HYDROSTATICS [For Competitive Examinations and B. A.; B. Sc. Students

ff of All Indian Universities]

Prof. P. N. CHATTERJI Head of the Department of Mathematics

D. N. (P^qs gradnatej Cdllege, Meerut.

1.1

Published by'

rajhonsprahashon tmandii* EDUCATIONAL PUBLISHERS

0«y?AyM 4i.0K RAMNA&AR, MEERUT (U.R) .

Published by : telegramYRAJHANS* Meerut. M. P. Giel , pimes . Office : 73258 Rajhans Prakashai* Mandir (Regd.) • Depot : 73358 Educational Publishers, • ' , - Press : 76595. Dharma-Alok, Ram Nagar, Pankaj : J 6 3 9 2 MEERUT (U. P.) " ' Resi. : 72201

^4// Rights* Reserved with the Author

Ninth , Edition 1985

Price : Rs. 14-40 only

iW't

>afili Wo

Priuted by : Rajee^Soel

* M. Cora. Rajhans Press, Rajhans Nagar, Garh Road, Meet*. Phone . 76595

i I

C"*&}H • *.

B r THE SAME AUTHOR ' " " ' Statics

Algebra Matrices namics

Trigonometry Real Analysis

Vector Algebra Vector Calculus

Integral Calculus Infinite Series

Theory of Equations Modern Algebra

Differential Calculus Set Theory & Numbers

Differential Equations Spherical Astronomy

Mathematical Statistics Co-ordinate" Gcometiy (2D)

Solid Geometry (3D) Spherical Trigonometry

PREFACE TO THE NINTH EDITION

It gives me great pleasure in Bringing, out the Ninth Edition of this book in a very short time.

The bock has been thoroughly revised and many new 'exa­mples selected from recent examination pgpers, have been added.

Besides giving due credits to the printers and publishers, I express my thanks to the professors and students for the apprecia­tion and patronage of the book. I am very grateful to Smt. S. Chaudiry who has revised the book and given it a new shape.

Suggestions for further improvement of the book will be highly appreciated. —Author

PREFACE TO THE FIRST EDITION

The present book comprising the subject "Hydrostatics" is meant for the students appearing at B. A.; B. Sc. Examinations of different Indian Universities and also various competitive Examinations. Efforts have been made to make the treatment logical and simple.

I gratefully acknowledge my indebtedness to various authors and publishers whose books have been freely consulted during the preparation of this book.

I shall be grateful to the readers for pointing out errors and omissions that, inspite of all care, might have crept in.

I look forward to the suggestions from trie* readers for the improvement of the book. '—Author

Chapters ' Pages

List of Formulae ny / j v \ 1. Introduction i _ 17

Jl. Fluid Pressure 18 72 3. Resultant Thrust on Curved Surfaces 73 125 4.-'Centre of Pressure 126. 192

^Floating Bodies 193 268 6. Atmospheric Pressure 269 304

SOME USEFUL* GEOMETRICAL RESULTS

Areas

Triangle, (i) If h be the height and a the length of the base, then Its area==|ah. •

(ii) If a and b be two sides and 9 the angle between them, then its area =|ab sta 6.

Hexagon. If a be the side of a regular hexagon, then its , „ 3^3aa

area= - ~ — .

Octagon. If a be the side of a regular octagon, then its area=2 (1+V2) a8. .

Circle. If r be the radius of a circle, then its area=nra and its circumference=27rr. *

Ellipse. If 2a and 26 be thelengths of the major and raior axes of an ellipse, then Its area=»jrab. •

Sector of a circle. The area of a sector of a circle of radius subtending an angle 20 at the cen're^it* d. „

Volumes and Surfaces.

Rectangular Solid. If a, b, and c be the length, breadth and height of the rectangular solid, then its volume -- abc ; • its whole surfaee=-2 (ab-f-bc+ca) ; and its diasonal=\/(aa~l~ba+c8)

•&

Cube. If a be the length of its

each edge, ihen its volume=a3,

its whole smface«=»6a2 and

its *diagonaI=a-y/3 <

\£ZA V

136/F/i

Hydrostatics

Prism. If the base of the prlsft •he a regular polygon, then

its volurae«< (area of the base) x height; -its lateral surface

—(perimeter of the base) X height.

Cylinder. If r be the radius of its base and /; be-its height, then its volum3=.jtr2h ; • its curved surfacera27trh and its whole surface ~2mr(h.-fr}

$-<? '

Pyramid.!

The volume of a pyramia e

=»JX(area of its basse) Xhaight and Its slant surface . - .

= 4 X (perimeter of its base) X slant height.

Cone. If h he the height; r the radius of its base, /its slant height and « its semi-vertical angle, then

. r<=h tan a, /*=h sec a Its vdlume^iw^^&jrb8 tan2 a, its curved surface ^jr/r^reh8 sec a ton a

Frnstum of a cone. If r{, rs be the radif of its circular ends and k be the perpendicular distance between them, the^its volume

T . . . " -i«k"(»»*+rir,+r^. , Sphere. If r be the radius of the

sphere,it9 yolurherafw3 and 'ts-curved surface=4nra.

+t~Uj~n

'Some Usesll Geometrical Results t'u • _

The volume of the j»ortioa.of & sphere, of radius r, included "between two parallel planes at distance*: hu Aa from the centre of the sphe:e

= l« ihi-hs) {3r*--tl»ia+M»-rl».*>} And the area of the zone of a sphere ,

or the area of the curved surface of ihz portion of the sphere included between two parallel planes => 2md, where t Is the radius of the sphere and d is the perpendicular distance between the planes.

Centres of Gravity.

C. G. of a cylinder (hollow or solid) is the mid-point of Ita axis.

C. G. of a solid cone. (Right circular) of ueigdt h is the polot 3h *

on the axis at a distance — from the vertex. 4 •

C. G. of the curved surface of a circular coae of height h is the

poinr on axis at a distance — from the vertex.

C. G. of (lis half cone. The distance of the C. G. ofihe half

cone from the axis of the whole cone<=> —, where r is the radius of

the base of the cone. C. G, of a Solid hemispiiere of radius r, is a point on its axis at

3r a distance -5- .from the centre of its basje.

o C, G. of • the curved surface of a hemisphere of radius 1; is a

r point on its axis at a distance -~- from tbe centre of this base*

C. G. of a semi-circle of radius r, is a point on the radius perpendicular £0 the bounding diameter at a distance 4r x~ horn the cntre of tbe ciiclc.

C. G. of a semi-ellipse bounded by the winor axis is a poir<t on the major axis whose distance from che centre ="4s/37t, where 2c s the length of the major axis.

J

Hydrosta|cs 136/F/i

C. G. of a sector of a circle.of radfus r subtending an angle 29 at the centie is a point, on the bisector of the angle 20, whose

' • 2 r sin 0 distance from the cen'tte= - — j — .

C. G. of circular are of radius r subtending an angle IB at the cenlre is a point on the bisector of the angle 20, whose distance from „ , . r sin 0 the cent i c ^ —;—.

CHAPTER I

introduction

§ 1"01. Definitions. States of Matter. Matter or substance existing In nature is

generally divided into two classes (a) Solids and (b) Fluids. Solid. A soiid is a substance which does not change its shape

or size without the application of some external force. Fluid. Fluids are those substances which are capable of

flowing, hence some external force is always required to maintain the size and shape os fluids.

Kinds of Fluids : Fluids are of two kinds — (a) Compressible" and (b) Incompressible. Compressible fluid, It is definsi as a fluid which can bs made

to occupy a small or large spa;e by• increasing or decreasing the pressure eg. air J n tubes or toy-baioons. Gases corns under this category. *

Incompressible fluids. It is defined as a fluid whioh under pressure cannot occupy a small or large spice e.g. water, oiis etc. Liquids are under tttis category.

Gases are als) call: 1 Elastic Maids as they can bs compressed or expanded. On other hand liquids are called Inelastic fluids as they can not be compressed or expanded.

§ 102. Generally the distinction between solid, liquid and gas is stated by authors as :

A solid has a definite size and definite shape. A liquid has a definite size but not a definite shape. A gas has neither a definite size not a definite shape. Experimentally it has beea seen that a substance cai be

changed from one of these three states to another. For example by increasing the temperature sufficiently, a mstal can be chwgeijfrom solid to liquid state and by the appllcati m of m >re heat it can bs changed Into gas.

§ 1*03. Hydrostatics. It is that branch of Mathematics^which deals with the equilibrium of masses of fluids acted upon by forces and of solids in contact with fluids at rest.

§ 1*04. Shearing Stress. Surrounding any point P inside a material substance imagine

a small element of area. Consider the two portions of the substance 136/1

•2 Hydrostatics

on either side of%hisarea, there is mutual action aud reaction between thein. This action and reaction Is called the nuitnal stress^ Let this stress be denoted by R, which can bs resolved into two components normally and tangentially to this plane vii. R cos 0 and R sin 6, R cos 0 is called the mutual'thrust or tension and R sin 6 (acting tangentially) is

i known as shearing stress. * (Fig. 1) § 1'05. Types of fluid. Fluids are of two types :,(a) Perfect and (b) Viscous. Perfect fluid. It is that ideal fluid for which, whether at rest

or in motion, the shearing stress is zero and it can exert an action only in the sense normal'to the surface with which it is in contact.

In nature no such fluid exists. If water in a vessel is set revolving, it .,comesi> to rest after some time due to the frictional resistance (tangential force/ between the vessel and water and diffe­rent elements of water. If this tangential force or shearing stress is totally absent, the water would*not come to rest.

In this book we shall deal with perfect fluids only unless other­wise stated.

Viscous fluid. When the particles of a fluid exert resistance to one another or it offers rsistance to the motion of a body in it then the fluid is called viscous. Honey, tar etc. are the examples of such a fluid.

§ T06 Pressure at a point. If a hole is made in the side of a vessel containing fluid and this hole is covered by an exactly fitting plate, then we find that the plate will remain at rest only if some external force is .applied to it, which shows that the fluid exerts some force on the plate which pushes it outwards. If P be the external force reqaired tolceepthe plate in positi n, then Pis th? measure of the fluid pressure or thrust on the plate.

Also by definition the force exerted by the fluid is always perpeadkular to each element of area of the plate

If the fluid pressure is proportional to the area i.e. it the force exerted by the fluid on each equal element of area be the same, then the pressure is called nDiform pressure. If p be the pressure ouan, area S la this case, then pressure at a point =plS.

If on the other hand, the pressure on each equal element of area is not the same, then the pressure is called varying pres§«e, If in this case pressure be Sp, on an element of area S5,' then press-

^ , Llm 8B dp u r e a t a p o n H - g ^ ^ ^

Irfi roduction •3

This is also defined as Intensity of pressur* at that point. _ Mean pressure. Trie* mean pressure on a plane area is the

uniform pressure on it which gives the same resultant thrust as the actual thrust on it.

**§ T07. The pressure at any point of a fluid at rest is the same in all directions. (M,ar, 76 > Mugadh 75 ; Mithla 82 ;

Muzzafarpur 80 ; Rajasthan 76 ; Banchi 82, 80, 76 ; Vikram 82,10)

Pioof Take three mutually perpendicular lines Ox, Oy and Oz at O, any point in the fluid.

Take short unequal lengths OA, OB and OC along these lines, Join AC, AB and BC and this will form a small tetrahedron (O, ABC) of liquid. Mass of the fluid within this tetrahedron<=»px volume of the fecrahedron, where p is the mean dens ty of the fluid in it.

«=--px$area of the baseX height =*9X\ABOCxOA. Let p be the thrust at any (Fig. 2)

point of A BOC. This thrust will be acting in the direction perpe­ndicular to A BOC i e. in the sense of OA or Ox. Hence thrust of the fluid over the area of A BOC in the sense of Ox

•=>p X ABOC. The thrust of the fluid over the areas AOB and AOC will be

zero in the sense of Ox. Let p' be the thrust at any point on A ABC and let 0 denote the angle/between A OBC and A ABC

:. The fluid thrust on A ABC^p'xA ABC and the resolved pait of this thrust in the sense of Ox=p' x A ABC cos 8

•*•/>' XABOC, since A ABC cos 6 = A BOC. • Let the fluid be under the action of some external field of

foice, such as grayity. Let X be the force acting on the fluid (per unit mass) along Ox. Then the component of the external fgrce on the fluid In tetrahedron (O, ABC) In the sense of Ox

•=<Xxmass of the fluid in the tetrahedron •=X~xiABOCxOA.P.

Resolving all the forces along Ox, we have p. ABOC—p' ABOC+X.^x ABOCxOAxp^O

or p-p' + i.XxOAx9=0 - (i) When every dimension of the tetrahedron is diminished, then

OA<=>o and p,p' become the pressure at O iwtwo different direc: tions and we have from (i) p—p'-*0 or p=>p' which proves the theorem for liquids at rest.

fctydfdstatlc t m~t6

*$1*08. Tranjmissbiiity of Liquid Pressure (Pascal's Law) \ An increase of pressure at any paint Of a Liquid at rest under

gi?ea external forces is transmitted without change to every other point of the liquid. (** / /*WM. 79)

Let A and B be any two points in the liquid. Case I. When the straight line joining A and B lies wholly in

the fluid. • . ,„ , „ .. With the straight line AB as axis construct a cylinder ot small

cross-section" a whose plane ends are perpjndioular to __.____.___•__- . --.. AB. Lst -p and p' be the* I ^ r r r ^ ^ ^ g r fluid thrust at any point on r _r___ - ~ ^ ' 3 the plane ends at A and B =- —* respectively. ' _:_

,lhe needle shaped __ cylinder of liquid is in =: equilibrium under tiie _.-j _ action of, the following rf^^fJJ^-j^gt^gr^^K ^ ^

(i) The fluid thrusts p% • r ^ ^ C ^ ^ ^ ^ p - ^ ^ ^ ^ ^ S : ? acd p'a on the plane ends ~ --"—"-"• at A and B respectively * N ' along AB, as shown in the (Fig 3)

.diagram. (ft) The pressure at -each point on the curved surface of the

cylinder acting at right angles to AB.. (iii) The resultant external forse on ths liquid within the _

cylinder, whose component along AB Is X (say). Resolving all these forces along AB, we have

px-p'xi-X^O —(1) Now suppose that an extra pressure of intensity P is applied

at the end A aad let the corresponding increase in pressure at the end B hi of intensity P'. Then as b.fo-c, resolving the forces along AB, the system still bein? in equilibrium, we hava

(p+P) a - ( /> '+P')«-rZ=0 - (2) .Substituting (1) from (2), we get _?«-_'«=() or P-=P' Hence increase of

pr.ssilre at A is equal to > J the increase of pressure 4

atB. Case II. When the

straight Hue joining- A, B does not lie wholly in the Moid.

Join A, B by means of straight lin^s AW. PQ, QR and RB each of which Uss wholly to the fluid. (Fig 4)

IntromuctioB

E=3 :

Then from case I above, we et increase of pressure at A & increase of pressure at P =increase of pressure at Q ^increase of,pressure at R "increase of pressure at B

Hence the theorem in this case. . *§ V09 Hydraulic or Bramah's Press. (Vikram 83) Hydraulic or Bramah's press are constructed osfprinciple of

transmissibility of liquid pressure. This machine consists

of two veriical cylitdeis communica'ing wilh each other by a tube near their bafes and one of the verti­cal cylinders is of wider cross-section than the other. Let the areas of their crosB-sections be A and X res­pectively as shown m the figure. Tne cylinders are fitted wiih water tight pistons and the space belo v {he pistons is filled with water. •

Let a downward forcep be applied by the piston of area A. This increases the pressure by (p!A) per unit area, which by the

principle of ttadmissibility of liquid pressure will be transmitted throughout the I;qu d And consequently, the resultant upward thrust on the pistou of areaXwlll be given by P= (p/A) X. or P A~p X or P : p=>X : A i.e. ttie thrusts on the pistons are pro­portion il to their cross sections.

Hence, by suitably adjusting the Tatio of the areas of CTOSS<-sections, any force however small can be made ro support a weight however large. This is called 'Hidro'tatic Paradox'. But in practice we find that th? strength of the side of (he vessel does not support an indefinitely great grmunt of fluid pressure. (Vikram 83)

Solved Examples on § 1 08 and § 1 09 Ex. 1. In a Bramah's Press the diameters of *he larger and

smaller pistons are 59 cm. end 4 cm. respectively. Find thjp mass which can be supported by a kilogram placed on the smaller piston.

Solution. We know Pressure on larger piston : pressure en smaller piston

=area of larger piston : area of the smaller piston Pressure on larger piston ^

(Fig. 5)

or , area of larger piston ^ 0 Q , m a l l e r p i s t Q n

area of smaller piston

Hvdrostafcs ifc

M^)*1 , i , 62.5 , e r , c "Vfl? X k , b g r a m~ faI56"25 kg- Ans.

Ex. 2. A cylindrical pipe which is filled with water opens into another pipe, the diameter of which ss tiiree times its owa diameter ; if a force of 20 lbs. wt. be applied to the water in the smaller pipe, find the force on the open end of the larger pipe, which is necessary to keep the water at rest.

Soluti.cn. Let 2r and 6r be the diameters of the two tubes. Then as in the last example, we have

Foice on the open eni of larger pipe area of larger pio* . _ „

•= T;—!—r— x force on smaller pjpe area of smaller pipe

=ii:ttry ^ w t _ r a l 8 0 l b s # w t f

•K (r)* An . Ex. 3. la a Bramah's Press the area of the larger piston is

100 sq inches, and that of the smaller one Ss £ sq. inch, find the force that must be applied to the latter so that the former may lift l t o a .

Solution. We know Pressure on smaller pisto» area of the smaller piston Pressure on lacgtr piston area o! the larger pistoa Pressure on smaller piston £ , . . .

° r " i"x224u ™100 [1 ion-2240 lbs. wt,] ii 2240x1 28 , , „ t .

or pressure on t ie smaller p is ton^™—2T = ->- ^ J ' O lbs. wt. An9. *Ex. 4. ht a Bramah's pr«s<? ihe piston can safely hear a

pressure of 1200 lbs. wt per square foot. What will be the greatest weight that can be placed on the smaller piston, its cros? section being 5 square inches.

Solution. Let t ie area of the larger piston be x square feet. Then the pressure on the larger piston=1205x lbs. wt.

*5 Also the area of the sma'ler piston <= 5 sq In =JT, sq. ft.

Now pressure on the smaller piston * Area of smaller piston ' , . — Xpressure on larger piston. Area of larger piston

» 125 * — * X 1200* lb. wt.=»^- «=>41§ lbs. wt. x 3 Aas.

Ex. 5. In a Bramab's press a total pressure of 1 ton is pro­duced when a pressure is applied to the smaller piston by means of force of S lbs. daced at the end uftfee lever If the diameters of the pistons are as 8 : 1, fiaii the ratim of the arms of the lever employed to work the piston *

Solution. Since the diameters of the pistons are in iheratia

8 : I, si their radii are also and 8 r. Let the pressure on the smaller piston be P lbs wt.

Now pressure on the smaller pise m

Area of the smaller piston"! Aiea oi itie larger piston J

Introduction , •

in the same ratio.* Let their radii be r F

i or P =

7TJ" XIX 2240 lbs wt, =

7r(8r)a

2240 64

Now for the lever,

=35 lbs. wt.

at (Fig. 6) A and B (he forces are of 35 lbs. wt. and 5 lbs, wt. respect vely.

.'. 35xFA*=5xFB or FA : FB=1 : 7. Ans. *Ex. 6. A vessel fall of water is fitted with a tight cork. How

is it that a slight blow on the cork may be sufficient to break the vessel ?

Solution Let A be the area of cross-sectlou cf the cork and S the area of the surface of the vessel. Let p be the intensity of the slight blow on the cork then as the water is incompressible so by Pascal's Law, the pressure {pi A) per unit art a will be transmitted to each point on the suiface of the vessel, tnd the total presswre thus transmitted will bs (p/A) S, which is >eiy large as S is large. This pressure on the surface of the vessel, being very large may bu sufficient to break the vessel,

Uence a slight blow on the cork may be sufficient to break the vessel. The cbanc-s of breaking increases if the surface of the vessel S is large and the area of the cork viz. A is small.

Exercises on § 1*09

E x . 1 . If the section of the cylinders of a Bramah's Press ba 18 sq. in, and 1 sq. ft. respectively, v?hat pressure must be applied to the smaller cylinder to produce a pr> ssute of two tons upon the larger ? (Ans 560 lbs. wt.)

Ex .2 . In a Bramah's Press the diameter of the larger,plston is £0" and that of <be smaller one is 20". What force must be applied to tne smaller piston to raise a weight of ,000 tons at the other piston? (Ans. 15*625 J^ns wt.)

§ I"10. Density. The density of a homogeneous substance is defined as Us mass per unit volume.

The mass of a cubic centimetre of water at 4°C is one gram. Hence in C. G. S, the density of water at 4°C is 1. Similarly the mass of a cubic foot of water at A°C is 162 5 lbs., hence in foot pound system the density of water at 4°C is 62 $

Homogeneous and Heterogeneous substances. If equal volumes,

8 Hydrostatics!

.however small, taken from B substance, have equal masses, then the substance JS said to be Hcmcgeneois (or of uniform density). And If their masses are different then the substance is called Heterogeneous (or of variable density).

§1-11. Weight in terns of density. JfXVbe the weight of a given substance in pounds, V its volume in cubic foot, P its density in pound pet^ cubic foot and g the ccceleration due to gravity in foot-second uni/s^tkenW=-V pg.

We know from Dynamics, W=Mg, where M is the mass of the^ubstance.

Also from the definition of density, we have Af«=>mass of V cubic- feel of substance

-^rxrnass of one cubic foot of the substance =FXP,

JL From (i) we get W=V9g ' § X 12. Specific gravity. Specific gravity or relative density of

a substance is defined as the ratio of weights OJ mass r f equal volu­mes of the substar.ees and a standard substance (which is generally taken as water at 4°Q. In short specific gravity is written as sp gr.

Note Specific gravity or relative density is a unrulier. § 1 13. Weight in terms «f specific graviiy. IfW be the weight

of a given substance, V its volumes its specific gravity and w the weight of unit volume of the standard substance, then W=>V's w.

We know, from definition of specific gravity. weight of a unit volume of the substance

weight cf a unit volume of the standard substance weight of unit volume of the substance or s=>

w or weight of unit volume of the substance<=*s.w. or weight of V units of the \oIume of the substance=V.s.w. or W<=>V.sw. Hence proved.

**§ 114. Specific Gravity «f Mixtures. (a) To find the spec fie gra\ Uy of a mixture ofgivtn substances

whose volumes and specific gravities an given Let Vx, V^ v3 • be the volumes and su ss, s&,... be the specific

gravities of the different substarces Then if w be trie weight of a unit volume of the standard substance ihe weights of the different substances are Vx sx w, V% sa w, V3 sa w,...

Also the volume of the mixture is Vl+Vi\-Vz+...

Let s be the specific gravity of the mixture, then the weight

of the mixture=>(Fi+ V2+V3+...) 1. w. ...(1) • Also the sum of the weights of the different substances is

ViSiw+Vesaw+VaSzW+... ...(2) These two weights must be equal.

Introduction 9

.*. Equating (1) and, (2) we get

flVf-P.+FVf-...) J w^ViStw+V&w+V&w-b •• n r - F'lJJi-l-Fy.+ T W - ••„ 'gyiJ.

In some cases chemical reaction takes place in the proeess of mixing the substances i.e. the volume "of the mixture Is different from the sum of the volumes of the different substances.

1 • Then let the volume of the mixture be - [Vi+Va+Va+...] x . c • * u - - V,st+l?is2+Vas3+.. nSVtsx

As before in this case , = — j — — J i ^ - - ^ (6) To find the specific gravity of a mixture of given substances

whose weights and specific gravities are given. (Magadh 73 ; Patna 82) Let Wx, Wv Wa, .. be the weights and sx, sa, st, .. be the

specific gravities of the different substances. Then if w be the weight of a unit volume of the standard substance, their respective volumes

. . . . Wt Wa Wa will be —\ —a, —?, ... sxw shw s3w

.'. If s be the specific gravity of the mixture, the weight of the

mixture= —1 + —J+-^-(-... 1 s w=\ — t —1-— r -• I s ...(1) \sxw sjv saw i \sx s2 sa }

Also the sum of the weights of the different substances is Wx+Wi\Wt+... - ( 2 ) '

These two weights must be equal, hence equating (I) and (2)

we get ( E I + 3 J ! + 2 J ! + \ J=WX+ W,+ Wt+... \ sx sa sa I

or ' 1 - WxW\Wz+... = IW1

Sx S2 S3 Sx

The ab,ove formula can be modified as in case (a) when some chemical reaction takes place in the process of mixing the substances.

Solved Examples on § 1-14. Ex. 1. The specific gravities of pure gold and copper are 19-3

and 8'62. Find the sp. gr of standard gold which is an allojwof gold and copper in the proponion of 11 : 1.

Solution Let the volumes of pure gold and copper be llv and v respectively. *

V s A-V s Now the sp. gr. ot the mixtures Xy\_y a

_(HvXl9-3)+Cvx8 62) a2l2 3 4 - V 6 2 a l 8 . 1 A n g

llp + v 12 D

Ex. 2. (a) What is the volume ofamfss ofwoodofsp.gr. 0*5 sp that when attached to 500 ozs. of iron of sp. gr. 7, the mean sp. gr. of the whole may be unify ?

10 Hydrostatys w

Solution. Let $v ozs. wt. be the weight of the wood which is ""attached to the given iron.' The sp. gr. «f water at 4°C is 1.

AW W ~Xjr—-

...see§ J -14 (b) Page 9 , 500+iv 500 . w cr.n / , 1 \ soo eftA

7 " V s or ii'=500x"? = a Y f l ozs- w t -

.*. If the required volume be Feu. ft. then from 'W^Vsw' \we have afl,aflXiV=7(0-5K62-5). V wt. of 1 cu. ft. of water«=62'5 lbs. or . F=(6/7) cu. ft. ' Ans.

Ex. 2. (b) How much water must be added to 20 c. c. of a salt solution whose sp. gr. is 1*08 so that the sp. gr. of the mixture may be I'd5 ? tMagadh 73)

Solution. Let x c.c. of water be added to the salt solution. Also the sp. gr. of water is 1.

W -\-W„ Now the sp. gr. of mixture- y 2 _See § M4 (b) Page 9

i-_i-)-__J

1-08 1

.or • ,(,<«-.)-» ( l - g ? ) - » Q 20x0-03 2000X3 llO „ 1 or x=3 —= = — =11 — c.c. ,

108X0-05 108x5 9 9 • Ans. Ex 3. (a"y A mixture is formed of two fluids. The density p of the

mixture, the ratio m : 1 of Che volumes, and the ratio n: 1 of densities are given. Prove that the densities of fluids are

E5±« and to+^) p. mn + 1 mn 4-1

Solution. Let the volume of two fluids be mv and v ,and their, densities be nor and a respectively.

Then weight of the misture=>(mv + v)x p. •• H) Sum of the weights of the two fluids=jwv./jor + \a. ...(2)

"These two weights being equal, equating (!) and (2) we have (mv + v) p=mnvo -{- va

or 0»+l) P=(w«-H) or or o-=(m + l) P/(w»-f-1) H?nce the densities of the fluids are no and o

n (/w+l)P . (m+l)P (wn-f-1) (mn+l)' • Hence proved.

Ex. 3 (b) The density of milk in a vessel is p and the density of pure milk is a. Prove that the ratio of volumes of water and milk

, in the vessel is °-~. (Ranchi 82) P - l

Introduction 1J

Solution. Let v and y' be the volumes [o# water and pure milk in tlie vessel. •

Then the density of the mixture

or v + v' ff=P (v-fV) or v'(or —P)=fp —1) v • or v//=lcr-p)/(p—1). Hence proved.

*Ex. 4. (a) Two liquids of specific gravities s Had s'and of volumes v and v', having been mixed, the sp. gr. of the mixture is found to be a. Find the volume of the mixture.

Solution. Let w be the weight per unit volume of the standard substance and V the volume of the mixture.

Then the weight of the mixture^Fw. ...(1) And the sum of the weight of two liquids

=VSII>+VVM;. ...(2) These two weights being equal, equating (1) and (2). we get

Vow~vsw + v's'w or 7=>(v.s + vV)/o. Ans. *Ex. 4. (b) The mixture of a quantity of liquid A with n kgs

of B has specific gravity s, with 2n kgs, of B a specific gravity s' and with 3n kgs. of B a specific gravity $". Find equations to determine the specific gravities of A and B. {Bihar 74 ; Muzzafarpur 82)

Solution. Let the liquid A be m kgs and the specific gravities of liquids A and B be st and j a respectively. •

Then according to the problem, we have m+" ...See § I'M (b) Page 9

- ( i )

..{ii)

Uii)

(m n \ or s 1— =/« -f n

Similarly s' ( — + — j =/w+2«

and ' - s"(~+3n)=m-{'!n \s1 s2 '

From (i) we have m | 1 ]«=« fl I

or m (s—Jj) s2=nsx (ja—s) ...<'lv) Similarly from (ii) and (iii) we have

m(,s'—s1)si<='2ns1(sa—s') ...(v) and m (s"—s-i) As^nSi (sz-s") - ...(vi)

Dividing (Iv) by (v) we get £ £ t = / ^ ^

Dividing (iv) by (vi) we get £ * - J * ^ _ ( ^

These (vii) and (viii) are the required equations.

12 Hydrostsjtics

*Ex. 5. When equal volumes of tw» substances are mixed, the sp. gr. of the mixture is 8 ; when equal weights of the same substances-are mixed, the sp gr. of the mixture is 6. Find the sp. gr. of the Substances. (Muzaffarpur 80)

Solution Let the required sp. gr. of the substances be sx and s2. When equal volumes, (say V) of the liquids are mixed, sp. gr.,

of the mixture is given by

When equal weigbis flsay W) of the liquids' are mixed, the sp. gr. of the mixture is given by

W+W 1 , 1 .

E^E OT^+*7=i -(2)

From (2), we get J - ^ = * or i l = i ^ f f o m ( I )

or ^ 3 = 4 8 . — <3) Solving (I) acd (3) we get ^-=12 and s2=4. Ans. Ex. 6. Two metals of which the sp. gr. are ll-22 and 7'25,

jvhen mixed in certain proportions without condensation form an alloy whose sp. gr. is 872 : find the proportions by volume of the metals in the alloy.

SoJ. Let the required volumes of the two metals be vx and va.

Then sp. gr. of the mixture^ v i ^±M?

or 8 -72 «=> faX",22>+fax7-25)

or 8-72 (v!+v3)=ll 22v1+7-25vs

or (872-725) va=(l 1-22-8-72) v, nr vt 8 72-7-25 1.-47 147 . 1>f, . ,«« M vl ~ IT22=872 =2 lo~2To ° r V l ' V*~U7 ' 250'. Ans.

**Ex. 7. Three litres "of the liquid whose sp. gr. is 08 are mixed with 5 litres of anothpr liquid whose sp. gr. is 104. Find the sp gr. of the mixture if there is a contraction of 5% on the joint v o l u m e i*Sra 77)

Solution. Let wbe the weight per litre of standard substance. Then the sum of the weights of the liquids >

. • =(3x0-8w) + (5xl 04w) < ~(1) Also the volume of the mixture after 5% contraction

=fo5o (3+5) points=i§ x 8=sB

a pints. (Note)

i. The weight of the mixture=-afi7n>, where s is the requ­

ired sp gr of the mixture. Since these two»weights are equal, hence we have

as J.H>=.(3X0-8H0+<5 X r04»)=7'6 w

In'tnpduction 13

or l=[(7;6x5)/38]=l Ans. *£x. 7 (b). A mixture is made of 7 en, cms. of sulphuric acid

(sp. gr =1843) and 3 ca. cms. of distilled water and its sp gr. when cold is found to be 1 615. What contraction has taken place l

{Ranchl 73) Solution. Let the volume of the mixture when cold be

i [ H 3 ] - I ( U , ) c a . e i M . _ ( i ) t N o t e )

Also ws know if j be the sp gr of the mixture rf volumes vi and v2 of two liquids ofsp. gr. Si anJ j a and if the volume of the mixture after contraction be (Vi+v2)//7, then

n (v^i-l v*s2) s==> .... VitV2 ~ ( l l )

.. see § r i 4 ( a ) Page 8 Here we are given J = 1 ' 6 1 5 : vx=7 cu cms. ; va="3 cu. cms.

si=1*84J and s2-=\ 0.

„•. From (n) we get 1 615«= ^—^ - — >—

or 10 (;'615)-n[12-90H--q'O] = l5-901 « 16 150 16151 '

or ii =a " " 15 901 159jl

.". From (i) we find that the contra "ted volume = foT^ ( ' 0 ) CU" CmS a ( 0 ' 9 8 5 ^ (I0) c u ' c m s-•=9 8,5 cu cms.

•*. The percentage of contraction

*E« 8. Three equri vessel* A, B and C are half full of liquids of densities plt pi ani Pa respectively. If now B be filied from A and then C fro a B, pro re that the density of the liquids now contained in C is I (Px f?2 >2P3) tie liquids being supposed to mix completely.

(Mithila 80) Solution. Let S be the volume of liquids in each of the three

vessels when they are half full. Let p bs the density of the liquid in B after it has been filled from A. ' •

Then' P = VPl + V9* - ^ ' men p= y+y 2 . Now the densities of the liquids in B and C are \ (Pi+P2) and

P., respectively Thsn C is filled from B I.e. now C contains the mixture of t»vo liqaicis earn of volum; V (as C w as half foil pre­viously) ani of densities J (Pj fPa) and P„. Letts be the required density, then

a K+K 4 Hence proved.

14 Hydrostafics

Ex. 9. A mmure is formed of e#qual volumes of n liquids the densities of which are in the ratios of t&e numbers 1, 2, 3,..., n, find the density of the mixture.

Solution. Let V be volume of each liquid. Let the densities of the liquids be p, 2p, 3P, . , np

• j • •* F.P+K.2P+F3P4- +Vnp Then the required aensityo „ y+y±_—T~p

" l + 2 ^ 3 . . . + / l 0 inOH-l) " x 1 . = — •—P™ p = . — — p. A n s .

" . n 2 *Ex, 10 Equal weights of gold, silver and alloy of gold and

silver are dipped successively in a cylinder of water to rise a, b and c inches respectively. Prove that the alloy contains gold and silver in the proportion b—c : c~ a by weight.

Solution, Let equal weights W of gold, silver and alloy be dipped in the cylinder containing water. Let A be the area of the cross-section of the cylinder.

V Volume of body^volume of water displaced. .*. Volume of gold, silver and alloy are Aa, Ab, and Ac

respectively. Let P1( Pa and P3 be the densities of gold, silver and alloy res­

pectively ' Thenwt. of go\d.='W-=>(Aa)x?xy.g or 9x=WjAag Similarly 9=WjAbg and pa<=aW/Acg Let Wx and W2 be the weights and Vx and Vs be the volumes

of gold and silver contained in the alloy respectively. Wj= Wx _^AaWx

'PiS iWlAag) g ^ W Similarly V^(AbW*)lW

:. Total volume of the alloy=F1+ V^^^+^f

or total weight of the alloy=>(volurne of the alloy) X p3sg lAaWl.AbWt\y, W , aW^bW,

-\-w+-w-rik*g-—c— But total weight of the alloy«»wt. of gold+wt. of silver in the

a l l o y s * »s . .'. " Wx+W^aWl+b^ or ic-a)Wx^W^{h~c)

or Wx : Wtf*=-{b—c) : (c—a) Hence proved. **Ex. 11. Show that the Bp, gr, of a mixture of n liquids is

greater when equal volumes are taken than when equal weights are taken, assuming no change in volume as the result of mixing.

(Mithila 81 ; Muzzafarpur 811 Ranchi 79, 77)

Then W1=V1p1g or V^

Introduction 15

Solution, Let s and s\bs the-sp gr. of the mixture when equal volumes and equal weights of the substances are taken respectively and s1} s2, s3, „., sB be the sp. gravities of the different substances.

Then we have, s~ Ji + J»+M-...+*B (S e e § 1-J4 Page 8) ..,(1)

and /«=>-

n n

or s'

J.+.i+-L+...+i- . . sl st ^3 ^n

s' n \s-s_ s2 s3 "" s„ I ...(2) Also we know that the Arilhnietic Mean of n unequal positive

numbers is greater than their Harmonic Mean. But A.M. of st, s2, s3,..., SnB*[s1+s2+-..-t So]/n

And H.M. of slt st,...t s„<=- : — < = - s ' . - + - + . . .+-Si s% sn

Hence s > s\ which proves the result. Ex. 12. If equal weights of two different substances be mixed

prove that the sp gr. of the mixture* fs the Harmonic Mean of toe pp. gr. of compound substances.

Solution. Let sx and s2 be the sp. gr. of the two substances. If s is the sp. gr. of the mixture and equal weights of the substances are taken, then from § 1*14 Pages 8-9, we have

2 1 , / 1 1 , 1

— -p.— s1 S%

or — = H 1-s* J

i e. s is the Harmonic Mean of sx and sz. Hence proved. *Ex. 13. Equal volumes of three liquids are mixed and the

mixture separated into three parts ; to each part is added its own volume orthe original liquids, and Abe densities so formed are in the ratio 3 : 4 : 5 . Prove that the densities of the liquids are as 1 : 2 : 3.

Solution. Let ps, p2 and pa be the densities of the three liquids. If ?! be the density of the mixture and volume Vt of each liquid be

taken, we have P - r&±h±& o r p_ft±Jji±P»

Now let the three parts of this mixture have the volumes Vi, V2. Fa. To first part of this mixture is added a volume Vx of the first liquid of density px. If 3a is the density of this mixture, then

w e h a v e 3 ^ F A + F - l ( ^ P ^

or 3or=i (4PJ+P2+P3) -V-'i

16- hydrostatic* ^ H

Similarly for *2nd and 3rd parts of the mixture, if we add liquids of volumes V2 of density p2 and V3 of density pa respectively, and the densities of (he new mixtures are 4or and 5c respectively; then we have

4tra P*±£±ri ...(2) and 5a= h±h±h (Note Densities \a, 4<r, 5<r are taken because of given ratio

3 : 4 : 5 ) . . Solving equations (1), (2 and (3), we find

Pi=2ff ; P2=4o,J pa 6<T

Hence P» : p2 : p3=2 : 4 : 6=1 : 2 : 3 Hence proved. **E< 14 The specific gravity of a mixture of equal volumes

of two liquids is a and thai of a mixture of equal weights of the same liquids is 6 Show that the specific gravities of liquids are given by X*-2ax+ab=0. (Ranchi 77, 75)

Solution. Let sx and s2 be the specific gravities of the two liquids.

Then if V volume of each liquid be taken, then sp. gr. of the mixture

ZVX V-tV 2 ° l S l v e n ' i.e. ' sx+s2=2a .. (1)

And if W weights of each liquid is taken, then sp. gr. of the mixture

"27 fly W+W

2 l i / J i )+( i / j , )

=fl (given)

i.e. - = _ - j - _ ^ J — a = — , from (i) •

°r SiSi^ab ...(H)

Also we know that the quadratic equation whose roots are Si and s2 can be written as

x2~(s1+s2)x^(s1s2)=0 (Note) (See Author's Algebra or Theory of Equations)

or x2—lax i ab=>0, from (i) and (ii). Hence proved.

Exercises on § I ' l l to § 114

Ex. 1. The sp. gr. of a cork being 0-24, find what volume of water weighs as mifcb. as a cubic yard of cork. (Ans. 6 48 cu. ft.)

136/2 Introduction 17

Ex. 2. When the equal volumes of two sulsf ances are mixed, the sp. gr. of the mixture is 6, when equal weight of the same substances are mixed, the sp. gr. of-the mixture is 4. Find the sp. gr. of the substances. Ans. 6±2\/3 (Vikram 83,78)

Ex. 3. When equal volumes of two liquids are mixed the specificgravity of the mixture Is 4, when equal weights are mixed the sp. gr. is 3. Find the sp. gr. of the liquids. {Bihar 75)

Ex. 4. Any alloy of zinc of sp gr. 7'2 and copper of sp. gr. 8 95 has a mass of 467 grammes. Its volume is 60 cu cmr Find the volume of each component. Ans. 40 c c ; 20 c.c.

Ex. 5 Having given the sp. gr. *p of a mixture formed tf equal volumes of two fluids, and also the sp gr. a of a mixture formed by taking a volume of one fluid double that of the other, prove that the sp. gr of the fluids are (3o—2p) and (4p -3<r).

Ex. 6. The density of mixture of equal volumes of three liquids Is A and that of a mixture of equal weights of the same three liquids is B Show that A > B. if there is no change in volume on m j x j n g # (Magadh 76, 74 ; Eanchi 81, 80, 76)

Ex. 7. The mixture of a quantity of a liquid A with x lbs. of B has a sp. gr. s, with 2x lbs. of B asp. gr. *', with 2x lbs. of 5 a sp. gr. s", show that 2 (s—sx) Q a - s ' H (.%-.?) (s'-sj, and 3 (si-s") '(s -sj^isz-s) . ( A " - ^ ) , where ^ and ss are the sp. gra­vities of the liquids. {Muzzafarpur 82)

CHAPTER II

Fluid Pressure § 2*01. A mass of fluid may be in equilibrium t.e, at res*

under action of a number of forces. In the present and the succee­ding chapters we shall regard the external force acting on the fluid as its weight i.e. force due to gravity, unless otherwise stated.

If the force due .to gravity is not to be taken into consideration! then the liquid is called light and if it is to be taken into consider­ation then the liquid is called heavy.

*§ 2 02, The pressure of a heavy homogeneous Maid at rest is the same at any two paiats in the same horizontal plane.

(Ranchi 80; Vikram 78)

Take two points P \ -and Q' situated in the ^-^0^^~^~~=^S^^-l~z^=k?^^3-same horizontal line in ^=3^^v-^i-_^^£ifH.^^j%^=_^?e-~-" the liquid. Join PQ. With ^p^S^^^^^j^f^?^ PQ as aais construct a = ~_?=4/^?r"~---~-~ ~ ~-~- ~-~--T--:nl/f>r-^ cylinder, whose cross- r£=ir^^?i^SSS^3l^^^ section is small and a 1~S^-^~-~€^--T-„^^J:SS'~1~:~Z. closed curve of any type whose area is a (say). (Fig. 7) Consider a portion of the liquid contained with the cylinder.

Letpt andpa be the pressures at P and Q.' Now pressure in general changes from point to point, but as the area of the cross section of the cylinder is small, so the pressure of the surrounding liquid on the bases can be taken as constant.

The liquid contained within the cylinder is in equilibrium under the action of the following forces :

(i) the thrusts p^o. and j?aa on the bases acting in opposite directions at right angles to them ;

* (if) the thrust at each point on the curved surface acting perpendicular to the line PQ and

_(iii) the weight of ths liquid contained acting vertically down­wards.

Hence resolving all the forces in the sense of the horizontal line PQ, we get » />ja—/?2a=>0 or p\*sp2. ^

Hence the pressures at Pand Q are equal. *§ 203. Pressure ia heavy homogeneous fluid. (a) In a heavy homogeneous fluid st rest." 1 he difference bet­

ween the pressures at two points varies as the difference of the ver-

Fluid Pressure [9

tical depths of these two points below the free surface.

Inside the fluid take points P and Q.

Case I. When the straight line PQ is vertical and lies entirely (Muzzafarpur 79)

•^ST?:~~'~~~-:i

mm&m^ (Fig. 8)

in the fluid. With PQ as axis, construct

a cylinder ef small cross-section of area a (say). Let PQ=z

Let /7, and p2 be the pressures at P and Q. jjCThe liquid within

. this small cylinder is in equili-b ium under the action of the following forces :

(i) The thrust pp., acting vertically downwards on the upper end at P.

(ii) The thrust p2<*, acting vertically upwards on the lower end at Q.

(iii) The thrusts at each point of the curved surface actiDg hori­zontally.

(iv) The weight of the fluid contained in the cylinder, whioh is z.xgp, acting vertically downwards, p being the density of the fluid

Resolving the forces in the vertical sense, we have p^+z «.gp-p&=0 or pa~Pi=zgp.

Hence the pressuie ,at the lower point Q differs from the pressure at the upper point P by zg. p, where z is the vertical dis­tance between the points P and Q.

Case IT. When the straight line PQ is not vertical and does not He entirely in the fluid.

Let the height of P above Q be z. Take two points M and N such that M and N are vertically below P and Q respectively. Also M and N are situated in the same hori­zontal level inside the liouid. Let the pressure at P, Q and MbepvPz and/73 respectively.

Now as M and N are sit­uated in the same horizontal line, so the pressure at Af= pre­ssure a{ N^pv

L<?<

•-M,

rS-

•

(Fig. 9)

M

N!

2

r—ir~J

20 Hydrostatics" ' 184-66

Again as M and P, N and (2 are situated in the same vertical Imes, A Pa-Pl-g p.PMl C a s e j and - Pz-Pa^g P 0 ^ 1

Subtracting we get Q>*-pJ-(Pt—pJ*=g? (PM-QN) or /'a_^i=^Pz« Hence the theorem.

*§ 204. Pressure at a depth z below the surface of liquid. v (a) When the liquid is heterogeneous. {Magadh 77)

Take any point P in the liquid at a depth z below the surface-From P draw a perpendicular PO to the sur- ,Q face and let OP be produced t o g , such that

With PQ as axis, construct a small 3I " cylinder, the area of whose cross-section 5s P Let p be the density of the liquid, which is varying from point to point.

Volume of the iiquld in the cylinder $ = a B z ; -•

.*. Weight of the liquid contained in (=±= )0 the cyllnder=a Sz <?g. Letpjsnd/i-f Bp be the intensity of pressures 'at P and Q respectively. - (Fig. 10)

The foices acting on this cylinder of liquid are t •(!) The thrust pa. on the plane face- at P , acting vertically

downwards. (til The thrust (p+ty) a on the plane face at Q, acting veiti-

cally upwards. (in)1 The thru5t due to the surrounding liquid at each point of

thecurved surface acting horizontally. (iv) Weight of the liquid contained in the cylinder viz. « Bz.pg,

acting vertically downwards. Resolving the forces vertically, we get

pa—(p+8p).a-pa.Bz pg"=0 or 8p=pg.8£ Bp Lim Sp dp ,

or v r = P g o r ~^n«r = ?g or -f- = P£ ' Sz $z-+0 8z ° dz (Mogadh77)

' Integrating, p=g J p dz+A, ...(1) where A is the constant of integration.

^ Let p = / ' (z) I.e. a futction of z. Then only (i) can be integr­ated and from (1) we havep=gf(z)+A.

(b) When the liquid is homogeneous. (Lucknqw 83) Let II be the atmospheric pressure and in this case p is also

constant, so as above we have dpjdz^pg. ;

Integrating, p=gpz+A, s . ; -(jj) where A is constant of integration.'

Fluid Pressure 2i

On the free surface, z=.0, p - I I , so from (ii) we get 11=0+4 or -4=11.

Substituting this value of A in (ii), we have p=*H+gpz Jf we consider that there is no atmospheric pressure, then

11=0 and we get p=gpz. *§ 2 05. To show that the surface trf heavy liquid at rest is

horizontal. JR<mchi78) Take two points P and Q in the

liquid, lying in the same horizontal plane - From P and Q draw PA and * A _._. QB perpesdiculars to the surface. y^M^^-~f^

'.' P and Q are in the same c-:;r-;^ll^r=f€S:

horizontal plane. ^^fMb^fffz^ „*. Pressure at P=Presiure at Q g H g ^ ^ - ^ ^

or Il+gpAP~n+g9BQ, ^ p M ^ ^ : : ...See § 2-04 (b) above ^ ^ = ^ = ^ ^ ^ ^ r S j =

where p is the density of the liquid, — z ^ i ^ ^ ^ Y ^ ^ S ^ or AP~BQ.

Hence A and B are in the same (Fig. II) horizontal level since P and Q are in the same horizontal level.

In this way we can prove that all points on the surface of the liquid are in the same horizontal leve 1.

Hence the free surface of a liquid is horizontal. *§ 2 06. In a heavy fluid at rest, the densities at any two points

in the same horizontal level are equal. Let A and B be two points

lying in the same horizontal line : - :^ : - ; sy- -~^>"-" : " -" -~ : ; i within the fluid. Let P and Q ce grj|'=---~~'-'I">-~-'-££'J:.?ggI| two points at a short distance ig-;?;] F£ K vertically below A and B, such !£^^^_-_--^-----^- _i_—M@i~ that AP~BQ. --""^^=*=r^Sr:^=Vita

Let Pi and pa be the mean " _ J densities of the fluid between AP (Fig. 12) and BQ respectively.

Then pressure at P—pressure at A=g PxAP *..j>l) and pressure at Q—pressure at 5=»g.p2 BQ .. (2)

"But pressure at 4=pressure at B^p' (say) * and pressure at P--pressure at Q<=p (say)

X from (I) and (2) we gtt p -p' =g?1AP, p -p'<=*g. p2 BQ whence we get £Pi. />«=»g,.Pa.BQ or Pi"=p2. since AP=BQ. Hence the theorem.

**§ 2*07. The surface of separation of two h^avy homogeneous liquids, which da aot mix, is a horizontal plane.

{Jodhpur 76; Bfijasihan 781 Ranchi 75, 73)

% Hydrostatic^

(Fig. 13)

Suppose that the surface of separation PQ is a carved surface. Take two points A and B in the F same horizontal lin*3 but In two ^ different liquids.

But as A and B are in the same horizontal line,# so the z-JzrA densities and pressures at A and B must bathe same. But here the densities at A and B are different. Hence the surface of separation PQ cannot be curved one. Henoe PQ must be horizontal.

**§ 208. To Snd the pressure at any point in the lower of the two liquids which do not mix. {Sanchi 74)

Let 9x and pa be the densities of the two liquids. Let the depth of the upper liquid of density w aa^O Px be h. Take a point' P in the lower liquid of density P2 at a depth z below the surface of separation of the liquids. Let IL be the atmos­pherics pressure. Through F draw PQ a perpendicular

, line fo the free surface. With the vertical line PQ as axis consider a cylinder of liquid of small crosssection a. Let p be the pressure at P.

This xylinder of liquid is in equilibrium under the action of the following

' forces: (i) weight of the liquid within the cylinder*which is the sum

of the weights of the liquids b6twecn QR and RP i.e. the sum of g$xti7. and gw*i e. (gPifa-f£p2za) acting vertically downwards.

. (ii) the thrust pa. on the base at P, acting vertically upwards, (iii) the atmospheric pressure II« on the base at Q, acting

vertically downwards. • (iv) the pressure due to the sufrounding liquid at each point

on the curved surface of the cylinder, acting horizontally. Resolving these forces vertically, we have

P^^gPihai+gp^+lIat or p = l l 4-gfth +gp2z or p=»H+w^ + w2z, where wt and ww are the weights per unit volume of the upper and lower liquids respectively.

If we are to neglect the atmospheric pressure, thcap^mji j wtf*

v. (Fig. 14)

Fluid Pressure 23

Corollary. If there ben liquids of densities plp p2, paf-..., p„ and of depths hit h2, h3,..., hn respectivelly,. starting from the uppermost liquid, then the pressure at any point on the bottom of the lowest liquid is given by ,p«II+gp1h1+gpA+-..+gp r thn, where II is the atmospheric pressure

or p="II+w1h1-f-w2ha+...4-wnh,j, where. wu M>2,..., wn are the weights per unit volume of th? liquids. (Magadh 75, 77; Muzzajarpur 82)

§ 2 09. Characteristics of liquids. From the theorems given above we find that :

(1) Surface of heavy liquid at rest is horizontal. (2) Surface of separation of liquids which do not mix are

horizontal.. (3) Liquids find their own level. (4) If there be two liquids, which do not mix, then the lighter

will occupy the higher position and the heavier the lower one. § 2'10 Eflective surface. We know that in a liquid of den­

sity p, the pressurep at a point at a depth z below the free surface is given by p=H-\-gpz, where II is the atmospheric pressure. Now if we suppose that there is no atmosphere and a stratum of the same liquid (of density p) of thickness tII/gp)=A (say) is added above the original liquid, then the pressure at a point at a depth below the original surface is g? lh+z)=gp [(H/gp)+z]=»n+gpz, which is the same as it was before the removal of the atmosphere.

The upper surface of this superimposed liquid (of thickness k) is called the surface of zero pressure or the Effective surface. So we find that the pressure at any point of a heavy homogeneous fluid is proportional to the depth of the point below the effective surface.

The pressure at a depth h feet below the effective surface is sometimes called as pressure due to a head of h feet of the liquid.

Solved Examples on fiuid pressure under gravity. Ex. 1. The pressure in a water pipe at the base of a building

is 38 lbs. wl. per sq. inch and on the roof it is 19 lbs. wt. per sq. inch ; find the height of the roof (1 en. ft. of water weighs 62*5 lbs).

{Agra 73) Solution. We know (see § 2 03 Page 18) that the difference

between the pressures at two points in a heavy homogeneous liquid is equal to g?h, where p is the density of the liquid and ftds the vertical distance between the points.

p<=«39 lbs. wt. per sq. inch=>39x 144 lbs. wt. per sq. ft. and/>'~19x 144 lbs. wt. per sq. ft. where p and p' are pressures at the base and the roof respectively.

Hence here (39xl44)-(19xl44)~gp.A, f «(1) where A is the required height of the building and gp is the weight per cubic foot of the water,

»

24 Hydrostatics , i

We are given that weight of 1 cu. ft. of water>=62,5 lbs. wt.

& From (1), (39x 144)-(19x 144)=62-5"A or A = ^ ~ 4 4 .

or A=45-08 ft AM. Ex. 2. Pressure at the bottom of a well, filled completely with

water, is three times the pressure at a depth of 3 feet. If the height of the water barometer be 30 ft, find the depth of the well.

iVikram 78) Solution. Let the required depth of the well be z ft. and p be

the density of the water in the well. Then the pressure at a depth of 3 ft.=(30+3) gp (Note)

and the pressure at the bottom>=(30+zj gp According to the problem, we have

(30+z) pg=3 ($0+3) g? or 30fz=>99 or z=>69 ft. Ans. Ex 3 Find the depth in water at which the pressure per aq.

inch is 140 ibs. assuming atmospheric pressure to be 15 lbs. per sq. inch and the weight of one cubic foot of water is 1000 ozs.

Solution. Let z inches be the required depth in water, where (he pressure is 140 lbs. per sq inch

Then 149=15+£p z, where gp is the weight per cubic inch of the water.

Now weight of water per cubic inch 1000 . 1000 ,. '

oz- w l ' a i , v i v , i i v u lbs. wt. 12X12X12 UX12X12X16 A w e h a v e . 4 0 - 1 5 + - U x l 2 x U x i 6 z .

o r Z c a1 2 5 x 1 2 * * 2 x j * x . ! i ~ l 2 x 12x 12x2 lnches~288ft. Ans.

*Ex. 4. Two liquids A and B da not mix and have different densities When A is poured in a vessel to a vertical height h the pressure on the bottom is the same as when A stands to a height hx and B to a height h2 above it. Shaw that the ratio of the density of A to that^f B i» h2/(h - h^ •

Solution. Let px and pa be the densities of the liquids A and B respectively.

. A the pressure on the bottom, when A stands to a height h oll+gPjfe, where il is the atmospheric pressure.

And the pressure on the bottom when A stands to a height hi and fito a height hz =11 • gPA+gPa/ia.

V these two pressures are given to be equal, so we get U+gPih^lI+gpA+SPA or P,A=PA+PA

or P A = P I (A—hJ or Px/Pa / ftA—Ax). Hence proved. **Ex. 5 Equal volumes of two liquids of densities p and 3p,

which do not mix together just fill a cone which is held with its axis, vertical and the vertex uppermost Prove that the pressure at any point of the base is 3 - 3 \ /4 times the pres&ure at the same point when the cone is filled with the lighter liquid.

{Agra 83; Avadh 79; Rmchi 74)

ftuid Pressure 11

23

Solution. Let h bs the height and a the semi-vertical angle of the cone. Let the hsavier liqu'd be filled in the cone to a depth h-h', so that the lighter oscuoles the uDper portion of height h . Since the volutna of the liquids in tte two parts are equal, so the volume of the upper liquid is equal to half the volume of the whole cone I.e. ^n(h' tan a)8 xh'

4x^( f i t ana 2 x/z o r h'*=W or A-A/21'8 (1)

point filled

Now the pressure at any on the base, when the cone is with the lighter liquid only

<=gP.h=p (say) ...(2) And the pressure at any point

on the base, when the c;ne is filled with two liquids as given in the problem

*~g9h'+g 3P (h-h')^gp (3A -2h')

-/i tcmol

(Fig. 15)

=>g?h [3-4M3-^h43-^] =g?h [3—22'*] =g?h [3-(4)'"] = [3-(4)Vsj Xp> f r o m ( 2 ) . Hence proved. Ex. 6. A layer of mercury (sp gr 136) 25 cm deep is coie-

red by one of water of the same depth ; above this there«is a layer 50 cm in depth of oil (sp gr. 0 9) ; fiod the pressure at the bottom in gros. wt. per sq cm. assuming the atmospheric pressure to be due to a column of 7o cm. of mercury.

Solution. Let p bethe.densty of the water, then gp is the weight of one cubic centimeter of water and is equal to 1 gm. wt. Hence flie weight of 1 cu. cm. of mer­cury and oil will bs U 6 g9 and 0"9 gp which will b& tbe same as 136 gm. wt. or < 9 gm. wt res­pectively. Hence the required pressure at

any point on the bottom

MERCURY

OIL

WATER k.~^^E^=.

76 cm-

(Fig. 16) 11+50x0-9 gPi-l5Xg9 + 25 X136 g?.

where If is the atmospheric pressure and is due to a column ofmercury/«. 11=76x1 >-6 gp. % • Reqd. pressure=76xl3'6gp+50xO-9gp+25 gp4r25xl3-6gp

= 1443-6 gp=> 1443-6 gms, wt, Ans,

26 Hydrostatics

, Ex. 7. Prove that if a parallelogram be immersed in any manner in a heavy homogeneous liquid, the sum of the pressure' at the extremities of one diagonal is equal to the SUM of the pres,ures at

(Magadh 77 ; Ranchi 78) the extremities of the ojher diagonal. Solution. Let ABCD be the parallelogram immersed in the

liquid of density p. Let the ' depths of the vertices A, B, C and p below the free" surface be zlt z2, z3 and z4.

Then the depth of the middle point of the diag­onal AC-=>\ (zt+za) and that of the m'ddJe point of BD is I (z2+Zi).

V diagonals of a * parallelogram bisect each other

Z. these mid. pts. of the two diagonals are the' same point '•«., i f o + ' a W (z2-K> or z1+z3^z2+zi .A

Also the sum of the pressures at A and C=gPz1+gPZ3

and the sum of the pressures at B and D-gpzJIfJ^Lu]) ^ . by the virtue of (i), these two sums of pressures are equal >

Ex. 8. Prove that the pressure at the centre of mtfalt t triangular lamina wholly immersed in a homogeneous li?„?d i„ anv manner, is one-third of the pressure a t the angular points. y

„ , . . r t ' Wgra74;Ra]astIian76) Solution. Let ABC be the

triangle immersed in a liquid of density P.' Let the depth of the vertices A, B and C below She free surface be zlt z2 and z. tively. Then the depth of C.G. of the AABC will be

J (zi+Za+zJ Z. the pressure at G •=gPl(Zi+za+za) ^[g?Zl + gPZ2^gPZ3]

r»spec« G.Jhe

(Fig. 18) * [Pressure at .4-{-pressure at 5+pressure at CJ

- i [sum of the pressures at ^vert ices] . Hence proved

BeSS, and the d L f y T f t t ^ I X T o t l T ^ t l T l M*-

uin stratum. dS<trbwl78;MmafarpurM)

fruid Pressure 27

Solution. Let the thick­ness of each stratum be h. >

Then the pressure at th? J f P \ v lowest point of the nth stratum

•=gph&g.2p./i+g3ph + ...+g.nPh

-ffl* [ l + 2 + 3 + ~ . + » ] ,y

*=>gph ti«(H+l)], "| " • ~ ft W»1K T 27B-*B (n+1)

—iffP*C«+J> A. Ans. * (Fig. 19) **Ex. 10. If a liquid he heterogeneous and of density pi/a at a

depth z, show that the pressure is II-KgPz2/2a),

M A\

%\

tf

•

•

•

p PP 3P

ft W ilP

atmospheric pressure. Solution. Let A be a point at a

depth z below the free surface.. From A draw ^40 perpenduular to the free surface. On OA take two points P and Q at depths x and (# i £A) below o.

With 0,4 as exis, construct a small cylinder of liquid whose area of cross seetion is a (say).

Then the weight of a small ele­ment of this cylinder with PQ as axis=(x.SxJ.(pA/o)j?. since (Px/a) is the density of the liquid at a depth x.

.'. The weight of fthe liquid within this cylinder

P *

{Lucknow 79 where II is the Muzzafarpur 78)

O

M

JSX j

/>*

a rfx. •1 a Jo n Jo 2A

(Fig. 20)

^ z 2

Jo 2fl „.([), This cylinder of liquid is in equilibrium under the action of

the following forces: (i) The atmospheric pressure II a on the upper face acting

vertically downwards. (ii) The pressure p.v. on the lower face acting vertically up­

wards, where p is the pressure at A, (iii) The weight of the liquid within th& cylinder viz. (apgz2/2a)

acting vertically downwards, and (iv) The pressure of the surrounding fluid at each point of

(he curved surface of the cylinder acting horizontally. Resolving these forces vertically, we get U.a.+(a.pgz>l2o) -pa^O or f«=H+P (gsa/«) Hence proved.

28 <A t I . J

Hydrostatics

Fx. 11. If p, p' be, the densities of fluids fp<p') and the lengths of the arms of a U-fube in which they meet be m and a inches respec-'tively, prove fiat in order that the tube may be completely filled the he'ght of the column of the lighter fluid above the horizontat plane in which they meet must be [p'(m nV(p'-p)] inches. (Magadh74)

Solution Given that-, AB=m" and CD=n": Lej£Fbe the plane df_ separ­ation of the two fljids. AE is the column of lighter fluid-of density p and we are to find length ol AE [=-z (say)]. Let R be the point in the column CD which is in level with EF. Since E and R are in the same hori- (Fig. 21) zontal level so the pressure at i?=>pressure at R. t e. - gP.AE~g'(> CR or g?.z=g?' (CD -RD)=g?' (n-EB) or * pz=p'[B-(/W—z)]=p'(n-m)f p'z

' o r p' (m -n)*=z (?'-p) or z<=(m-n) P ' / (P ' -P ) . Hence proved Ex. 12. (a) In a/U-tube of uniform cross-section there is some

mercury ; water is then nonred into one limb and it occupies a length 9 inches. Ifthesp. gr. of mercury is taken to be 13 5. find the dis­tance through which the mercury level in the other limb is raised.

, (Garhwal 81) Solution Let ini­

tially the height of mer­cury in each tube bey inches. Then water added to the arm AB of the tube which ( causes the level of mercury to sink through- x" in the arm^B Shd the level of mercury in tbe arm EK would rise. Also as the cross-section of the tube is uniform, so the level of mercury in the arm EK would also rise*'. (Fig. 22) Hence in the position of equilibrium, In {he arm AB the water

BL

c

0

1

3 i

Fluid Pressure 29

occupies a length AD*=9' and mercury occupied a length DB*>(y—x) Inches and in (he arm .EX"the mercury occupies (x-\-y) inches.

Also B and K being in the same horizontal level, the pressure at 5=pressure at K

or n+g?.9igxl3-5p(y-x)~Il+gxU-5?(y+x), where p is the density of water and II is tie atmospheric pressure.

or 9+13 5 (y-x)^l3'5 (y+x) or 9=27* or *=&•=»£ inch. Ans. **Es. 13. The lower ends of (wo vertical tubes whose cross-

sections are 2 and 0"2 sq. inches respectively are connected by a tube. The tubes cm tain mercury (sp gr. 13 6). How much wafer must be poured in (he larger tube to raise the lei el of the mercury in the smaller tube by 2 inches ?

Solution. Initially let EF and KP be the levels of mercury in the two limbs (EF and KP are in the same horizontal level), such that EB*=PD=>y inches (say).

Let z inches of water be poured in the tube AB, so that level of mercury in this tube is pressed to QR, such that EQ=x" and the mercury so pushed from the tube AB enters in the tube CD whi:h causes-the level of mercury in the tube CD to rise through PC<=>2" (given"*. (Fig. 2i)

Since the volume of mercury pushed from tube AB is equal to the volume of mercury entered in tube CD, therefore,

2xEQ=02xCP or 2xx= .02x2 or x^O'2 in. Hence finally we have z inches.of water and (y—x) inches or

(y-0'2) inches of mercury in the tube AB and (y+2) inches of mercury in the tube CD.

Also B and D being in the same horizontal level, pressure at 5=pressure at D

or Il+gpz+gX 13'6P 0>-0-2)-II+#X 13*6p (y+2), where p is the density of the water and II is the atmospherffpressure ' ot z f l 3 6 ( j - 0 2)=13 6(y+2) or z=13-6 [2+C-2]=B'6x2-2=29-92 inches. Ans,

**Ex. 14. A U-tube of uniform section, open at both ends, contains 27 inches of water and such a quantity of mercury that the water is entirely in one of the straight portions of the tube. The straight portions are connected by a fine horizontal tube which passes

30 Hydrostatics

A

•8

p

p

t i

R

i

-——

Q

c*

IT

D*

- * > •

(Fig. 24) 'H+py (13-5Xl*5).

between them half an iuch above the surface of separation of the water and mercury, the passage through which is originally closed. If this passage is now opened ; find in what direction flow takes place through it, and the volume of Ihe liquid which passes before equilibrium is again established, (sp. gr. of mercury being 13"5). >

Solution. Connecting tube is very fine and hence its capacity is negligible. Before the passage is opened ;

Let the height of mercury in the tube AB be x inches. •

V The pressure at 5 = the pressure at D.

L IH-gXl3-5P* = I I + g P.27, where p is the density of the water and II is the atmospheric pressure x,"

27 or x = ^-_=2 inches.

Also the pressure at Q ~lI+gp(27-l)=IH-£P(26£)

and the pressure at P=II+£X 13*5P (x Hence it is apparent that

the pressifre at Q is greater • than the pressure at P.

After the passage is opened :

As the new passage is opened, the water will flow into the other arm of the tube through the fine tube PQ towards P from Q. And the pressure of the liquid will increase in the arra AB, conse­quently the mercury will fbw through the tube BD into the other arm CD. nThe level of mercusy in the "watertube CD" will be at least half an inch beCause at P and Q (Fig. 25) (which are the points in the same horizontal line) the densities will be the same. Let z" be the height of water which has entered in the other tube AB.

Now the tube AB contains z inches of water and (2—f) i.e. If inches of mercury vvhereas the tube CD contains (27—z) inches of water and \ inch of mercury.

i K

E

R

P\

M

F T

Q

(27- ft

T;

1 D

Flul3 Pressor* 31

Now the pressure at Polity gPz+gX 13«5PX ( l j - i ) (Note) and pressure at 0=*II+£p (27-z) + gx 13'5P. (J-JV

The two pressures being equal, we get Il+gP z+gxl3'59X1^11+gp(27-~z)

OT z+13M5-=>27~z , or 2z=>27—13'5 or z=$ (13'5>= 6*75 Inches. Ans.

**Ex. 15. In the lower half of a uniform circular tubs, one quadrant is occupied by a liquid of density 2p, and the other by two liquids of densities p and 3p- Prove that the volume of the lower of tbe two latter liquids is twice that of the other.

{Lucknow 82, 80 ; Magadh 76 / Rajasthan 78, 75 ; Ranchi 74)

Solution. The quadrant BD is occupied by the liquids of densities p and .IP and the quadrant AB by the liquid of density J P.

Let CF be the surface of separation of the liquids of densities p and 3P. Let the radius OC be inclined at an angle 0 to the horizontal A diameter AD. Let a hi the radius of the circular tube Then OE=a sin 0 and

EB^OB- OE=a—a sin 0 Now pressure at B due to

the liquid in AB ' ~ g 2P.OB~2gPa.

And pressure at B due to (Fig. 26) ' the liquids in DB

=g?.OE } g.3?.EB=g?.a sin 0 j.g.3?.(a~a sin 0). For equilibrium, these two pressures must be equal, hence we -

have 2g?a=g9a sin 9+°-.3p (a -a sin 0) or 2=sin 0 f-3 (1— sin 0) or 2sin0«=<l or sin 0=4 or 0=»3O°, i.e. f_ DOC=:0°

L. CaB=>9Oo-0=»9O°-3G°=>6O°«=2Z_ DOC :. Arc BC=2 arc CD.

.'. the volume of the liquid in arc BC =•2 (volume of the liquid in arc CD). Hence proved.

**Ex. 16. -A small uniform tube is bent into the form of a circle whose plane is vertical; ,equal volumes of two fluifls whose densities are pB cr fill half the tube. Show that the radius passing through

3 2 \ Hydrostatics 136/2

the common scrface Inakes with the vertical an angle 6, given by tan fl«[(p-a)/(P+q)J.

(Bhopal 83, 80; Jodhpur 78; Lucknow 75 i Mogadh 75 ; - Ranchi 81 ; Rohilkhand 81, 79 ; Vikrom 83, 78)

isolation. BK is the sur-face of separation of the liquids, such that OB makes an angle 9 with the vejticaJ. , ^ C is <he lowest point of the Jw^TTa 1F

tube and so n-ust be occupied "^ ^ by ' the heavier liquid * of ^ . \ i n density p Let a be the radius f J \ i v of the tube. ^ <r

Then O.E=asin fl= OF, I l k / ^ - ^ N ^ FC=OC-OF-=a-a sin 6, ^Wh§/ N'f

ON<=a cos 6 and r^^^-—\, NC~OC- ON* a-a cos 6. " ^ ^ ,

Now pressure at C due • , to liquids in tube ABC (Fig. < 7)

=go EN-i gP NC =g<r (EO+ON)+g? NC

i =•#<* (a sin 6+a cos 0)+gP ( a - a cos 5). v And pressure at C due to liquid in tube DC

"=g? FC=gP (a—a sin 6). For equilibrium these two pressures must be equal, hence we

have go (a sin 6+a cos 0)+g? (a—a cos 0)=gp (a—a sin 6) or o (sin 0-j-cos 0)-f p (1—cos 0)=P (l=sin 0) or (o-f P) sin 0^(9-a) cos 0 or tan B=W—o)l(?+o-)]

Hence proved. , From here we can find that

s j n a <»- g) ' cn<! B=a ' <P+°) '

sin 2fl=>2 sin 6 cos 0 « ^P~g ) (P4-Q)

V[(?-°)*+(P+O)'Y V[(P-^ a+(P+g)23 2 ( P 2 - «ra)

""(P-o^Hft+ff)*" (Mogadh 75) **Ex. 17. In a uniform circular tube two liquids are placed so

as to subtend 90° each at the centre. If the diameter joining the two free surfaces be inclined at 60° to the vertical, prove that the densities of the two liquids are as v / 3 + 1 '< V3 - 1 .

\Bhopal 73 ; Lucknow 81,77 ; Mlthila 81 ; Rajasthan 74 ; Ranchi 79 ; Vikram 83,78,77, 75)

136/3 Fluid Pressure M

Solution. AD is the diameter joining the two free surfaces. Let the densities of two liquids be a± and os, where °a > °i Then B, the lowest point of the tube, must be occu­pied by the heavier liquid of density cr2. Let O be the centre and a the radius of the tube. Then, OE=a c o s ( 0 ° = | a = l OF j OK=-a cos 30°=lfl\/3; FB=OB-OF<=>a-la=\a ', KB=OB~OK=a-a ( ^ 3 )

- J a ( 2 - t f 3 ) . Now pressure at B due to liquid in tube AB

j ~g<72 FB^gOv (\o). •• And pressure at B due to liquid in tube BCD

~gorx lEK)+ga% KB ~gox(EO + OK)+g<*zKB *=g°i ty+WSj+g*, la ( 2 - ^ 3 ) . For equilibrium these two pressures must be equal, hence we

ga2 \a=go1 \a (l+V3)+g<r2 \a ( 2 - ^ 3 ) •

(Fig. 28)

«V-»it t+tf3)+ffB(2-V3)

< T 2 : ^ - ( V 3 + 1 ) : ( V 3 - 1 ) . Hence proved.

have or or or

Ex. 18. A circular tube of fine uniform bare is half filled with equal volumes of four liquids which do not mix and whose densities are 1 : 4 t 8 : 7 and is held with its plane vertical. Show that the diameter joining the free surfaces makes aa angle tan -1 2 with the vertical.

'Solution. Since the volume " \ of each liquid/ls the same, so they subtend equal angles at the centre.

Hence L_ AOB=-L BOD ~LDOE=LEOF~45°. Let the densities of the

liquids be p, 3P, 8p and 7P res> pectively. The lowest point C of tue tube will be occupied by the heaviest liquid of density 8p.

Let a be the radius of the circular tube and AF, the dia­meter joining the free sui faces, be inclined at an angle 0 to the vertical. (Fig. 29)

34 Hydrostatics

Then l_FOK=t± AOC=B; /_BOC= 0-45°; LCOD=>£.BOD- ABOp^W-B.

Hence OK^a cos 0 ; OL=a cos 0 OM=acos L^EOC^-a cos (/_EOD+l_DOC)

= a cos (45°+9O°-0)=.-fl sin (45°-0) ; (9N=a cos Z_BOC=-a cos (0—45°) OP=o cos f_DOC=a cos (90°—0)=a sin 9.

Pressure at C due to the liquids in tube ABO =>g.7?LN+g 8p.2vC=g.7p.(0iV- OZ)+g.8P (OC-ON) -g.7p.a {cos (0-45°)-cos 0}+g.8P {a-o cos (0-45°)} And pressure atJC due to the liquids in the tube FEDC ~£P KM+g 4P.MP- g.W.PC =gP (OJST+0M)+g.4p (0P-0M)+g.8p (OC-OP) **g?a {cos 0-sin (45°-0)}+£.4P.« {sin 0+sin (45°-0)}

+ g 8p.fl (1-sin 0). For equilibrium these two pressures must be equal, hence we

have gJp.a {cos (0-45°)-cos 0}+g.8pa {1-cos (0-45°)} «=gPa {cos 0—sin (45°—0)} * +g 4p.fl {sin 0+sin (45°- 8)}+g 8P.e (1-sin 0)

or 7 {cos 0 cos 45°+sin 0 sin 45°—cos 0} + 8 {1-cos 0 cos 45°-sin 0 sin 45°}

•={cos 0— sin 45° cos 0 - cos 45° sin 0) +4 {sin 0+sin 45° cos 0—cos 45° sin 0}+8 (1—sin 0)

or (7/^2) {cos 0+ sin 0 - ^2 cos 0}+(8/\/2) {</2- cos 8 - sin 0) ~ ( W 2 ) {\/2 cos 0-cos 0+sin 0} +(4/V2) {^2 sin 0+cos 0-sin 0}+8 (1-sin 0)

or 7 cos 0+7 sin 0—ly/2 cos 0 - 8 cos 0—8 sin 0 = ^ 2 cos 0—cos 0+sin 0 + 4 ^ 2 sin 0+4 cos 0

- 4 sin 0+8\/2—8\/2 sin 0 -or ( 7 - 7 ^ 2 - 8 - ^ 2 + 1 - 4 ) cos 0

=(1 + 4 ^ 2 - 4 - 8 ^ 2 + 8 - 7 ) sin 0 or ( - 8 \ / 2 - 4 ) c o s 0 = ( - 2 - 4 V 2 ) s i n 0 or 2 cos 0=sin 0 or * tan 0=2 or 0=tan_ 12. Hence proved.

**Ex. 19' A fine circular tube in the vertical plane contains a column of liquid of density 8, which subtends a right angle at the centre, and a column of density S' subtending an angle a. Prove that the radius through, the common surface makes with the vertical an angle tan"1 [ (g- §'+S' cos a)/(S+ S' sin a)]

{4gra 74 ; Bhopal 82 ; Jodhfus 761

Fluid Pressure 3S

Sol. Let OC, the radius through the common surface make an angle 0 with the verti­cal. Let the portion AC of the tube be occupied by the heavier liquid of density 8 and the portion CD by the lighter liquid of density 8'. Let B be the lowest point ol the tube.

Let a be the radius of the circular tube. Then OK=a cos 0, 0^=0 008(90°—6)<=>a sin 0 j OE=>a cos (oc 4 0).

Pressure at B due to liquid in the tube AB

*=>g8FB*g l.(OB-OF)**g 8. (a—a sin B) And pressure at B due to liquids In the tube DCB

» g 8' EK+g S KB~g $'.(OK-OE)4>g S (OB-OK) *=>g 8' a {cos 0-cos (x+B)}+g $.a (1—cos 0).

For equilibrium these two pressures must be equal, hence we g 8 a (1-sin 6)=>g8' a {cos B- cos (x+B)}+g 8 a (1—cos 6)

—8 sin 0=8' {cos 0-cos a cos 0-f sin « sin 6}— 8 c<Js 0 ( - 8 , - 8 ' sin a.) sin 0~(8'—S' cos a-S) cos 6

(Fig. 30)

have or or

or tan0= g—S'-t-8' cos a 8-j-S' sin a or 0=tan' i! -S'+Scos« } 6+8'sin a

Hence proved. *Ex. 20. A fine circular glass tube of uniform bore is comple*

tely filled with equal volumes of thrre fluids which do not mix and whose densities are p, 2p and 3P. Proi e that when the tube is held in a vertical plane, the surface of separation of the liquids of densities p and tfp is an extre­mity of the horizontal diameter of circle.

Sol.V Let the liquids of densities p, 2P and ?p occupy the portions AD, DC and CA of the tube respectively. From D, C and A draw DK, CN and AM perpendiculars to the vertical diameter EB Let AO. the radius through the surf >ce of separation of the liquids of densities p and 3p, make an angle 0 with the horizontal diameter PQ. (Fig. 31)

36 Hydrostatics 184-66

(6Oo-0)~6O°+0

Then £. AOB=9Q°TB 'J L COB^UO0—(9O°-0)=3O°+0,

Z. COe=90o-(30°+fl)=60°-fl ; /_ QOD=>L COD-A CO0=>12O° £. D0£=9Uo-(6Oo+0)= ,3O°-0 ; & Pressure at the lowest point B, due to the liquids in the

tube EPB •=g? ffM+g 3P MB=>gP (EO+OMHg 3p.(OB-OM) «=> g p {a+a cos (90°- 0)}+g.?? {a - a cos (90°- 0)}, where a is

the radius of the tube *=>gpa ( l+sm0)+£.3pa( l - s in0) And the pressure at B due to the liquids in the tube EQB «=*.P EK+g 2p KN+g4?.NB =>g.? (OE-OK)+g2p (OK+ON)+g.3p (OB-ON) «=g.P {a-a cos C30°—0)}-f£.7p {a cos (30°-0) <

-fa cos OO°+0)H# 3p {a -a cos (3O°+0)} •=gP.a {1-cos 30° cos 0—sin 30° sin 0}

+g 2p a {2 cos 30° cos 6}+g.3P a {1-cos (3O°+0)} These two pressures are equal for equilibrium, hence we have

gp a (1 + sin 0)+£.3p.3 (1-sin 0) «=g.p.fl {1—cos H0° cos 0~sin 30° sin 6}+g.2P a<tf3 cos 0

-f g 3p.s {1 - cos 30° cos 0+sin 30° sin 0} or sin 0 - 3 sin 0

•=— W$ cos 0- -J sinx0+2V3 cos 0—1^3 cos 0+§ sin ft or - 3 sin 0=—2^3 cos 0 + 2 ^ 3 cos 0=0 or sin 0=0 or 0<=O, which shows that the surface of separation of the liquids of densities p and 3P is an extiemity of the horizontal diameter of the circle.

**Es. 21. A circular tobe wilh centre O is filled with three fluids of densities Pu p3., p3 (in descending order of magnitude) and placed in a vertical plane. If 2a, 2p» 17 be the angles subten­ded at tbe centre by the fluids and P be the point on tbe circumference midway between tbe ends of tbe lightest fluid ; then tbe angle 0 which OP makes with the vertical is given by Pa~ $i„ sin a sin (6 - 0) p i ~ Pa sin(s+0)" sin p

Sol. We are given that P1E>p2>p3 and 2a, 2(1, 2y are the angles subtended by the fluids.

Let a be the radius of the circular tube.i The portions (Fig. 32)

Mlaid Pressure 37

BD, DA and AB of the tube are filled with fluids of densities p^ pa and pg respectively. Let C be the lowest point of the tube and different angles are shown in the figure.

Pressure at C, due to the fluids in the tube EABC ~g P3 EK+g?a KF+g9l.FC ~g p3 (OE-OK)+g92 (OK+OF)+g%?1 (OC-OF) *=-g P3 {

fl~a °os (y~0)}+g P2 [a cos (y-0) +a cos {2«-(rc—6—Y)}] +gPx [a—a cos {2a—(n-6—y)}]

And pressure at C, due to the fluids in the tube EPDC <=SP3 Eff+gPi NC<=gps {OE±ON)±g Pl (OC-ON) =gpi ( a + a c o s («—0-y)}+£Pi {«-« cos (TC-fl-7/}. For equilibrium the two pressures are equal,, hence we have gPi {a—a cos (7-0)}+gP2 {a cos (y-6)+a cos (TT—2«-0-y)}

+£Pi {«—a cos (w— 2a—0 -7)}<=g? {a -a cos (5—y)} -f «Pi {a+a cos (fl+y)}

P3 {-cos ty -e)} + p2 {cos (y-0) -cos (2a+0+y)} •+ P2 {cos (2a+0+y)}=.p3 {-cos (0+y)}+p, {C0S (0+y)} _

(Pi-Pn) cos (y-»)+(Pi-P2) cos (2a-f S+y) +<P3—Pi) cos (0-f-y>=»O

<Pa-P3) cos (y -ff)+(Pi-P» + P» - Pa' cos (2a+0+y) +(P3 -Pi) cos (P+y) =.0 (Note)

(P2-Pa> [cos (y -0)-cos <2a+0+y)] =(Pi-P2) [cos (0+y)-cos (2a+0+y)]

(Pi-Pa) [2 sin (a +y) sin (a+0)] - C I - P O [2 sin (a+0+y) sin «]

Pa - P3^ sin (q+fl+y) sin a „sin {0+frr-P)} sin a Pi-Pa sin (a+y) sin (a+0) sin (re - p) sin (a+0)

V 2«+2p+2y=.2rc or a + y = m - P •=[sin (p — (9) sin a]/[sln P sin (a+0)]. Hence proved.

**Ex. 22. Three fluids whose densities are in A.P. fill a semi­circular tube whose bounding diameter is horizontal. Prove that the depth of one of the common surfaces is double that of the other.

' (Lucknow 83, 76 ; Mithil a 82) Sol. Let the densities

of the fluids be p -o, p and p+o respectively Let C be the lowest point of the tube. Then C will be occupied by the heaviest fluid density (9+0). The portions AB, BCD and DE of the tubs are occupied by the fluids of densities p -o , p-f<r and p respectively. (Fig 33) /

Let OF=x and OK=y and the radius of t"he circular tube be <z. TheaX?(?=a, FC**OC- OF^a - x and KC<=> OC- OK<=a -s\

or

or

or

or

or

or

ft O

38 Hydrostatics

Pressure at C„due to the fluids in the tube ABC H?(P-») <>F+g(.9+o) FC~g (p-o) x$g (p+o) (p-x)

And pressure at C, due to the fluids in the tube EDC

~gp OK±g (p+o) KC=>gp y+g (p+o) (<»-j>) For equilibrium these two pressures are equal and hence, we

have g (p-o) x+g (p+o) (a -*)=gj>p+g (P + o) (fl-;p) or (p— o-'p—o) *=»(p—p—o)j or l°x=-oy or y=>2x

{6. OK=20F. Hence depth of cne common surface is double that of the other. •

*Ex, 23, A closed tube in the form of an equilateral triangle Contains equal volumes of three liquids which do not mix and is placed with its lowest side horizontal Prove that if the densities of the liquids are in A. P., their surface of separation will be at the points of trisection of the sides of the triangle. (Rojastkan 77)

Solution. ABC is the given closed tube. Let a be the length of each side of the tube. P, Q ^and R are the points on the surfaces of separation of the liqu ds. Let the densities of the liquids be (p-o), p and (p + o). The portions PAQ, QCR and RBP be filled with liquids ot densities P—o, p and p+o respectively.

Let AQ^x^CR^BP, then AP~BR=>CQ=a-x.

Pressure at B, due to the liquids in the tube APB

=g (P—<0 AEJrg{H o).EF ~g (P~o) AP cos 30°+g (P+o) BP sin 60° "g (P~CT) («-*) 5V3+g(P+o) x.WT

And pressure at C, due to the liquids in the tube AQC *=>g (.?-") AQ'cos30°+g.<? gCsin60° * -«• (P-o) x iV3+^.P (h-x) iV3 .

Since B and C are in the same horizontal line, so the pressure at 5=pjessure at C

or g (P-o) (a-x) W3+g (P+ ») x . | ^ 3 = g (P~ <T) X J V3 - rg .P(a-*) i t f3

or (P-o) ia~x)+(? +o) x =(p_o) *+p (fl_^) or 3o.x'=a.fl or x = Ja Hence proved.

**Ex. 24. A fine gla-js tube in the shape of an e qui lateral triangle is filled with equal volumes of three liqnids which <to not mix, whose densities arc in A. P. Tlie Jube is held in a vertical plane and the side that contains portions of the heaviest a«d lightest

Fluid Pressure 39

>CP-(Fig. 35)

»a—».

then and

liquids makes an angle 6 with the vertical. Show that the surfaces of separation divide the sides in the ratio cos (J—0): cos (%n+0).

Solution. Let ABC be tri­angular tube and let its each side be a in length. Let the densities of the fluid be P - a . p, P+a. Let A be the lowest point and B {he highest point of the tube, then A will be occupied by the heaviest liquid of density 9±a and B by the lightest liquid of density P - a . The portions RBP, PCQ and QAR are occupied by the liquids of densities P - a , P and P+a respec­tively. Draw RN, BD, PP, CM and QK perpendiculars from R, B, P, C and Q respectively on the vertical line DA through A.

Let BP~x~CQ~AR, then AQ=BR Given l_DAB=B, i_DAC=\n-6t V each angle of the triangle ABC Is in

LCDA=-U+0\ AK=AQ cos (£rc-0)~(a-x) Cos (in-0) MK^CQ cos (in— B)=x cos (in—6) PM~CP cos (in+OWa-x) cos (Ja+0) DF=BP cos (in+0)=x cos (JTC+0) AN=AR cos 0=*x cos 0; DN=BR cos 0=(a—x) cos 0

Equating the pressures at A, we have g (?+o).x cos 0 +g (P-a) ( a -x ) cos 0<=g (p-o) x cos ftff+0)

-f gP (a—as) cos (&7r-f 0)+gP.» cos (ire-0) +8 (?+") ( a -* ) cos Qre-0)

x [(p+o) cos 0 - (p -a ) cos ( £ K - H ) - P C O S QJC-0) ] =*(a-x) [p cos (Jrc^0)-f (P+a) cos (Jre—0)—(p-a) cos 0]

X [p {cos 0,-cos (i7r-(-0)-cos (£re-0# + a {cos 0-fcos CJje+^)}l = ( a - x ) [P {cos (£n+0)+cos (in -0) -cos 0}

-fa {cos ^Jt—0)-f cos 0}J x [p {cos 0 2 cos in cos 0} + a {2 cos (i?t-f0) cos in}]

=>{a-x) [P {2 cos in cos 0-cos 0}^a{2 cos \n cos (Jre—0)J x [p {cos 0-2.1 cos 0} + a {2 cos (£n+ 0) cos JTE}]

= ( a - x) [P (2. J cos 0 - cos 0}-$ P {2 cos fw cos (\n-4>)\ X [P (0)+a.2.COS (\n+6) COS Jn]

= < a - x ) [p.(0)+a.2 cos \n cos (Jw—0)1 x ^ 2 cos (fo - 6) cos JTT^COS (ftre—0)

a—x = 2 cos (J7t + 0,j cos fra cos (Jrc+0) x : ( a - x ) = c o s (Jw-0) : cos (Jre+0). , Henc6 proved.

i*fix. 25. A cycloids! uniform tube contains equal weights of

or

or

or

or

or

or

01

kb , llydrosta'tjci

two liquids occupying lengths a and b ; if it be placedjvith its axis vertical, prove that height of the surfaces of ileitis above the vertex of the tube are as ?3a+b)2 to (3b+a)2.

Solution. Let the densities of thejwo liquids be pi and p2 andvlet these occupy the portions ABC and CD of the tube res­pectively. • Given that ABC=>a and CD=b. (Fig

"7 the weights of liquids are equal ; •\ flPig=»*Paff or Pi/p2=6/fl. ...(0 Let the vertical heights of A, C and D above the vertex B be

3Waand;P3. Then pressure at B due to the liquid In the tube AB—gpx yx . And the pressure at B due to the liquid in the tube BD

Equaling these two pressures, we have SPi5'i=£PiV2+£p2 Oa ~ J'z)-

Pi (^i-j2l = p2 O'a -J's) Or PJ/PJ,- 0'a - JaVO'i- Tz)- - (2) Also if a be the radius of the generating circle of the cycloid,

8aj', where s is the arcural distance from the vertex of a point, whose height above the vertex is y.

Let arc BC=s, then we have arc AB^a—s and arc BD" Hence we have (a -s)2=>kaylt

s*=%oy2 (b+s)*~Zay3

Subtracting (4) frcm (3) we get aa - 2as<=>%a (y1—iy2) -Subtracting (4) from (5) vie get b2 f 2fo=8fl 0>a—JV2)

or

we know s?

and

>b+s ...(3) ...(4) ...(5) ...(6) ...(7)

2as yj - j», p2

or or

Dividing (6) by (7) we get g—^^fz:

b (a2-2a.?)

a I

6 --a{b*+1bs) or 4fl^=a6(fl

Now from (3) and (5) we have

...from (I) and (2)

(3a+6)2 "*)J

...(8)

...from (8)

(3 b +- o)a" • Heace pro ved. ^""Ex 26 A Cube in the from of a parabola held with its ver­

tex downwards and axis vertical, is fiiled with two different liquids of densities 5 aad S'. If the distances of the free surface of the liquids from the focus be r and r' respectively, show that the distance of their common surface ftom the focus is (rS r'§')/(*- 8').

- i {Lucknow 75)

• Fluid Pressure k\

'-b

Solution. Y Is the ver­tex, S js the focus and FD is the directrix of the parabola. Portion AYB and BC of the tube ate occupied by the liquids of densities 8 and 8' respectively.

SA=>r and SC=r' (given) ' Let the required focal

distance of the common sur­face viz- SB be R.

Also we know that for a parabola, the focal distance of any point on it=itsdis- (Fig. 37) tance from the directrix.

Hence AS**AE=r ; BS^BE ~R and CS^CD^r' Let VZ=.b, then AK^AF-b^r-b ; BL~R-b;CM=

and CN=CM-CL=(r'~ b)-~ (R-b)=r'- R Now pressure at V, due to liquid in the tube AV

~g.§AK~g.§.(r-b) And pressure at V, due to liquids in the tube VC

=g 8 6L+gB' CN=g Z.iR-b)+g§'Xr'-R) For equilibrium, these two pressures are equal, and we have

g8(r b)*=g%.{R~b)+gr(r'-R) or R(B'-B)=B'r'~Sr or RMrS-r'8')/(8-B').

••Ex. 27. A closed tube in the form of an ellipse with its* major axis vertical is filled with three different liquids of densities Pi, P2> P3 respectively. If the distances of the surfaces of separation from either focus be rlt ra, ra respectively, prove that

*i (P» - P8) -i r2 (P3 -Pi)+ra (Pi-pa)=>0. Solution. S is ihs

focus and KM the directrix of the ellipse. &4=ra, SC=r3 and SD^^

Let die portions AC CD and DA be occupied by I-quids of densities P^ Pa and p3 respectively.

Also wc know that for an ellipse, the focal distance of any point on St=e times distance from the direc rix, where e is the eccentricity.

Hence AS=r.AK or AK^!le)AS^rJe Similarly CM- rs/e and DL^tiie (Fig. 38)

4£ Hydrostatics.

Let BZ=>b, then AP=AR-b~ir2le)—b. DQ={rJe)-b and CE=*(rJe)-b.

Let BE=2a, then EN=EB -AP=>2a-(r3le) + b and EF=EB-DQ=>2a-(rxle)+b,

FH=DQ -CR~(rxle)-(r3le)=>0le) (ri-ra) .'. The pressure at the lowest point B, due t o the liquids in

the tube EAB * a -gPg £tff gP, ^P=£P3 {2a-(r /e)*ft} + £Pi {(^/e)-^}

And the pressure at B, due to the liquids in the tube ECB ^ ~g?aEF+gp2FH+g?xCR

=gP3 {2fl-(ri/e)+*}+£-P2 (1/e) (ri-r,)+CT! {(r3/e)-*} These two pressures being equal, we have gp3 (2fl - r,/e+W + gPi {rJe-b)~gP3 (la-rje+b)

+^P.(l/«)(ri-r1)+*P l(rJc-6) P3 (-r , 'e) + Pi (ra/e)=P3 (-ii/e)+P, (l/e) (rj-ra)+P» (r,/e)

—rjPa+r.Pi—rA + r lP2-r3P2+ r3Px

>i (P«-P«)+',« (p3-pi)+r8 (Pi-P2)«=0. Hence proved p **£x. 28. A fine tube bent in the form of an ellipse is hele

with its plane vertical and is filled with n liquids whose densities art Pi. Pa, p3. •••> P" t a k e Q i n , o r d e r r o u n d t h e e I 1 ' P t i c tuDe- I f rl> rif •••» r» b e

the distances of the paints of separation from either focus, prove tba ri (Pi-Pa)+ra(Pi-Ps) * - + r » (P.-P,)=0.

Solution. S is the

or or or

focus and MN is the i directrix of ellipse. Let the portions AXAS, A&A3„ A3A„ .. bs occupied by liquids of densities Pa,

M_ M, <W> hh

etc. Then SAX*

SA. 'a»

f i . SAa*= rn From' Alt A2,A3, ..

draw A t Mx, A% Af2, .. perpendicu-the directrix

A3 M3, lars to MN.

Then as in the last example, we have

' SA^e AxMxt

where g is the eccen- (Fig. 3) trioity of the ellipse or rx=.e ^xMi or AxMx=rx\e

Similarly AaMa=r2le9 AdM3=r3le, ..., AnMn*=>r„le. Let the pressures at A x , A i t A 3 , „ be p x , p i t p 3 . „ respectively. Then p 2 -Pi"="pressure at A 2 - pressure at Ax

„,. =.gPa.42Ma - g9»AxM, =gP 2 (A^It-AlMl) =>gPa (rje-rile) or P3-P1V (gPa/e) (ra— rx)

- Similarly p t -p»=(g*ile) ( r a - - ra) ,

iljluid Pressure 43 '

h~Ph-i={g?hle) (rm—rM) , Pt Pn^gPi/e) (ft-r,,) Adding these results, we get 0=(g/e) [p2 Cra-ri)+P3 (r3-r2)-l-~+Pn <rr-r,ua)+Pi (r,- r„)]

or Pa (/-a-^O + Ps fo -»"a)+—+Pn ( ' ' « - ' » - I ) + P I fa—l^O or ri (pj-Pj+rj, (p2-p3)+...+r„ (Pn-pJ-^0. Hence proved.

Exercises on Fluid Pressure Ex 1. The sp. gr of mercury is 13 6. At what depth in

mercury will (he pressure be equal to that at 500 meters in water ? Ans. 36f ? meters.

Ex. 2. The pressure at the bottom of a well is four times that at a depth of two feet. What is the depth of the well if the pressure of atmosphere be equivalent to that of 30 ft. of water ? Ans. 96 ft.

Ex. 3. The atmospheric pressure at the surface of a lake is 15 lbs. wt. per sq inch. Find at what depth tde pressure will be 45 lbs wt. per sq. inch, the weight of a cubic foot of water bding taken to be 1000 ozs. Ans 6912 ft.

Ex. 4. (a) The pressure in the water pipe at the basement of a building is 35 lbs wt. to the square inch, whereas at the third floor it is only 19*375 lbs. wt. to the squire inch. Find the height of the third floor. (Taking a cubic foot of water to weigh 62*5 lbs.)

{Magadh 75) Ex. 4. (b) Tae pressure in the waterpipe at the basement of

a building is 3 S lb. wt. to the sqjare inch, and at the upper floor la 1875 lb wt. Find the height, if one cable foot of water=6251b. wt>

"" (Patna 82 J Ex. 5. If a liquid be heterogeneous and of density kz at a

depth z, show that the piessure is U+igkz2, where II is the atmos­pheric pressure. \

Ex. 6. If a liquid be heterogeneous and of density pzja at a depth z. show that the pressure is II -\-{pgz%l{2a)}t where II is the atmospheric pressure. (Rajasthan 77)

Ex. 7. A glass tube of uniform bore is bent into the form of a U-tube jwhose arms are vertical. • It contains mercury of specific gravity 13-5. Water is poured into one arm to occupy a height 40 cm. of it Find the rise in the level of mercury in the other arm.

Ex. 8. The two arms of a U-tube are close together. In one arm there is water and in the other mercury ; so that their common surface is at the lowest point. One quarter of the water is taken out and is pourej into the other arm over the mercury. Proijp that in the new equilibrium position the difference of heights of the upper surfaces Is one-half of what it was formerly.

*Ex. 9. A circular tube of uniform thin bore is half filled with equal volumes of three liquids of f pecific gravities 3, 4 and 6, and is kept with its plane vertical. Prove that the diameter joining the two free surfaces makes an angle 8 to,the vertical such that tan0=19/V3-

tk •" ' Hydrosfaticb*

Ex 10. Three liquids whose densities plt p2, p, (pj > pa" > p ) are in A.P , completely fill a circular tube in a vertical plane and Gccupy length of the arcs subtending at the centre angles 2a, 2P, 27 respectively. If tfce radius to the point midway between the ends of the lightest liquid makes with the vertical an aDgle 8, prove that

cot 0=»cot a+2 cot p. Ex. 11. • Liquids of density p and a occupy areas subtending

angles a and P respectively at the centre of a fine vertical circular -tube. If the surface of separation be at the lowest point, prove

that sin Jp==V(P,or)sinK Ex. 12. A fine circular tube in a vertical plane contains two

liquids of densities p and cr which subtend angles a and p respec­tively at the centre of the tube. Find the angle which the radius through the common surface of the liquids makes with the vertical. (Given p > a) (Magadh 73)

Ex. 13. Prove that in\ a fluid at rest under gravity horizontal planes are surfaces of equal pressure and equal density. (Avadh 81)

Ex. 14 A fine'circular tube in a vertical plane contains two liquids of densities p and a (p > a) js^bich subtend angles a, p res­pectively at the centie of the tube. Find the angle which the radius through the common surface of the liquids makes with the vertical.

{Patna 82)

*•§ 2*11. Whole Pressure. (Bhopal 81 / Jodhpur 76 ; Rojasthan 78)

• When a surface is in contact with a fluid, the pressure exerted by the fluid on each element of it is in the direction of the normal to the element. If the surface be plane surface, the direction of thrust at each element being normal to the surface is the same everywhere, these thrusts thus form a system of like parallel forces which can be compounded into a single force called total thrust or resultant thrust of the fluid on the plane surface. Its magnitude will be the arithmetical sum of the thrusts on different elements of the plane surface.

But if the surface be *a curved one, then thrust? at different elements of it will not form a system of parallel forces and hence tbeir resultant will not be equal to their arithmetical sum. But whatever be the surface9 plane or curved, this arithmetical sum of the thrusts on each element of the surface is called the whole pre­ssure and is defined as '.—The whole pressure of a fluid on a surface is the smt of the normal thrusts exerted by the fluid on each element of the surface.

**§ 2"I2. If a plane surface of a'aa S is immersed in a heavy homogeneous liquid, the whole pressure or the resultant thrust of the liquid upon this surface is given by v? z S, where z is the lieptfa of the centre of gravity of the area below the free surface and w is the weight per unit volume of tfce liquid.

(Bhopal 81; Jodhpur 76; Magadh 74; Rajastkan 78 ; Vikram 73)

Fluid Pressure 45

Let the area S be divided into a large number of small ele­ments of areas au aa, «3,... and let the depths of these elemen­tary areas below the free sur­face be zv z2, zs,... respec­tively.

.*. The pressure at a depth Zi below the free Miiface Is Zi iv, where w is the weight per unit volume of the liquid, therefore the pressure on the element of area aL is ax zx w. Similarly the pressures on elementary areas og, <xs, ... are as Zt w, a3 z3 w,... respectively.

.*. whole pressure on the area S= arithmetic sum of the pre­ssures on each element of the area S

«=«i Zi w>4-«2 zi w-\-*a Z3 H>+... "=w («J 2i+«2 Zi+«3 z3 f. . .) ' ...(1)

Also, from Statics, we know that if z be the depth below the free surface of the centre of gravity of the area S, then

«i+«s+«s+ — (See Author's Statics) . But «i+aa+a,4r ...=»sum of the areas of different elements

= 5

(Fig. 40)

<tyZx-\- «2?i,+er8z34-.

S or S z=a1z1+aaz24 «%^Jr •

A From (U the whole pressure on the area S=w z S. Note. The above theorem may also be stated as :— 1. Thf whole pressure of a liquid on a surface in equal to the

weight of a column of liquid whose base is equal to the area of the given plane surface, and whose height Is equal to the depth of the centre of gravity of the given plane surface below the surface of the liquid.

Or - 2. The resultant thrust on a plane area which ts immefted in a

liquid at rest is equal to the product of Its area and the pressure at its C.G. ' (Mogadk 73 ; Mithila 81 ; Patna 82)

Solved Examples on § 2 11—§ 212 (whole pressure) Ex. 1. Determine the total thrust on one side of a rectangular

vertical dock ga'e 50 it. wide, immersed in sajt water to a depth of 25 It. having given that one cubic foot of salt water weighs 1026 oza,

46 Hydrostatics

Solution. The area Immersed in salt water (f.e. S)=50x25 sq.ft.

The depth of C. 6. of the area below the free surface (I.e. z)=^2

5 ft. the required

thrust on the dock gate •="W z S

= J ? § a . ¥ . (50X25) •=1001953 | lbs. wt.

so Ft..

(Fig. 41) Ans.

*Ex 2. The width of a rectangular vertical dock gate is 40 ft. and on one side there is sfllt water (sp. g. 1*026) to depth of 24 ft. On the other side there is fresh water. Find its depth if the thrusts on the two sides are equal.

• Solution. The area immersed in salt water=40x24 sq. ft. The depth of the [C G. of the area below the free surface of

the salt waters s2

4=12 ft. - ~X The thiust due to salt water on one side of the dock gate

- w s.5=(l-026xx^a).12.(40x24) lbs. wf. where *f §* is the weight in lbs. wt. of 1 en. ft. of water.

Let x be the required depth of fresh water on the other side, then the thrust due to fresh water on the other side of the dock-gate

•=w.S5=2 f«a . lx . (40xx) lbs. w' .

Since these two thrusts ar equal, so we have

' 1fs--.^(40XJc) =(r026x1£!f t).12.(40x24)

or . *2<=1 026X2X12X24 or *=24xl'0129~24-309ft.

* Ans. Ex. ,3. Find the thiust on a

rectangular board, 3 ft. by 2 ft. immersed vertically in water wiih the smaller top edge horizontal and at a depth of 24 5# ft below tie (Fig. 42) surface of water, the atmospheric pressure being equal to a column of 34. ft. of water. _ '(See.Figure 42 above)

Fluid Pressure 4?

Solution. Area of the boards 3 x 2=6 sq. ft. The depth of the C. G. of the board below the effective surface

=34+(24 ,5+l,5)=60ft. (Note)' £. the required thrust

=w z.5=>(62 5)X60X6 lbs. wt. =625 X 36=225000 lbs. wt. • Ans.

Ex. 4. In the vertical side of a water tank there is a square plate whose upper edge is horizontal and 8 ft. below the surface of the water. The depth of the plate is one foot, find the resoltant pressure on the plate, taking the weight of 1 cu ft. of water to be 625 lbs.

Solution. Area of the plate=l X1 sq. ft. The depth of the C. G. of the plate below the free surface

- 8 + i ( l ) = ¥ f t . .*. The required pressure on the plate=B> Z S

=.(62 5) . '^ 1=J(1062-5)=531.25 lbs. wt. Aas. Ex. 5. Find the whole pressure on a plane area, in the form

of a trapezium immersed vertically in a heavy homogeneous liquid with one of the parallel sides, of length a in the surface and the other, of length b, at a depth h below the surface.

Solution. Let ABCD be the given trapezium. Join the diagonal BD, If AD is in the surface. AD=a and BC=b ; vertical distance between AD and BC=h. Thrust on triangle ABD=(\ah) {\h) w [ V Area of the A ABD=\ ah and depth of the C G. =. \h] Similarly thrust on triangle BCD^b h) (f h) w. A Whole pressure on the trapezium

=thrust on A ABD -(-thrust on A BCD » t aft»w + 5 M 8 w=| Ifiw (a+2b)~i h*(a f 2ft) w, Ans.

Ex, 6. Find the whole pressure on a triangle, the depths of whose vertices are a, (3, 7, the liquid being homogeneous and S being the area of tlte triangle.

Solution. Area of the triangle= S. Then depth of the C,G. of the triangle below the free surface

- i '«+P + y). Let w be the weight per unit volume of the liquid. Then the required whole pressure=Hz.5=w4 («I £ - fy^ .

Ans. *Ex. 7. A vessel containing some water rests on a horizontal

table, ii a person dips his hand in water without touching the vessel, how is the pressure on the table affected ? (Rojasthan 76)

Solution. Let the water be filled in the vessel to a height h and w be the weight.per unit volume ofttev^ter. Let 5" be the area of the vessel which is in contact with the table. Then the whole pressure on the table=w.A S. - _(1J

»

At Hydrostatics 136/3

When a person dips his hand in wafer without touching the vessel, S and w remain unaffected whereas the level of water in the ves sel rises and thereby h increases.

Hence from (1), we find that the whole pressure on the table increases.

Ex. 8. An ellipse is placed with its minor axis OH tbe surface of water ard its plane vertical. A circle is described OB the minor axis as diameter. Ffnd tbe total pressure of v> arer on the portion of the area enclosed between the ellipse and the circle.

Solution. AB is the minor axis iof the ellipse lying on the free surface. /

Arcs ACB and ADB are of the (Fig. 43) circle and the ellipse respectively. Let AB=2b and OD*=a

Then the area of the semi-ellipse^' $itad"=% Xnab. The area of tbe semi circle="j7trs"=ljria. Let Gx and G3 be the centres of gravity of the semi-circle and

semi-ellipse respectively. v Then OGi=46/3re and OG^>=4a/3-n:. Let.w be the weight per unit volume of water. Then pressure of water on the semi-ellipse

And pressure of water on the semi circle *~"w .z.S" ~w.(4bl37z).hnbfl~%baw.

'. Required pressure on the portion of the area enclosed between the ellipse and tbe circle

•^pressure on the semi-ellipse—pressure on the semi-

•= §a*bw - f&8M"=J b («»- P) w. Ex. 9. Find the thrust on a rect­

angular plate whose sides are a and b, the side 'a being horizontal and at a depth c below tbe surface and tbe plane of tbe rectasgle being inclined at an angle 6 to the vertical.

Solution. ABCD is the rectangular plate,- whose side AB=b and BC<=a. The side AD is at a depth c below tbe free surface, Fis the middle point of the side AD and G is the C. G. of the plate ABCD. Then GE<=>$b and

EC= EG cos 8*° %b cos 6.

136/4 Fluid Pressurs 49

Let GL be the perpendicular drawn from G to the free surface, then GL=AM+EK=e+lb cos 6.

Also area of the rectangular plate=axA. A, The required itarust on the plate

=,V2.S"'='ve {c-i lb cos 0} ah, where w is the weight per unit volume of the liquid. Ans.

Ex. 10. A square wbose edge is 8 inches, i- immersed its water, its upper edge befog h«ri7o»tal and at a depth of 12 inches* below the surface of the plate when It is inclined at 45° to the horizon ; the mass of a en ft of water being 64 lbs

^ Relation. Proceed as in the above example. Here c=»12"=l ft., a=>a=>8"«=*i ft. ; fl^Jrc and w=64 lbs. wt.

Ans 35*148 lbs wt. Ex.11. Find the total thrust of water on a semi-circular area

of radio' 6 inches immersed in water w'th its diameter in the surface atsd the plane inclined to the vertical at an angle of 60°.

Solution. ACB is the semi-circu­lar f rea. O is the middle pokt of AB and OC is the radius perpendicular to the diameter AB Let G be the C. G. of the semicircle and GL the

drawn from G to the surface. perp. (Fig. 45)

Then OG* "4a" 4x6 8 S

'=— inches=7=- =5— ft. 3JC 3rc

.-. 'z'~GL°=OG cos 60o=(2/3rc)XJ- (1 _/3«) ft Also the area of the semi- circle«*"£ .*. the required thrust

„., 1 n 62 "

:i«r*"-ijB Ci)a sq. ft.

-(62-3).^. y — 5 4 lbs. wt.= 29

• 2 s l b . . w t Ans.

*Es 12 A square is placed in a liquid with one side in the surface Show. how to draw a horizontal, line in the square dividing it into two portions the thrust on which are the same. (Avacth 80)

boSution. ABCD is the square with its side AD in she surface, Ga is the C G. of the square ABCD. Draw a horizontal line £Fat a depth x below the free surface, such U&t the pressure on the portion AEFD and EBCF are equal or the pressure on tne square ABCD is double of the pressure on the portion AEFD.

Let (?! be the C. G. of the por- : -K=>33p§^v tion AEFD. Let the side of the square - - - = •• be a. Then the depth of Ga and Gx below the free surface are ia and J»

(Fig. 4 $

30 Hydrostatics

. Also the areas of the square ABCD and (he portion AEFD are a2 and, ax respectively.

.'. The thrust, on the square ABCD=-''w.5S"—w. J.«r a* and the thrust on the portion AEFD=*w\x.ax, where w is the weight per unit volume of the liquid.

V The thrust on the square ABCD=>2 times the thrust on the portion AEFD. .

A fk$a.aa»=2.w % x ax or 2xa=a1

Ex. 13 A rectangular lamina is immersed with one side in. the free sur­face of a liquid. Show how to divide it into tw» parts by a straight line drawn parallel to this side, so that the pressures on the upper and the lower parts are in the ratio 4 } 5. iRajasthan 75)

Solution. Let ABCD ba the rect­angle, whose sides AB=a and BC<=b. Lets' the line EF divide AB into two parts, so that the pressu es on the upper and the lower parts are in the ratio 4 : 5. Let AE=x.

or x<=a{</l •

x A L

~f=fz

m

.-a.-.*

I

r i i i

i h i ! !

i^--.S----&J-^--r--r--. "2

4ns.

0 ?=lf

JLyi

w d§

or

(Fig. 47) P/essure on rectangle AEFD =>" wz. S'' •=• w.\x.bx Pressure on rectangle EBCF^w [xH-J (o - x)] (a—x)b

pressure on rect. AEFD 4

pressure on lect. EBCF"* 8

w \x>bx A $xa

Given that

or - • = #

or x—ia w [x+i(a-x)] (a -*) b * %[x+a){a-x)

or *a/(aa-jc2)=f or 5Aa«=.4aa-4A;* or x* . & a - » = . a - f a « J a

A AE : EB=x : at-x=>\a : \a=l :1 . Ans. **Ex. 14. A square lamina ABCD which is immersed in water

has the side BA in the surface. Draw a line AB to a point E in CD, such that the pressures on the two portions into which it divides the lamina, may be ^qual. (Bhopal 83 ; Bihar 76)

Solution. Let DE=x. Then CE=a—x, where a is the side of the square We are given that the pre­ssures, on the two portions ABED and BEC are equal i.e. the pressure on the square ^4BCD=2 times the pressure on A BEC (Note)

A L

SF/i

V=~

Z.-Z.-ZT.

j

G/

bzY ... /.

it t

\ /: A '

3 TZ ~

:=rz.

W, hl-lI-iE^t=^Ay^^:c^-;i

(Fig. 48)

Fluid Pressure 51

or or or or

w.G-L.aa=2 w. Gi Lv{i (a—*) a} iv.|a.fl2=2 w la.i(a-x) a a=f (a—x) or 4x=a or x=-£a DE-=l.DC Ans.

*Ex. 15. Divide a square immersed vertically in the fluid with one side in the surface by a straight line parallel to the diagonal so that the thrusts on the two porticos are equal. m (Mithila 82\

Solution. EFis the line drawn parallel to the diagonal BD of the square ABCD, such that BE=x=DE and £C«= (a—x)>=CF, where a is the side of the square ABCD. GZ=depth of G, the C.G. of the square below the free surface=£a and the depth of Glt the C.G. of A ECF^GyL^DN

=DF+FN=x+% (a-x)=i(2a+x) Now pressures on the two por­

tions aie equal, or the pressure on the A ECF «=£ the pressure on the square ABGD (Fig. 49) or w.GiLy area of AECF<=% w.GL area of the square ABCD or w & (2a+x). £ (a—x) («—*)= \ w.\a.a*

6aa«-|-as=»0, which gives or 2 (2a3-3aax+xs)<=*3a3 or 2xa

the value of x. Ex. 16. ABCD is a rectangle immersed in a homogeneous

liquid wiih AB in the free surface and AD vertical. Show how to draw a straight line through A, dividing the rectangular area into two portions equally pressed.

Solution- Let AE be the line drawn through A, dividing the rectan­gular area into two portions equally pressed.

Let CE=x, so that DE<=a—x, where AB=CD=a. Let AD=BC<=b. I

V The pressures on the two por- : tions are equal, =

^ The pressure on A ADF I =>£ the pressure on the rect. ABCD or w.G^Lx area of AADE

=\.w GL. area of the rectangle ABCD. or w.*b.$b (a—x)=\.w.\b.ab or 4(fl-*)=3a or x<=\a or EC=la~lJ)G Ans.

*£x. 17. A triangle is immersed in a liquid with one vertex in the surface and the opposite side horizontal. Neglecting the

(Fig. 50)

32 Hydrostatics 284-66

atmospheric pressure, find Ibe ratio in vihicb the mediao' through the vertex will be divided by a horizontal line vibich divides the triangle into two parts on which the pressures are equal. {Ranchi 82)

Solution. Let EF be the horizontal line which divides the A ABC into two parts 'the pre­ssures on which are equal. From the vertex A draw AL perpendi­cular to side BC ceeting EF in M. le t AM-=*x and AL^h.

V As ^ C and AEFsxs similar.

. EF AM BC

x "h

or

AL

EF=(xlh).BC.

Also in similar As ARM and ADL, TK^-TT ^ r AD AL, n

Depth of C.G. of A ABC*=% AL^lh and the depth of C.G. of A AEF=* AM<=\x

A Pressure on A ABC=i'w.z.s"'~w-lh.{\ BCAL) «=4w h\BC.

And the pressure on A AEF=.wz.S*=w.%x.(l.EF.AM) *=\wx.{xlh\ BCx.

V The pressure on A -<lEF=»the pressure on the figure EBCF ;. The pressure on the A AEF~>\ the pressure on A ABC

or 4w.(tf/A)JK?-H.wA'£C or 2x3<=/23 or x=/?/21/3

x J_ AD 21'8

/2ra2'/8 o r AJT'\ 21'3 . AD-AK 21/3_i KD 2 1 ' 8 - !

— | or =

From (2), - ^

or AD . TK~1'

•1 o r or AK 1 AK' 1 Ans. *Ex. 18. The side AB of a triangle ABC is in the surface of

a Buid and a point D is taken in AC, such that the pressures on the triangle BAD and BDC are equal ; find the ratio AB

Solution. From D and C draw DK and CL perpen- __/f K L diculSrs to AB.

Pressure on A ABC ='w.z.S' maW.lCL \AB.CL - i AB (CL\Z w = | ABAC2cm* Aw,

since CL—AC sin A.

DC.

B -r?p-

1 1

v I

rsH-' y \

V

1 1 1 I „

- '1

1 "* 1

V '

\ ' * •

c Similarly pressure on A ADB (Fig. 82)

Fluid Fiessu're 53

•w& KD.{hAB.KD)^\.AB.<KDf w

or or

<=%AB.ADZ sin2 A.w, since KD=°AD sin A. Now the pressure on A^-05=pressure on ABDC

the pressure on /\ADB=\. pressure on A ABC I AB.AD* sin8 A.w<=*\ k.AB.AC* sina A.w

AC or 2 ADa=>ACz or = - „ = V or

or ^ AD

AD

«1 or w

V2 1 42- 1 or AD

1

Z>C = 1 :<y2 - l )

t

k B

T x i D

r if

T

I

Ans. **£*. 19. 4 rectaogalar area is immersed in a heary liquid

with two sides horizontal, and is dirided by horizontal lines into strips on whish the total thrusts are equal. Prove that, if a, b, c are the breadths of tnree consecutive strips, then

a (a+b) (b -c j=c (b+c) (a-b.?. (Bhopal 52 ; Lucknow 81, 79)

Solution. ABCD is the rectangle immersed in the liquid with sides AD and BC horizontal.

Let AD<^BC<*d and AD be at a depth x below the free surface.

Then the depth of C. G. of the strip AEFD below the free surface=#+£a.

Similarly the depths of the C. G. of strips EKMF and KBCM are (» \-a \-\b) and (x+a+bY $c) respectively.

Let/?,,/?a and/7a be the thrusts on the strips AEFD, EKMF and KBCM respectively. Then/>i=/>2=7>3 (given)

Now p1='itw.zS"'=>w.{x+la) ad; p2

and par=>w{x+a-tb+%c).cd V P!=ps, / . w(x+la).ad=w%x+a+%b)bd

ax+la*=>bx+ab+bb* or x (a ~b)^ob+W-W Again j ^ P a . s ° we n a v e

w (x+a+J6).W=w (x+a+b+fc) cd bx+ab+ibs=cx-{-ac+bo+icz

x {b-c)=*ac-\-bo+\c%-ab-lbz

DM*. (.) by (2, we ge« H - a e + g + g ' i r ^ ' Cross-multiplying and simplifying, we get

aac -f ac* - a2b ~ ab^-b^c—bc^+labc^Q Rearranging,, we get abc—btc-fccia-bi}a=>—gbe+aidJ7baa—a2c

o? o (a+ b) (p~c)**c (b+c) (o—b). Hence proved,

(Fig. 33)

'W.(x+a+$b).bd

or

or or

CD

(2)

u • Hydrostatic

Ex. 20. A conical vessel, filled with jplane base on a horizontal table. Prove that the resultant thrast of the liquid on the base of the cone is three times the weight of the contained water.

Solution. Let h be the height and r the radius of the base of the cone. Then the

v\veighc of the contained vvater=$wJA w, where w is tBe weight of the water per unit volume.

Also the thrust on the base of the cone="w z S"\=w.h.itr2<=3 (far* hw) •=3 (wt. of the water contained in the cone).

Hence proved. *Ex. 21. The sides of a cistern are vertical

horizontal regular hexagon each side of which is V3 ft. long Find the depth if, when it is full of water, the thrust on each of its sides is the same as on its

water, stand with the

(Fig. 54) Its base is a

(Magadh 74) be the depth depth of the cistern below

base. Solution. Let h ft.

of the cistern. Then the Henfh of ths 1 1 1 i l l * C.G. of each side of the the free surface=|A ft.

£v The thrust on the side of the cistern=»"vvz 5"'=w \h (A\/3), where w is the weight per unit volume of the water. > ^

Area of the base (Fig. 55) «=area of a hexagon of side ^ 3 = 6 (area of an equilateral triangle of side V3) «=6xJ tf3 V3. h / 3 = f V 3 sq.ft.*

[Because area of a triangle<^\ab sin 6] A Thrust on the base=)v./2.J.9<^3 It is given that the thrust on the base=thrust, on each side

or w.h.\Sy/^-=w.\h.hy/Z or h=9 ft. Ans. **Ex. 22. (a) A hollow weightless hemisphere, filied with

liquid is suspended freely from a point in the rim of its base, show that the thrust on the plane base is to the weight of the contained liqaid as 12 : y/p3). (Lucknow 77 ; Rajasthan 75 ; Ranchi 81, 80, 75)

Solution Let G be the C G. of the hemisphere full of liquid (solid hemisphere) and O the centre of the base of the hemisphere. Then OG^^a, where a is the radius of the hemisphere. Let the

Fluid Pressure

hemisphere be suspended from the point • A. Then the hemisphere (full of liquid) is in equilibrium under ths action ot two forces viz. wt. of liquid acting at G and the forces at A. He ace A and G must be in the same vertical line. Let OA i.e. the bass of the hemisphere bs inclined to the vertical at an angle 9.

^ , „ OG 3«/8 3 Then tan 0= TTA^-=-=** OA a or tan 0=3/8. ...(1)

Also the depth of C. Q. of the base BMOL=AO cos 0=a. cos 9.

.*. The thrust on the base of the hemisphere="iv.2.S" =w.0L n<P=>wa cos O.-na*, where iv is the weight per unit

volume of the liquid. Also the weight of the liquid contained«= And from (1) we get

1 ^ 1 , 1_ •V/(secs0)'= "

(Fig. 56)

=§ 7caa.w.

cos 9 V(l+tan*0> VO+-A) V(?3) . The thrust on the plane base^w a cos 9 ngz 3 cos 9 " wt. of the contained liquid f itasw ~~2

_ 3X8 _ 12 ""2^(73) \?(73) Hence proved.

**Ex. 22 (b). A hollow weightless hemisphere, filled with liquid is suspended freely from a point in the rim of its base, prove that the whole pressure on the carved surface and that on the base are in tha ratio 13 : 8. (A?ra 74 ; Bhopal 80 ; Garhwal 81 ; Vikram 73)

Solution. Refer Fig. 56 above. Proceed exactly as in Ex. 22 (a) above. Here OGt^\a, where Gt is the C. G. of the curved sur­face of the hemispherical shell.

The thrust on the base of the hemisphere =na3w cos 9, as in Ex. 22 (a*) above

and tan 9=3/8, where 9 is the inclination of the base to the vertical. Now the depth of Gx below the free surface

•=OA cos 0-f-OGi sin 9 (Note) = a cos 0+£a sin 9

.'. The whole pressure on the curved surface <=>wzS=w (a cos 9+ia sin 9) (2nfl8)=7to3w (2 cos 0+gfh e)

-r, . , whole pressure on curved surface .*. Requised ratio =• -T—T- —

^ pressure on the base ncPw (2 cos fl+sin 9) KCPW cos 0

-2+1=19/6.

•2+tan 9

Hence proved.

58 fiydros'ta^cs \

*Ex 23. The incliaatioa of th« axis of a submerges! solid cylinder to the vertical in two different positions are complementary to each other. If P and F' be (he difference between the prej&nres on the two ends in the two cases, prove that the Weight of displaced ' fluid is equal to <P2

TF'2) i a

(jodapur 76; Lucknow 7&

(Fig. 57)

centre of the lower end below the free

Solution. Let h be the height iand r the radius ot the base of the cylinder In the first position, let the axis of the cylin­der be inclined to vertical at an angle B. Also let 0 the centre of the upper end of tbe cylinder, be at a depth x below the free surface ie. OL=x,

Tin n the depth of the surface—0'L'=x-t h cos 0.

.*. The thrust on the upper end="iv.z.S"=n'.x„7rr2

And the thrust on the lower end—w.(x+h cos 0).rcra

A, P*° the difference between the pressures on the two ends.' *=w (xi-h cos 6) xra—w.x.nr2'=wJi cos 6,xra

In the second position, the axis of the cylinder is inclined to the vertieal at an angle complementary to d i,e. 9a°—Q.

Then P'^w h cos (90°- 0).nrs^w.h sin S.nrz

« {Pa+P'z)We={(w>h cos 0.rcr2)a+ (w.h sin 0.W2)2}1'2

*=>w.nr2!t{cos2 0+sin3 oyP^w-Tzr^h' "^weight of the liquid displaced by the cylinder,

\ Hence proved. Ex. 21. A holJow weightless cylinder ailed with water is

suspended freely from a point oa the ram of ifs piane end. If the height of the cylinder be twice tbe radius of plane circular ends, show that tfc thrusts oa the ends are in the ratio 3 : 1 .

Solntloa. The cylinder is suspended from the point A and the G.G. of the cylinder full of water is G Let the radius of the base of the cylinder be r, then its height=2r (given).

L The cylinder is in equilibrium under the action of two forces viz, the weight of the cylinder full of water acting at G and

\

r?lu!d Pressure 5 /

the force at A. So the line of action of the force at A is also verti­cal and t&ersfore A aad G are in the same vertical line.

AO In A AOG, tan 6' OG ll A L

i M \ \

\ \ X \ \

1^45° ;

6/45° N ^ V

0 /

. A

a

e a X ^ i where 6 is the r

inclination of the axis of the cylinder to the vertical.

.'. /LOAL*=0=4i°and AG=OG.seo 45°=r^2.

.*. The depth of O, the C G. of the upper end below the free suiface *=*OL~AO sin 45° «r ,y/1.

And the dep*h of O', the C G. of the lower end^O'Z/ •=*0'M+GA=rcos 4? + ry/2

«(r/V2)+rV2="3r/V2 Also area of each end^nr2. A Thmst on the upper end

*="w2S"^w.OL nr2=w (rtfl).™*. And the thrust on the lower end =>w.O'L'.•Kr*<=w.(3r/<tfZ).nr2

Thrust on the upper end _7twrs/&2 __x

Tarust on the lower end Reqd. ratio

*Ex. 25 (a).

3xwr*WZ d' Hence proved.

A cone fall of water, is placed on its side on J horizontal table, show that the thrust on its base is 3 sin a times the weight of the contained fluid, where 2a is the vertical angle of the cone. • (Avadh 82 ; Vikram 74)

Solution V is the vertex ot the cone and VB is the generator in contact with the horizontal table. A is the highest point of the cone and the horizontal pane passing through A is the level of (Fig. 59) the free surface. Let « be the serni-veitical angle of the coae, then i_ AOL^ a. Let h be the height of the cone, then the radius of the base of the cone^A tan a.

Also the depth of O, the C. G. of the base of the cone below the free surface=OL^AO cos «=>h tan a cos *>=>h sin «

/ . Thrust on the base of ths cone %

«=»"if .2,5'' *=>w.Ol.it {h \m «1*

$&- hydrostatics

=»j«v./j sin oc.ft2 tan a~3 sin a [luh3 tau8. u.w] = 3 sin a [weight of the liquid contained in the cone].

Hence proved^ Ex. 26\ A cone full of water is placed on its side on a hori­

zontal table. If the thrust on the plane base is l^times the weight of the contained liquid, show that vertical angle of the cone is 60°.

Solution. As in the last example, let h be the height and « the semi-yertica*l angle of the cone.

Then the thrust on the base . ' =>3 sin a [weight of the liquid contained]

01 weignt of the liquid contained

o r sin (*=>£ or <x=>3Q°. .*. The vertical angle of the cone=»2a>=»60o. Hence proved. *»Ex. 27 (a). A hollow weightless cone of semi-vertical angle

a and of height h, is filled with liquid and freely hung from a point on the rim of the base, show that the thrust of the water on the base is

47thV fan8 « sin « V a + 1 5 sin8 a) »

(Agra 81 : Jodhpur 78 ; Rohilkhand79)

Solution. Let V be the vertex, " a the semi-vertical angle and h be

the height of the cone. Let O be the centre of the base of the cone. Then CM ~ A tan « Let G be the C G. of the cone full of liquid, then OG=>lVO=lh. The cone, full of liquid is hung from the point A on its rim and is in equili­brium under the action of two forces viz., the weight of ths cone full of liquid acting vertically downwards through G and the y force at A Hence these two forces must have the same line of action - (Fig. 60) f e. A and G should be in the same vertical line. Let £ OAG^O.

* , , 0 0 ? jh _ 1 Then tan 0 - O A ^ J T ^ ^ " ^ ^ ' ...(«)

• Depth of O, the C. G. of the base, below the free surface W=LOL<=AO cos 0=/z tan a cos 0

Area of the ba$e of the cone=>"w2"=n/ta tana a ; & thrust op the base of cone «=• "w%S"

Pressure & '

=w.h tan a.cos O.nh* tan8 a=mv/z8 tan3 a cos 0. ...(2)

Also from (1), sec2 0=>l+tan2 f a l + -j—i- = 1 + ^ , , 16 tan'oc 16 sin2 a

16 sin* «4 cos* <*. 15 sin" a+(sin2 a + cos* a) 16 sin" a 16 sin2a j

_ l + 15sin2« 16 sin2 a

4 sin a a U 16 sin2 a 1 _ or COB •- iy tr+i5ita^r } -tf(l + 15sin8a)

X • from (2), the thrust on the base of the cone ia A « 4 sin a

••JCH'«', tan8 a. ,,, , t c » - • TT „ v ( l + 1 5 sina«j Hence proved. •Ex. 27 (b). A hollow weightless cone of vertical angle 2ce is

filled with a fluid and suspended freely from a point on the rim of its base. Show that thrust on the base is to the weight of the fluid the ' cone would contain as 12 sin2 a : cos 0^(1+15 sin8 a) {Agra 82,77)

Solution. As In Ex. 27 (a) above we can find that the thrust 4 sin oc

on the base of the coLe=nwhs tan3 a -77—, . , . ,„ . ... V(i+15sin2a) -.(1)

Also the weight of the fluid the cone would contain = liz.h3 tan1 a.vi> ...(ii)

, . x. mvh* tan3«4 s'n a 3 A required ratio - - ^ — ^ ^ - ^ . ^ - ^ - ^

from (i) 12 sin2 a TT

"cos«V(l + 15wn^0- H e n c e p r o v e d -•Ex. 28. A closed hollow cane is just filled with liquid, and

is placed with its vertex upwards aad axi j vertical, divide its curved surface by a horizontal plane into two parts on which the whole pressures are equal. (Agra 81, 73)

Solution. Let h be the height, a the semi-vertical angle of the cone whose vertex is Kand the centre of its base be O.

Then OB<=>h tan a and VB=h sec a. Let the plane PQ divide the curved surface of the cgpe into

two parts, the pressures on which are equal Let this plane PQ be at a distance h' from the vertex V i.e. VO'=h'. Then O'Q—h' tan a and VQ=h'. sec a.

- Let G and G' be the C G.'s of the7curved surfaces of the cones {V, AB) and {V, PQ) respectively. Then VG-\h and FG'•=§£'. Also the curved surfaces of the cones (F, AB) and (V, PQ) are

•nJi sec «./i tan « and •a.h' sec « h' tan «,

'60 ftjdrosta&rf

(Note i the area of the curved surface of a cone=7t/r)

A The thrust on the curved surface of the whole cone (V. AB) i="wzS"=w.%h WJ2 tan a sec a.

'And similarly the thrust on the curved surface of the cone (F„ PQ) 'BW f h'.nh'2 tan a sec a."

<& The'thrust on the curved surfaces of the two parts of the eone (V, AB) are equal.

.*. The thrust on the curved surface of the cone (F, AB) ' (Fig. 61)

=2 times the thrust on the curved surface of the cone (V, PQ) or , w i/z.a7z2tan a sec a.=:2Xw.fh .it A'2 tana sec a or /z3-=2/z's or A'=A/2]'S, which gives the required position of the plane PQ.

*Ez. 29. The side AB of a trimgle ABC is in the surface of a fluid and points P, E are takes in AC, suck that the pressures on the

triangles S4D, BDE, BEC are equal; find the rafio , A L M N _ 8 AD : DE : EC.

Solution. From D. E and C draw DL, EM and CN perpendiculars to AB I.e. the free surface. Then

DL—AD sin A„ Hi -) 4 / /2 ~ EM<=>AE sin A J 01 by (Fig. 62)

and CN=>AC sin A ...(1) .*. the depths of C.G.'s of. As ABD, ABE m& ABC are

JDZ,, $Z?M and \CN respectively. Also the areas of As ABDt ABE and ABC are \.AB DL

iABEM and lAB.CN respectively. A The pressure on A ABD^'wzS'^w \DL.\AB.DL

<=*\ ABDUw=>\.AB ADa.s'ms A.w, from (1). Similarly the pressures on As ABE and ABC are

v J AB AEa sin8 A w and & AB AC2 sin8 ^ JV respectively. Now St is given that the pressures on As ABD, DBE and EBC

araequll i.e. the pressures on As ABD, ABE and ABC are in the ratio 1 : 2 : 3 . or | ^B ^Da sin2 A.w : \AB AE* sin2 A.w : |/4S.^Ca sin8 A.w

<=1:2 S3 ^ C 2 - l : 2 : 3 or uD : AE ; ^ C = l ; ^ 2 ; ^ 3 ^Da M£2

or AD

or - r e AC* AD AE-AD AC-AE ^3 or 1 ^ 2 . - 1 tf3-Y?

Fluid Pressure 61

AD_ DE _ EC 0 r 1 i?2-l=Vs-^' Ans.

**Ex. 30. A triangle ABC is immersed fn a liquid with the vertex C in the surface and (he sides AC, BC equally inclined to the

^surface.' Show that tie vertical through C divides tfce Hiargle into i two others, the fluid pressures on which are as b8+3ab8 : a3+3a2b

(Agra 77 ; Bhopal 81 ; Lucknow 82, 74 ; Sojosthan 78, 76) Solution. Let the sides AC

and BC be inclined to the free , surface at an angle a. Let CD

be the vertical line, through C. From A and B draw AL

and BM perpendiculars to the free surface. Then

AL*=*AC sin a=6 sin a and BM<=>BC sin a=a sin a.

Also l_ACD*=90°- a =/_BCD

or the line CD is the bisector of /_ ACB .'. AD : DB=AC: BC=b i a >

(Fig. 63)

Also AL, CD and BM are parallel.

CD* b (BM)+a (AL) b (a sin a)+a (b sin a) lab sin o>\ a+b4 b+a b+a

The depth of C G. of A ADC below the surface «=4 [depth of A\ depth of D +depth of C] =4 [b sin a-!- CD4 0j -£ [b sin a^{(2a?"sin a)/(a+£)}]

And the depth of C,G, of ABCD below the surface » £ [a sin «+{(?«£ sin a)/(a-f-&)}]

The fluid pressure on A ADC="wzS" •=w h [b sin a+ {{lab sin a)/(a+i;}].^D.C!E

Similarly the pressure on AABC-, •*w J- [a sin a f {lab sin a)/(a-f-£)}UfiZ>.C£

™ . , pressure on A AD? I. The required ra t io- pressure on A BDC

I", . 2aZ)sina1 . - . ^ E , r * j 2fl*

w.£ L a sin a -

**Ex. 31. A being vertical and the circumscribed

lab sin a

" as+3as5*

].MD.« [^.2*]

Hence proved. triangle ABC is immersed in a liquid, its plane

the side AB in (he surface jwi O be the centre of ciicle of A ABC, prove that the pressure on

AOCA : pressure on AOCB=sin 2B ; sin 2A.

62 Hydrostatics

or

(Fig. 64)

Solution. Join OD and OE where D and E are the mid points of the sides AC and Be respectively. Let Gt and (?2 be the centres of gravity of As AOC and BOC respectively.

Then OG^f.ODandOGg^f.OS, OGi _OD OG2 OE

Hence from A* OGfis and ODE we find that G,^ is parallel to DE,

Again BE being the line joining the mid points of tbe sides DC and BC, DE is parallel to the side AB.

:. G,G2 is parallel to the line AB i.e. the depths of Gx and G2 below AB are equal say x.

V O is the circumcentre of A ABC. J. OA'=>OB=OC=R, the clrcumradius. Also /_ AOC-=2 L ABC=2B.. angle at the centre is twice the

angle at the circumference. Similarly £_ BOC^IA Area of AAOC^AO.OC sin AOC*=\R.R sin 7B=\ R?sin 25 Similarly area of A BOC<=\R? sin 24. pressure en A ^OC_^.xj i? a sin ?5^sjn 25 pressure on A ROC W.X.\R2 sin 24 sin 2A Hence proved. *Ex. 32. A parallelogram is immersed in a homogeneons liquid

with one side in the suiface ; show how to draw horizontal lines dividing it inton portions the thtusts on which are equal.

Solution. ABCD is the parallelogram immersed irilhe given liquid with side AD=a (say) • • • " ~ - * A . . , „ in the free surface. Let the horizontal lines dividing jt into portions the thrusts on which are equal be at depths, Aii h> A3>...respectively.

Then the area of the first strip=ahx

area of the second strip

Area of the third strlp=>a (Aa-A2), etc. (Fig. 65) Depth of C.G. of the first strip^fAj, Depth of C.G. of the second strip=$ (Aj-f h2), Depth of C G. of the third strip=ilha+ha), etc. .'. The thur^t on the first strip«=»w. J//x oh^

the thrust on the second strip - w 4 (A!+A2).a (A2-A0=»| *w (tf-hfl,

D

iowh2t

FluSd Pressure /

63

or

the thrust on the third strip -w.§ (M-A,) a (A3-A8)«=J aw (h3

z-h22)

Since the thrust on the strips are equal; JL I awhf^l aw (hf-hi*) or *,"=*„•-*»'

2A,«-V or Aj/A2~l/\/2 Also 1 flwA^-i aw (*,•-*»"> or V V - V

.(1)

or or

V = V - 2 A i ' . . . - V V-2Ai" 3V=V or Ai_ 1 At. Similarly — ^ ^

2

*i/A,= l / ^3 ...(2)

6 V5 ' ' " ' A nr a V «

.*. h1:h2iha:hi:...:hn*='l:\/2i^3:j/4:...:<i?n Ans. ••Ex. 33 (a). A semi-circle is immersed vertically in a liquid

with the diameter in the surface ; show how to divide it into n sectors, such that the thrust on each of ihem is the same.

Solution. O is the centre of the semi- AH,M N2, N$ i N* circular arc, AB is the diameter in the surface, AOPv PxOP2.P*OPs ... are the secto-s, the thrusts on which are equal. I. There are x sectors in AOPx Let [_AOPx=2Q and Kbe the mid point of the aro APX, then

/_AOK<=>6. (Fig. 66) Let G be the C. G. of sector AOPx on the line

a sin 6 OK, then

OGo e

, where a is the radius of the semi-circle.

or

or

.'. The depth of G, the C. G. of tbe sector AOPx

* «=GL=.0G sin 6=(2a sin3 6)138 And area of the sector AOPx=i a a.20-=as9 '.' The thrust on each sector is the same /. The thrust on one secfor=>(l/«) thrust on the semi-circle

the thrust on sector AOPx which contains x sectors •= (x 'n) thrust on the semi-circle.

iv Gt,. area of the sector AOPa=(xjn) w. (depth of C. G. of the semi-circle), area of the semi-circle,

2a sina 0 „ „ x 4a 1 „ or H. —„-.—.a2 0 = — .WX-.}TZO?

30 n 3TC n or sin2 Q=x\n or 2 sina 6 or a (1—cos 20)=2ax/n or

=2x/« or 1—c&s 29« o—a cos 2fl«="2flx/n

(Note) '2xjn

64 Hydrostatics 136/4

OB a—0JV>»2as/n, V l_AOP^=29 and PXNX Is perp. to A3 or ANx*=*2axln, V 0.4=-a and ANx=-OA~ONa^a—ON*

Putting ««=>!, 2, 3, ... we have AN1^2aln,AN2'='4aln, AN3^6a/n,...

& N1Ns^AN2—AN1^(4aln)—(2alri)=2alf3, NiNa^ANa~ANi=(6aln)-(4afri)^2alnt...,

.*. AN1**N1N2*=>N2Na=>NaNt=>.~<z=2aln. ,& Divide the diameter ^S into » equal parts ANV N-jNn N%Na,

Nz Nt,... From Nu 2V2l Ar8, draw iVYPi, iV2P2, iVaPjj,... perpendiculars

to AB meeting the arc of the semi-circle in Plt P2, P%, ... Join these points P l8 P2, Ps ..with O, then the sectors AOPlt PxOP2t P^OP^... are the sectors the thrusts on which are equal.

*Ex. 34, A cylindrical vessel m a horizontal circular base of radios as is filled with a liquid oU density w to a heig&i h, If now a sphere of a radios a and density greater than H' JS suspended by a thread so that it ss completely immersed Sad the increase of pressure om the base of the vessel, aud show ifeat ihe increase off the whole pressure on curved surface is (8rc/ a) wca [a+(2c3/3aaj] g.

Solution. Let the level of the liquid stand at PQ before the suspension of the sphere. In this position the pressme on the base

«="w.z S"=wg.h.na? And the whole pressure on the curved surface

=»"w.z.S" => wg.lh.2nah After the complete immer­

sion ©f the sphere m the liquid, let the level of the liquid rise to MS, suoh that MS is at a height x above PQ. S. The height of the liquid in the cylinder in this case is (h+x).

;. After the immersion" of the spheie,

the pressure on the base B»**W z S"<=wg (A+x) 7taa .. (3)

and the whole pressure on the curved surface= "w.z S"

•=-wg.hih+x).2iza i/j-f-x)

•CD

=7tfl (k+x)Z Wg •(4) (Fig. 67) Also the volume of the liquid

displaced by the sphere-volume of the sphere or the volume of the liquid PQMS=volume of the sphere or jtaa*=frtC3 or x<=*4c3j3as (5)

and <3), Hence the required increase of pressure on the base from (1)

'Wg {h+x) raa3—wghjzd *wasxwg

136/5 Fluid Pressure 65

=jraa.(4c3/3a?) wg, ...from (5) =$KCz.wg Ans.

and the increase of the whole pressure on the curved surface from (2) and (4) =na (h+x)2 wg—naPwg^Tza (2hx+x2) wg

= « « (2h+x) wg=-3ra.(4c73a») [2h+(4c3[3a*)] wg

= (87t/3a) wcs [ft+(2c3/3aa)] g. Hence proved.

Exercises on § 212

Ex. 1. An ellipse is placed in a liquid with its majoi axis in the surface of water and its plane vertical. A circle is deicribed on the major axis as diameter. Find the total pressure of water on the portion of the area enclosed between the ellipse and the circle.

Ex. 2. ABCD is a rectangle, AB being equal to 2BC. It is immersed in water with its plane vertical and AB in (he surface. Show how to divide the area into two parts by a straight line from A such that the pressures on the tw3 parts may be equal.

Ex. 3. A conical vessel 10 inches high on a flat oireular bass of 5 inches radius, is filled with water. Calculate the vertical thrust on the base when the vertex is upwards. (Garhwal 78)

Ex. 4. A cone full of water is placed on its side on a hori­zontal table. Show that the thrust on the base is 3r/i^(ra+hs) times the weight of the contained liquid, where h is the height and r is the radius of the base of the cone. (Ranchi 76)

Ex. S. A semi-circular area is immersed In a liquid with diameter in the surface. Show how to divide it into three sectors, so that ihe thrusts on the three are the same. (Magadh 76)

Ex. 6. Taking a cubic foot of water'as 1,000 ozs, find the pressure of the water arising from its weight on a side of cistern 7 ft. wide and 8 ft. deep, if the cistern is filled with water.

Ans. 14000 lbs. wt. Ex. 7. Find the thrust on a square board whose side is 1 foot

when sunk in water to a depth of 20 feet, the board being horizontal and the atmospheric pressure at the surface being equivalent to a height of 30 inches of mercury, (sp. gr. of mercury being 13'59j.

Ans. 3373-4375 lbs. wt. Ex. 8. A rectangular plate is immersed vertically in a liquid

with one side in the surface. Divide it into two parts by a line parallel to the surface such that the pressure on the upper half is double the pressure on the lower half.

Ex. 9. A rectangular lamina is immersed vertically in a liquid with one side in the tsurface. Show how to dras? a horizontal line dividing it into two'parts such that the ratio of the thrusts on the upper and the lower parts is 9 :7. (Mithila 81)

66 Hydrostatics

Ex. 10. An isosceles triangle is immersed vertically in a fluid with its vertex in the surface and base horizontal. Determine how it must be divided by ,a line parallel to the base so that the pressure on the upper and lower portions may be in the ratio m : n.

Aas. Depth on the line below the vertex^m^m+n)]1'3./!, where h is the height* of the triangle.

§ 2 13. Layers of different liquids. Case I. There are n liquids of densities plt P2, p3, p4, ..,, pn

filed one over the other and of thickness hlt h2, ha, ..., hn beginning from the top and there is a plane surface which is in contact with the lowest liquid only. To find the whole pressure on the plane surface.

Let Zj be the depth of an element »x of the plane surface below the lowest surface of separation. Then the pressure on this element o •=£PA«i+£P2Vi+ — +g?f>-\ hn-i«i+g?nZi*i. (Note)

Similarly pressure on different elements «2, aa, oe4, ... can be obtained. / v

.*. The whole pressure on thcplane surface =£PiAi («i4 cc2+*..•) + gP2M«i+«2+ - ) + ...

•-.+g?n-i /*„-! (ai + a a f . . 0+gP« (21K1 + Za«2+—)

where S is the area of the plane surface and f is the depth of the C. G. of the area of the plane surface below the lowest surface of separation. Also z1cc1+z2a2-f...'==z..S (See § 2" 12 Page 44)

.'. The whole pressure on the surface t=>(gPih1+gPaha+...+g9n_1h„_1+g?,2) S

™= (pressure at the C.G. of the sarface)'x area of the surface.^ (Note) Note. If the atmospheric pressure II is to bs taken into consi­

deration, then the whole pressure on the plane surface ~(II+*P1Al+Sftft,+ ...+ff>ii-l Vl+gPn§)tf.

§ 2*14 A plane area in contact with more than one fluid.' If a plane area be in contact with more than one fluid, then

divide «he plane area into several parts such that each of these parts be in contact with one fluid only. Then the pressure on each part ean bepcalculated with the help of the formula'given in § 2*13 above and the sum of these pressures gives the whole pressure on the plane area

Solved Examples on § 2 '13- § 2*14. Ex. 1. A cubical vessel is filled with two liquids of densities

f aid P', the volumtf of each being the same. Fiad the pressure on the base and on one of the sides of the vessel. (Rajasthan 74)

Fluid Pressure 67

(Fig. 68) Ans.

on one of the sides of tbe

Solution. Let a be the side of the cube. T the volume of each liquid is the same, so the height of each liquid is \a.

Let the densities of the upper andthe lower liquids be P' and P respectively.

.'. The pressure at the C. G. of the base=gp'.$a +gP.£a

.'. The pressure on the base «=> (pressure at the C.G. of

the base) x area of the base

_ =ia3g(P'-!-p).~ Now we are to find the pressure

vessel. The pressure on the upper half of a side of the vessel

="vv.z.5"'=gP'.ia.(fl.ia)=.|fl8P'f And the pressure on the lower half of a side of the vessel

=>(pressure at its C.G.) x its area =teP'-ia+.?P.ia)X(o.|a)=|a8

< 2 P ' + P ) # Ans.( .*. The pressure on one of the sides of the vessel

«= pressure on upper half+pressure on lower half. - M > ' g + | a 3

(2P '+P) g~kas (3p'-M g. Ans. Ex 2. A cylindrical tumbler, half filled with liquid of density

P, is filled up with a liquid of density p' which does not mis with the former one. Show that the pressure on the base of the tumbler is to the whole pressure on its curved surface as 2r (p+p') to h (p#3p'), where h is the height and r the radius of the base of tumbler.

{Lucknow 76) Solution. Pressure at C.G. of the base

=«P'.|A+g.Pift .'. Pressure on the base of the cylfader ^(pressure at the C.G. of the base)

X area of the base •=(g9'hh+g9 P).Jira

- I W ^ P ' + P ) gh. ...(1) The whole pressure on the curved

surface of the cylinder=»the whole pressure on the upper half-f-the whole pressure on the lower half of the curved surface.

The whole pressure on (he upper half of the curved surface=(pressure at the C.G. of Ntis upper half) X area of the upper half

=rSP'.}A)x(-2nr.iA)'=iirrA«gP' * - (2 ) The whole pressure on the lower half of the curved surface

68 Hydrostatics 2§4- -oo

Hence proved.

•it

Jf

=(pressure at the C. G. of this lower half) X (area of this lower half) ~(g?'-lh+g? lh)x{.2nr.lh)~lTzrhz (2P '+P) g ...(3) A From (2) and (3), the whole pressure on the eurved surface ~lnrhVg+l*rh* (2P'-f P) g ^rh2

(3P '+P) g ...(4) . „ ,IN A (A\ Tf le whole pressure on the base . . from (1) and (4;, ,=- -,—: -r -r ?

The whole pressure on the curved surface lfrr a(P'-fp)gft _2rfp '+P)

lnrh*09'-t?)g A(3P' + Pj Ex. 3. A cylinder is filled

with equal volumes of n different liquids which do not mix, the density of the highest is P, that of the next is 2P, and so on, that of the lowest being wP; show that the whole pressure on different por­tions of the curved surfaces of the cjlindar are in the ratios

1» ,"2" ,3V. . . . I I» . Solution. Let x be the

height of each liquid and r the radius of the base of the cylinder. Consider the liquid of density pp. Ltt & bo the C. G. of the portion •f the curved surface'of the cylin­der in contact with the liquid of density pp.

Then the pressure at G , (Fig. 70) "-gP.x + g.lp.x f £.3P.x+... + g (p-1) p.x+g.pp.(x/3t

'- X The whole pressure on the portion of the curved surfac* in eontact with the liquid of density p?

—pressure at Gxthe area of this portion of the curved surfaee *-[g?x+g.2?x +... + g.Cp-1) Px+gppjx] x [2Kr.x] ~2itgv** [1+2+3 f... + (/»-l)+ip] •=*7nrxsgP, [l(p-1)p+ip], V S i - i n (*+1) ' -2wx«gPfo>[(p-l)+U •=nrxsgp p*-=>Pp*, where P>=Ttrx2gp. Puttingp=>l, 2, B...(n—1), n we find that the whole pressure

different portions of the curved surface are r

P.la , P.2\ P.3a, ..., P (»-!)% Pns which are in the ratio

cn-np

on

Is, 2\ 3\ ..., (« - l ) a , n\ Hence provid. ••Ex. 4. A vessel contains n different liquids resting in hori­

zontal layers and of densities PM P2 p8, .., P„ starting from the highest liquid. A trianglefis held with its base in the upper surface of the highest liquid and with its vertex in the nth liquid. Prove that if A

Fluid Pressure 69 ,

be the area of the triangle and hi, ha, ..., b„ be the depths of tha vertex below the upper surface of the 1st, 2nd, pectively, the thrust on the triangle is

1 (SA/VJ {Pi (b18-ba

s) + P2 ( V ~ I n ­solation. Let ABC

bt the triangle immersed in the given n liquids -with its base BC in the upper surface of the liquid of density pj. The vertex A of the triangle is in tht liquid of den­sity «p.

AN "hi, ANi-h,, AN^hz, . . (given)

Also A = area of the A ABC^xBCxAN or

, nth liquids res-

+ P„b«8}.

c

\ BC.hx

BC^Afhi Again from similar triangles ABC and ABjCt, we have

h „, r>n ** R r „ ^ s 2A

h «i «i • 2 A W

Similarly 2?aC2=2 A W , 5 | Q = 2 A W . -.•. Areas of As ^5C, J-BjQ, -4-&2Q, 58C3...etc.

A, i^Cj ^JVi, i52Q.^JV2, ±BaCt.AN3... etc. are

A / 2A/i8 •ha, i 2AA,

. / j 3> etc.

er ..etc. A, AV/V, AA,W» AV/V,-Also the required thrust on A-dSC ' «=»The tbrust on A ABC in a liquid of density px

+rhe thrust on A^AB^Cx in a liquid of density (P»-pi) +tke thrust on h.AB&Ct in a >jquid of density Ps-{(p3+Pi)-Pi}, i-e. (Ps-P2) and so on. <Note)

«rPi.i/J1A+g(Pa-P1)^a.(A/iss/V)+g(P»-Pa).Pi-(AA8

a/V) + .--f g (Pn-Pn_j).|An.(A/ln,,//l1

8) - t (^A/VJfPiV+(p8-Pi) V + (P.-fc) V+-(P--Pn- i)A. 'J - 4 ( M / V ) [Pi (V-A.")+Pi (* . , -V)+. . .+p .* . , I . .

Hence proved. **Ex. 5. The lighter of the two liquids of density p rests oa

the heavier of density a to a depth of a inches. A square of side b is Immersed in a vertical position with one side in the surface of the upper liquid, if the thrusts on the two portions of the square in con­tact with the two fluids be equal, pio^e that Pa (3a - 26)=o ib-af.

(Mcknow 80, 78, 75, 74)

* 70 Hydrostatics

Solution. ABCD is the given square with side AD in the surface of the liquid of density P.

Let Gt and G2 be the C.G.'s of the two portions AEFD and EBCF .respectively.

.'. The thrust on the portion AEFD=?gla{bXa) "...(I)

And the thrust on the portion zl^rBz^=tli^fzTpPiQ EBCF=*(the pressure at (72) " t " ~~-

xareaEBCF (Fig. 72) »fep.«+Sw{H*-«)}Jx* (*-«)• * ' ..(2) V The thrust on the portion AEFD

=thrust on the portion EBCF. ,'. From (1) and (2), we have

gP.ia (a.b)~[gPa+go ft (b -a)]] b (b~a) or paa«{2Pa+a (b-a)} (b-a) or P [a*-2a (b-a)]=a (b-a)s

or Pa (3a ~ 26) =• a (b - a)3. Hence pToved. *Ex. 6. A square of side a is dipped vertically in two liquids

of densities P and a with the upper side in the free surface. The depth of the upper liquid of density P is b (< a). Find the thrust on the square.

Solution. ABCD is the given square (Fig. 72 above) with side AD in the free surface of liquid of density P.

Let EFbs the surface of separation and AE=b (given). Thrust on the portion AEFD=?g.\b (bxa) ...(1) and the thrust on the portion EBCF

=(pressure at the C. G. of EBCF) x Area EBCF = {g?.b+g°{h(a-b)}]a(a-b) ...(2)

.'. Thrust on the square =thrust on ^£FI>-Mhrust on EBCF " =?g.\b.(bxa)\[g.9 b+ga ft (a-b)}] a (S-b) ~lP.g.Pa+{g.p.ba(o-6)}+g.<r.ia.(fi-byi

=iga [PZ»a+2p6a- 2p.ba+o.(a-b?] =|a& (2a- b) 9g+%a (a -by ag. Ans.

Exercise on § 2 1 3 - § 2'14 ,Ex. A rectangular box is half filled with water and remaining

half with oil. If the oil be half as heavy as water, show that the total thrust on aside of ihs box is one fourth greater than it would be if the box were filled with oil only.

SOME IMPORTANT SOLVED EXAMPLES *Ex. 1. The central board of a yacht is in the form of a trape­

zoid in which site parallel sides are 3 and 5 ft. respectively, and the side perpendicular to these two is 4 ft. in length. Assuming

Fluid Pressure 71

that the last named side is parallel to the surface of the water at a depth of 1 foot and that the parallel sides are yertical. find the total pressure on the board.

Solution. ABCD is the central board of the yacht, with side AD (—4 ft) one foot below the surface of the liquid and sides AB—3' and £>C=5' are vertical. From B draw BF perpendicular to CD. Let Gx and Gs be the C G.s of the rect. ABFD and triangle BCF respec­tively. From G.t draw GaL perpen­dicular to BF, then G9L=i FC

- 1 (5-3)—f ft. ' & The thrust on rectangle

ABFD="w.v S" = w.(l + l$i (3X4) «=(62i)Xixl2=.125xl51bs.

And the thrust on the triangle BCF

="»'.z.S"'=w.(l +3+f) . ( i4 7) - w ( ¥ ) (4)=(62J) <*.*) ( 4 ) - | (125X28) lbs. & Required thrust on the board =thrust on the rect. ABFD f thrust on LBCP

-.1

(Fig. 73)

>(125x25)+& (I25x28)=i (125x73)= 3041 § lbs. Ans.

Ex. 2. A quantity of water which just fills a cone, whose* height is h and the radios of whose base is r, is ponred into a cylinder of radios r Compare the thrusts on the two bases the axis of each vessel being vertical. <Asfa S3»

Solution. In the case of the cone, the thrust on the ba<e of the C9ne="u»z5"^ w h 7rr2-=>7rrahw, •• 0) where w is the weight of the water per uni t volume.

Again the v o k m e of the water that fills the cone=ȣjrra h. This water"} s poured 'n to a cylinder df radius r Let a height h' of the cylinder be occupied by this water. Then we must have .

Vol. of water in the cylfnder=VoJ. of water tha t fills the cone te. •Kr1h'~%itr*h Ue. h'«=\h •»(»)

.". The thrust on the base of the cylinder » « H « S " = W h'.nr* m (Note) «=H> \h izr*, from (ii) <=*%itrahw ...(Hi)

Hence from (i) and (iii), the required rat io <=nrafiA>: %nr*hw=l : $ - 3 : 1. Ans.

Ex. 3 . A parallelogram is immersed in a fluid with a diagonal vertical, one extremity of which is in the* surface of the liquid.

72 Hydrostatics

Tbrough this point lines are drawn dividing the parallelogram into three equal parts. Compare the pressare on three parts, and if P2 be the pressure on the middle part and E1( P3 those on the other two proTC that I6P2«11 (Pi+P^.

Solution. ABCD is the para­llelogram with diagonal DB verti­cal. Let the depths of A. B and € below the free surface be zu z% and ?3 respectively. Then if O be the centre 6f the parallelogram, then OD^iza.

;

Also OD-=\{AK+CN)

Zl^ZiJ-Z3 . . (1) Let P and Q be such points

AB and BC respectively so that area of A-<4.DP«= area of fig. DPBg^areaof ADQC=*A (say)

& Area of the parallelogram ABCD=3 A.

Also AL and PM are perpendiculars to BD. Now area of ADPB^enea. of AABD—area of AAPD

=i(3A)~A^hA=iiU) area of A DPB<=% (area of AABD)

IBD.FM^ (hDB.AL) or PM=>\AL X In similar triangles ABL and PBAf, we have

BM PM t . .,. j ^ = | , from (2) or

or or

on

or or (2)

BL or -*i)

or

or

BM^IBL or ML=>f&L=f (5£>-^iS:)=| (*,-.'. Depth of P below the free surface

. =DM=AK+ML=z1+% (it%-zd-l (zi+2z2) .'. Depth of C. G. of A APD

=1 (depth of A\depth of P+depth of/)) =H*iH"Wr-2*)-r-0}-$ (2Z1+Z2)

/. P^thrus t on A^P»=e,w.2.S"'«=w.f (2z1+z8J!/i Similarly P , = thrust on ADQC*=w.l (2za+zt) A. And P,«=> thrust on the parallelogram ABCD-(PX+ Pa)

= w . K (3A)-[w § (2zx f z8) ^+w f (2z*+z£ A] <~w.A [ ^ a - l {22r,+2za+2z9}]«w.^.[fz2-fzsi], from (1)

J W J w Azs .-(3) Hence thrust P,, P2 and P3 can be compared. Also Pl+Pn**w.$ ilz^+zj A+w.% (2zs+22) A

^fw^.fo+Za+ZsHI w./4.z8 ... V Zj+zB=z2 = 1 [H*.]. from (3)

11 (P1+P3)=16 Pa. Hence proved.

CHAPTER III

Resultant Thrust on Curved Surfaces § 3 01. Thrus* on a curved surface.

Suppose curved surface is in contact with a fluid. Take a point P on the surface and consider the small element of area surrounding this point P. The thrust of the liquid on thiselement will be act­ing normally because the fluid is a perfect fluid. Similarly the thrust on the other elements of area of the surface will be acting normally. Thus these thrusts form a system of forces which are neither in the same direction nor generally In the (Fig. 75) same plane, and such a system can be reduced in general to a single force together with a single couple and not to a single force. But fibis totality of pressure is in fact equivalent to certain forces in the horizontal and vertical directions which can be obtained in the following manner.

Let the fluid thrust on each element be firstly resolved into horizontal and vertical components. These vertical components being parallel will form a system of parallel forces, which is equi­valent to a single force say Z. This single force Z is called the Resultant Vertical Thrust of the fluid on the surfaee. {Rohtikhemd 79)

Next let us consider the horizontal components of the fluid thrusts on different elements. The direction of these horizontal thrusts are nflt parallel. Therefore ctioose two tixed directions Ox and Oy at right angles to one another and res Vive the horizontal components of thrusts on every element along these two perpendi­cular directions. Then all the components resolved parallel to Ox will form a system of parallel forces which can be compounded into a single force, say X, acting parallel to Ox. Similarly the com­ponents resolved parallel to Oy can also be compounded into a single force, say T, acting parallel to Oy. These resultants X and Y which can be thus determined in magnitude and direction are called the Resultant Horizontal Thrusts of the fluid on the surface parallel to Ox and Oy respectively. (Rohllkhand79)

Thus the resultant fluid thrust on the curved surface is equi­valent to three forces X. Y and Z. If in any particular case these three forces X, Y and Z are concurrent i.e. they pass through the

7d Hydrostatics

£ £." rr_ z~zz -~z . — 1

• w 9 5 5 ^

~-=-i? ~ ~ —_r_*^=r

^Hil

« ^Z. — —Z-

~~~ -.z. —-rci—z.

Lr^rt.—

-----_ —

E ^ f ~

M^/MSS^

& rE: r_v ™ . —

n : —-2—-

w (Fig. 76)

the curved surface of the cylinder acting

same point then thev can be compounded into a single resultant force of magnitude y/(X*+ Y2+Za) which may be called Resultant Fluid Thrust on the curved surface. _ (U.P.P.C S. 78)

*§ 302. To obtain the Resultant Vertical Thrust. {Agra 78, 74 ; Magadh 77; Rajasthan 78)

Let the surface on which the thrust is to be calculated be bounded by a curve PQRS. From every point of this curve draw vertical lines to meet the fre'e surface (or the effective surface) of the fluid in a curve P'Q'R'S'. These lines form a cylinder. The fluid in this cylinder called the superincumbent liquid, is in equilibrium under the action of the following forces :

(i) the weight of the fluid in the cylinder acting vertically downwards,

(ii) the mutual action and reaction between the liquid and the curved sur­face under consideration acting normally,

(iii) the thrust of the liquid sur­rounding the cylinder ©n horizontally.

Resolve all the forces vertically, then the weight of the fluid In the cylinder will be equal to the algebraic sum of the vertical components of all the reactions on different elements of the curved surface under consideration i.e. equal to the vertical components of the resultant reaction, which is nothing but the vertical component of the resultant thrust in magnitude.

Hence ths vertical thrust of the fluid on the surface PQRS is equal to the weight of the fluids in the cylinder PQRS, P'Q'R'S', which may be described as the "superincumbent fluid" and acts through the centre of gravity of the superincumbent fluid.

§ 3-03. Special Cases. Case I. When the fluid presses on the surface from below i.e.

upwards. Let PQRS be such a sur­

face as shown in the adjoining figure. Then the pressure at each point being due to the depth "below the free (or effec­tive) surface, it is clear that if we draw vertical lines as before to meet the plane of the free (or effective) surface in the curve P'Q'R'S', then the thrust will act in the upward direction and its vertical component will (Fig. 77)

/

Vertical Thrust on Curved Surfaces 7»

be equal to the weight of the superincumbent fluid which would fill the cylinder PQRS, P'Q'R'S'.

Case I\ When the fluid presses on surface partly upwards and partly downwards.

In the figure below, PQR is the surface in which PQ is pressed upwards and QR downwards.

(Fig. 78^ .". Pressure on PQ

=weight of the superincumbent fluid filling the cylinder PQQ'R' and acting vertically upwards.

And pressure on QR «=>weight of the superincumbent fluid filling the cylinder

RQQ'R' and acting vertically downwards .*. Pressure on Pg/?=Pressure on QR—pressure on PQ

('.' they are in opposite directions f =(weight of the fluid in the cylinder RQQ'R')

—(weight of the fluid in the cylinder PQQ'R') «=» weight of the fluid in PQRP.

.'. The thrust on a surface part; pressed downwards and partly pressed upwards is equal to the weight of the fluid contained.

Solved Examples on Vertical Thrust. *Ex. 1. A conical wine glass is filled with water and placed

in an invented position upon a tab|e ; show that the resultant vertical thrust of the water on the glass is two-thirds that on the table. (Agra 75 : Bhopal 83 ;

Jodhpur 76 ; Luckaow 83) Solution. Let h be the height and

r the radius of the base of the cone. The water contained in the cone is

pressing the curved surface of the cone upwards and tha resultant vertical thrust of the water on the curved sur­face is equal to the weight of the 'super­incumbent liquid' which would fill the space between the cone VAB and the circumscribing cylinder ABCD. (Fig. 99)

76 Hydrostatics

*\ The resultant -vertical thrust on the curved surface of the cone =[The vol. of the cylinder ABCD-the vol. of the cone (VAB)]g? •=(nrah—itcr2h) gp*=*l-nr*hgp, where p is the density »f the water.

Also the thrust on the base of the cone •=• weight of the superincumbent liquid =(The volume of the cylinder ABGD) g9=*itrVi.gp

:. The resultant vertical thrust on the curved surfaee of the cone—f jrrsAgP=%f (the thrust on base of the cone). Hence prored.

Es. 2. A conical vessel filled with water, stands with its plane base on a horizontal table. Prove that the thrust of the liquid oa the basa of the cone is three times the weight of the contained water. How do you explain the result ?

Solution. As in the last example if A be the height and r the radius of the base of the cone, then

the weight of the liquid contained= (JwraA) gg. Also the thrust on the base==weight of the superincumbent

liquid Explanation of the result. The water in the eone is in equili­

brium under the action of the following forcss : (1) th«. reaction ef the base en the water aitinj vertically

upwards, (2) the reaction «f the curved surfats ©f ths tone BJJ»B the

water acting vertically downwards, and m (3) the weight of the liquid %nr*hg?, acting vertleally dew»-wards.

Whence we conclude that the reaction of the base™ the reaction of the curved surface-f weight of the liquid contained.

i e. the reaction of the base or the thrust of the wa'er on the base > the weight of the water contained in the cone.

Ex. 3 A bucket in the form of a frustum of cone is filled with' water If the top and the bqttom ends be of radii o and b (b < a) and the height k, find the resultant ver­tical thrust on ttee curved surface

Solution The vertical thrust on the curved surface of the bucket -=»the weight of the superincum­bent liquid. , •='the weight of the liquid filling the space between the frustum ABCD and the cylinder ABFE =• (volume of the frustum ABCD —volume of the cylinder ABFE) gp <=[fck (al+b*+ob)-*ndVi] gp.

Vertical Thrust OH Curved Surface 11

(Fig. 81)

=4?cA (<J2+afl-2dB) gP. Ans. Ex. 4. A hemispherical vessel filled with water is placed in an

inverted position on a horizontal table, find the resultant vertical thrust of the water on the vessel. If the water is on the point of escaping, find the weight of the vessel. (Agra 75; Magadh 77; Ranchi 73)

Solution. Let r be the radius of the hemispherical vessel AEB. Resultant thrust of the

water on the curved surface of the vessel acting upwards -=«WGight of the superincum­bent liquid which would fill th« space between the hemi­sphere AEB an J the circums­cribing cylinder ABCD*a(vol-ume of the cylinder ABCD —volume of the hemisphere AEB).gP =(wa.r— §nr8).gP°-lnrs gpt

wh«re p is the density of water. If the water is on the point of eseaping, then the wiigkt ©f the

vesssl must be equal to the upward thrust on the vessel / e. weight of the vessel=j7i:r*.£p. Ans. [Also the thrust on the table™,8gpz1S'"'-"gp.r.Tcra, v the depth of the C.G. of the plane base belew E is r. Hence the resultant vertical thrust on the surface of the hemi-

sph«8-4TCr3gP—\ (thru»t on the table). • Vera M. 5c 75)] **Ex. S. A hollow cone is placed with its vertex upwards on a

horizontal table and liquid is poured in through a small bole _ in the vertex ; if the cone begins to rise when the weight ©f the liquid pou­nd in is equal to its own weight, prove that its weight is to the weight •f the liquid required to fill the cone, as 9—3\/3 : 4«

Solution. Let (V, AB)\K the cone of* height A and the semi-vertical angle «.

Let the cone begin to rise when the height of liquid in the eone is (A—A').

Then the weight of the cone =weight of the liquid contained

(given) =[vol. of cone (V, AB)- Volume

of cone {V, EF)].gp <=(!*&" tan8 a—fak" tan2 a) g?, where p is the density of the liquid •»lw(A ,-A ,»)tan ,a.£p. ^Fig. 82)

Also the vertical thrust on the curved surface of the cone in contaet with the liquid, acting vertically upwards.

ihrh")! £=£=z=± L"JL-z45:

n Hydrostatics

JirA8 tan3« gp

=weight of the superincumbent liquid that would fill the space bet­ween the frustum ABE and the circumscribing cylinder ABCD. =(vol. of cylinder ABCD) gp—weight of the liquid contained (Note)

={7r (h tan «)a. (h-h')} gp—^r (ha-h'a) tan3 a gp =4w tan2 a {3h* (h~h')-ih3~h'a)} g? =£71 (h-h') tan* a (2h*-hh'-h'a) g?

The cone, will begin to rise when the weight of the cone»=»the upward vertical thrust of the liquid on the cone t e. \n (h*-h'3) tan8 agp^rc {h-h') tan3 a (2h2~hh'-h'2) gp or (h3-h'a)=*(h-k') (2h2-hh'—h'2) or tf+hh'+h'^ltf-hh'-h'*, V h~li ^ 0 or 2h'*+2h'h—tii=0 or h'~l [-2h±i/(.4h&^m)]^l [-2/i±2\/3AS=4 (V3-1) h,

as h' cannot be —ve „ weight of the vessel ^jn (/z8—ft'3) tan2 a gp

e n C e weight of liquid required to fill it

= 1-1 (V3- l ) B <=l - t [3V3-3.3.1+3.V3.1-1] = | [8—(6^3—10)]=^ [l8-6«^3]-K9~3V3). Hence proved.

' Ex .6 . A vessel in the shape of a hollow hemisphere surmounted by a cone is held with the axis vertical and vertex uppermost. If it be filled with a liquid so as to submerge half the axis of the cone in the liquid, and the height of the cone be double the radius of its base, show that the resultant downward throst of the liquid on the vessel is 15/8 times the weight of the liquid that the hemisphere can hold,

Solution. LQt r be the radius of the base of the hemisphere or cone. Then the Tieight of the cone=2r

(given) i.e. VK<=>2r or VL=LK*=>r Also from similar triangles VLF

and VKB, we ^j^yf^^ "i

cr LF<=\KB<=*\n .'. The volume of the frustum

ABEF^'fr h W + r . ' + i y , ) " s = ^ r [ r a + ( 4 r ) H r ^ r ) ] — A w 3 .

And the volume of the hemi-sphere =>|wrs.

i. weight 0f the liquid con­tained in the vessel

Vertical Thrust on Curved Surfaces 79

a(isnra + lnrs)g?c=lnr3gP, where p is the density of the liquid. Now the resultant vertical thrust on the vessel, which is partly

pressed upwards and partly downwards =weight of the liquid contained (see § 3'03 case II Page 75) = ! w 3 g p = 1

86 ( 5 f r V ) (Note)

•= V (the weight of the liquid that the hemisphere can hold). Hence proved.

Ex. 7. A conical cup whose weight is |tn of Tthe freight of water which would just fill it is placed vertex upwards on a smooth table and water is gradually poured in through a hole made in the top. Show that the cop will be on the point of rising from the table when the water reaches half the height of the cup. (Lucknow 74)

Solution. Let A be the height of the cone and a its semi verti­cal angle. Then the volume of water which can just fill the cone

«=>i7iA8 tana a [See Fig. 82 Page 77] .'. The weight of the conical cup.

= | (the weight of water which would just fill it) =»| [Wz8 tana a gp], where p is the density of the water =•!& rc/i8 tan2 a gp (!)

Let (li—h') be the height of water in the cone, when it is on '• the point of rising from the table.

The upward thrust on the curved surface of the cone la contact with water =>the weight of the superincumbent liquid which can fill ihe space between ABFE and the circumscribing cylinder ABCD.

~[(vol. of the cylinder ABCD) -(vol. of the frustum ABFE] g? =•({* (A tan a)a (h-h')}-{&& tan2 a - ^ / j ' 3 tan2 a}] gP - i * tan2« [3A« (h-h')-h*+h'*} g? Now the cone will begin to rise when the weight of the cup

•^upward thrust on the curved surface of the cone i e. -Arf tan" a gp=ln tan2 « [3h* (h-h')-h3+h'3] g? or f A3=.3/za (h-h') -h3^h'^2h3~3h2 h'+h's

or 5A3=. 1 6 ^ - 2 ^ " h' + W* or ll/*3-24/ja/j'+8/i'8=0 or ih-2h') (nha-2hh'-4h'2)**0 or h—2ft'=0, other values of h' being inadmissible. or ti>=*\h or h—h'^h—ih^ih i e. the water rises to half the height ol the cup. Hence proved.

Ex. 8. A pyramid with a square base and with sides which are equilateral triangles is placed on a horizontal plane and filled with a liquid through an aperture in the vertex, find the pressure on one of the side. {Agra 73)

If the pyramid has BO base, find its least weight consistent with its not being raised.

Solution. Let V be the vertex and ABCD be the base of the pyraaaid. Let h be its height and a the length of each side of its base. Then BZ)«VC>a+a2

l)=flV'2.v Also if O be the mid-point

80 Hydrostatics 136/5

or or or

the

of BD, then OB=i BD^\a^2^a(^2. Also in A VOB, VO=>h, OB=-a\y/l and VB<=°AB*=,a. And in right angled triangle VOB, VBs=-VOs+OBa

aa«=,Aa+(a/V2)a

A-=(«/tf2) ...(1) Now suppose we are to find

pressure on tbe face F5Cof the pyramid. Then the depth cf C G. of &VBC below the level of

:. Thrust on the face VBC t=*"wz.S"~g?ih.(ia.a sin 60°), where p is the density of the liquid = \ha\l</Sg9=>WS (a/V2) a* g?

...from(l)[/z~a/V2 ="iW6a3gP. Ans.

Also the vertical upward thrust on the slant faces of the eumbent liquid which would fill and the circumscribing prism.

=(vol. of the prism

(Fig. 84) pyramid=weight of the superin-the space between the pyramid

(Note) vol. of the pyramid).gp

<=*[{a\h)-{\a\h)\g9=lcPh gp-§a» (a/V2) gP~iV2a3 gp. If there be no base of the pyramid, then the pyramid would

not rise if the weight of the pyramid=avertical thrust of the liquid •n the pyramid=>it^2a* gP. Ans.

Ex. 9. A hollow cone filled with water and closed, is held with its axis herizontal; find the resultant Tertical pressure on the upper half of its curved surface. Find it on the lower half as well.

{Lucknow 78 ; Rajasthan 76 ; Rahilkhand 78 ; yikram 74) Solution. Let V be the

vertex of the cone and O the; centre of its base. Let h be Its height and r the radius of its base. The horizontal plane VAB diyidesthe cone into two halves.

&, Vertical thrust on the upper half of the curved smface of the cone=welght of the superincumbent liquid, which would fill the space between the upper haff of the <. (Fig. 85) cone and the circumscribing prism on A VAB as base=[(vol. of prism on A VAB as base)—(volume of upper half of the cone)] gP, where p is the .density of the water.

136/6 Vertical Thrust on Curved Surfaces 81

=[(area of A VABx 0 0 ' ) - * (hnr*h)] g? . N -[(*2r.Axr)-(Si.r»A)l«i-(l-iw)r*AgP . Ans.

And the vertical" thrust on the lower half of the cone=>weight of the superincumbent liquid which would fill the lower half of the cone as well as the prism on A VAB as base.

«= [(volume of lower half of the cone)+(volume of the prism on A VAB as base)] g?

-[<i.JiwJA) + (*.2r.A r)] gP~(Jro-,A+r2A)gP~<l-r*7c)',iiAgP. Ans. Ex. 10. A solid right cone, height A and radius a (of the base),

is complete!? immersed in a liquid with its axis horizontal and at depth b. Compare the vertical thrust on the two halves into which the carved surface is divided by a horizontal plane through the axis.

Solution. Let VAB ba the horizontal plane through the axis VO, which divides the cone in tw i halves. Also the cone here is solid and liquid surroaads i t The free surface is at a height b above-the plane VAB.

c (Fig. 86)

/. Vertical thrust (acting downwards) on the upper half of the curved surface of the cone=weight of the superincumbent liquid, which would fill the space between the upper half of the cone and prism on A V4fi a s base and of height b • D£P [(volume of the prism on A VAB as base and height b) —(volume of upper half cone)], where p is the density of the liquid -#> [{area of A VAB) b}-{l&o*h}) «=gP [{hh.2a.b}-Qna2h)]~g9 [hab-\nO?h]~lhag9 [6b—no].

Also the vertical thrust (acting upwards) on the lower half of the curved surface=welght of the superincumbent liquid =gP [(volume^pf the prism on A VAB as base and height b)

+(volume of lower half of cone)] ~g P [{J h.2a &}-HH7rasA}]=i/ifl 8? (66-f-wa)

.*. Required ratio of the vertical thrusts on two lialves ^\hag? (6b—-na) : lhagp (66-f7ra)=>(66-wfl) s {6b-\-ita) Ans.

Ex. 11. Find the resultant vertical fluid %ust on the lower half of the curved surface of a cylindrical pipe of length A and radius r when the pipe is full. {Agra 73)

~82 Hydrostatics

(Fig. 87)

Solution Let ABCD be the horizontal plane through the axis which divides the cylinder in two halves. Then toe downward vertical thrust on the lower half . , of thecylinder=weight " » Q O of the superincumbent liquid which would fill the lower half of the cylinder as weJl es the prism on ABCD as base and whose height is A A' («=/•)= [vol. of the lower half of the cylmder)+(vol. of the prism on ABCD as base)] gp, where p is the density of the fluid <=.[(|.7crs/2)-Harea of rectangle ABCDxAA')] g? <=*[(knrsh)-t(2rhXr)] gp«=-[(irc+2j r!h gP]. Aas.

*&x 12. The shape of the interior of a vessel is a doable cone; the ends being open and the two portions being connected by a minute aperture at the common vertex. It is placed with ons circular rim fitting close upon a horizontal plane and is filled with water. Find the resultant vertical thrust upon it, and prove that if it be zero,, the ratio of the axes of the two portions is 1 : 2.

Solution. Let hv and hz be the heights of the upper and the lower cones respectively. Let« be the semi-vertical angle of each cone. The vertical thrust on the curved surface of the upper cone acting downwards=weight of the super­incumbent liquid i e. weight of the liquid in this cone=^7sAj3 tan2 a gp, where p is tae density of the liquid.

And the vertical thru'st on the curved surface of the lower cone acting upwards •^weight of the superincumbent liquid whicD would fill the space between the cylinder of height {h2+h^ on the base AB of the lower cone and the lower cone.

<=g9 [{Vol of cylinder on the base AB asd of height (^i+/22)}—vol. of lower cone]

=gP [{« (A. tana)8 fo+ft,)}-***/ tan2 «] =\ng9 tan2 a [3ft/A1+3Aa»-A«'M>OT tan2 a (lhfhx+lh£) ,\ Resultant 'vertical thrust on the vessel acting down­

wards = (vertical thrust on the upper cone) —(vertical thrust on the lower cone)

(Fig- 88) J

Vertical Thrust on Curved Surfaces 83

Ans. =0

or or or

= ( W tai,2 a g?)-{§ng? tan3 « (3!h%+2h23)}

~ i « tan2 a [V-3VAi-2A«1 g? If the thrust is zero, then we have {h^-3h^h1~2hz

s)= ih-2h) ( V + 2 A A + V ) 1 3 0

(^-2Aa) (A,*W'=*0 or Ax-2Aa=0, V Ai+Aa^O ;2,=.2/z2 or A 1 :A 2 =2:1 . Hence proved.

**Ex. 13. A doable funnel is formed by joining two equal hollow cones at their vertices anil stands on a horizontal plane with the common axis vertical ; liquid is poured into the cone until its surface bisects the axis of the upper cone. If the liquid be on the point of escaping between the Sower cone and the table, prove that the weight of either cone is to that of liquid it can hold as 27 s16.

Solution. Let a be the semi-vertical angle and h the height of each cone.

The vertical" thrust on the curved surface of the upper cone VCD, acting downwards =»weight of the superincumbent

liquid . •^weight of the liquid contained

in this cone = 3^ i\hf tan2 a,gp, where p !s

density of the liquid. •=Tfcph* tan3 a gp.

And the vertical thrust on the curved surface of the lower cone VAB, acling upwards =weight of the superincumbent liquid which would fill the space between the cylinder on base AB and of height (h + \h) I.e. lh and the lower cone* (Fig, 89) =#P [vol. of the cylinder on base AB and of height fA)—(vol. of the lower cone)] «jgP [<JB (h tan «)«.f A}-{inA» tan8 a}] = »rAs tan2 <*[£—$] gp=%nh9 tana « g?

J. Resultant upward thrust on the surface of vessel •=» (Thrust on the surface of lower cone)

•—(thrust on she surface of tae upper coae)N

=(£rcA8 tan2 « gp)-faith* tan2 a gP)«=»§7r/z3 tan* a gp if the liquid be on the point of escaping between the lower

cone and the table, then the weight of the vessel (z.e.donble cone} must be equal to the upward vertical thrust on the v%sse]

I.e. 2W=%nhs tan8 « gp, where W is the weight of each cone or W=f&nh* tan8 «..gp.

84 Hydrostatics 284 -66

Jl weight of either cone W

Inh3 tan8 « g? weight of the liquid it can hold •ftnh* taoa a gP = ? 3 L

a Inh* tan8 a g? 16 Hence proved. **Ex. 14. A hollow sphere of radius a is just filled with water,

find the resultant vertical thrusts on the two portions into which the surface is divided by a horizontal plane at depth c below the centre.

Solution. O is the centre 1 of the sphere which Is just filled with water and hence the free surface of the water is the hori­zontal plane patslng through the highest point P of the sphere. Let CD be the horizontal plane, at a depth c below O, which divides the sphere into two portions. AB is the horizontal plane passing through the centre O.

Also ND~-tf(OD*-OW)

and PN*=PO + ON<=>a+c

The downward vertical thrust on the portion CQD of the curved surface of the sphere (Fig. 90)

=weight of the superincumbent liquid C'CQDD' •= [(vol. of the cylinder CDD'C)

-{-(vol of the spherical segment CQD)] gp, where p is the density of the water

=[{«r (NDf.PN}+{fr (a -c) (3a2~aa-flc-ca)}] gP (Note) '[{it (aa-<:a).(a+c)} + \K (a -c) (2a2-ac-o*)] gP >\TC (a-c) [3 («+c)a+(2aa-so-ea)] gP b(a-c)(5fls+5fl<H-2c2)gp - l l )

The vestical thrust on the portion CAPBD of the curved surface of the sphere is comprised of the upward vertical thrust on the curved surface APB and the downward vertical thrust on the spheri­cal zone ACDB.

Now the upward vertical thrust on the portion APB >=weight of the superincumbent liquid which would

fill the space between the hemisphere APB and the circumscribing cylinder ABB'A' =[(vol. of the cylinder ABB' .4')-(vol. 0f the hemisphere APB)] gP

>[(jraa.a)-(f7rfl3)]gp= 4rea8gp .(2) And the downward thrust on the zone ACDB

=weiglft of the superincumbent liquid =[vol of the cylinder ABB'A'+vol of the cone ACDB

-vol . of the cylinder CDD'C] gP (Note)

Resultant Horizontal Thrust 85

=[(7ra2.fl)-B* (c) IW~CS)-TZ (a2-c2) (a + c)] g? **\K [3a3+3asc -c3-3a3+3ac»-3a2c+3c3] g? «iTC(?ac2+2c8^P. ..(3)

Hence the resultant vertical thrust on the curved surfact CAPBD of the sphere will be the difference of (2) and (3)

I.e. & (3ac»+2cs) £P~4roJ8gp or fr f(3ac"+2c3)~as] g? ...(4) which would be upwards or downwards according as (2) is greater or less than (3), depending upon the values of a and c.

Ex. 15. A hollow sphere is just filled with wfiter, prove that the resultant vertical pressures on the upper and lower halves of the internal sarface are \w and f w respectively, where w is the weight of the water (Asra 8°. 73>

Solution. Hint : w=|7ra3gp Putc~OinEx. 14 Page 84 Equation (1) gives the vertical thrust on the lower half

=»f;rfl8gp=fw Equation (4) gives the vertical thrust on the upper half

= | r t A8 £p=£ji>. Hence proved. Exercises on Vertical Thrust.

Ex. 1. A conical vessel 20 cm. high on a flat circular base of 10 cm. radius is filled with water. Calculate the vertical thrust on the base when the vertex Is upwards

Ex. 2. A hollow closed vessel in the shape of a cylinder sur­mounted by a cone is filled with liquid If the axis of the cone be 3 times as long that of the cylinder, prove that the resultant thrust on the surface of the cone will be the same in the two positions in which the vessel can be placed with its axis vertical

*Ex. 3 A vessel in the shape of the greater segment of a sphere of radius a cut off by a plane at a distance c from the centre rests with its base on a table. If the segment is filled with water, show that the resultant vertical thrust on the curved surface will be an upward rfoce if c < J«.

!§ 3C4. To obtain the Resultant Horizontal Thrust. (Jodhpur 76 ; Mithila 82 ; Rojasthan 78 ; RancM 81,76; Vikram 74)

Let VLBM be the curved surface on which we are to obtain the horl- " - - y ^ 7 - — zontal thrust of the fluid in an assigned direction, the direction of AA', say.

Draw horizontal lines in the assigned direction I.e. parallel to AA' through every point of the ==f5^™^€-==>Si5i^ boundary of the surface -^---~~-^------^.------ABGD, and take a vertical \ cross-section A'L'B'M' of (F|g. 91)

Lti—— z. - ~ ~ :

#6 Hydrostatip

the cylinder formed by these parallel lines Then A'L'B'M' is the projection of the surface ALBM on a~vertical place.

Consider the equilibrium of the fluid in this cylinder bounded by these parallel lines, the curved surface ALBM and the vertical plane A'L'B'M'. The forces acting on the fluid, within this cylinder in the direction of its generator i.e. the horizontal direction AA' are

(i) the horizontal component of the reactioa pf the curved surface ALBM on the fluid in the assigned direction, and

(ii) the thrust of the plane end A'L'B'M' acting at right angles to this area i.e. parallel to the assigned direction.

In equilibrium these forces must balance one another. Also the horizontal thiust of the fluid on the curved surface ALBM is equal to the horizontal component of the reactioa of the curved surface ALBM on the fluid Hence the required horizontal compo­nent of the thrust on the curved surface ALBM is equal to the thrust on the plane end A'L'B'M'.

Thus we obtain the following rule for finding the Resultant Horizontal Thrust :—

The resultant Horizontal ThrHSt on given surface in contact with a liquid in an assigned horizontal direction is. ia magnitude and line of action, equal to the whole pressure on the projection of the surface upon a vertical plane perpendicular to the assigned direction.

[Important Note. In the above theorem, it has been assumed that each of the horizontal lines drawn parallel to AA' cuts the surface at one point only. If however lines parallel to AA' cut the surface at more than one point, then the given surface should be divided into two or more parts so that the above theorem may be applicable to each part separately. The horizontal thrust in the assigned direction on each of these parts should be found separately, which after being compounded will give the resulSaut horizontal thrust on the whole of the curved surface in the assigned direction.

§3 05. Resultant Thrast In § 301, Page 73, we have dis­cussed how the resultant thrust on a curved surface can be obtained in terms of X, Y and Z We have shown in § 3 02 Page 74 and § 3'04 P. 85 how the vertical component Z and the horizontal component, say X, in an assigned direction can be obtained. In a similar manner as explained in § 3-04 P. 85, we can obtain the resultant horizontal thrust. on the curved surface in a direction perpendicular to the previously assigned direct'on and call it Y. In this way we can completely determine the Resultant Thrust on any surface in contact with a fluid, by determining the vertical and horizontal thrusts on it.

In case of certain symmetrical bodies, these three forces X, Y

Resultant Thrust on Curved Surfaces 87

and Z are found to be coplanar aDd can be compounded into a single force <\Z(X*-\-Y&-\-Zz) in magnitude, which is called the Resultant thrust.

*•§ 3.06. Resultant Thrust on a solid immersed in fluid—Prin­ciple of Archimedes.

Principle of Archimedes may be stated as follows : When a body is wholly or partially immersed in hea<*y fluid at

rest the resultant thrust of the fluid on the solid is equal and opposite to the freight of the fluid displaced by the solid and acts vertically upwards through the centre of gravity of the fluid displaced.

(Bihar 74 ; Lucknow 77 ; Vikram SO) Proof i— Case I. When the body is wholly immersed in a heavy homo­

geneous liquid Take a point P on the body

and consider the small element of area surrounding this point P. The thrust of the liquid on the element will be acting normally because the fluid is a perfect fluid. Similarly the t hrusts on th e other elements of the body will be acting normally. The direction of the thrust on each element bsing different, it is difficult to find the resultant thrusC by compounding all the separate thrusts on different elements. Hence we proceed as follows :—

It is necessary to observe that the resultant thrust of fluid on the (Fig. 92) solid depends only on the shape of the solid and the position which It occupies, and is clearly Independent of the material of which the solid is composed of. The thrust of the fluid on all bodies which would exactly fill the same space would be the same. If then we imagine the solid to be removed and the space which it occupied to be filled up with the fluid of the same kind as the surrounding fluid, this mass, »&ich we may call, "the displaced fluid" would be in equilibrium under the action of its weight and the thrust of the surrounding fluid upon it Hence the resultant thrust of the fluid upon the solid which would fill the same space must be a force equal and opposite to the weight of the flald displaced and act upwards through the centre of gravity of the fluid displaced. The force Js called the force of buoyancy and the centre of gravity of the fluid is called the centre of buoyancy.

Case II. When the body is completely Immersed in the heavy heterogeneous fluid.

88 Hydrostatics

When the fluid of different densities are" at rest under gra­vity and do not mix, the surfaces of separation are the hoiizontal planes and the fluid pressure is the same on each separate surface at each and every point.

Now replace the body by - the actual fluffi surrounding the body i.e. if we pass a horizontal plane through the fluid wbkh replaces the solid and the sur­rounding fluid, the pressure at

' each and eveiy point of this plane should remain the same. . (Fig 93) Then the fluid which replaces the body will be in equilibrium under the action of two forces viz. the resultant thrust of the surrounding fluid and its weight acting vertically downwards through the centre of gravity of the fluid which replaces the body.' Hence the resultant thrust will be equal to the weight of the fluid displaced acting verti­cally upwards through the centre of gravity of the fluid displaced.

Case III When the body is partially immersed in a heavy liquid,

In this case also the resul­tant thrust of the liquid over the portion of the body immersed must be equal in magnitude to the weight of the body to maintain equilibrium, and should act \ert i- , cally upwards through tbe centre of gravity of the body.

§ 3-07. Centre of Pressure. We have already seen that when a plane surface is In contact with a fluid, the fluid exerts pressure at eaoh,point of the plane surface acting normally to it. The magni­tude of the thrust at each point of the plane surface depends upon the depth of the point below the free surface and hence differ from point to point but the direction of all these thrusts beitjg parallel', these form a system of like parallel forces which can be compoanded into a single force acting at some definite point of the plane surface. The single force is the Resultant Thrust or the Resultant Pressure of the fluid on plane surface and the

t

--_-j=^3

/ /

*"*--

\ \

>

6 I . - --Z

t-_:

(Fig 94)

Centre of Pressure 89

point on the plane surface where it acts is called the Centre of Pre­ssure of the plane, area.

Dafinition. The Centre of Pressure of a plane area in contact with a fluid is the point of the area at which the Resultant thrust on one side of the area acts.

Below we give the position of the centre of pressure for a few simple standard cases, the proofs of the same will be given in the next chapter on "Centre of Pressure". In all the cases given below atmospheric pressure has been neglected.

0.

M^^^Mf^^M (a) Centre of Pressure for a Rectangle. ABCD is the recfan- ^J^SS^KrJr^SSS~^=---^--~^-gular area immersed Jn a homo­geneous liquid with the side AD (Fig. 95) In the surface. E and F are the middle points of the sides AD and BC respectively, Then the Centre of Pressure of the rectangle ABCD will be at P In EF, such that

EP-=>IEF i.e. EP : PF">2 : I. (b) Centre of Pressure of a

triangle with its base in the sur­face.

ABC is the triangle immer­sed in the homogeneous liquid with its base AC in the surface of the liquid. E is the middle point of the base AC. Then the centre of pressure P of the tri­angle ABC is the middle point of BE.

(c) Centre of Pressure of a triangle with its vertex in the surface and base horizontal.

ABC is the triangle immer­sed in the homogeneous liquid with the vertex A in the surface and base BC horizontal. D is the middle point of the side BC. Then the centre of pressure P of the triangle ABC is the point on AD, such that AP<=*\AD i.e. AP i PZ)=3 : 1.

i

(Fig. 97)

90 Hydrostatics

(d) Centre of Pressure of a circle. ABCD is the circle, of radius a with

centre Ot totally immersed in the homo­geneous liquid with it's plane vertical and centre O at a depth h below she free sur­face, so that OL<=*h and AC is the vertical diameter.

Then the centre of pressure P of this circular area lies on AC,, suck that

or LP' az

(Fig. 98) Solved Examples on Horizontal Thrust and Resultant Thrust. Ex. 1. A right circular cone filled with liquid is held with its

axis veitical." Prove that the horizontal thrust on half the carved surface Cut off by a plane through the axis, when the vertex is up­wards is twice that when the vertex is downwards.

Solution. Let h be the height and r the radius of the base of the cone, whose vertex is V,

If the vertex V is upwards. If X and Y{ be the thrusts in two mutually perpendicular

directions OA and OD respectively, then X=> horizontal thrust In the direction of OA. =»thrust on the projection of half the

curved surface VABD on a vertical plane at right angles to OA

=thrust on A VBD =>fP-fA.(i.2r.A)=|rAagp, where p is the density of the llqcid.

Also y=horizontal thrust in the direction1 of OD

=0s since the surface 5s symmetrical about a plane VOA perpendicular to OD. (Note)

Hence in this case' the resultant horizontal thrust

If the vertex V is downwards. Let VABD be the half curved surface of the cone under consi­

deration. If X' and Y't be the thrusts in two mutualiy perpendioular

directions OA and OD respectively, then

-aWiagP

Resultant Thrust 91

^^Z

(Fig. 100) Hence proved.

X'=horizontal thrust in the direc­tion OA

=thrust on the projection of half the curved surface VABD on a vertical plane at right angles to OA «thrust on A VBD

and Y'=horizontal thrust in the direction 0C=O, as before by symmetry.

„.'„ Resultant Horizontal thrust in this case =tf{X'*+Y'*)~\rhi g?. ...(2)

Hence from (1) and (2) we get that the horizontal thrust on half the curved surface VABD of the coi,e when the vertex is upwards is twice that when the vertex is downwards.

Ex, 2. A hollow cylinder closed by a plate base is filled with liquid and held with its axis vertical ; find the magnitude and the line of actios of the resultant fbrusr oo half the cylinder cut off by a vertical plane through the axis.

Solution. Let the vertical plane ABB'A divide the cylinder Into two halves. Let h be the height and r the radius of the cylinder.

If Zand Fbe the horizontal thrusts in two mutually perpendicular directions OC and OA respectively, then

Z=horizontal thrust in the direction OC

>=• thrust on the projection of half the curved surface on a vertical plane perpendicular to OC

=tbrust on the rectangle ABB'A' *=°gp.%h(2r h)=2rh?gp,

where p is the density of the liquid, and Y«=» horizontal thrust In the direction OA

(Fig 101) «=0, as the curved surface is symmetrical about the p'ane

OCC'O' perpendicular to OA. If Z be the vertical thrust on half the curved surface under

consideration, then Z=welght of the liquid contained in half the cylinder

•=> $ [nrsh gp] => \Tzr2hgp ;. Resultant thrust on half the curved surface of the cylinder

- ^a 2 +y 2 +Z a )=V{(W^p) 9 +(0)4 (^ r%P) 2 } =ir/jgpV{4AH^r2}.

&2 Hydrostatics

( nr 2h

Also if & be the angle which the resultant thrust makes with the horizontal, then

. „ Z inr7hg? srr „ . .

Ex. 3. A right circular cone is divided into two parts by a plane through its axis. One of these portions is just immersed vertex downwards in water. Find the resultant thrust on its carved surface, and show that it is Inclined at an angle tan-1 (£rc tan a) to the hori­zontal, where a is the semi-vertical angle of the cone. {Lucknow 75)

Solution. Let h be the height of the cone and V Its vertex. Let VBD be the vertical plane which divides the COQC into t\ro halves. Let VCDB be the half under consideration.

I f Z a n d T b e the horizontal thrusts on the curved surface of this half in two mutually perpendicular directions OC and OD respectively then

X= resultant horizontal thrust in the direction of OC

=thrust on the projection of „,' the curved su face under con- ' \ sideration on a vertical plane perpendicular to OC

«= thrust on A VBD "=£P \h (i 7h tan a h), where p is

the density of the liquid «»£/z3 tan a g? ~.(1J

and r=resultant horizontal thrust in the direction of OD

=0, as surface is symmetrical „ about the plane VOC perpen- ^ dlcular to OD. (Fig. 102)

Let Z be the vertical thrust on the curved surface under consi­deration, then Z=>vertical thrust on the curved surface

= weight of the superincumbent liquid which would fill half the cone

~ * (inh9 tan2 a gP)=\TZ1I3 tan8 a g? ...(2) Also X acts normal to the plane VBD at its centre of pressure

which lies on VO, hence line of action of X lies in the plane VCO. And Z passes through the centre of gravity of half the cone

under consideration, which also lies in the symmetrical plane VCO. Hence X and Z lie in one plane and they must meet • each

other at some point, as they are not parallel. £. Resultant Thrust«=a<v

/cr8+rs-r-Z2)

^^{iW tan oc g?)'+(Q)*+tfrh8 tan2 a #>)•}, from (1) and (2) ^ ~4/r* t a n a g P ^ l - f J tan8a) . , Ans.

Resultant Thrusts on Curved Surfaces 03

Also if Obe the inclination of the resultant thrust to the .. , , , . Z J r f tan8 <x ;eP , ,

vertical, then tan 0 - y ° i / t a t a D K g p - i « tan «

or flratan"1 (iit tan a). Hence proved. Ex. 4. A hemisphere of radius a is immersed in a liquid of

density <r. The plane of the base is vertical and its centre at a depth a\/5 below the surface. Show that the resultant force on the curved

its direction makes with ttfeh {Lucknow 80}

surface is inaga3 and its direction makes with th*e horizontal an angle 6, where tan 0=»2/V45.

Solution. OA*=>a^S (given) From Archimedes Principle, we have resultant hrust on the

entire hemisphere =*the weight of the liquid displaced mafltcflga. ^

This force acts vertically upwards at the centre of buoyancy G, where 0(j=fa.

The thrust on the plane area will be normal to the plane and acting at the centre of pressure vertically below the centre of gravity O of the plane and In magnitude

=the horizontal thrust on the curved surface=Z (say).

There is no vertical thrust on the plane area. Hence the two components of the thrust on the curved surface in the vertical and horizontal directions are

Z=1JT a*ga and X^itas\^5ga Hence the required resultant thrust

=^(X»+Z») »V{(fT0V) i+(7 taV5^)a} w=7tasgo-v/{»+5}=|3ras g°

Also if the resultant thrust makes an angle 0 with the horizontal, then

Z _ %Ka*ga _ 2 _

(Fig. 103)

tan 0=- Hence proved. X jraV5gcr 3V5 V(4S) Ex. 5. A solid hemisphere of radius a is placed with its centre

at a distance h below the surface of water and has its plane face vertical Find the horizontal thrust on the curved surface. Find also the resultant thrust on it.

Solution. If X and Y be the resultant horizontal thrust in two mutually perpendicular directions EO and AO respectively, then

Z=horizontal thrust in the direction of &0=thrust on the projection of the curved surface perpendicular to EO

94 Hydrostatics

=thrust on the circle ACBD =gp.h 7ia\ where p is the density

of the liquid. F=horizontal Ihrust in the direction of

AO^O, since the surface is symme­trical about the plane DEC. Hence the Squired horizontal thrust

on the curved surface ^V(^a+Ysi'=nashg9 Ans. Also if Z be the vertical thrust on the

curved surface, then Z=>weight of the liquid displaced

(Principle of Archimedes) ~ =?frcas£P-

.'. The required resultant thrust on the curved surface

- VGP+ Y*+Z*)~ tflfraVigpf+O+(fra5g?)s] «=7:a

3gP<(A2+fa3)'=^n,aV(9/ja+4fl2) g? • Ans.

then tan 8*

Also if this resultant thrust makes an angle 0 with the horizontal

'X na*hgp 3h o r " " t a n \ 3 A ; Ans *Ex. 6. A hemispherical bowl is filled with water, find the

horizontal fluid pressure on one half of the surface divided by a verti­cal diametral plane, and show that it is lfr of the magnitude of the resultant Quid thrust on the whole surface.

i . . (Garhwal 78 : Lucknow 76 ; Rojastkan 75 ; Vikram 7S)\ Solution. CED is the verti­

cal plane dividing the hemisphere into two halves. Let r be the radius of the base of the hemis­phere.

If X and Y be the horizon­tal thrusts in two mutually per­pendicular, directions OA and OC respectively, then

X=horizontal thrust in the direction of OA. (.Fig. 105)

=thrust on the projection of half of the curved surface urider consideration perpendicular to OA

=thrust on semi-circle CED' =£P (4r/3rc) |Tcr2«f r3gp

' ...(depth of C. G. of the semi-circle=«4r/3n) and Y<=* horizontal thrust in the direction of OC

Resultant Thrust on Curved Surfaces 95

«=0, since the surface is symmetrical about the plane AOE which is perpendicular to OC.

". Resultant horizontal thrust on half of the curved surface of the hemisphere•=tf(Xz+¥a)=|r8 #>-(l/n) l%nr»gp]

«=•(!/«) [the weight of the liquid contained] =(l/re) [the resultant fluid thrust on the whole

surface]. Hence proved. Ex. 7. A hemispherical bow] with its lowest ppini downwards

and the plane base horizontal is filled with water. The water is poured into a cylindrical rambler, radius of whose base is equal to that of the hemisphere. Prove that the horizontal thrust on half of their carved surfaces in which they may be divided by vertical planes through their axes are in the ratio of 3 : 2.

Solution. Let r be the radius of the base of the hemisphere and cylinder. If the water stands to a height h in the cylinder then the volume of water in the cylinder being equal to the volume of water in the hemisphere, we have 7r/-r/z=i|7cr8 or /z=a|p. ...(I)

If AT and Y be the horizontal thrusts on half of the curved

(Fig, 106 (A) (Fig. 106 B) surface of the hemisphere in two mutually perpendicular directions OD and OA respectively, then

Jf= horizontal thrust in the direction of OD ^thrust on the projection of the curved surface on a vertical

plane perpendicular to OD =thrust on semi-circle ABC=^g? (4r/3rc) |re/'a='|r3^ps

where p is the density of the liquid. and r=» horizontal thrust in the direction OA=>Q„ by symmetry.

.*. Resultant horizontal thrust oa the half curved surface of the hemisphere=^(ra-f-F2)=fr^P=i71 (say) ...<2)

Similarly if X' and Y' be the horizontal tbrusts on half of the curved surface of the cylinder in contact jfrith the liquid in two mutually perpendicular directions OR and OP respectively, then

X*«= horizontal thrust in the direction of OR * =>thrust on the rectangle PQQ'P, the projection of the

96 Hydrostatics 136/6

curved surface under consideration on a vertical plane perpendicular to OH

=gp.ih.2rh=rtiig9='r &r)a gp, from(l) ~$r»gp

and y = 0 (by symmetry as before) ^. Resultant horizontal thrust on the curved surface under

consideration of the cyllnder=V(X'a+ r 2 ) = f ra gp=# 2 (say) ...(3) Hence the required ra t io^ i^ : Hz

•=|r3 g? : f r3gp, from (2) and (3) = 3 : 2 . Hence proved.

Ex. 8. A solid circular cylinder is divided in two equal parts by a plane through the axis. If it is held just immersed in a liquid with the axis horizontal and tbe plane section vertical, what will be the resultant horizontal thrust on the carved surface ? C/

Solution. Let the vertical plane ABCD divide the cylinder into two parts. Let h be the height of the cylinder and r tbe radius of Its base.

(Fig. 107)

If X and Y be the horizontal thrusts in two mutually perpendi­cular directions EO and OO' respectively, then

Z=>horizontal thrust in the direction of EO => thrust on the projection of curved surface of half the

cylinder on a vertical plane perpendicular to EO •= thrust on the rectangle ABCD=gp r {2r.h) * •=>2rzhgp, where p is the density of the liquid

Y= horizontal thrust in the direction of OO' =thrust on the projection of the curved surface of half the

cylinder on a vertical plane perpendicular to OO' F«="thrust on the arc AED

=Q, as the area of the arc AED is zero. (Note) .*. Required resultant horizontal thrust on the curved surface

~l/(X*+Y*)=2r*hg?. Ans. *Ex. 9. A hollow right circular cylinder is filled with liquid

and held with its axis horizontal, find the magnitude and the line of action of the resultant thrust on half the curved surface cut off by a vertical plane through the axis. (Agra 80)

Solution. As in the last example, we can find the resultant

and

or

136/7 Resultant Thrust on Curved Surfaces 5»7

horizontal thrust on hal f the curved surface cut off by a vertical plane through the ax i s=2r 2 / i gp=># (say). '

(Fig. 108) Let V be the vertical thrust on this curved surface, then V

=weight of the liquid contained in half the cylinder=ȣ [nraftg?]. .'. If R be the required resultant thrust making an angle 0

with the horizontal, then R=\/(H2+ F»)= «Jrafep\/{16-fna}

and t a n f l = 4 = ^ 2 y = - J " or * = tu- . - , ,. . H 2r2/y*p 4 \ 4 / Ans,

[If the resultant thrust R makes an angle <j> with the vertical then we have

= V{(2»"VzgP)a+(i:cr'tep)2j

( * ) •

tan =r» — *= 2r3hgP

or (f> =tan-(-l). V inr'tfigp n r \ rc / ' Ans .

• • E x . 10 . A cylindrical pipe of a circular cross-section i s half . full of water. I f the pipe be imagined to be divided into two halves by a vertical plane along the middle, show that the water wil l tend to push them asunder horizontally with a force (Win), where W is the

Show t h a t . the resultant thrust o f the pipe makes with the vertical an angle

1\ i

weight of the water contained, water on either half of the cof1 (?r/2).

x Solution. Let the len­gth of the pipe" be / and r the radius of its base. If W be the weight of the water contained in half of the pipe, then

W~\ (w»).fe?, ...(]) where p is the density of the water.

The vertical thrust on D ° the^ curved surface [BOC, (Fig. 109) B'O'C') of half the pipe=weight of the superincumbent liquid which Is exqual to the weight of the liquid contained —Z (say)

«=•* {\*r*)lg^l™Hg?~\W ...fronwOK Also if X and Y be the resultant horizontal thrusts o n the

98 hydrostatics

curved surface of half the cylinder in the directions of OC and 00' respectively, then

X<= horizontal thrust on the ourved surface of half the cylinder in the direction of OC

=thrust on the projection of this curved surface on a vertical plane perpendicular to OC

=thiust on rectangle OBB'O' *=g?.l-Kr.lr=>\lr2g?*aW[T: , ...from (1)

or X^Wjiz ..(2) And F=horizon<aI thrust on the above curved' surface in the

direction 00' =•(), by symmetry. Hence the resultant horizontal thrust on the curved surface

under consideration=^'(2r2+ F3)= Wjit. ' Also if 6 be the angle which the resultant thrust makes with

.the vertical, then tan 0=>X/Z

or cot 6=-^'='—-'='— or ^=cot_1 I -y I: Hence proved.

Es. II . A closed cylinderical vessel with hemispherical eods is filled with water, and placed with its axis horizontal. Find the resultant thrust on each of the ends and determine its line of action.'

(Garhwal 79) Solution. Let r be the radius of the base of the cylinder. If X and Y be the resultant horizontal - thrusts on each end in

two mutually perpendicular directions OE and OC respectively, then

(Fig. 110) .T= horizontal thrust on the curved surface of the hemi­

spherical end in the difeetiou of OE *=thrust on the circle ABCD, which is the projection of the

above surface on a vertical plane perpendicular to OE. ^gP.r [nr2)=7rr9gp, where p is the density of water,

and r=horizontal thrust on the curved surface under considera­tion in thedirection of OC=0, by symmetry about the plane BED.

Also if Z 8e the vertical thrust on the above curved surface, then Z=»weight of the liquid contained=fw3gp.

Resultant Thrust on Curved Surfaces 99

.'. If R be the resultant thrust on the above curved surface, making an angle 6 with the horizontal, then!

J?= tf(X*+Ya+Za)=tf{(nr3gp)H 0+(f w V)2} = I <(/(13)nrsgP

and * n z tan0= ^-' 3™sgP__

Kr8gp " or 0=tan_1 (§).

Ans. *Ex. 12. A right circular cone is just immersed in a liquid with

its axis horizontal. Find the resultant horizontal thruSt (i) on half the coue cut off by a vertical plane through the axis, (ii) on lower half of the cone cut off by a horizontal plane through the axis.

{Bhopal 73 ; Magadh 74) Solution. - Let r be the radius

of the base and ft the height of the cone.

<i) If X and Y be the horizontal thrusts on the curved surface of half the cone in two mutually perpendicular directions BO and VO respectively, then

Z=horizontal thrust on the above curved surface In the direction of BO

=>thrust on A VAC, which is the projection of the (Fig. I l l ) above curved surface on a vertical plane perpendicular to BO.

or X^gpr {lh.2r)i=>r2hgp, where p is the density of the liquid, and r«=>horizontal thrust on the above curved surface in the direction

oiVO =>thrust on semi-circle ABC, which is the projection of the

above surface on a vertical plane perpendicular to VO. =ugp%,Snsgp.r &nr2)=>litrz gp £. The required resultant horizontal thrust=<?(X2+ Y*)

•=\/{(r V+(i*r 3gP)*}~rB£P tf(A«+!«•'•/. - -Ans. (ii) The horizontal plane VBD divides the cone into two

halves. I f Z a n d r be the resultant horizontal thrust on the curved

surface of the lower half in two mutually perpendicular directions BO and VO respectively, then

Z=>the horizontal thrust on the above curved surface in the direction of BO.

=>0, since'the above surface is symmetrical t« the plane VOC and r= the horizontal thrust on the above surface in the direction

of VO

ISO Hydrostatics 284-66

(Fig. 112)

>= the thrust on semicircle BCD, which is the projection of the above surface on a veitical place perpendicular to VO.

-gp [r+(4r/3*)](lw»), (Note)

the depth ©f G. G. of the semi-circle BCD below the

-frte surface being equal to [r+(4r/3n)]

= i^pr3{H(4/37r)]. Henee the required resultant horizontal thrust=<^(Ta j- Yz)

«£7cgpr3{l+(4/3TC)}. Ans. *Ex 13.' The end of a horizontal pipe is closed by a sphere of

tke same radius as the internal section of the pipe. The sphere is kinged at its highest point. If the pipe is just full of liquid of density t, prove that the moment about the hinge of the liquid pressure on the sphere is gpw&\ (Lucknow 79 ; V. P. P. C. S. 77)

Solution. Only the curved surface of the hemisphere ACB is ia •ontaet with the liquid, hence the fluid thrust will act only on this surfaea

If X and Y be the horizontal thrusts on this surface in two Mutually porpendicular directions CO and EO, then ^

Z—horizontal thrust on surface uader consideration in the direction of CO

=»thru«t on the circle ADBE, which is the projeetion of the above surface on a plane perpendicular to.CO

=»' gp z5"=g-p.fl.Tfla

(Fig, 113)

Resultant Thrust on Curved Surfaces 101

=>7tfl3gp, acting at P, the centre of pressure of the circle ABBE, such that AP~>a+(aal4a)~(5al4)

and F-= the horizontal thrust on the above surface in the direoti«m of EO

•=0, since the surface is symmetrical about the plane A®B& Also if Zbe upward vertical thrust on the above-curved

surface then Z«=<weight of the liquid displaced •

•^fito^P. acting through G, the centre of gravity of the liquid displaced, where AO^Qaji) ;

.*. The required moment of the liquid pressure about the hinge at A<=>X AP—Z OG-=>nasgP.ia—$na3g94« =7raV-

Hence proved. *Ex. 14. A spherical shell

formed of two halves in contact along a vertical' plane is filled with water ; show that the resul­tant pressure on either half of the shell is \/(13)/4 of the total weight of the liquid (Lucknow 77 71:

Rajasthan 75; Vikram 77) Solution As in Ex. 11 Page

98 we tan prove that

Also the total weight of the liquid in the shell = fvrr'gp

A .R-iVf<13)(frrrs«p) -==£^(13) (total weight (Fig. 114)

of the Hquid in the shell). Hence proved. Ex. 15. A weightless sphere is divided by a vertical plane into

two halves, which are hinged together at their lowest point and it is just filled with water. Show that the tension of a string, which ties together the highest points of the two halres, is three-eighth of the weight of water that the sphere would contain.

Solution. Let a be the radius of the sphere. ACBD is the vertical plane dividing the sphere into two parts , which are hinged at the lowest point B

If Zand Y be the horizontal thrusts on the curved surface of the hemisphere AEB in two mutually perpendicular directions OE and OD respectively, then

X— horizontal thrust in direc­tion of OE

«=»thrust on the circle ACBD

102 Hydrostatics

="£p.2.S"=g-p.a.7raa

=7ta8gp, acting through P, the centre of pressure of the circle ACBD, such that AP=a+(a*t4a)<=>%a and 5P=2a- | a= . fa

and F=> horizontal thrust in the direction of OD =9, by symmetry of the surface about the plane AOBE.

Also iT Z be the downward vertical thrust on this curved sur­face, then Z^weight of the liquid contained

"frea^p, acting through G the centre of of gravity of the liquid contained, where OG=*%a.

If The the tension of the string at the highest points of the two halves. then taking moments about B, in order to avoid the reaction, of the hinge at B, we have

T.BA-X.BP-Z.OG~Q /or r . 2 f i - J . f a~Zfa=0 . or T.2a=X.ia+Z.$a=*Tiasg9.la+%Ka!igp.la

<=*la,nasg? (1-|-|)=f a (iitaPgp) or 7V»! (fnePg?)

=Ȥ (weight of water that the sphere would contain). Hence proved.

**Ex. 16. Two closely fitting hemispheres made of sheet metal of small uiiform thickness are hinged together at a point on their rim, aad are saspended from the hinge, the rims being greased so that they form a water-tight spherical shell, this shell is MOW filled with water through a smali aperture near the hinge. Prove that the contact will not give way if the weight of the shell exceed* three times that of ths water it contains. (U. P. P. a s. 79)

Solution. Let r be the radius of the base of each hemisphere Let the hemispheres be hinged at A and suspended from the same point.

If Zand F be the horizontal thrusts on the curved surface of the hemisphere AEB in two mutually perpendicular directions OB and OD respectively,, then

X«=>the horizontal thrust on' the surface in the direction ofOE

=thrust on the circle ACBD - ="g,P-^.5"=gp r w a

«7rrsgP, acting at P, the centre'of pressure of the circle ACBD, where AP

aar+(raJ4r)=>r-t |r«=»|r

Resultant Thrust on Curved Surfaces 103

And F«=» horizontal thrust on the surface in the direction of OD. •=>0, since the surface is symmetrical about the plane AOBE.

If Z be the downward vertical thrust on the surface, then Z=weight of the liquid contained**! rcr3 gP, acting through G, the C.G. of the liquid contained, where <X?=*3r/8.

Also if Wbe the weight of the hemispherical shejl AEB, then Reacts downwards through G\ the C. G. of the shell such that OG'=*lr.

From the above figure it is evident that the force X tries to separate the two halves while forces Z and W prevent this separa­tion. Hence the two parts will not separate if the sum of the moments of Z and W about A > the moment of X about A

i.e. W.OG'^Z.OG > X.AP a or W.lr+lnrsg9.^r > irr*g?%r or W§r > fw^p—lnr*gP or W>2nrzgP or 2W>4srr3gp or 2W > 3 (fjrr8 g?) or the weight of the shell > 3 (weight of the water contained).

Hence proved. **Ex. 17. A solid sphere of density p is placed at the bottom

of a vessel which is horizontal, and a liquid of density a ( < f) is poured in so as to cover up the sphere. The sphere is then cut along the vertical diametral plane. Prove that the two parts will not separate if P < 4cr.

Solution. ABCD is the vertioal plane dividing the sphere of radius r (say) into two halves.

If X and Y be the horizontal thrusts on the curved surface of the hemisphere AEB in two mutually perpendicular directions EO and DO respectively, then

Z=horizontal thrust in the direction of EO

=thrust on the circle ABCD ^"gP.z.s" ^go.r.Ttr^uaKT^g*,

acting through P, the centre of pressure of the circle ACBD, where

-AP^Sr/4. and r = horizontal thrust * In the

direction of OD=0 (by symmetry) (Fig. 117)

Also if Z be the vertical fluid thrust on the surface of this hemisphere, then Z=>the weight of the liquid displaced.

104 Hydrostatics

or Z=f nr*go. acting vertically upwards at G, the O. G. of the liquid displaced.

.*. The forces acting on the hemisphere AEB are (i) the horizontal thrust Z«=nrV» acting through P,

where BP=»2r—fr=f r. (ii) the upward vertioal thrust Z = | w r V . acting through G,

such that or?=fr. (iii) the weight W=>1 w s gP of this hemisphere acting verti-

eally downwards through G, and (iv) the reaction at the lowest point B Out of these forces. X and Z try to keep the hemispheres

together whereas the force W tries to separate them Hence the two parts will not separate if the sum of the moments of X and Z about B is greater than the moment of W about B, where B is the point about which the hemisphere can turn in case of separation i.e. X.BP+Z.OG>W.OG or rtrjga |r+ftrV.fr>§7rr3gP.fr or o+ i 0 , >5 p or 4<r>p. Hence proved.

*Er. 18. A hollow right circular cone is divided into two parts by a plane through the axis, and the two parts are hinged together at the vertex, the edges being greased so as to be water-tight. The vessel is tben hung up by the hinge and filled up with water through a small aperture near the hinge. If the water does not flow out what must be the least value of the vertical angle of the cone ?

Solution. Let h be the height and a the semi-vertical angle of the cone.

The vertical plane VCD divides the cone into two halves. Let X and Fbe the horizontal thrusts on the curved surface of

the semi-cone (V, BCD) in the directions of OB and OC which are mutually perpendicular, then

X=horizontal thrust in the direction of OB =thrust on A VCD-Ug& j"-gP.fh & (2h fan «)}. (Note)

= f hs tan K gp, acting through P. the centre of the pressure of A VCD, where VP<=>1 h

and F«=»horizontal thrust in the direc­tion of OC

•=0 (by symmetry). If Z be the vertical thrust on the

surface of this semi-cone, then - Z<=ths weight of the liquid in this

semi-cone=| (£n;/j8 tanaagP*3 = JTCA3 tan8 cc.gp. acting verti­

cally ajosrawards through G, the C.G. of the liquid contai­ned, where the horizontal dis-

Resultant Thrust on Curved Surfaces lOo

tance of G from the axis of the cone=[(A tan »)/n] (Proof of it is given in £he next example)

Now the water will not flow out, if the moment of Z about V > moment of X about V. inh* tan3 «.gp. [(h tan «)/«•] > § h3 tan a gp.'A

(Note)

i tan2« > I or tan « > ^3 or a > 60° or 2a > 120° i.e. or A The required least value of the vertical angle/rf the cone=120o.

**Ex. 19 A solid circular cone of uniform material and height h and of vertical angle 2«, floats in water with its axis vertical and vertex downwards and a length h' of the axis immersed. The cone is bisected by a vertical plane through the asis and the two parts are hinged together at the vertex. Show that the two parts will remain in contact, if h' > h sin3 a or tan2 a < {h'7(h-h')].

{Lucknow 82 ; U. P. P. C. S. 80)

Solution. Let P and a be the densities of water and the material of the cone respectively. VAB is the vertical plane which divides the cone in two halves.

Consider the equilibrium of the semi-cone (V, BCD), OC is perpendicular to BD. Let Gx be the C. G. of the semi-circular base of the semi-cone.

TU n^ "4r" 4/ztanot. Then OGx=-w- •=—= lit J7C

Join VGU then the C. G. of this semi-cone will be at G on this i-_-_-_r>;r-3iV-:fc line VGU such that VG=l VGt. L;

From G draw GL perpendicular to the axis of Jhe cone. (Fig. 119;

In similar &S.VLG and VOGlt we find y^r***r7~—=* '-; OGt KGi 4

nr 8 fxn a 4A tan a h tan a or GL=\.OG^=\. —-z => 5tt it

The cone is immersed upto EF, such that the height of EF above the vertex Fis h' i.e, VO'<=*h'.

If X and Z be the horizontal and vertical thrusts on the curved surface of the semi-cone in oontact with water, then

X=horizontal fluid thrust on surface in the direction of FQ' «=thrust on AVB'D'-="gp.z.S"=g9.yi'. {\h'.2h' tan a) =»£A'3 tan agP, acting horizontally at P. the centre of pres­

sure of the triangle VB'D', such that 0'P'<=-W or VP<=\h'. And Z=the vertical fluid thrust' on the surface

" 106 Hydrostatics

=the weight of the liquid displaced^! (inh's tan2 «) gP =|TTA'3 tan9 agp, acting upwards at G\ the C. O. of the water

displaced where the horizontal distance of G' from the axis ofthecone=G'L'=.(/j' tan a)/w

Also the weight of the semi-cone=§ (in/ts"tana a) go t=>\izh2 tan3 a ga^W (say), acting vertically downwards at G%

the C. Q of the semi-cone0 where GLi=>(h (an «)/re. Hence tlie forces acting on the semi-cone.(F, BCD) are a (i) The horizontal thrust X=4 h'a tan « gP, acting at P, where

TT-Jfc'. <ii) the vertical thrust Z=>|n:/i'3 tan8 a.gp, acting vertically

upwards at G', where G'L'^(h' tan cO/rc and (iii) the weight of the semi-cone=> W>=*\%hz tan2 a ga, acting

vertically downwards at G, where GL<=>(h tan «)/Tt. Out of these three forces, X and Z help to keep two parts together whereas the force W tries to separate them. Hence the two parts will remain in contact if the sum of the moments of X and Z about V > the moment of IF about Vi.e X.VP+Z.G'U > W.GL

or \h'z tan a gp. TT+WI'3 tanBa.gP. —&>$rc/i3 tan3 agcr _^5_? 2 7E 7T

or ^ A'*p+/2'a tan" a p>A4 tan3 a » or h'1 (1+tan2 a) P>/i« tan2« a or A'4 P>A*(sla2 a} a. ...(i)

Also for equilibrium, the weight of the body must be equal to the weight of the water displaced.

.'. %nh3 tan3 x.go^lnh'8 tana a gP or A3crra/['3p or P=(/J3O/A'3)

.*. From (1), we have the required condition A'*.(AV^'8) > h* sin2 a a or A'>A sin2 a. Hence proved

/.e. sin2 a < (h'/h) . or tan8 a < [h'IQi-h')] **£x. 20. A hollow right circular cone, filled with water is

held with its axis vertical asd vertex downwards. Find the resultant pressure on the portion of its surface contained between two vertical planes through its axis, and show thai if the inclination of these planes to each other be 20 and the vertical angle of the cone 2a, the direction of the resultant pressure makes with the vertical an angle equal to tan - ' [(sin 0)1(6 tan «)]. (Bhopal 80)

Solution. Let h be the height of the cone and r the radius of its base. Let (V, OACB) be the portion of the cone lying between two vertical planss VOA and VOB. It being given that AAOB=*28, if C be the middle point of the arc AB, th/sn

L.AOC^d=LBOC. Also l_OVA<=>a.'=>t3in~i (r/h), V r*»h tan « for the cone. Area of the sector 04C»=»«,Jr,0*,-»ir>.20«-r,0.

Resultant Thrust on Curved Surfaces 107

Let Zand Ybe the resultant horizontal and veitical thrusts on the curved surface of the portion under consideration. Then

Zrahorizontal thrust on the curved surface of the portion in the direc­tion of OC

•=thrust on the projection of this curved surface on a vertical plane

\ perpendicular to OC. -thrust on A VAC*=*'egP.zS" •=>gp.\h. area of A VAC -gP.iMlA AC)=gp.^h.(lh.2r sin 6)

(Note) =4/j3rsin Bgp.

Here, we must note that the horizontal thrust in the direction perpendicular to OC is zero by symmetry. And Z = the downward vertical thrust on the

curved surface of poi-tioa under consideration.

«*ttie weight of the liquid contained =»gpX [volume of the portion

(V, OACB) of the cone] =gp (\ area of the sector OACBh)=>gP.% ra O.h.

A Required angle ^ (say) is given by tan <f>~XIZ

or i * _i / X\ * -i rift9'' si° 6§P~\ t ^(hsme\

r<=> h tan a for the cone

Hence proved.

•tan^KftsfnflyCMtancOI... =tan-x [(sin* *)/(« tan a)],

Ex. 21. The stem of a fsen el has a radius a and its month has a radius b, the height of the stem is h and that of slanting part is /. The mouth is placed on a horizontal plase (being greased so as to remain water-tight) and the funnel is filled with wates to the top of the stem ; prove that the upward pressure on the funnel is equal to •KW (b—a) {h (a+b) 4-%l (a-f 2b)}, where w is the weight of unit volume of water. Also find the horizontal thrust on half of the curved surface of the funnel, which is bisected by a vertical plane through the axis.

Solution. The upward pressure on the funnel =the upward pressure on the curved surface of the

frustum PQBK =weight of the superincumbent liquid which would fill

the space between the cylinder of height (h + l) on PQ as base and the funnel,

108 . Hydrostatics

*H> [vol. of the cylinder \PQQ'S'—{vol. of the cylinder - MKLN+vol. of the frustum PQLK}]

=w fab2 (h+l)-{kazh+§Kl ( a » + J H # ] •JW [h (bz-aa)+I {&-% (a2+68+a6)}]

it-t1- -!l

(Fig. 121)

=jiw [h (6a-a2) HI (26a -aa-«ft>] ~TCW [A (62 - a a ) + £/(&- fl) (26 \-a)] = n (6 -a ) w [h (d+a)+%l (2b+a)]. Hence proved.

Also horizontal thrust on the curved surface of half the funnel on the right of the vextical plane through the axis.

•=*the horizontal thrust on the curved surface of half the cylinder (ENF, CLD) + the horizontal thrust on the curved surface' of the frustum (CLD, ~AQB), in the direction of OQ.

•=(the thrust on the rectangle CDFE)+(th@ thrust on the trapezium ABCD), these being the projections on the vertical plane perpendicular to oQ

[The horizontal thrust in the direction perpendicular to OQ is zero by symmetry and the depth of C.G. of the trapezium ABCD below CD is {£ (a+2fi) l}l(a+b)]. ^

Exercise on Horizontal Thrust Ex. A solid right circular cone of uniform material and

Thrust on Gurved Surfaces Bounded by Plane Area .109 '

height h and of vertical angle 90°, floats in water with its axis vertical and vertex downwards and a length h' of its axis immersed. The cone is bisected by a vertical plane through the axis and the two parts are hinged together at the vertex. Show that the two parts will remain in contact if h' > \h. {Rohilkhand 77)

§ 308. Thrust on a carved surface bounded by plane area. (Rohilkhand 82 ; Vikram 73)

Suppose a solid body bounded by a plane area.is completely immersed in a liquid. Then we know from the Principle of Archimedes that the thrust of the liquid on the whole body acts at the centre of buoyancy in the upward vertical direction and is equal to the weight of the liquid displaced. So we find that the thrust on the body as a whole is vertical and there is no component of the thrust on the whole body in the horizontal direction.

Let X' and T be the components of the resultant thrust on the curved surface in the horizontal and vertical directions respectively. Let X and Y be the components of x/ the thrust on the plane area — » -in the same two directions respectively.

Then since the resultant thrust on the body as a" whole is vertical, so we have (Fig. 122)

X'~ X°»0 (in the horizontal direction) and Y'—Y~°Vg? (inthe vertical direction), where V is the volume of the body and p the density "of the liquid.

From these two equations, we get X'—Zand Y'^Y+Vg?, which when compounded gives the resultant thrust on the curved surface. * Solved Examples on § 3*08.

•Ex. 1. A solid cone whose vertical angle is 2a, is immersed in a liquid with its vertex in the surface and axis vertical. Prove that if P be whole pressure on the curved surfacc-and base and P ' the resultant pressure, then P : P' : : (2+3 sin a) : sin a.

(Agra 78, 73 ; Rajasthan 7S) Solution. Let h be the height of the cone The whole pressure on the base of the 'cone

~"#P zS" ~gP.fi % (h tan a)a=7cA3 tana % gp, where p is the density of the liquid.

And the whole pressure on the curved surface. «="gP2„y"=,gP.f& (nfr)~PfA (nh sec a h tan «) = ITC/J3 tan a sec a g<? '

Then P<=>the whole pressure on the curved surface and the,

110 Hydrostatics

base=f?r/i3 tan a sec « gP -f^/j3 tan3 a gp _ «irc/i3 tan « (2 sec a-1-3 f a n » g? ^^=-=^=r^

and P'=the resultant pressure on the surface of the cone

=>tbe weight of the liquid displaced =4n/z3 tan2 a gP

P _£rc/'» tan a (2 sec a+3 tan «) gp P' £TC/J3 tana « gP

(2 sec a+3 tan a)

PV

(Fig. 123) tan «

=(2+3 sin «)/(sin a) , Hence proved.

•Ex. 2. A hollow weightless hemisphere with a plane base is filled with water and hung up by means of a string, one end of which is attached to a point of the rim of its base, find the inclination to the horizontal of the resultant thrust on its curved surface. (Lucft'now 79)

Also find the resultant thrust on the curved surface. (Agra 82) Solution. The hemisphere full of liquid is in equilibrium under

the action of three forces viz. (i) the weight of the hemisphere full of liquid acting vertically" downwards through G, the C.G. of the hemisphere of liquid such that OG=f r, where r is the radius of the hemis­phere, (ii) the downward vertical thrust of the water on the surface of hemisphere, acting at G, the C.G. of the liquid contained and (iii) the tension at A, (Here we should note that on the hemisphere as a whole, there will be no horizontal thrust). Since out of these three forces two are vertical i.e. parallel and passing through G, the third force viz. the tension £t A- should also be vertical and passing through G i.e. G and A should be in the same vertical line. (Fig. 124)

.*. If the base AB of the hemisphere makes an angle « with the horizontal, then [_ AGO*=>x.

. t AO r 8 » t a n "-OCT(37/8) " 3-whence sin a ^ S / V ^ ) and cos a=3/<v

/(73) If P be thrusj on the plane end of the hemisphere, then

.-(1)

P«»"gPz,s"=>gpr sia a (ffr?)=mr3 sin a gP, acting normally to the plane base (P being the density of water).

Thrust on Curved Surfaces Bounded by Plane Area 111

If X' and Y' be the components of the thrust P in the hori­zontal and vertical directions, then X'^P sin oc and Y'*=>P cos a.

Also the resultant thrust on the vessel is vertical and equal to V, say. Then F«=»welght of the liquid contained=f7r/-3 gp.

.'. If Zand The the components of the thrusts on the curved surface in the horizontal and vertical directions respectively, then we have . X^X'^P sin a' #

and Y—Y'=V, V V acts vertically downwards or Y=V+Y'-=*V+P cos a,

L If ^ be the required inclination to the horizontal to the resultant thrust on the curved surface of the hemisphere, then

Y V-\ P co? a %w3 gf+7rr3 sin a gp cos a t a n <£«=» T^=' j ; D — s—: —r

Y X P sin a 7cr sin a gP sin a 2 f 3 sin a cos «_2+3 (8/^73) (3/^73)

" 3 sin2 a 3 (8/\/73)a — • f r o m {1)

. , (2X^3)+72_218_109 /109V or t a i i r f - 1 - ^ ^ 192 ~96 0 r * = t a n ( T S ) Ans.

Also 2?=^(Z2+72)=. V[CP sin «)«+(F+/> cos K)a] = ^ [ P a s»n2 « 4-I/a+-P2 cos2 a -f-2 VP cos a] -=><^(Fa+Pa+2 VP cos oc) ~^[(f7cr3 gp) s+(^8 sin a #P)a

-f2 §w3 gP.w3 sin a gP. cos a]

' " ^ W [ 1+^ « 4 ^3)- 3)]

-* nr3 S? J[^flY34 «'* gP [^73)] Ans. ••Ex. 3. A hemisphere, radius a, is entirely submerged in a

liquid of density p, so that its diametral plane makes an angle 6 with the horizontal and has its centre at a depth h. Prove that the resul­tant force on the curved surface is rca2 gp (|aa-f h2±$ah cos 6)1".

(Agra 74) Solution. Case I (base upwards). Let P be the thiust on the plane of the hemisphere acting at

the centre of pressure of the base. Then P=»"gPS"=gp h mi2

Let X' and Y' be its horizontal and vertical components, then Z ' = P sin 0 and r ' = P cos 6 (acting vertically downwards).

Also the resultant thrust V on the hemisphere in acting verti­cally upwards and equal to the weight of liquid displaced acting through G, the C. G. of the liquid displaced. Thei; r=§7co3 g?

If X and Y be the components of the thrust on the curved surface of the hemisphere in horizontal and vertical directions

' 112 Hydrostatics 136/7

respectively, then resolving hori­zontally and vertically all the forces acting on the hemisphere as a whole, we get

' z = Z ' = P sin 6 ...0) and or

Y- Y'=V<=%ita»gp Y-lna*g9+Y'

~frca3gP+P cos 6 ...(2)

Ifi? be the resultant thrust on the curved surface making an angle $ with the hoiizontal, then from (1) and (2) we get

Re*i/(X'+Y')-{(P sin 6)* +(|™^P+Pcose) a} ] / a (Fig. 125)

•=[P* sin8 B+tmWgY+P* cos2 0+%Tta3gp. P cos 0]1'8

•=[P*+§nW?a+fra2g?.Pcos 6]1'* >=>[(no*hg?)a+§nsQ6g2ei+%™8gP (na'hg?) cos 0\V\

P=7caa/jgp

...from (1) and (2) 2Q+lh cos 8

3A sin 6

=.7taagP [/za+fa2 + |a/* cos 0]1'3

A n r l t f l n , Y _ jnePgP+P cos 6 And tan ^ - ^

f7ra3gp+7ra2/?gP cos $ = 7ra2/igP sin 0

Case II. (base downwards). In this case resolving the forces

acting on the hemisphere horizontally and vertically we have

X^X'^P sin 0 ...(3) r — 7 = F or r = F ' - F (Note)

or Y=>P cos 6~lna3g? ...(4) Also as in case I, if R be the

resultant thrust on the curved surface .making an angle (f> with the horizon­tal, then

R~4(,X'+ Fa) =[(P sin fl)z

+(P cos 8-$na?g9)s]1'*

=na3gp [AH£<Ja-f«/j cos e]1'2 (Fig. 126)

and tan $=•=?= . 8fl—2r«=—^-r-;——-.

^ Z Psmfl 3h sm 0 *Ex. 4. A solid hemisphere is immersed in a liquid with highest

point of its plane 'base in the surface and the base is inclined at an angle tan-1 2 to the horizon ; show that the resultant thrust on the

136/8 Thrust on Curved Surfaces Bounded by Plane Area 113 x

curved surface is equal to twice the weight of the displaced liquid. (/. A. S. 73 ; Jodhpur 78 : Rajasthan 74 ;

Sohilkhand 781 Vikram 80)

Solution Let a be the radius of the sphere. AB is its, diameter, inclined to the horizon at an angle a, where tan oc<=2 (given), whence sin «'=2/\/5'| and cos oc^l/VSJ ...(i) If P be the normal thrust on the plane base, then

P="gPz.S"=£p.asin area8

= 7ra3s]na#p, ...(2) where P is the density of the liquid. (Fig. 127)

If X' and Y' be the components of P in the horizontal and vertical directions, then.X'=-P sin a and 7'=»P cos a, acting verti­cally downwards.

If V be the resultant vertical thrust on the hemisphere, then V=the weight of the liquid displaced or V'='li«**gP, acting verti­cally upwards.

If X and Y be the horizontal and vertical components of the thrusts on the curved surface then resolving horizontally and verti­cally the forces acting oa the hemisphere, we have

X=X'«=Psina ...(3) T- Y'=V or T - F + r = f n o 3 g P + P cos a ...(4)

A If R be the resultant thrust on the carved surface, then R~tf(Xz+ Y*)={(P sin a) a+(!w3gP+P cos a)1}1"

~{Pa sin2 a+freVgV+P" cos* a+frcasgp. P cos a}1'" e={P*+iK*a°gy+i™3g?P cos «}V3 •={(na3 sin a.g-p)3 f f7tVgaP2+f izazg9jzcPg9 sin a cos a}1'3

«=7ta3gp {sin* a + t + f sin a cos a}1'*

and

>*a»gP { | + | + | - J 5 - Ts} 1 ' 2 . from (1)

~™3g9 {4+t+A-}V,=t«tf»gP«2 (fnasgP) •=2 (weight of the liquid displaoed).

Also if this resultant thrust R makes an angle 6 with the horl= zontal, then

Hence proved.

tan d Y

'X* fm^gp + Pcos a

P sin a

rca3gp sin8 a ' v ' =(2+3 sin os cos a)/(3 sin* a)

ADS,

114 Hydrostatics

Fsnd the magui-Ex. 5. A spherical shell is filled with liquid, hide and the line of action of the resultant pressure on each of the hemispheres divided by any diametral plane. {Agra 83)

Solution Let the diametral plane AB make an angle 0 with the horizontal Proceed further as in Ex. 3, Pages 111-112).

(Here h=a). Ex. 6. A cylindrical vessel full

of water is held with its axis inclined (Fig. 128) at aa angle of 45° to the vertical Find the magnitude to the pressure on the ends and show that the resultant pressure on the curved snrface will equal the difference between the pressures on the ends

Solution Let k be the height and r the radius of the base of the cylinder, then if Px and Pa be the thrusts on the lower and•> upper ends respectively, we have

P1="g?zS,,=gp (h cos 45"+r cos 45°) rer2

*=(l/\/2)gP (A fr) w a , where p is the density of the liquid and P,=*t,g&S"g9 (r cos 45°) 7ira=(i/V2) nr*gp

V. Pi-P^iHy/1) g? (A+r) w»-(l / t f2) ™*8P -(BI-VV2) hg9 ..(!>

Let X and Y be the resultant horizontal and vertical thrusts on the cur­ved surface of the cylinder.

Then resolving the forces acting on the whole cylinder horizontally, we get Z+P 2 cos 45°=Pi cos 45° or Z=(P , -P 2 ) cos 45°

=(w'/*2)Agp.(i/v2), x from (i)

or X=>\nrVi g? . . (2 )

Again is V be the resultant vertical thrust on the cylinder as a whole, then from Principle of Archimedes, we know

F=» weight of the liquid contained aeting vertically downwards through G, contained.

.'. Revolving the forces acting on the whole cylinder verti cally, we get V*~Y-{-Px cos 45°-P2 cos 45°.

R Coil 5' (Fig. 129)

<nr2hgp9

the C. G. of the liquid

Thrust on Curved Surfaces Bounded by Plane Area J13

or Y= V-P% cos 45°+P2 cos 45°=7ir2/;gP-(P1-P8) cos 45° «7rr2%p-(Iir

2/\/2) hgp.(lf</2), from (1) =4;rra/igP ' ...(3)

.*. If R be the resultaut thrust on the curved surface, we have ; ? = ^ ( X 2 + r 2 ) = = v ' { ( § w W H ( £ w 2 W } J from U) and (3)

=i7rr 2 ^ P ^( l + l)=.(]/V2) Trr*hgP~P1~P2, from (1). Hence proved.

**Ex. 7. A cose floats, with its axis horizontal in a liquid of density double its own, find the pressure on its base and prove that if 0 be the inclination to the vertical of the resultant thrust on the curved surface, and a the semi-vertical angle of the cone, then

tan 8 =>(4/TC) tan «.. (Lucknow 77, 76 / Rajasthan 78, 74)

Solution. Since the density of the material of the cone is half that of the liquid, so half of the cone will be immersed in the liquid i.e. the horizontal plane OBD passing through the axis of the cone will be in the free surface. If ft be the height of the cone, then the radius of its base will be h tan k.

If V be the resultant thrust on the whole surface of the semi-cone immersed in the liquid, then

7 = weight of liquid displaced™ £ (JnA3 tan2 « g?) «=J7C/J3 tan2 a. gP, acting vertically upwards through G, the

C. G. of the liquid displaced. If X and Y be the

horizontal and vertical components of the thrust on the curved surface of this semi-cone, then we have in the horizontal and vertical directions : — X=*P, the horizontal thrust on fife semi-circle BCD

where r<=>h tan « g?h3 tan3 a ...(I)

And Y~V (Fig. 130) (V there is no vertical thrust on the semi-circle BCD)

or Y<=* JTIA3. tan* a.gp ' .. (2) Hence if R be the resultant thrust on the curved surface of the

semi-cone making an angle 6 with the vertical, then R=> s/(X2+Y2), which can be evaluated from (h) and (2)

f/»3tan3agP' 4 and tan B =. i. tan a. Y §7i/i3tan'agP n »Hence proved.

•*Ex; 8. A hollow cone without weight, closed and filled with

116 Hydrostatics 234-66

water is suspended from a point in the rim of its base j if <f> be tbe angle which the direction of the resultant pressure makes with the

*• 1 .V 1- 4.U 4. A. J. 2 8 C O t « + C°tS «

vertical, then show that cot 9 = To , a being the semi-vertical angle of the cone. (Lucknow 81)

Solution Since tbe cone filled with liquid is suspended from a point A>on its rim, so in the position of equilibrium A and G, the C.G. of the cone filled with liquid must be in the same vertical line. Let h be the height of the cone, then the radius of its base=A tan a. Let 0 be the angle which the base of the cone makes with the ver­tical. Then in A AOG, we have

tan 6=OG/OA

•(1) or tan 0= ih t . — =»i cot a h tan a 4 PCoiO

whence cos 0=

and sin 0=

V(16+cota a) cot a

V(16+cotaa) - (2)

Now if P be the thrust on the base of the cone, then P*="gP2s"-=>gp. (r cos 0) w a , where r=>h tan a (the radius of the base) (Fig. (131) or P^nh3 tan3 a. cos 0 gP ...(3)

If V be the resultant vertical thrus>t on the cone as a whole, then F^weight of the liquid contained. (Principal of Archimedes)

=»into? fan8 a gP ...(4) acting vertically downwards through G.

I f Z a n d T b e the horizontal and vertical components of the thrnst on the curved surfaceof the cone, then resolving the forces acting on the cone in the horizontal and vertical directions, we have

X-=P cos 6<=nkz tan3 a cos2 0 gP .. (5) and V=> Y-P sin 0, " V is the resultant of Y and P sin 0 or r=» V+P sin 0=Jn/z3 tan3 « gP+rc/i3 tan3 a cos 6 sin 0 gP,

from (3) and (4) «=>£n/i3 tana a gP U + 3 cos 0 sin 0 tan a)

Hence if 54 be the angle which the resultant thrust on the curved surface makes with the vertical, then cot 4>*=>YIX

jrcft3 tan2 «gP (1+3 cos* 0 sin 8 tan cc) TC/J3 tan3 a cos8 0 gP

'1+3 cos 0 sin 0 tana i tan a cos8 0

or cot $=

Thrust on Curved Surfaces Bounded by Plane Area H»

or cot <f>*

* , „ / 4 cot a \ .

\ 16+co tMj

, from (2) 3 tan a

j e + c o t M 16+cot2 «+12_28 cot «+cot8 a

P'/Xt*

48 tan« 48 ' Hence proved. •EX. 9. (a) A solid right circular cone of vertical angle 2a is

just immersed in water so that one generator is in the surface of the liquid, prove that the resultant pressure on the curved surface of the cone is to the weight of the fluid displaced by the- cone, as V ( l + 3 sina a) : 1 and that it is inclined to the axis of the cone at an angle cot"1 (2 tan a). {.Rajasthaa 771 Rohllkhand 83)

Solution. Let h be the height of the cone tben the radius of its base is h tan oc.

If V be the resultant thrust on the cone as a whole, then

F=the weight of the liquid displaced

•=\ith3 tan2 ce.g-P, acting vertically upwards through G, theCG. of the liquid displaced.

I fPbethe thiust on the plane base of the cone, ' (Fig. 132) thenP="gp2S"=£P.(0.4 cos «).(wa), where r=h tan a

<=gPnh3 tan3 a.cos a, acting at right angles to'the plane end. If JT and y be the horizontal and vertical components of the

thrust on the curved surface, then resolving theforces horizontally and vertically we have

X=P cos oc=»gP 7u/z3 tan8 a cos a cos a«=7t&3 tan a sin3 a g? and 7 + P sin a ^ or Y^V-P&ina or Y^ith3 tan2 a g?—nh3 tan3 a cos a sin a g?

=i-rch3 tan8 a gP (1 -3 sin3 a) A If R be the resultant thrust on the curved surface making

an angle 6 with the horizontal, then R~tf(X2+Y2)

=v^(jtA3 tan a sin2 «gP)sf{lnh3 tan3 « g9 ( 1 - 3 sin* a)}1] =nh3 tan a gP^fsin* a+'£ tan2 a (1+3 sin2 a)2] =^v:h3 tan «.gpi/[9 sin* cc+tan2 a (1+9 sin* ot-6 sin1 a)] • ^ J j r ^ t a ^ a g p ^ s i ^ o c c o t 2 a + ( l + 9 sin4 « - 6 sin" a)] =^/i8 tan a «.g?^[9 sin* a (cot2 « + l ) + l - 6 sin2 «] =\nhs tan8 «.gp<<(9 sin4 « cosec2 a+1—6 sin2 a}, =V (1+3 sin" a), where F«=inA8 tan"« g9

11? Hydrostatics

R tf(l + 3 sin2 a) A Required ratio

Also tan 0 =

F 1 7 37c/zstan2agP(l~3s!nsa)

'X u/z3 tan « sin2 a gp •=tan a (I—3 sin2 a)/3 sin2 a

or ran 6^ \ tan a (cosec8 a—3)=§ tan a (1-j-cot8 a—V) «=& tan a (cots a - 2 ) = ( l - 2 tan3 a)/3 tan a.

V. Required angle=angie between i? and the axis of the cone =»(*+GO-tarT1 [tan (0-f a)] (Note)

f l - 2 tan3 « •} , f tan fl -f tan « 1 . , ' 3 tan a I

^ t a n ! r ^ n - ^ l a n T J = t a n <| 7" / ' - 2 tatf«l f f

(. 1-(-rsinr)tB,,aJ -tM-1 [2 t i f ^ i «-l-tanJ I ^ W 1 <* cot«> =cot - 1 (2/cot a) = cor 1 (2 tan a). Hence proved. *Ex 9 (b). A solid right circular cone of vertical angle 60° is

just immersed in wafer with a generating line in ths surface. Find the angle which the resultant thrnst on the curved surface makes with the vertical

Solution. Proceeding exactly as in Ex. 9 (a) above we can get j 2 tan2 oc

tan 6=—- ', where 6 is the angle which the resultant thrust 3 tan a °

makes with the horizontal and 2a is the vertical angle of cone. Here7a=60° or a«i0° or tan ««=>tan 30°=l/\/3 . „ . . i - 2 n / < / 3 « l - f 1 ... Here tan « - _ _ £ _ . - _ - _ _ Now we are required to find the angle which the resultant

thrust makes with the vertical. Let it be 9$. ' Then 0=90°- 6 (See Fig. 132 Page 117)

or cottf=cot(9O°-0)=t»n0-l/(3\/3) _* -or tan<fi=3<&3 or ^ t a a " 1 ($</*) 'Ans.

*Ex. 10 A" cone whose vertical angle is 2a, has its lowest generator horizontal and is filled with liquid, prove that the resultant pressure on the curved surface is V d + l S sin8 a1*. W, where W is the weight of the liquid. (Agra 83 ; Jodhpur 76 ; Lucknow 78 ;

Ranchl 78 : Rohilkhand 81) Solution. Let h be the height of the cone, then the radius of

its base=ft tan «. I fF be the resultant thrust on the whole cone then from

Principle of Archimedes, we have ' F=>tbe weight of the liquid contained

= Jjc^tan aagP /

Thrust on Curved Surfaces Bounded by Plane Area 119

X (Fig. 133)

resolving forces horizontally and veiti-

~W (given), ...(1) acting vertically downwards through G, the C G. of the liquid contained. ^

If P be the thrust on 'the plane bass, then P^'gpzS"

•=gP.(AO cos a) % (AO)a, where AO<= h tan a

<=gP,7th3 {an8 « cos a ...(2)

Also if X and Y be the horizontal and vertical components of the resul­tant thrust on the curved surface of the cone, then cally, we have

X=P cos a^nh3 tan3 a cos8 a gP, from (2) and Y-P sin « = F or F = F + P s i n « or Y*=M$izh3 tan8 a gp+nh3 tan3 a cos a sin a gP

*~inhs tan8 Kg? (1 + 3 sina a) A If R be the resultant thrust on the curved surface making

an angle 0 with the horizontal, then R=\/(Xz-\- F2). or R-tf[(ith3 tan8 a cos2 a gP)1+{***' tana a gP (1 + 3 sin2 a)}8]

=|rc/i3 tan* a gPVl9 tan8 a cos4 a +1+9 sin4 a+6 sin8 a] •^Inh3 tan2 a gP\/[l+9 sin8 a (cos1 « f sin8 a)+6 sin" a] ~fah3 tan8 a gPV(l+15 sin8 «) •=W^\/(.l+l5 sin8 a), from (1> Hence proved. •*Ex. 11. A solid cone is just immersed with a generating

line in the surface ; if 8 be the inclination to the vertical of the res­ultant pressure on the curved surface, prove that \

(1—3 sin8 a) tan 0=3 sin a cos a, 2K being the vertical angle of the cone.

(Agra 79. 75 ; Lucknow 83 ; Rajasthan 76 ; Rohilkhaud 82), Solution. Proceeding exactly in the same way as in Ex. 9 (a)

Pages 117—118, we have X<=nh3 tan a sin8 a gp and r=47rA3 tan8 « gP.( I - 3 sin2 a)

The resultant thrust on the curved surface makes an angle 9 -with the vertical, therefore tan 0—X/Y

or

or

or

tan 0<

tan0«

ith3 tan a sin2agP 3 sin8

\nh3 tan8 « gP (1—3 sin2 a) 3 sin a cos a

tan a(l-3sina<x)"

1—3 sin2 a ^(1—3 sin8 a) tan 0=3 sin « cos a. Heuce proved **£x. 12. A right cone of semi-vertical angle a» is just imner-

120 Hydrostatics

sed with s slant sWe of length / in the free sarface. Show that the resultant thrust on the curved surface will cut this slant side at a

distance j—^— . 3 from the vertex, and find the magnitude of 4 (X—3 sin tfj

this thrust. Solution. Let h be the height of the cone, then

/=ftseca or fc=*/cosa, -"(l) Proceeding exactly in the same way as In Ex. 9 (a) Page 117,

we have J»*»the thrust on the base of ihe cone=^p.7rA8 tan3 a cos a, acting normally to the plane base through C, (he C. P of the cir­cular base, when 0'C'=(aa/4h), 'where 'If denotes the distance of the centre from the effective surface in the plane of the circle.

I.e.

(Fig. 134)

O'Ccs (A tan « ) ' „ ^ tan eg. ra4 {h tan *T 4

i?=the resultant thrust on the cutved surface =4TTA8 tan2 a \ / ( l+3 sin2 a) g?„

making an angle 0 with the horizontal and cutting OA at-£,&uch that OE=x (say) x

and F=the resultant thrust on the whole cone=»j7t'A3 tana a gP, acting vertically upwards through G, the C. G. of the liquid, where OG^lh.

V V is the resultant of R and P .'. Taking moments about the vertex O, we have

V.OG cos a^R.OEsme-P.O'C * (Note) $nh3 tan2 a gp.f A cos Q8=j7rA3 tan2 a-\/(l+3 sin2 a) g? x sin &

—nh3 tan3 « cos a. g? \h tan a lh cos a=x \ / ( l +3 sin2 a) sin 0—3 sin a.£/z tan a

x \ / ( l * 3 sin2 a) s!n *x "" — • • " " " * - " - 3h

or

or

or

or

2h , 3h sin3 a s-r- cos a + -j • 4 • 4 C03 a 4 cos a

3A 4.CQS a sin 6^/(1+3 sin2 a) * ••(2)

Thrust on Curved Surfaces Bounded by Plane Area 121

Also we have proved in Ex, 9 (a) P. 117 that fan fl= 3 t a n g '

1-2 tan2« sin 6'

or sin 0—

VK1-2tan a«>2+(3tana)a] 1-2 tan2 a

V(1_r 4 tan" a 4-5 tan2a) , 1-2 tan2 a 1-2 tan2 a

VK1 + 4 tan9 *)(l-ttm* a;] sec av/(i+4 tan* a) 1—3 sin2 a , . . A . , " . . . , .

5 3 ^ ( 1 + 3 sin'aj' c h a n 8 i n S int<> sinefonly. (Note) .*. From (2)7 we have required distance

3h B* . (1—3 sin2 a) .., , , . „ .

4 c O S a - ^ ( l + 3 s i n 2 a ) - ^ 1 + 3 s i n ! , g ) 3A 3 / cos a ~ , , .

or J C ^ T 7 i — 5 — ~ 9 ~ \ = ' ^ ;—=—••—*—r-> from (!) 4 cos 06 (1—3 sin2a) 4 cos a (1—3 sin" a) ' v '

«=3//[4 (1—3 sin2 a)] . Hence proved. E x . 13. A hemisphere, totally immersed in a homogeneous

fluid, rests with i ts plane base in contact with a rough fixed incliiied plane. Show that the reactions between the plane asd the hemisphere are equivalent to the following forces, viz a force \wh a normal to the base and a vertical force \wa a both acting at the centre, and a third force equal to the weight of tbe.body acting at its centre of gravity, where o is the area of the curved surface of the hemisphere, h the depth of its centre, a its radius and w the weight of a unit volume of the fluid. (Af. T.)

Solution. The forces acting on the hemisphere a re :— , (i) Its own weight acting a t G, the centre of gravity of the

hemisphere. (il) Resultant fluid thrust

on the curved surface of the hemisphere, passing through O, the centre of the base, since the thrust a t each point of the curved surface acts normally to the surface hence passes through the centre 0. (Also the pJane base being not exposed to ihe liquid, there -is no fluid thrust >v * < V on it. k " (Pig- 135)

(iii) ^Normal reaction at tne centre O of the base, and \ (iv) 'Force'of friction a 'ong the inclined plane. ^ These forces keep the hemisphere in equilibrium, hence we can

say tha t the resultant reaction (resultant of friction and normal •reaction) is equal and opposite to the resultant of weight of the body and the resultant fluid thrust on the curved surface.

i

122 Hydrostatics

Also we are given that <r= >area of the curved surface.

\ Let the magnitude of the fluid thrust on the curved surface be denoted by T. This acts at the centre O of the base.

In order to find this thrust on the carved surface draw diagram supposing that there is n"> inclined plane with which the hemisphere is in contact. |

Then thrust on the plane baStt="gP.z S"<=Mi>/i.7ras=4 who, acting normally to the plane area asjshown'in the figure. Also the resultant thrust on the whole hemisphere=weight of the liquid displaced=|7ra3w=io,atf, from (1). acting vertically upwards through G, the C.G. of the liquid displaced. |

The thrust T on the curved surface is the resultant of the two forces viz. $ who acting normally to the plane base and $a<jw acting in the vertical direction.

Hence the required reactions are the forces which are equal and opposite to the forces of which T is the rusultant and the weight of the hemisphere acting vertically downwards, or they are equivalent to the foices viz. a force £ who actinlg normally to the base and a vertical force \.wav both acting at thje centre together with a third force equal to the weight of the body acting at G, the centre of the hemisphere.

*Es. 14. A solid octant of a sphere is immersed in water with a plane face in the surface ; prove that the resultant pressure on the curved surface is [ 1 -f (8/TC2)]1>2 times the1 weight of water displaced by the octant. I

» I Solution. (What is octant ? If up pass

two planes mutually at rigt angles through the axis of a hemisphere, the hemispher'e will be divided into four equal parts. Each is called the octant of the sphere).

Let a be the radius of the sphere. ; Entire thiust on the octant =weight of

the water displaced=i (ffQ3<?P)» acting (verti­cally upwards through the centre of gravity of the liquid displaced. \~

This entire thrust (or the resultant thrust on the body as a, whole) is vertical and hence there is no horizontal thiust on the body as a whole. The vertical plane faces of octant are immersed in the water. "'The thrust on these will be purely horizontal; whereas on the curved surface will be horizontal as well as vertical thrusts.

Thrust on face OAB= "gpzS" \ • =gp.(4fl/3re) (iia2) . (Note) •=\<P g-p=ithrust on the face OBC

part

(Fig. 136) there

Thrust on Curved Surfaces Bounded by Plane Area 123

These "thrusts, on the faces OAB 'and OBC will be horizontal and normal to the respective faces and they will be equal and opposite to two horizontal thrust on the curved surface. Let these two horizontal thrusts which are in two mutually perpendicular directions, be denoted by X and Y. Then X=*Y=\ az gp.

Also there being no vertical thrust on the plane faces, the vertical thrust on the entire body will be due to the curved surface. Let this vertical component of thiust on the curved surface be denoted by Z, then Z=> J (f TA3 go) •=• imfgp #

A The resultant thrust on the curved surface - V(*J+r»-Fz*)=[(ifl3 g?)*HW gpy+(i™sgpWi\ <rlaa g9 [1 + l+(W]vaCTJTO8 gP [i+(8/»i»jp/a =[l+(8/« a)]" ! X weight of the liquid displaced.

Hence proved. *Ex. IS. Portion of a sphere cut off by two planes through its

centre inclined at angle \n is just immersed in a liquid with one face in the surface. Find the resultant thrust on the carved surface and show that it makes an angle 0 with the horizontal such that

tan 0=Jre- l . (I.A.S.74) Solution. ABD atid

ABC ara the two plane faces of tfte solid. The plane face ABD is in the surface of the liquid. Let a be the radius of the sphere.

Then the thrust P on the plane face ABC

=^p.(0(r, sin£7c),$7ra', where OG1=Aall-K

(£' ^ ) * *=W W?) gp, - 0> acting normally to the plane face.

Also the resultant thrust V on the body as a whole =the weight of the liquid displaced

vol. of the solid Jrc — _ * 2 S — ,

vol. of the sphere In =>f£vol. of the sphere) gp«=»|(*7ta3) pg=>\n^gP, acting vertically upwards through the C.G..of the liquid displaced. ' /

Let the horizontal and vertical comnonents of the thrust oa the curved Surface be Zand Yand their resultant be Jt inclined at an angle 6 with the horizontal. Then as the resultant thrust on the

'gp. kna'

(Fig. 137)

'(volume of the solid) x^P, where- •£- (Note)

124 Hydrostatics

and • Y tan 0=» Y*

body as a whole Is the resultant of LR and the thrust P on the plane face/15C, so we have X<~P cos frc=4oV(2) gp (1/^2), from ( 0 or X="|a3gp. and 7 + P sin ^7i=resultant thrust F on the body as a whole or Y=V~P sin £7r=ijra3gp -iaV2^P.(l /^2)-ia»gp ($TC-1)

A Resultant thrust i? on the curved surface

£a3 gp 2 " Hence proved. *Ex. 16. A solid is formed by the revolution of a semicircle

of radius a about its bounding diameter through an angle a, and the solid is immersed with one plane face isa the surfece of a liquid, prove that the magnitude of the resultant thrust on the curved surface of the solid is fasgp {(a—sin a cos a)a + sin4 a}1'8.

Solution. ABC and ABD are the two plane faces' of the solid, the plane face ABD be'ng in the surface of the liquid. Let a be the radius of the sphere, v ^

Then the thrut P on the plane face ABC

' =gp (OGj sin a) $ita3, • where OG1=>4al3it

=fa3 sin a gp, acting normally to this plane facel

Also the resultant thrust V on the solid as a whole •=the weight of the liquid displaced

(Fig. 128)

vol. of the solid " a vol. of the sphere ' 2n (Note) «=»(vohime of the solid) Xgp, where

= [(«/27t).f7rti3] gp=fo3agP, acting vertically upwards through the centre o' gravity of the liquid displaced.

Let the horizontal and vertical components of the thrust on the curved surface of 8ae solid be X and Y and their resultant be R. Then as V is the resultant of R and P, so we have

X=P sin «•=§o3 sin a gP sin a=f a3 sin8 a gp and Y+P cos a = V or T=-K—P cos a or 7=fa3agP—fa3 sin « gP cos a<=jja3 ,'a—sin a cos ec) gP.

.*. Resultant thrust R on the curved surJace of the solid •sin a cos a)gp}aJ,/a

Hence proved. . \ / (Z a+r a)=[(Io3sin s ocgP)a+{fa3 (a-"fasgp [sin8 «<K«-sin a cos a)8]1'*.

Thrust on Curved Surfaces Bounded by Plane Area 125 (

Exercises on Thrust on Curved Surfaces Bounded by Plane Area Ex. 1. A sphere of radius a is filled with liquid of d. nsity o.

Determine the resultant horizontal thiust upon half the curved surface of the sphere divided by a vertical plane through its centre.

Ans, na2og. Ex. 2. Find the direction and magnitude of the resultant on

the curved surface of a hemisphere of radius .>* placed with its base vertical and centre at a depth of 6" below the free surface of a liquid of which one cubic inch weights w grammes.

Ans. tan -1 (£J with the horizontal, 178 88 w gms. Ex. 3. A closed cylinder, whose base diameter is equal to its

length, is full of water.\ and hangs freely from a point in its upper rim, prove that the vertical and horizontal components of the resul tant thrust on the curved surface are each half the weight of the water. (U.P.P.c.S 78)

Ex. 4. A solid cone is just immersed in water with a genera­ting line in the surface ; prove that the inclination to the vertical

( 3 tan a \ . , , — g - j ,

where 2a is the vertical angle of the cone. Ex. 5. A solid right circular cone of height h and semi-

vertical angle a floats-just immersed in a heavy homogeneous liquid with a generator in the surface. Find the resultant thrust on its curved surface.

Ex. 6. A right circular cone of vertical angle 60° has Its uppermost generator in the surface of a liquid. ^ Find the resultant thrust on its curved surface if a be the radius of its base

* (Rohilkhand 77) Ex. 7. A solid right circular cone is placed with its axis hori­

zontal and at a depth A below the surface of a l.'quid. Find the horizontal thrust on the cone cut off by a vertical plane through the axis.

Ex. 8. "A right cone of semi-vertical an^le 30° has its upper­most generator In tbe surface of a liquid. Show that the resultant thrust on its curved surface passes through the centre of its base.

Ex. 9. A right chcalar cone closed by plane base, is held with its axis horizontal and is full of water. Find the resultant thrust on the lower half of the curved surface. (R<mchi82)

(Hint i See Ex. 7 Page 115). Ex. 10. A hollow weightless cone of vertical angle 120° is

filled with water and is held with its lowest generator horizontal. Prove that the resultant pressure on the curved surface is 3 J times the weight of the water contained. {Agra 81)

(Hint: See Ex. 10 Page 118).

•

CHAPTER IV

Centre of Pressure 4"0l.% Centre of Pressure. (Lucknow 79; Vikram 83,81)

If a plane area be Immersed in a fluid, then we know that the pressure on every element of area will be acting normally to the plane and in magnitude would be - proportional to the depth of the element below the free swface of the fluid. Then all these pressures on one side of the plane area will constitute a .system of like parallel forces. "The magnitude of their resultant will be the arithmetical sum of the magnitudes of these fwrcea and is called the Kcsultant fluid thrust The point where this resultant fluid thrust acts is known

. as the centre of pressure and written as C. P. Definition. (Luckhow 74 ; Rajasthan 75)

- The centre of pressure of a plane area immersed In a fluid is that point of the area at which the resultant thrust'of the fluid on one side of the plane area qcts.

J ^ ' 0 2 . To find the Centre of Pressure or C. P. Let the plane area be verti- Q ^ y.

' cal and equal to S. Take the line' of intersection of this plane area with the fr'ee surface as ^-axis, and a line perpendicular to it and5*" lying in the plane of the area as

• x-axis. . ,

(Fig. 139) the required centre of

Take any point P (x, y) in the area. Surrounding, P, consi­der an element of area ds. lip be the fluid pressure at P, then' the fluid thrust over this 'element' of area=/> ds and the total thrust on the whole plane area=»27 p ds.

Let C (x, y) be the coordinates of pressure, then the total thrust on the whole plane area acts at C. _.

".*. The moment of the Ihrust on the whole area about y-axis is equal to the .algebraic^, sum of the. moments of thrusts on each element about j>-axis..

*. Takirig moments about y-axis, we get xH,p ds<=*2 xp ds

- 2 xp.ds J xp ds 27 p ds J p-ds or (if integration is used)

Centre of Pressure 127 '

The limits of integration should be such that the whole area is covered.

Similarly by taking moments about x-axis, we get - 2 yp ds J yp ds y o — i i o ' J pds J p d s '

If double integration is used i.e. if we take ds=>dx dy, then - J Jxpdxd? - J J yp dx if

ffpdxdy , a y ~ JJpdxdy ' limits of integration covering the whole-area.

If toe liquid be homogeneous then, pressure at P is given by p=>g Px, where p is the density of the fluid.

• v = J * # P X dS „nrl « lygPxds • . * = • ,- j - dull .y=" —f j —

or - J x2 ds - ' xy ds x^1 7-rr-n and y — — Jxds " " " J J x d s *

403. Position of the centre of pressure of an area immersed in a floid remains unlatered relative to the area, if the plane of the area is turned about its line of intersection with the free surface.

Let ABCD be the y plane area, immersed in a y" fluid, whose inclination to - ' the vertical is 6 and let OY be its line of intersection with the free sui face. Take a line perpendicular to OY and lying in the plane. f the area as x-axis. Take aoy point P (x. y) in the area ABCD. Surrounding P consider an element of area ds. From P draw PL perpendicular to the free surface, then PL<=*x cos 9.

Z. Ifp be the fluid pressure at » P, then p*=g? PL=gpx cos 0, where P is the density of the fluid.

Also from § 4*02 above i yp ^ I p ds

Substituting p=g? x cos 0 in these, we get

, — f xp ds , — we have x= -,—j— and j * i p ds ' (Fig. 140)

J x gp x cos 0 ds j gp x cos 0 ds

and y* f y gP x cos $ as J gP JC cos 6 ds

1 8 Hydrostatics lft/8

or - j" x2 ds n - - f xy ds x= -, _,_• and y*~

r-*-\

w---:\

iz&y

J x ds """ ' J * ds which are. independent ofTl.e. the position of C. P. relative to the area remains unchanged when 6 changes i.e. when the plane of area is turned about the line Oy.

Note. Hence we can always suppose the plane of the area as vertical while finding out its centre of pressure without any loss of generality. *

_4-4%4. Geometrical method of finding Centre of Pressure. Let a plane area ABCD, not \-^££^~j^jry^y-^-jrjr=£z-.^

in a vertical position, be immersed r w-a--=^=^= -~ in a liquid. Draw vertical lines through every point of its bounr dary to meet the free surface of the liquid in the curve A'B'C'D. Consider the equilibrium of the superincumbent liquid in this cylinder. The forces acting on it are :—

(i) Resultant vertical thrust on the plane area ABCD acting through P, the C. P. of this area.

.(iij The weight of the liquid within this cylinder acting verti­cally downwards through G, the C. G. of the superincumbent liquid, and

(iii) The'thrust of the surrounding liquid on the curved sur­face of the cylinder, acting horizontally.

(The plane end A'B'C'D' being on the surface cf the liquid, there is no thrust on it.)

Resolving all these furces vertically we find that the upward vertical thrust on the plane area ABCD passing through its centre of pressure is equal in magnitude but opposite in direction to the weight of the liquid in ihe cylinder acting vertically downwards through the C. G. oftheifquid. ' r

Hence the vertical line through the C. G. of the superincum-bentiiquid in the cylinder will also pass through the C. P. of the plane area I.e. G and P are in the same vertical line.

Hence it follows that the centre of pressure of a plane area immersed in a liquid lies vertically below the centre of gravity of the superincumbent liquid.

sf*§ 4*05. To prove that the depth of C. P. of a plane area immersed in a liquid is greater than depth of its C. G.

(Avadh 83 ; Gorakhpur 77 ; Rajasthan 75) We know tflat in a heavy homogeneous liquid, the intensity

of pressure increases with the depth. If the plane area under con-

H6/9 -Centre of Pressure 129

sideration fs horizontal, then only the pressure at every point will be the seme and consequently the resultant thrust on the plane area will act through the C G. or centroid of the area. In all other cases when the plene area is not horizontal, the pressure on all the points of the lower portion of the area is greater than those in the upper one ; heme the centre of pressure, which is the point uhere the resultant of all these piessures on different elements of area acts, will be in the lower portion ard ttus below the centroid of the area.

/*§-4"06. Centre of pressure in some standard cases. ^ (a ) C. P. of a rectangle immersed in a homogeneous liquid

with one side in the surface. (Garhwal 79 ; Mithila 81; Ranchi 79) Let the side AD of the rect­

angle ABCD be in the surface. Let AB=a and AD*=>b.

Let EF be the line joining the middle points of AD and BC.

jen by symmetry the C. P. of the le lies on EF. Take an elementary strip

PQRS of thickness Ix at a depth x and parallel to AD.

Then the area of the strip ='8.s'=6 Sx and the pressure at

A E ,

7 1

T_-_-_-_«t-

X

Sx

0

-Q—Z-.

*gpx, (Fig. 142)

any point of the strip (whose thickness is very small) =>'/»' where p Is the density of the liquid.

.*. If 8 be the depth of the C. P. of the rectangle below AD,

thenx=S*p£s, } 8 0s

x gp.xb dx \ x2 dx - j J

or a Er

n&7 ', kjnce El V. P. of a triangular area immersed in a homogeneous

liquid with one side in the surface. (Agra 73 ; Bhopal 82 : Gorakhpur 75 / Lucknow 79; Masadh 73 /

Mithila 82 ; Muzoffarpur 81; Ranchi 82, 80, 74, 73'fRohilkhand 83)

Let the side BC of the triangular lamina ABC be in the surface.

D is the mid-point of BC. Let BC=a and AD=d. Let AD make an angle 8 with the side BC.

Take an elementary strip PQRS parallel to the side BC. Let DM =*x and MN^Zx,

so that the thickness of the strip=.FK"g>gjg sin fl and the depth of PQ below the free surface <=>EK<=x sin 6

Hydrostatics

From similar As ABC and APQ, we have FQIBC=>AM/AD

PQ d—x. -_ a(d— x) or —=-|—'or Pg«= , . a d d Lr Area of the strip PQRS

Q 8s=FQ Sx sin 6 a, (d - x)

sin 5 S*>

and the pressure at any point on the strip«=p^_sin-fl=ge, where P is " (Fig. 143) the densityTfTtelicTuid. Also due to symmetry, the C. P. will He OH the median AD. Hence if x be the distance of the C P. of the A ABC from D along DA, then

_ "J xp ds" j p as

\ (d-x) Jo

r as.x stBn&TF. -2 (d-jeysffir&dx-

\ xshr&gp: ^d^xTsitt-0-dx Jo

x*dx

{d- x) x dx

dx* 3 4 J,

-V .2 3 J,

.* ft-i)r --id

OT\X= \.DA. ].e. CXuflii&g^ ^BC is the mid-point of the median. ^ " ( c ) C. P. of triangular area imraersed in a homogeneons

liquid with its vertex in the surface and base horizontal. (Bhopal 83, 81 ; Garhwal 81 ; Gorakhpur 79 ; Mogadh 75 ;

^ .' - 'Muzoffarpur 80, Patna 82, Rohiikhand 77 ; Vikram 75, 73)

A

1€C-J3L

Let ABC be the ,trJanpIe with its vertex A in the surface^and base -z.-z=.-z^.-_~-. BC horizontal D is the. mid-point y^^s-l-y. of BG. Let BC<=a and ,the 'median L^-^r_r_rr AD<=d. Also AD makes an? angle 8 —£"£"£"-/* witbBC. \'K

w

Take an elementary strip PQRS parallel to, the j'slde BC. Let Afi£=x and MN=8x. •

.*. The"thickness of the strip T <=FK<=°$x sin 5 (Fig. 144)

and the depth of PQ below the free s\nface=iAF=x. ,sin 8 Also from similar As APQ and .4BC, we get

PQ AM , PQ x. _ - <JX

J. "" are^ of the strip PQRS= Ss*=PQ 8x sin e^(ax/d) sin 0 8x, and the pressure at any .point of the strip=»7?'«=gp x sin 9, where p

Centre of Pressure m Is the density of the liquid. Also due to symmetry, the C. P. of A ABC will lie on the median AD. Hence if x be the distance of (he C.P. of A ABC from A along AD, then

J xpds ax xgQ x sinfl. -j-sin

ax gp x sin 0. - j sin 6 dx

edx \lX'dx id* 3d

Jo Xs rf*

or |lx= J^(d) C. P. of a vertical circle of radios a wholly immersed in

a homogeneous liquid with its centre at a depth h below thesarface. (Agra 83, 81, 75 ; Avadh 79 ; Bhopal 80 : Bihar 74 ; Gorakhpur 80, 78 ;

Jodhpur 78 ; Luckriow 83j 81,74 ; Magash 76, 74 J Muzaffarpur 82 / Rajasthan 74; Ranchi 78 ; Rohilkhand 82; Vikram 83,81)

The centre O of the circle is at £ a depth EO-=h below the free surface.

Take an elementary strip PQRS of thickness gx at a depth x below 0, ^ - 5 ^ = = ^ ^ . such that PQ is horizontal.

Then the area of this strip •='Sz'~PQ Ix^lMQ lx ~2f(0&-OM*) 8x

And the depth of PQ below the surface=(A+x), so the pressure at any point of the strip«='/>'»=£P (h+x), where P is the density of the liquid. By symmetry the C. P. of the circle will lie on the vertical diameter AB.

= - T p :

Hence, if jc be the depth of G. P / of,the circular area below O,

^Eigv

/ ^ I

it_ _ f xp ds thQ"x-\7dT'

Mr t" xg4(h+x) 2:Ma)'-x*)dx J-o r rfftf' •• f" ^Af*)2.$$af-*a)<to

{" X(h+xWiM$X?)dx

s:

m

(h+xWia^-x^dx^;

/J(Note)

fw/a

l - ir/a a sin 0 (/z+a sin 0),a cos 0 a cos 0 do

(A+fl sin 0) fl cos 8.a cos 0 <#

putting ««<;

' 132 Hydrostatics 284—66

sin & cos2 6 d&+ a41 s?n2 B cos8 9 dB -it/i ' - T T / 2

a2h \" co^edd+a3 P " sin0cosa0d0 J - i f /2 J-w/2

sin" 0 cos8 6 d9 «= Y Ta » applying the property

» 2a2A cos2 6 d$

Jo

J /(*> *e~2 ]/<*) dx or 0, according*as f(x) is an even or odd function of*.

a m 2 2 i

Hence the depth of the C. P. of the circular area below the free surfacc^ff^Ha^hJ^

2§f4 07. C. P. of sum or difference of two areas. (a) C. P. of sum of two areas. If P1 and P2 be the thrusts on

the two parts of a given plane area and xt and xa be the depths of > their centres of pressure belov/ the free surface, then x, the depth of the C.P. of the whole area below the free-surface is given by

P1S1-fP2x3

Px+Pa * The generalised form of the above formula is

r.^ P^4P2x2+P3x„-i-.. , P i+P 2 +P 3 i - •• '

when the plane arta is divided into more than two paits. rt>) C.P. of the difference of two areas. If P and Pi be the pressures on the whole plane area and one

part of it respectively and x and xt be ihe depths of their centres of pressure below the free surface, then x2, the depth of the C. P. of the

. • • i P x - P i X , remainSng part is given by x a = - p — X--

From the above two cases we can observe that the method of determining the C. P. of sum or difference of areas is similar to that of determining C. G. of scm or difference of areas.

(See Author's Statics) The difference in these is that the place of weight in C G. is

taken by pressure in C. P. /Solved Examples on C. P . of reef angle, triangle etc.

v / l x , 1. Tfae gate of a dock is 10 ft. wide and 18 ft. deep and it has the pressure, due to 15 ft. of fresh water on one side, 10 ft. of

Centre 01 FieSsurb 133

sea water on the other side. Find the magnitude and position of (he resultant pressure on the surface of the gate.

(Fresh water weighs 62 3 lbs. per cu. ft. and sea-water neighs 64 lbs per cu. ft.)

Solution. Let AB be the side elevation of the gaie. Let Xj and X2 be the thrusts due to sea-water and fresh water respectively on either side of the gate, acting at, the centres of pressure P% and P2.

Then X^thrust due to f 0' sea-water on one side of the gate, which is 10 feet wide (Fig. 146)

-*'«S.S , ,-fPi (V) (10x10), 1 where £px is the weight per cubic ft. of sea water

=64 x 5 x 100 lbs , a; ting at P„ where AP^S- ft. And Xa=» thrust doe to 15' of fresh water on the other side of

the gate, which is 10' wide. ="g?zS' -^gPt ( ¥ ) (15 x 10), where gPa is the weight

per cubic ft. of fresh water •=(62-3) xi X150 lbs., acting at P.,, where AP^^^y,

.'. The resultant pressure on surface of gate^/J (say) =X3-X4«{(62 3)xl5x7j}-{64x5X100} = 123 {560-7-256}-125x304-7~38087-5 lbs.

Let this resultant pressure R act at a height of x ft. above the lowest point A then taking moments about A, we get

R.x=>Xi.APi-Xx,APi

(X2~Xj) x=-5X2-\aXu sinoe ^ P . ^ 5 ' and 4P i—? «•

15AV-10A"! _15 (623X 15x751-10 (61x5x100) *"* 3a (Zg-Xi) 3 (62-3X15X75-64X5X100)

=»6-4 ft. nearly. Ans. &&. 2. In the vertical side of a vessel containing water there

is a square trap-door, opeaiag freely outwards about a hinge in its upper edge which is horizontal. The length of a side of the door is 3 cm. and water rises in the'vessel upto the level of the hinge. Find the least force in grammes that will keep the trap door closed.

Solution. ABCD is the trap door whose upper edge AB Is hori­zontal. The thrust on this trap­door «"gp2S'! (Pig. 147>

or

or

4-£- 5-A

134* l9ydro§talics

«=l.f.(3x3) gms. w t . ^Y gms. wt.„ acting at P, the C. P. of the square ABCD, such that PE=\ EF=2 cm.

Let X be the least force applied on any point in CD to keep the trap door closed, then the moments of Zand thrust on the trap door about^tie hinge AB should be equal.

.*.yX.3=a^.2 or Z=9gms. wt. Ans. \JSx. 3 Prove that the horizontal line through the centre of

pressure of a rectangle, immersed in a liquid with one side in the surface divides the rectangle in two parts, the liquid pressures on which are in the ratio 4 : 5 .

N Solution. ABCD Is the given rectangle with side AD in the surface. Let AB=a and -AD=b. Let EF be the line joining the mid­points of BC. Then centre of pressure P of the rectangle lies on EF, such that 2?P=>fa and FP=>ia. LM is the horizontal line through P. Let P, and P2

De the thrusts on rectangle ALMD and LBCM respectively.

Then P,«=thrust on rect. ALMD ="gPzS"=gP ( | § ab) (%ab)^as bg?

Also the depth of C. G of rect LBCM

:. Px~the thrust onject. LBCM^'gezS'

2/3a

B

D

P

7 'A*

M

F,

(Fig. 148)

='A a*bgP. I or Pj

•gtWdob)

Hence P a ; P2~§aatep : -*da2figp~2 : f or Pt : P8 : : 4 : 5.

_^*Ex. 4. A rectangle is immersed vertically ia a heavy homo­geneous liquid with two of its sides horizontal and at depths a and b below the surface. Find the depths of the centre of pressure. v. , (Agra 79 ; Bihar 76; Rajasthan 78 ; Ranchl 79)

Solution. ABCD is the given rectangle with side AB and DC at depths a and b below the free sur­face. Take an elefnentary strip of thickness 5x at a depth x below the free surface. Then the area of this strip=i"Ss"=rf Sx where AB=d, and the pressure at any point of this strip=tp,'=gP.x, where p is the density of the liquid. IfEF be the line joining the mid points of AB and DC, then the C. P. of the rectangle lies on EF.

7 . If S be the depth of C. P. of the rectangle below the free surface then

fa A

h

0~ f £ E

•

F

X

I X

6 /

Sx

C

(Fig. 149)

\xp ds 1 xgpxddx | &2.dx _ ^ — — •=> -- 1

[/><&• f g9X.ddx \ xrf*

Centre of Pressure 133

Ex. 5. P is the centre of pressure of a recta'igSe ABCD. the side AB being in the surface. Prove that Hue through A and P divides the area into two portions the thrust on which are equal.

Solution. Let AP produced meet the A side CD in K. Let AB=>a and AD=0

Then £P~f ft and PF^\b In similar As AKD and PKF. we get

FK_PF_\AD KD AD^AD"* FK=\KD^{DFi-FK)

\KF^\DF=\t\a) or FK^la or or

D£=.aF+ jFX=ia+la«fa

(Fig. 550) .'. Thrust on AADK^'g? z ~S" .

=gp I AD\\AD.DK)=>\b\lag9, V AD=b\DK=la

•=3fl6agP» where p is the density of the liquid. And thrust on rectangle ABCD="gp z,s»

^gp.lb.ab^lao* gp=2 (thrust on A ADK). :. Thrust on A ADK and figure AKCBA are equal.

/ ' Hence proved. Ex/6. A triangle is wholly Immersed ia a liquid with its base

in the surface. Show that a horizontal straight liae drawn through the centre of pressure of the triangle divides it into two parts, toe pressure oa which are equal. i^sra S3)

Solution. Let the side AC of the triangle ABC be in the surface. Let DE be the horizontal line through P, the centre of pressure of A ABC. Then the depth of the line DE below AC is ih.

In similar triangles ABC and DBE, DE_BN_\h

~=2

Also the depth of C.G. of A DBE below DE«*?. ,A = £/i. the depth of C.G. of A DBE below AC-=>lk + $h=*U.

.*. Thtust on A DBE<=>"g9.z.S"<=gf>.2h.iDE.AN DgP Ih 11 AC.\h ^h^.AC.gp, where p is the density of

the liquid. And thrust on AABC-=t',gp.z.S"-=>g9.lh.l AC.h**lh\ACg?s

A. Thrust on figure ADEC =thius| on A ABC-thiust oa ADBE *

ne Hydrostatics

Huh

-IP. AC gP-ikh*.AC g?~?ih\AC g? •=»thnist on A DBE. Hence proved.

*Ex. 7. ABC is a triangular lamina vertically immersed in a Ibid with A in the surface and BC horizontal. M and N are the mid-points of AB and AC. Prove that the depth of the centre of pressure of the area MNCB is f frf, where d is the depth of B.

Solution. Since MN is the line joining the mid-points of the sides AB and AC.

:. MN is parallel to BC and MN=\BC.

Let P and Pt be the fluid thrust on A ABC and A AMN respectively.

Then P="gP.z.S"=.£P.f</.£ BC.d, -where density of the fluid=p

or ' P=°tdaBCge And P ^ ' g P 2S"~gP | ft*) (WN.W).

=.-aV d2BCgp, since MN^BC Also if x and x^ be the depths of C. Pvof As ABC and AMN

below the free surface, then x**\d and xx=l (Jrf)«-f J. .*. If x2 be the required depth of the C. P. of the figure

MNCB. then * . ! * * = J ^ - ^ ^ W -J&* BC *i * 15<f

(Fig. 152)

»'A-da MiV.gp .(1)

P-Pa i4 ,5c.«p-tf^.^c gp — i_^ A» 64 ~ 7l=Sf''- Hence proved.

^/**E*. S. A lamina in the shape of a quadrilateral A BCD has the side CD in the surface, and the sides AD, BC vertical and of lengths a and P respectively. Prove that the depth of centre of pressure is £ [(a«-p*)/(a8 - &•)].

{Bhopal 81 ; Magadh 75 ; Ranch! 80; Rohilkhand82) Solution. Let the side AB be produced to meet the free

surface m O, such that OC-=>x (say). In similar As OBC and OAD, we get

OD_AD_« o f c / ) « o C = « OC BC p

IfP>and Px be the thrusts on A OAD _ and A OBC respectively, and P«=>thrust on A OAD >="gPzS"=>g?.i« \OD a •=\«?.OD g? *=W.{*I$) rgP = | (a8/p) X gp and Px=thrust on A OBC -"gP2.S"=gp£p.ixp - t o " X gp •

p p ... D C<— & >£

«x

• • — • . — - - . — — • -

A (Fig. 153)

tTentre of Pressure 13t

Also if z and zx be the depths of C.P.'s of AOAD and A OBC below the free surface, thenz«»£» and za=jp.

A If z2 be (he required depth of the C. P. of the quadrilateral ABCD, then

aP.z-P.7lrJ (a3/fi) x gp. ja- IS2 xg? jp 2 /»-/», i(a8/PJxgp-|32XgP 4

[The above value of z2 can be further simplified and written as («+j8) faa+Pa)

(a8+ap-f-p2) by factorlsing num. and denominator • 1 (Bhopal 81 ; Muzaffarpur 82 ; Ranchi 80)

^*Ex. 9. A quadrilateral is immersed in water with two angular points in the surface and the otiier two at depths a and b. If x and y are the depths below the surface of the centres of gravity and pressure respectively, show that 6*j> <• ab=3x (o-f-6).

(Gorakhpur 78, 74 ; Lucknow 822 74 ; Mogadh 74)

Solution. Let ABCD be the given quadrilateral with two angular points A and B in the surface. Let DC produced meet the free surface in O.

^0

(Fig. 154) If Ax and A2 be the areas of As OBC and OAD respectively,

then Ax=\ OB.b and A&*=% OA.a. ...{1) Also the depths of C. G. of these »areas below the surface are

\b and \a respectively. A^a—A^b

A x=>depth of C. G. of the quad. ABCD*

_\OA_a.ka-kOBb.\b x a\QA—ba.OB = 3 ' a.OA-b.OB

-Ax

or

or

\OA.a-\OBb 3x (a.OA-b.OB)=ai (OA)-ba (OB), cross-multipjying

Oxa-a*) ( * - ( * * - * , OB or ^ - g £ ^ _ ( 2 )

Also if Px and Pa be the thrusts on As OBC and OAD respec­tively, then P1-"gPg5 , ,=gpi 6.J OB.b=»\W.OB g9 and i W g P z S ' ! =gpJa \OA.a^\a\OA Jp.

•138 Hydrostatics

And if xx and x2 be the depthsi of tbe C. P. of As OBC and OAD below the free surface xx*=>\b and xa=»|a.

=the depths of C.P. of the quad. ABCD below the free P^-Ptxx _ > * OA gp.\a-jb* OB g? \b

' Pt-Pi &.OAg9-lP.OBg? . a*OA~b3OB

surface

or or or

'"<*• OA-b* OB 2y (cf.OA-b* OB)~(ar> OA-bs OB), cross-multiplying

OA {2a*y-as)=t2bay-bz) OB OA W{2v-b)

i OB a

From (2) and (3) we get

3 <2y'-a) (3x-A) b b* <2y-b)

•(3) \3x-a)a- a2 (2y—a) or ~ (3* - b) a {2y -a) ~(3x-a) b (2y-b) or 6xya-3xa* -2aty + a a 6 = 6 ^ - 3 x 6 a -2aby+ab* or r &c iy(a-*)-3A:(aa-A2)+fl6(a-6)=0 or - / 6xy—3x(fl + 8)+fl6=0. Hence proved.

\^?Es 10. ABC is a triangular lamina immersed vertically in water with C in the surface and AB horizontal Show how to divide the area by a horizontal line PQ info portions on which the pressures are equal, P and Q being points on AC and BC respectively.

If h be the length of the perpendicular from C on AB prove . that the height above AB of the centre of pressure of the areai APQB

in the above case is \h (33^4—4). Solution Let the depths

ofifBand PQ below the free surface be h and x respectively. Let AB=c. Then in similar As CPQ' and CAB, we have

AB h

« . * - i " - i - ••<•> (FIg.,55) Since the thrusts on the two portions viz. A PCQ and figure

PABQ are equal, therefore the thrusts on A CAB=2 times the thrust on A CPQ.

or or

U. g?%h(\AB.h)* » ' h* AB**2.x2 PQ • h3c=2xsc/h or *3=£A3 or *=(/i/V2)

'2g9^x. (I PQ.x) ..using gpzS •2.x2 (x/h) c, from (1) and AB=c

& • The line PQ should be at a depth (fi/V 2), where h is the depth of AB below the free surface. v Ans.

Let P be thefhrust on ACPQ, then the thrust on A CAB=2P, •iace the thrusts oa two parts, made by PQ ace equal. Also if xA

Centre of Pressure / 139

and x2 be the depths of C. P. of As CAB and CPQ, bslow ths free surface, then x^lh and x%-=>l (A/V^), from (2)

X The depth of the C. P. of area APQB 2Px1-Pxa . .(3h\ 3h 2>h I 1 \

- J A [ 2 - l V 4 ] = f A [ 4 - V 4 I - J!. Required height of the C. P. of area APQB above AB

£ > A - i / ( 4 - V4) -" i* (3 .V4-4) . tience proved. £« **Ex: 11. Prove that the depth of the centre of pressure of a trapezium in water with side a in the surface and the parallel side b at a depth c below the surface is £ [(a-|-3b)/(a+2b)I c.

{Agra 74 ; Garhwal 81, 78 ; Lucknow 75 ; Muzaffarpur 81; Rohilkhand 80)

Solution. ABCD is /\ a,~ L D the given trapezium with side AD in the surface and AD=a, BC^b and CL^c. Join AC. Then AC divides trapezium in two triangles viz. A ABC and A ACD.

IfPiandPa^e the fluid thrusts on A* ABC and ACD respectively, then Pt

and P2

Also if^ and *2 bs the depths of the G. P.'s of As ABC and ACD below the fres surface respsctively, then »i=»4C and x 2 =|c .

A The required depth of O. P. of the trapezium ABCD „Plx1+P2x2 ^jbctgP fc+lac3 g? jc'

P1+P2 hbc* gp+frc* g?

/ W F T (45+S) "tfS+STi" Hence proved* *££ 12. A quadrilateral ABCD, with slides AB, CD parallel

(an3 AD, BC equal) is immersed in a hDm^geaeous liquid with AB in the free surface. Show that the centre of pressure will be tlie inter­section of the d iagon als if A B=tf3 CD. '

Solution. Let d be the depth of DC below the free surface. Let two diagonals AC and BD meet in P, such that the depth of P below the surface=P£'=z (say). Then PF**{d -z) and from simiiar As ABP and PDC, we have '

EP^AB ._ _J_t=AB n d-z_DC d-z~DQ

(Fig. 156) >"gP%S"<=>gP lc ibs=ibc2g? >"gP.zS"-=gP \c.\ac*=> $ac2gP

PP DO or or z AB

146 Hydrostatics

d .DC -

^d DC, , AB+DC

or z=d.AB/(ABi-DC) ...(1) Also^ the area of the

rectangle ABCD is the sum of the areas of A ABD and ABDC. Let A and A be ^ F the fluid thrusts on As ABD ' (Fig. 157) and BDC respectively and xlt xg be the depths of their C.P.'s below the free surface. Then x1'=^d ; x2^=ld

/>!-« gt.sS"=gf,y l$AB.d) = ld* AB gP and Pa=£p.fd.[\DC d)= \d\DC gP.

A The depth of C P. of the trapezium below the free surface „ P i * i + i V 2 _ ^ 8 ABg? \d+\d* DCgp.jd

Pi+P, Id* AB g9+%d» DCg? ^ f & ABH DC) d_(AB+l DC) d " QAB-t^DC) {2AB + 4DC)

If 0 . P. of the trapezium be P, the point of intersection of the , A (ABhWC)d

diagonals, then z - ^ — _ ^ o r (4B-r-/)Ci=

<2-iB+4/)Ci ' f f ° m ( 1 )

or 2AB(AB±IDC)<=>(AB±DC)(AB±WC) or 2AB*+4DC.AB='AB,i-h3AB.DC+DC.AB-\-3DC* or ABl=3DC* or AB=^3 DC. Hence proved.

Ex. 13. A Sat circular plats, of radius a, lies on a plane inclined at 30° to the horizontal and is subjected to water pressure on one face. The centre of pressure is at a distance (a/Id) from the geometric centre. Show that the geometric centre is at depth 2a below the free surface of the water.

Solution. Let d be the depth of the geometric centre below the free surface Let AB be the diameter of the circle along the line of greatest slope. Produce BA to meet the free surface In C. Then ,

OC^dcosea3ii°^2d Now if the plane of the

chcle be rotated so that it becomes (Fig. T5 ) vertical and the centre O of the circle be at a depth CO^ld below the free surface, tben the depth of G. P. of the circle below the centre O is "aa/4/i". (See § 4 Qi Page 127 and* § 4"06 (d) Page 131). i.e. OP~(a*(4.0C) or <a/16>=(aB/4.2d), V OP~(a/16) (given) or do>2fl. Hence proved.

Centre of Pressure 141

- *Ex. 13. (b). D etermJnc the depth cf the centre of pressure of a circular area whose p lane is inclined at an angle 6 to tbe horizontal.

i ( Rohilkhdnd 78)

Solution. Refer fig. 158 Page 140. Let d be the depth of the geometric centre O of the circle

below the free surface. Let AB be the di3n eter cf the circle along the line of greatest slope. Produce BA to meet tbe free surface in C. Then we have '

OC^d cosec 9. Now if the plane of the circle is rotated so that it becomes

vertical and the centre O of the circle will be at a depth=C0 «=</ ooseo 6» below the free surface, then the depth of the C.P. of the circle below the centre O is (aa/4#). i.e. 0P=»as/(4.C0), where a is the radius of the circle, or ' OP=a*l[4d cosec 0]

' A, CP=CO+OP=d cosec 0+[a"/(4d cosec 0)] & The required depth of the centre of pressure P below the

free surfaoe=CP sin B =•[<* cosec 0+{a2/(4d cosec 6)}] sin 0 =,d+{(a* sin2 0)/4d} Ann.

_JEx. 14. A square lamina, of side a, has a portion of it in the form of the inscribed circle removed from it and tbe remaining figure is immersed vertically in water with one side of the square in tbe surface. Jfr-oie that the depth of the C. P. is

&a(64-14ir)/(4-re). Solution. ABCD is the

square lamina from which the portion in the form of the Ins­cribed circle with centre O is removed. The side AD of the square is in tie surface. The radius of the inscribed circle

<=>i the side of the square

I f P and P, be tbe fluid thrusts on the square ABCD and the inscribed ci'de and x and *i be the depths of their centres of pressure below the free surface, then P«=the thrust on square ABCD*=>gpzS'=>gPf\a a2 ;

and

(Fig. 159) i ^ t h e thrust on the inscribed circJe«=»£P.$a.£7«ja 3 x=the depth of C. P. of square ABCD^\a x ^ t h e depth of C. P. of the inscribed circle

142 Hydrostatics

'A+<aW« ' 2 T 4(J f l ) a , a Sa

i — - 4 - _ £=a

2^8 8 * Hence the depth of the C P. of the remaining figure _ P . * - P i x_ _fo3gp fa~ _ita3gp \a__ f a—__na

P-Pi baag?-fra*gp (l-fcc)

& 24 /* \ 4—it ) Hence proved. _$&k. 15. A plane lamina consists of circular disc (radius a and

centre O) from which a circular portion (radius £a and centre P) has been cut. The lamina is completely immersed in a homogeneous fluid with its plane vertical and P vertically below O. If OP is equal to Ja

1 and the centre of pressure of the lamina is at O, prote that O is at a depth (11 a/8) below the surface of the fluid. (.Agra 80)

Solution. Let h be the depth of O below the free surface. Then the depth of P^LP^LO+OP= h+\a. If P and P_ be the fluid thrusts on the circles with centres O

and P respecthely and x and xx be the depths of their C.P. below O, then P=>thrust on circle with centre O _______ L ——

«."jpzS"«gPA.7caa ; P_=thrust on circle with centre P

=gp (h+_a) JZ (Jc)a

=£7taa(2/H-a)gp; x=the depth of C. P. of circle

with centre 0=(a2/4h) and the depth of C. P. of cjrcle with centre P below free surface - " A + ( f l W *

- ( *+§ ) - ( W -4(A+*fl)

or Jfji

+ 2 i S 0 + « ) ' =the depth of C. P. of circle with centre P below O

(Fig. 160)

, , a • a" i " 2 8 (2A+o) -A-la+

then x.

8 (2A+f l)" If xa be the depth of C. P. of the remaining part below 0, P.x—P_xx

_ i _ T..,- „,„ „.„„. ™_. „..- . .. -„ .„_ remain-P-Pt

ing part coincides with O, so that #a

og rca1

•4/f~

; but we are given that the C. P. of the

0

or P.*-P_.x_-0 o r r f i ^ . £ F - ^ ( 2 * + - ) ^ { | + I - ^ ) } - C

or •.7-»^*^n?y-«

Centre of Pressure 143*

or 16fl=>8A+5a or h<=>Ma/8. Hence proved. •Ex. 16. A circular area is immersed in a liquid with its plane

vertical and its centre at a depth c. If the area is lowered slowly so that the centre is at a depth f (t) at time t, show that the velocity of the centre of pressure at that instant is f [1—(a8/4Pj], where a is the radius. / (Vikaram78)

Solution. The dep*h of the C. P. of the circular area of radius a with its centre at a deoth c, below the free surface

= " k + (aa/4/j)"=»c+(aa/4c). It is given that the depth of the centre of circle at time t is f(t). .*. If x be the.distance of the C. P. of the circle from the free

surface, at time /, then x«=/(0+[oa/{4/(0}]. Dlfferentiatinng both sides with respect to t. we get

or the velocity of the centre of pressure at time t—f [1—(a2//3)]. Hence proved.

Exercises on § 401—§ 4 07 Ex. 1. The depth of water on one side of a rectangular lock-

gate is 16 ft. and on the other side is 4 ft., the breadth of the gate is 8 ft. Find the magnitude and the line of action of the resultant thrust on the gate.

Ex. 2. A rectangular area of sides a and b has the sides of length a vertical and the centre of the rectangle at a depth h below the surface of water in which it Is immersed. Prove that the C. P. is at a depth (aB/i2h$ below the centre.

**§ 4 8." Effect of further immersion. A plane area fs immersed tn.a homogeneous liquid and the depths

of its centres of gravity and pressure are a and b respectively. If the whole area is now lowered to a further depth h, without rotation to find the new position of the centre of pressure of area.

(Lucknow 80, 79; Raj as than 78, 76, 75) Let a plane area be Immersed in a liquid with its centre of

gravity G and centre of pressure P at depths a and b respectively below the free surface. If S be its area, then the thrust on this area is gpaS acting at iVwbere p is the density of the liquid.

When the area § is lowered to a further depth # (without rotation), let G' and P' denote the points G and P in the new position. »„ The depths of G' and P' below tlte free surface are (a+h) and (b±h) respectively.

144 Hydrostatic* 136/9

Due to this further Immersion, the pressuie^at every point of S will be increased by an arngijnjLgpJ_,and this-mcreaseH pressure is uniform over the whole area. Therefore_gpjjS, the total thrust as increased dueXo. further immersionjwjnjs_ACiing-at GH—Thes-inntEeTewTposition we

'fiave two tbfrusts (1) gphS acting at G' and (2) ggaS acting at P'. These two thrusts are parallel can be com­pounded into a single force acting at Plt a point on the line joining G\ P' and such that

gPaSxPj'^gphSxPfi'

or

z ^

d ID

PjG' " a If the depth of the point P^ (below the surface) is z, then

u PoS (b+ h) -J- gghS fa+h) a (b+h)+h {a+h) {a+h)

or

Corolla? a-j-h 7e have depth of Px-

•(2)

depth of P', f«ij) hl+2ah+ab , . , M h*+2ah+ab -(a+h)(b+h) • ° - ' a+~h {b+h)- V+h)

which is always negative since b>a. 'C-P

-ffii Hence depth of P' > depth of Py he. Pt is a higher level

than P' or due to further immersion, the centre of pressure is raised through the'distance [{.b—a)l{a+n){h.

Corollary 2. We have depth of Pj-depth of G' h2+2ah+ab {a+h)* tf+lah+ab-ia+h? ab—a2

'(a HO' a+h * ' (a+h) which varies inversely as (a+h) i.e. the depth of G'.

It follows that greater the value of h, smaller is the difference between the depths of Pa and G' Le. the centre of pressure P, and the centre of pravity G' come closer and closer as h is increased" and when a is infinite P^ aniG' coinage. (Magadh 77)

Corollary 3. If the C. P. of a given area is known when the atmospheric pressure is taken into account.

If II be the atmospheric pressure Its effect Is equivalent to supposing that the given area has been further lowered thiough a depth Ilfep, since a height Il'gpof the liquid would produce the same pressure II as due to the atmosphere.

136/10 Centre of Presstsrs 143

Sphed Examples oh § 4*08 \J EXv"l. Find the centre of pressure of an Isosceles triangle imraersetf with its plane vertical and its base horizontal half as far below the surface as vertea.

Solution. Let^flCbs the given triaagle.jof altitude h and side £C—a. We may\regard that previously the triangle was with side BC in the sur­face and now it is lowered to a further^ depth h.

The depths of C. G. and C. P. of the triangle below the surface in the position when BC is in the surface are \h and \h respectively.

/ . The required depth of the C. P. of the triangle when BC is at a depth h below the surface

J'WArlrii + ab'".

^_ r a 11* fA *" 6 AiA » ' ' Ans.

A rectaugle is immersed vertically in water with two

»}A and «d'=.p

...see § 4'08 Page 143

sides horizontal and at depths h and a j- h respectively below the effective surface ; prove that the distance of C. P. from the upper side Isja[(3h+2a)/(2h + a)].

Solution. We may regard that previously the side AD of the rectangle AB.CD was in the surface and then it has been lowered to depth h.

In the previous position when the side AD was in the surface, the depth of C. Q. of the rectangle below the surface = | a and the depth of C. P. of the rectangle below the sur­face =fa.

' J. The depth of C. P..of the rectangle when AD is at a depth hB+3a?i4ab"

(Fig. 163)

h below the surface=X=>

hs+2 (§a) h a+h

I4a+A) 2 jW+ldlA-a2)

3 {a+2h)

Hydrostatics

,\ ThedlstanceofC. P. fromtbe upper side AD

- • - A - 2 ^ ^ . f t = g j g , ' Hence proved. > -v 3 <a+?h) " 3 (a+2/») ( ^ V F X 3. A square lansiua is just immersed vertically in water

andis the" lowered through a depth h, if 'a' is the length of the edge of the jtqasre. prove that the distance of ibe centre oi pressure from thfc ceiitre*of the square is a2/(6a-f-l2h).

(Garhwal 78 ; lucknow 76 ; Vlkram 75) Solution. Let ABCD be the given

square, the length of whose edge js a and one of it« edges AD,i$ in the free surface. In this position the depths ofC. G. and O P . of this square below the free surface are ia and fa re&pectively.

Now let the square be lowered through a dep'h h. then 'he depth of new r-osition of centre of gravity viz. G' below the f>ee suTfaces=£G'=A+fo.

•» - -- And the depth of P', the "new posi­tion of the centre of pressure below the free suiface«=J5P' (Fig. 164)

, where V=>%a, 'V^la in this case _,h2+2ah+ab'

„/i2-J-2.£g./H jfl.fa _Aa+flft+£«a _J* (W+lah+a*)

A G'P'=EP'-FG'<=>

3(a+-2/0

6o.

i3 (a+2/i) 12ff»+12aft+4fla-12flft_12ff»—3fl'

. " 6(a+2fe) =fla/(6a-}-12/j). Hensepaoved. ^ 4. A rectangular area of a

height h is immersed in a liquid with one side in the surface ; show how to

' draw a horizontal line across the area so that the centre of pressure of the parts of the area above and below this line of division shall be equidistant from it.

Solution. Let ABCD be the given rectangle with side AD=a in the sur­face. Let LMbjS ihe.horizontal line at a depth z be.ow AD, such that ifPj and P2 be the C. P,'s of th,e rectangle

D

L

3

N P,

(Fig, 165)

i i i i i

h

k i i i i i i

Centra of Pressure 147

ALMS and LBCM tben NP^NP* Now EP^depth of C. P. of rect. ALMD below the

surface=fz.

Now LBCM is a rectangle whose two horizontal sides are at depths'2 «ana A belo>v the surface and LB=h—z. t

Also in Ex. 2 Page 145 we have found that the depth of the C. P. of the rectangle, whose horizontal sides are at depths a and a—h below the sutface, below the upper side is

^a[(3hi-2a)!(2h-\-a)). Putting a=h - z and h<^z in the re<mlf, we get

NP*~ r l2zH»-z)\ irlzTh J

or z(z+h)^(h-z)(z+2h) or z* + hz-h2**0 or z~\ [-h±\/(h*+4tis)] or z=$/i (\/5—1), neglecting the negative value. Ans.

Q) *%Xy4. A triangle is immfrsed in a Hqoid wish the base hori­zontal and vertex in sbe sorface. If the atmospheric pressure is equivalent to a heai! of H feet of the liquid, find through what height the centre of pressure of the triangle is raised in tbe plane of the triangle. {Bhopai 83; Rohilkhand 83)

, , Solution. Let the depth of A_ base BC of the triangle ABC, whose vertex A is in the surface, be h. When atmospheric pressure is neg­lected the depth of its C. G. and C. P. below the surface are f h and \h respectively. .

Now if the atmospheric pre- ,_, ssure which is equivalent to a head ^_ ,. D of H feet of the liquid, be taken (Eig. 166) into consideration, then the tffectofthis atmospheric pressure is equivalent to supposing that the triangle ABC has been further lowered through the distance H.

.". When atmospheric pressure is taken into consideration the depth of C. P. of the A ABC below the (effective) surface] "

'W+lah+ab". Hs+2qh}H+(U)($h) a+h ™ (|A)>ff

f(2/z-f-3tfj 2(2h+W) The depth of C. P. of A ABC in the case below A

" ' i i & + 3 f l r " 2 (2ft+3ifr (Note) A The height through which the C. P. will be raised

3ft 4ftff-4-3ft' ^ hH \<& / ~ 4 2 ( 2 A + 3 i ? ) °4(2ft4-3J7) Ans.

£,»~ fvJEx. 8. A triangle has its base in the surface of liquid and its vertex downwards, if atmospheric pressure be eqnivaleat to a height h of that liquid, prove that g D C the centre of pressure will be higher by a distance £ hd/fh+d) than it is when the atmospheric pressure fs neglected, where d is the distance of the centre of gravity of the triangle below the surface of the liquid

Solution. The side BC of &ABC is in the surface If atmos- (Fig. 167) pherio pressure Is neglected then the depth of G, the C. G. of t\ABC below the surface in given to be d.

_*. The depth of the vertex A below the surface is 3d. (Note) Hence the depth of P, the C. P. of A ABC below the surface

= * Qd) Now if the atmospheric pressure, which; is equivalent to a

head ft of the liquid, be taken into consideration, then its effect is equivalent to supposing that the &ABC has been further lowered through a distance ft.

.*. When atmospheric pressure is taken into consideration, the depth of P', the C. P. of &ABC below the surface (effective)

_"hz+2ah+ab'* h* + 2dh + d ($d)_2h*+4lidl-3d' , Z+h a d+k ' 2 (d+h)

,or the depth of P' below the side BO I J t ' * W + ^ L 2ftd + 3<?a

2 (d+ft) ft=,2(rffft) A The required distance^ depth of P- depth of P'

,s~\ 7^ 2 2 (d-i-h) a 2 (</+ ft)' Hence proved. ( J \ / *Ex 7 When the depth of the liquid is increased by an amount a, the^oepth of the centre of pressure is found to increase by y\ and when instead, the depth »f the liquid is increased by b, that of the centre of pressure is found to increase by z. Show that the depth of tbe centre of gravity of the area in the original state of the liquid is

• ab (b - a •+• y _ z)/(az—by).' Solution. Let xavddbe the depths of C.Q and C, P, of the

area in the orrgrnal state of the liquid.

Centre of Pressures 149

When the depth of the liquid ris increased by an amount a, then we are given that the depth of C. P. is increased by y.

d+y~"h2+2ah+ab°' i.e a + h

x-\-a or (d+yXx+a^aP+lax+xd r f - I -^ , or yx+ay+ad^as + 2ax ...(1) =

Again when the depth of § the liquid is increased by an z-amount b. we are given tbat the Zrl Q depth of C. P. is increased by z, ~~

. , "P+2ah+ab" ! „. d+za _ ; b*+2xb+xd '

(Fig 168)

cr or

or or

x+t> (x+b) (d+z)=b3+2xb+xd xz+bz4-bd<=b*+2bx ...(2)

Multiplying (1) by b and (2) by a and substracting, we have bpx-\- bay- axz—abz=a?b - abz

x (by—az)<=>a*b—ab*-bay+abz=ab (a—b—y+z) x<=ab (b—a+y—z)/(az—by). Hence proved.

Ex. 8. A plane area of any shape is completely submerged in a fluid of uniform density. Its plane is vertical. The depth of immer­sion is varied without rotating the area in its own plane. Show that the product of (i) the depth of the centre of gravity below the free surface and (ii) the depth of its centre of pressure below its centre of gravity is constant, the atmospheric pressure is to be neglected.

Solution. In § 4-08 Page 143 we have proved that if a plane area be immersed In a liquid with its centre of gravity G and centre of pressure P at ^depths a and ft respectively below the free surface and the whole area is then lowered to a further depth h, without rotation, the depth of new position of the centre of pressure below

the free surface^ 4n.— a n d t n e d e P £ n of C. G. of the area a+h In the new position is (a+h) below the free surface. -

& The depth of its C. P. below its O. G. In the new position h2+2ah+ab . . . . ab-a* . ,._ t.

° {a+h) ' ' - < * + « = (^HJf » o n amplification. .

£, The required product of the depth of C. Q. and depth of

« _ • «afl6-.al, which does not de* C.P.MowC.G.«(«+ft)x

H>drosfac(ics

pend-upon fa hence is constant for different depths of immeisiofi. Hence proved.

Ex, 9. A vertical circular area is immersed in water, h [he depth of its centre is doubles!, proie tbat the distance between its centre and the centre of press­ure will be halved.

Sohstion. Let a be the rad=us of life circular area. Let h be the depth of the centre C0 of the circular area below the free surface i.e OC0^=h. In this position let Pt be tne centre of pressure, then

CPi^a'hh. ...(0 (See § 4-06 d Page 131)

Now the circular area is lowered such that the centre C of (Fig. 169) the circular area is*at a depth 2/i below the free surface i.e OC^lh in this case. Let P2„be tue centre of pressure in this position.

Then O P , - ' - ^ ^ * — " . see § 4'08 Page 143.x

Hence 'h'-^h ; 'a'^h and 'h'^hr(a2[4h) &+2n.h+h[h+(ePi4h)] „V+2A 1 +AVia»

h + h 2/» J. CP,-OP,-OC=iI2A+(a«/8A;]~(2A)

CP%~aVih ...(ii) From (i) and (ii) we find that CP2=4 CP:. Hence proved. *** 4 09. Depth of tfas C.P. of a triangle in terms of tae depth's

or OP.

or

•»+e

of its three vertices. {Agra 79 ;

Gorakhpur 81; Luckm w 78 ; Rajaslkan 77 ; U. P. P. C. S. 81; Vikram 82) Let ABC be a tri­

angle,' the depths of whose vertices A, B and C below the free surface are a, (J and 7 respec­tively.

Let us first regard th i t the vertex A of the triangle ABC is in the free Mirface an J the depths . (Fig 170) of B and C beloar the iree surface are b and c respectively.

Let BC produced meet the free surface in 0, stun that OA=x> MPX ?nd P2 be the thrusts on As OAB*ad. OAC jespecfively

and ?j, Zi oe tne depths of taeir centre* of pressure,

Centre of Pressure is r

ihen

and

then

i 3 ^ thrust on A OAB = "gp7sS"=gp,%b.$x*<e>lxb2 g? ; / ,

2=thrust on A OAC^"gp3 S"-^gp.\c lxc=%xc2 gp ; Zi=>the depth of C.P of A OAB*>\b z2~tbe depth of C.P. of A OAC=\c.

. . If 2.be the depth of C.P. of A ABC below the free surface, ^ 1 g l •» 2 g 2

* ~|ft'sgP-ftcfogP ft'—V).

• ( 6 a —c a j °

(^.era 5i ; Gorakhpur 74 ; (Mithila 80 ; Rohiikhund8l, 79)

Now suppose that this tri­angle ABC is lowered through a distance « Without rotation. Then the depths of B and C will be A-fr-a and c-f a. But these (Fig. HI) depths have been taken as p and 7. Hence 6+«=P and e/.-fa=y or fl=»3-aandc=y—cs -^ * „.(2)

Also before lowering of the triangle, depth of C.G. of A ABC below the free surface was \ (b-j-c). (Note)

A The depth of C. P. of A ABC in this new positional!© § 4*08 Page i43).

k ib+c)-t*

, from (i) m « a +§ (6+c) «-f£ 16+c^ {(63+&o4-c»)/2 (b+c)}

^Cec'-Hs (5+c)+(63-<-fc+ca) & 2 (&+c)+6« ^ 6«a+-4oc ( f t -g+r-oO-H(P-«V»-!(P-a) (y -cQ4-(y-«)a]

2 (^—a-f-y—a)4-6« • - ^ 6a2 + 4a(Hy-2K)+(&2+3oC

a+y2 -3ap-3y«+py) ' " 2 tf+y+a)

« a + p a + y 9 + a p + p y + y a " M?n» 7P ; Gorakhpur 81 ; Lutfawto 78 ;

2{fl+ f+7) Rajasthun 77; U. P.P. C. S. 81)

~" Corollary 1. The above expression can also be written as

which Shows that .the centra of pressure coincides with the ceutre of three parallel forces acting at the middle points of the sides and

ibi Hydrostatic!

in magnitudes proportional to the corresponding depths. Corollary 2. The above expression can also be written as

k (2a4-S+y) « + fc (a-f 2j3-|-y) p+fc fa+ fi + 2y) 7 *(2a-|-p-t7JirAc(a-r-2p-hy)tA:(a+pf2y) ; '

which shows that the centre of pressure coincides with the centre of parallel forces acting at vertices A, B, C (i e. at depths a, 3 and 7 respectively) and in magnitude proportional to (2a-f-Pfy), (a-f-2P-j-7) aad (<t + $-f-2y) respectively.

Solved Examples on § 4*09

(Ex. 1. A plane triangular area is immersed in a liquid of uniform density with its plane vertical, one side horizontal and the opposite corner downwards. Its vertical attitude is h and the horizontal side is at a depth h below the effective surface bhow that Us centre of pressure is at a depth Y1 h below the surface.

Solution. ABC is the given (Fig. 172) triangle with its side AC horizontal and at a depth h below the free surface and the vertex B at a aepth h below AC or 2/i below the free surface. Hence the depths of A, B and C below the free surface are h, 2h and h respectively.

.*. The required depth of the C. P. of A ABC _ N

"aa+pa+y8+«P-i-Py+ya" 2(a-r-p-t-7)

y+(2k?+h*+h (2li)+(2h) h+h h 2 (k-t-Zh^h)

ll/:a lift

- [?N, Hence proved. £^ - -^•Esc^S. Find the centre of pressure of a square lamina isn-

meesedWa a fluid with one vertex in ihe surface and the diagonal Vertical. (Avadh 82, 80- Mithila 81, 80; MuZaffarpur 80; Rtmchi 78, Vikram 74\

8u the given

vertical Solution . ABCD is

' square with its diagonal DB and vertex D in the surface.

The lamina is symmetrical about the diagonal DB and hence the C P. of this square lies on DB. Also the depth of its C. P. is the same as that of A ABD.

Let,the depth of the vertices A and C below the free surface be d, then depth of B is 2d. .

, A, The depth of C. P. of A ABD <Fig. 173)

Centre of Pressure 153

the dep

_"«2 f p34-ya 4-g3+(fr+r«" 2(a+Pfy)

_Q + a!i+(2d)*+0.d+d2d+2d.Q_7da_7d B 2(Otd-i-2d) a6d'=' 6 '

If P be the C. P. of the square ABCD, then DP=*%d PB=>2d~l&=%d Ip:PB=id:$J~7i5. Ans.

3. Shaw that the centre of pressure of a triangular lamina 5s of ?vh»se angular points are «, (3, y is at a deptfi

A K2+Pa+r i-[3y-r«-«ft

below the centre of gravity of the lamina. Solution. If a? p, y be the depths of the vertices of the triangle

below the free surface, then the depth of C. P. below the free «a+P2+yaipy+y«+aP

2(«-r-P-r-y, and the depth of C.G. below the free surface^ (a+p+y).

,\ Required depth of C.P. below C.G. _aa > p»+y*+«P+Py+y« _ «+P+y

^ («+p+yj ^ _^ <q»+p»+-y«+aP+Py * 7a)-2 fa+ft +y)«

6(«-i-£J<W) 3 ( c

2+f i24-ya^ap+py+y«)~2 (da+P^-y2-r-2aP+23y+2yot) 6 («+p+y) "

«s+p3+y9-«p-py-y« Hence proved.

surface

*'% 6 («+£+->)

EX. 4. An equilateral triassgle ABC of altitude h has oae vertex fixed at a depth 2h below the surface of the liquid. k\®m the position in v isich BC is horizontal above A, the rria.igfe is turned tilt BC is again horizontal hut below A ; the plane of the triangle always remaining vertical. Show that the shift of the centre of pressore relative to the triangle is i.h/16).

Solution. Lee the veites A of A ABC be fixed and at a depth h below the side BC, which is horizon­tal and at a depth h below the free surface. .'. The vertices A, B and C of A 45C are at depths 2ft, h and A respectively Mow the free surface, (Fig, 174>

Ui ' , , hydrostatics

A In the case when A is below BC, the depth of the C. P. of A ABC below the free surface

"" 2 (ce-t P+7J r_(2A)a-4-A*+tf4-(?A) A+A.A4-A (2h)r_lU*_1JA " 2(ih-fh+k) ^ s-A = 8

or the distance of C. P. from A=2h-\xh = IA (Note) Again in the -second position w&en A is above BCB let ^ B ' C

be the new-position of A ABC. Then the depths or the vertices A, B' and C below the free surface are 2A, 3A and 3A respectively.

A The depth of C. P. of A ^ 5 ' C below the iree surface in . . . ' (2A)a+(3A)2+(3A8)+f2A) (3A)+(3A)(;A)-f (3A) (2A)_43,, tnis case= 2 (2h+3h + 3k)^- 16*"

/ . the distance of C.P. from A in this case*=f %h—2A (Note)

A ' The shift of C.P. relative to the triangle«iJA-gA (Note) , / v ^A" A. Hence proved.

•^^•yEjeCS. A quadrilateral immersed vertically haying two sides of lengths 2a and a parallel to^tae surface at depths A and Ih respec­tively. Show that depth of the centre of pressure is pi. /\Agra 80, 77)

Solution. The area of quardrilatera! ABCD «=the area of A ABD+the area of A BCD.

Let Pi and P2 be the thrusts — j ;.— . on A AMD and A BCD rsspec- - ; * ;

tively and zv ?a ^e the depths of «; !

their C.P. below the free surface. A z«- > n •> Then V ^ J^7 24

jP.^thrast on A ^5£>. \ !«• , , ~ ' ' / °"g^$" ^ A.?-''' - / ' ! «>gP(A+iA).(t2fl.A) ^ a C *

< B.f*»<«P, fFig. 175) Pa=thrust on'A BCD=*e°gp.z.S"~gP (hi-$h) (iah)-=Usag9> -

, .u P ^ o P A ,»n Aa-|-(2/zl2 + Aa+//.2A+2A.A + 2A A 4-fcpffa of C P. of A ^ - - - i f ^ ^ p j ~ - r -

*= VA and / ,. **, n « A A™ A8-H2A)2-!-(2A)a4-A.2A+2A.2A 4-2* A

x.-depth Of C. P. of A ACD-- 4 l A T ^ T 2 A r ~ — . l i t

A The icquired depth'of C . P . of the quadrilateral ABCD below the free surface- - , - & - - —^^s?)H^qg?) —

e=f A. Hence proved,

Centre of Pressure 355

**Ex^T A parallelogram has its corner at depths hx, b2) h3> h4 * below the surface of a liquid; and its centre at a d-pth h. Show tSaat the depth of its centre of pressure Is (b2

a

+fa22f5»3

s+V+8«ia)/(i2h'. {Agra 78 ; Lucknow 81)

Solution. ABCD is the parallelogram whose verti­ces A, B, C and D are at depths hu h2, h3 aad A4 respectively.

4 the depth of O, the middle point of AC and BD is A, so we have (Fig 17o)

\{hi\-h.i)=h^k{h-\-ll^ or /il+//B'=2/i«//,+A« ..(1) Let Px and Pa be the thrusts on A ABD and A -SC£> respectively

and Zi, z2 be tae depths of their centres of pressure below the free surface. Then

P^thrust on A ,4#Z)-=,sirp.z.S"=»gP £ Vh+Ih+hJ.S, where S is the are* of triangle; ABD^^gp (Aa+2A) .S ...from (1)

Pg=thrustoii A-B^D •="gp.s S '^gP. i (/22+/!8+W.S, where S is the area of

ABCD [T AABD~ABCD\ <~g9.\{2h\h3\S, ...fiom(l)

Zl=.depth of C. P. of A ABD V f A,' + At»+KK *• h &+ hjix

h?+hj f V+A x (A, +A4) f AaA4

2 (Ai+Aj+A*)

or Z . = -

.. J'xorn (I)

and r,-depth of G.P. of A #CI>= ^ H V l A ^ W ^ \ 2 ^Aa-f-As+AJ

* * T .—

I (llx -t-Jl)

G.P. of A >

2CA2+A4-rA3)

2(2A"+A.) - ...from(l) The required depth of the C. P. of parallelogram ^5CZ)

m

ifP Vh+2h) S

+l&(fh + 2h)S

2 (A!+2A)

2(2fe-t-ft,>

•4

i*P(AH52A) S>**P </23+2/i) S (fti'+V+V+2ftfti+M4) + (V+V+V+2ftft,+ft,ft«)

2ft)^(/23+2/2) (fti , (V+V

'2 * + V" -y)+(v+y+2 w + 2 * (&,+*,)

3* 2A+^ - f r o m W

Qv •A

6/z *(fc,' + V + A."-r *.'+»*•).

A2-t-A4=t2A

Hence proved. N ' "XiJE*. ?• A parallelogram whose plaae is vertical and centre at depth h below the surface is totally immersed. Show that if a and b are the lengths of the projections of its sides on a vertical line, then

aa4-ba

the depth of its centre of pressure will be h-f .

(/. A. S. 75 ; Lucknow 79) Sofatson. Through A, the uppermost vertex of the parallelo­

gram AECD draw a horizontal line PQ meeting the vertical lines drawn through B and D in P and Q. Also it is given that BP=*^

, and DQ<=>a. Hence BD is not horizontal.

\ .\ From the figure, it is evident, O being the mid-pout of BD, that ON=i {BP+DQ)

i •=»$ (b+a). Let hu fta hs and ht be the

depths of A, 3, C and D respec­tively below the free surface.

Then h^AM^ OL- ON<= h—i (c-f b), /22=depth of B=AM+b

t=>h—l (a 4 b)i b<=h - \ (a- 6); fc8~depth of C^OL+ON (Note) (Fig. 177)

aad ft4«=»dep$h of Z)=4.ftf+fl=ft-$ (a-f-d)-f a=/z+$ (a—b) A From last example, we have the required depth of G. P.

of the parallelogram ABCD K

12/i m nh after simplification of ( V + V - W + V )

iwA-t-Ktf+aWAJ Hence proved.

4 A parallels

gram ABCD is immersed in a liquid with A in tbe sur-

* face and BD horizontal Prove that the centre of pressure P lies on AC such that AP : AC=*7 :12

(Agra 89)

Solution. Let O be the point, where the diagonals AC and BD meet. The area of the parallelogram=>sum of the areas of A^-BCand A BCD.

IS?

Fig. 178)

Thus centres of pressure of both these triangles will be on their median ie. en AC Hence the C. P. of the parallelogram will also be on AC.

Let P1 and P3 be the thrust on A ABD and A BCD and zx and za be depths of tbe C. P.'s of A ABD and A BCD respectively.

Then P^thiust on A ABD=*"g(5.S"= gPffi.S, where A is the depth of BD below the free surface and 25" is the area of the paiallelogram ABCD or S !s the area of triangle BAD and BCD!

[V AABD=*&BCD]

P ^ thrust on LBCD^g? (A + 1A) S-=*$gphS, si—depth on C. P. of AABD*?3h/l

and za=depth of C. P. of &BCD ^ft3 s-(2hY+hs+M2h)+2h.h+h.h lift "" 2{.h+2h+h) D 8 '

.'. The^depth of Pt the C. P. of the parallelogram ABCD

below the surface • /Vi-4-PaZa

Also CM*=>2h. J. In similar triangles APL and /4CMS we have

AP PL (7A/6) » 7 — - r - — • .

AC CM 2A 12' Hence proved.

^S^rfl). A square whose side is 2a is completely immersed in a homogeneous liquid in a vertical plane with its centteat a depth d. Prove that the centre of pressure is at a distance a2/3d below the centie of tbe square, whatever be tbe inclination of tbe sides of tbe square to the vertical.

Solution. Let ABCD be the given square with its centre Q at depth d below the free surface," Let AM«=• A^and 0 be the inclination of^iO to the vertical.

Hydrostatics

and

or

or

or

The depth of S -- K (say)=BP=BK+AM *=>2a sin 0+ht;

/i4=>depth of D=DQ =>£2V4 AM<*>2a cos 0+ht

7;3<= depth of C=EQ =*£>Q + DE =>(2a cos & + A1)+2fl sin & •=2a (sin 0-j-cos 0H^i .

HowOL^CBP+KQ) - t (*.+**)

rf-i[(2asln0-«-At) ^ +<2o cos fl+Ai)

=fl(sinfl-)-cos^+A1

h^d-a (sin 0+cos 6)

...(1) Now V + V - ( M - 2 a sin 0)H-(A,+ 7a cosfl)3

=2V+4o2+4fl/j, (sin 8+cos 0) *i"+ ^ M V=?*i"4 4fl2+4flAs (sin 04cos 0)

Also Aaa= [Aj+^o (sin fl+cos 0)]a

j«=»V+4fla (sin 04cos fl)J+rfflA1 (sin 0+cos 0) » i V + V + fta"+*4,-3Ai,+4ft,+4flA1 (sin0-fcos 0)

4 V 4 4aa (sin 04 cos 0)a-f 4tf^ (sin 04 cos 0) —from 0 )

=•4/^4 8e&i (sin 04- cos 0)4 4a2 (sin 0+cos 0)24 4a2

= 4 [d-a (sin &-\ cos 0)ls4-8tf [«"-a (sin 0+cos fl)](sin 04cos0)

(Fig. 17S)

(2)

4 4a* (sin 0+cos 0)2+4a?, from(l) «=4d248fi2 (sin 0+cos 0)s-8a«'(sin 0+cos 6)

+ 8 a i (sin 0+cos 0)-8aa (sin 0+cos 0)a+4as

=4d»44fla. „A From Example 6, Page 155, we have the depth of C. P. of

the square below the free surface ~[h* f V 4 - V 4 hf+Sd»]ll2d (Note) _(4«»-Mg»)+gg» „12<P+<^ rf, £ .

l rf *" 12d " + 3d* A depth of C.P. below O, the C Q. of the square '

This result being free from 0, this distance will remain the same whatever/be the inclination of the side of the square to the vertical.

^Y'v \jj$Ex. 10. If a quadrilateral area be entirely immersed in water, and a, P„ J», S be deplbs of its fonr corners, and h that of its centre of gravity, show that tbe depth'of its centre of pressare is

Centre of Pressure 339

Solution. Let ABCD be the. given quadrilateral with Its Vertices at depths a, P, 7 and S below the free surface. Join AC.

Let Aiand A a be the aieas of triangles ADC and ABC respectively.

Let zx and z2 be the depths of centre of gravity of the triangles ADC and ABC respectively. Then

Zi=k (a-iyi S) and 2a=> J («+p+y).

Also h being the depth ofC.G. ofrhe quadrilateral ABCD, we have

AiZi+AaZ-g A i+A 2

(Ai+Aa) ^=-Ai?i+A2?a A1(A~ri1 = AaU2 A). A i „ z 8 - A A a A—r,

H « 4 P + T ) - A

A=

or or or

,tt+p4y-3A 33A-«-y- 8 ...(2) (Fig. 180)

/gain if *x and *2 be the depths of the centres of pressure of the triangles ADC and ABC respectively, then

8l»j.y»+8«+ay+y&+8* A a9 + P2+ya+«3+pv+-y« * 2 ^ + ^ + 8 ) a n d ^ . 2 ( .+Pi -y) .'. If Px and P2 be the -thrusts on triangles ADC and ^5C

respectively, then P1^i'gesS '=gp % (<*+y4. 8) Ai

and ^ iWpH<H-p4-y> Aa Hence the requiied depth of C.P. of the quadrilateral ^ M ^ I T " a * a "

fa2 4-yg+83+ay4-y8+5cQ_ 2(a+y+S)

i w i e i N o A (a a4-P3+y2^«P + Py4-ya)

+*(«+?+ y)*p A,. 2 ( a + p + ^ r .

H « + y + 8 ) Aigp.

i(«-

• * • •

T-y + 8) A, gP+i(« + P+30 A2£P (a2+y2f'Sa+«r+yS + s*) Ai

4-fa2 4 |324-y94ap+By4-y«) A3

AiZi + Aa^a (Note) A i

4. (a2 + yHSHay+y8+s«) . " 1 +(«HP a +y a +«3+pyfy«)

"A'KAt/AO + ll . . Substitutirg tie value of A 3/A a from (2) and simplifying we

get the required result.

Exercisei on § 4*09

*Ex. I. A rhombus is Immersed in a liquid with a vertex in the surface and the diagonal through the vertex vertical. Prove that the centre of pressure divides the diagonal in the ratio 7 : 5.

(Agra 75 ; Bihar 75 / Raechi 75 ; V. P. P. C. S. 77) **Ex. 2. A rhombus is Immersed in & liquid with a vertex in

the surfaccand the diagonal through the vertex vertical. Prove that the centre of pressure divides the diagonal in'the ratio 7 : 5.

(Agra 81, 75 ; Rohilkhand80)

[Hint, i Do as Ex. 2 Page 152] *Ex. 3. Show that the depth of the centre of pressure of a

rhombus totally immersed with one diagonal vertical and its centre at a depth h, is h+(az/Z4h), where a is the length of the vertical diagonal. (Lucknow 74)

Examples on Centres of Pressure by Integration.

*Ex. 1. A rectangle is immersed vertically in a liqnid whose density varies as the depth btlow the surface If the two horizontal sides of the rectangle are at depths h asd k below the surface, find the depth of the centre of pressure. (Agra 82 ; Patna 82)

Solution. Let ABCD be the rectangle whose sides AD and BC are horizontal end at depths k and h below the sur­face. Let AD=b. Let PQRS be elementary strip of breadth Sx at a depth x below the sur­face. Also the density P of the liquid at a depth x below the the surface=/t'x, where k' is constant.

(Fig.J81) Then the area of the strip PQRS^ 8s=fl8x, Also the thrustp at a,depth x-=gP x*=>g.k'x.X'=>k'x2g

' A If 55 be the depth of C. P. of the rectangle, then

\xP ds j* x.k' x2g.b dx r *8 dx i (x* y \p ds V k'x3g>b dx I* xa dx i (x3 j

(,

A

i -S 1 1

'M W//////M,

x !

y/////////;wl

9

<0

n

QR5X

c

W-k*)_z (h+k)Qi3 +Jfca) {h*+hk + k2)

the depth of the C. P.

Ans. ~ 4 (A3-/c3)" Ex. 2. Prove that the depth of the C. P. of a parallelogram

whose two sides*are horizontal and at depths h and kbelow the surface of a liquid whose density varies as the depth below the surface is

136/11 Centie of Pressure 161

3 h3+hak+kah+k3

4 ha+kh+ka • (Avadh 82; Gorakhphr 76 ;

Lucknow 78 ; Mithiia 82) Solution. Let ABCD

be the parallelogram whose sides AD and BC are parallel. Let AD=a. Consider an elementary strip of thickness 8x at a depth x below the sur­face.

Then &y«=area of the strip=aSxandp=pressure (Fig. 182) at a point at a depth x=gPx=g.^x x, since p=A*, A is const..

.*. If x be the depth of C. P of the parallelogram below the

\xp ds 1 xgkx2.odx 1 xa dx i (** ) surface, then x <=> j =• i | — =. ' * — —

\pds \ g\x*.aJx I x1

3 (£«-&«) 3 (k+h) (&a+A2)

2 dx i j x3 )

4(fcs-/J3J 4(fta+ifcAi-Aa) * kz+kh+hz ' Hence proved.

Ex. 3. Find the depth of the centre of pressure of a parallelo­gram immersed in a homogeneous liquid with a side in the free surface (Magadh 77)

Solution. Prove exactly as in Ex. 2 above. Here the side AD of the parallelogram being in the free surface, we shall get

\xp ds 1 xgP.x.adx

^pds - 4 ? -r

Jo

g?x adx

wherep==gPx, AD-=>a and depth of BC is h.

j* x* dx ( i x3)*

o> \ -^ i** M! it3

ADS.

'4. An ellipse is completely immersed with its minor axis horizontal and at a depth h ; find the position of the centre of pressure. (Avadh 83; Gorakhpur 77)

Solution. Referred to. major and minor axes as axes of refe-xz s>a

rence the equation of the ellipse Is -g- + rj-«=l. / n

162, Hydrostatics

where AA'**2a and BB'<=>2b. Consider an elementary

' strip of thickness Sx at a depth x below the centre C of the ellipse. I

_ Then &y=area of the stifp ~2?Sx=(2£>/a)V(oa-xa) 8*. sincey=*(d/a)</(az-x

2), from (1) and/?=pressure at any point at

a depth x - below C - g p {h+x).

If x be the depth of C. P. of ~ the ellipse below the centre of the

ellipse, on the major axfs (due to symmetry), then

— /

Ixp ds \: xg?(h+x). — i/(a*-xs) dx

ipdS \° i9 (h+x). 2±<na*_-x>) dx J - 0 /•»

/?J° xtf(oa-x2) dx+[° xVte 2 -* 3 ) dx

J" </(a2-x2) <fa+

Jo

x V ( a 8 - x ' ) dx

2h j° V(a2-^a) <*« , other integrals vanishing

I since J

* !

f{x) dx=0, iff(x) is an odd function of*

sins 0 cosa 0 dB

he? \ cos2 8 d8 Jo

aa r$n i

-, putting x=*a sin 9

4" «a

**Ex. 5. An ellipse is just immersed m water with its major axfs vertical. Show tbat if the centre of pressure coincides with the lower»focos, the t ccentricily of the ellipse msjst be J. '

(Agra 80 ; Avqdh 79 ; Bhopal SO / Lucknow 80 ; Rajasthan 77,75; Ranchi 81)

Solution. Referred to major and minor axes as axes of refer-x2 v2

snee, the equation of the ellipse is -g- + ^ - => 1,

Centre of Pressure 163

where AA'=2a and BB'=2b. Consider an elementary

strip of thickness 8x at a depth x below the centre of the ellipse.

Then Sj=»area of the strip =2?$x ~(2b/aW{ifi-x*) Sx,

from (1) and h =»pressure at any

point at a depth x below the centre of the ellipse

=g? (a+x). „ .'. If* be the depth of

C.P. below the centre of the ellipse, (Fig. 184)

then x \xp ds

= 3 - = S 3

[' x.gp (a+x). Qtf(o2 -JC2) dx J-a a

J p ds \" g? ( a+x) .—V(f l* -*V*

J -a a

aV JC\/(fl8-x*) dx-h\" x V ( a a - * J ) **

flf-0 ^a*~x2) dx+\" xV(a2-x2) dx •=>ia, integrating as in Ex. 4 Page 161

But we are given that C P. coincides with the focus i.e. the depth of C. P. below the centre of the ellipse«=>ae.

\ ae=\a or e=h Hence proved.

^**Es. 6 (a) A semi-ellipse bounded by its minor axis is just immersed in a liquid the density of which varies as the depth ; if the minor axis be in the surface, find the eccentricity in order that the -focus may be the centre of the pressure.

(Gorakhpur 81; Lucknow 82 ; Rohilkhand 83, 79)

Solution. Referred to major and minor axes as the axes of reference, the equation of the ellipse is

•a) where CB=b and CA=a.

L64 Hydrostatics 384-S6

Consider an elemen­tary strip of thickness $x at a depth x below the free surface, on which the centre C of the ellipse lies.

Then 8*=area of the strip=2j8x =2 (blaW(az-xa) dx,

V from CO we have y=>{bla)<tf(a*-x*)

and /?«=>pressure at any point at a depth x

*-gpx, where P=fc* *kgx\

B' '

\

\ & C \

C

X

•y

A

y

V

B y^

I.e. p (Fig. 185) A If x be the depth of C. P. below the free surface, then

fxp ds \" x.kgx*. — </(az-x2') dx

f r« 26 \ p ds \ kgx*. —-y/fa2-*a) <**

i/v(i fla-JC2) OX fl5 J !

Jo sin3 6 cosa fl </0

* V (°a—*s> <fc fl* [ * *sin8 *cosS 6 de Jo

putting x<=a sin 6 r2rjU 2f3 2.1 32a 2 f | " r | r f " " " t . J . i V M V « "i5ie"

But we are given that C.P. coincides with the focus, i.e. the ', dep<h of C. P. below the free surface=>a« -

M, „*. a<?=.32fl/(15re) or e=32/(157r). Ans. x / ^ E x . 6. (b) A' uniform elliptic lamina, whose ayes are 2a and 2b, is half immersed in water, (be axis 2b being in the surface. Find tbe centre of pressure. (Bhopal 82 ; Rajasihon 76)

Solution. Refer Fig. 185 above. Referred to major and minor axes as axes of reference, the

equation of the eliipse is i

' ere CB=\ BB'~\ {2b)*=b and CA*=o. ^Consider an elementary, strip of thickness Bx at a depth x

oelow the free surface, on which the centre C of the ellipse lies. Then 8s=area of the strip=2j8#"

and p«=pressure at any point at a depth x=gpx. .'. If * be, the depth of C.P. below the free surface, then

Centre of Pressure 165

j xp ds t)x_n

ipds I" gPx.lydx

x'ydx. h^ a_ 0 , where ? = ———

-x*)

\ xy dx

from (i)

1° x V ( a a - ^ ! ) dx 2a\ h := 1

'ol sin2~0 cos2 6 dd o

1° xy/(aa-x2) dx aa V"* sin 6 cos2 0 <#

?rfrfx_2rf putting x = a sin 0

2r3 "n r f

2.1 = 1 6 4ns. **Ex. 7. A semi-circular lamina is immersed in a liquid with

the diameter in the surface. Find the centre of pressure. {Agra 82, 78, 74 ; Avadh 81. 80 ; Gorakhvur 79, 76 ;

Garhwal 79 ; Vikram 78, 77)

Solution. Let the p)ane of the lamina be vertical with the diameter BOB' in the surface.

Let OA be a radius of the circle perpendicular to BOB'.

Choose OA and OB as x and y axes respec­tively. Then referred to these axes, the equation of the circular arc is

a* f / = a 2

where a is the radius of the circfe. Due to symmetry C.P. of the lamina lies on OA. Consider aQ

elementary strip ol thickness S# at a depth x below the surface. Then Sj=area of the strip=2y 8jc=.2\/(fla-xJ) Bx, from (1)

and /»«= pressure at any point at a depth x^gPx. .; If x be the depth of the lamina below the surface,

i xpds \ ' x gpx.2<tf(a2—xl) dx tnen JC= «=» -—

/ ? '

1 \ 0

y X

X

y

5...

) /<x,yy,

*Flg. 186)

.(1)

\"pds [" gpx.2tf(aa-xz) dx J0 JK-O

166 Hydrostatics

[" xV(a a -x2) dx a4 ["* sin2 8 cos* 6 dB Jo , Jo

! xV(a2-xa) dx a ir/2 »

•3 5 sin 6 cos2 0 </0

5' 0

\ x

B

y

putting x=a sin 6

r f r f y j z i ^ i^ i fv /u 3A7C = f l l 2r3 r i r f " 2.1 16 • Ans.

' Ex. 8.- A semi-circular area of radius a is immersed vertically with its diameter horizontal at' a depth b. 'If the circumference be below the centre, prove that the depth of centre of pressure is

. 3rc(a2+4b6j+33ab * 4a-f-3rcb

Solution. As in the last example, let OA and OB be the x and y axes respectively. Th,en referred to these axes, the equation ot the circle is

xi+yz^a!i ...(I) 1 Consider an elemen­

tary strip of thickness S* at a depth x below the .diameter B'OB.

' Then Bs=area of the strip~2y Sx=2V(a* -x2) Sx, from (i) (Fig. 187) and /»=pressure at any point at a depth x below B'OB

=gp (b+x) & If* be the depth of C.P. below the diameter B 'OB, then

\xpds [° x.g9(b+x).2tf(a*-xz)dx x~± - i 2 l

\pds \' g? (b + x) 2V(aa-xa) dx

\" xtfiaP-x*) ^ + ( ° * V ( o a - x a ) dx , ' ' ^ Jo Jo

b V ^<a a-x a) dx +• (° xtf(a*-xa) dx Jo Jo

, fwfs fir/a < a*b I sin 6 cos2 6 d6&a* \ sin2 6 cosa 8 dB

fifl2 l cosa 0 dfl+a3 \ sin2 0 cos2 0 </0 Jo Jo

putting x<=a sin 0

Centre of Pressure 167

or _ _ 16a& f 3rcaa

* 4(36TC+43) The depth of C. P. below the surface*

l€ab-{-3na2 , , 3TC <e?+4P) • 32 a6 , ' JC+*

+Z>« Hence proved. 4(3foH-4fl),"r"-"" 4( 0;e+4a)

Ex. 9. Frfive that the depth of pmswe of the area bonoded by (wo concentric semi-circles with their common bounding diameter in the surface is -fen (a4- b4)/(a3- bs), a and b being their radii.

Solution. Let Px and P2 be the fluid thrusts on semi-circles of radii a a»d b whose centres of pressure are at depths zx and za below the free surface

Also the depths of C. G.'s of these semi-circles below the surface are (4fl/3rc) and (46/37e) respectively.

Then Px=thrust on semi-circle of radius a

*=."gP.z 5 " =gP (4a/3rc) faa* •=-§a"gP 5

Ps=thrust on semi-circle of radius b=%b3gP, as above and z,=-A-fl7r. z_

SUL

>fgbn

(Fig. 188) ...see Ex. 7. Page 163

X. If x be the required depth of C. P. of the area bounded between the two semi-circles, then

_ P.fr-.rV^lfl 'gP.-AflTC- i&';gP.&far_3ir ( f l»-y) Px-P, VgP- l*3^P 16 (a 3 -6 3 ) '

Hence proved. °*Ex. 10 (a). A quadrant of a circle is just immersed verti­

cally in a heavy homogeneous liquid with one edge in the surface. Find the centre of pressure.

Solution. QAB is the quad-drant of the circle with radius OB in tie surface. Take the radii OA and OB as ,x aiid y axes respec­tively.

Then referred to these the equation of the circle is

(Gorakhpur 75)

-e—Y

j t»- j ->s

axes,

•(1) where a is the radius of the circle. Consider an elementary

strip of thickness 8» at a depth x below the free surface. ' (Fig. 189)

Then SJ=> the area of the strip =ySx«=^(aa—x2) 8* ; p=»pressure at any point at a depth x=gf>x.

168 Hydrostatics

, The coordinates of C. P. of this elementary strip are (», Jp) (Note)

Hence if (*, j>) be the coordinates of the required C. P. of the'

\xp ds \ x gPx. V <Ja—xz)dx quadrant, then *•=•? ,= yjj

I pds jgpx.\ /(a2-x z) dx

[' x>\/(a*-xa) dx a* y sin2 0 cos2 6 d0, > s D l « =,_-!» i

V xV(a 2 -* ) dx a8 \ sin 0 cos3 0 dd Jo Jo

putting x=>a sin 0 r iTf 2Tf. a.W»-t W * 3^«

= 2f3 r i r f"" 2.1 ° 16

or

-and i yp ds -.gpxi/W-x*) dx — - ^

Ua*,y ["gP*vf(a* *')*?*

\Q x(eP—jP)dx T JO

("xV(a9-x a)

S ir

0

dx , since^=V(aa-x2) from (I)

sin 0 cos8 0 d3 , putting x=»o sin 0

A8 1 sin fl'cos2 0 <*0 Jo

a riTl 2 T f „ a U _ 2-f.JV™ '« " 2 ; 2f3 r i r f 2 2.2.1 14 < « " 8

Hence the coordinates of C. P. of the quadrant are (&an, fa). Ans.

**Ex. 10. (b) A quadrant of a circle is just imaiersed vertically, ' with one edge in the surface, in a liquid whose density varies as the. O depth, find the coordinates of the centre of pressure

(Agra 77'; Rohilkhand 77) Solution. OAB is the quad- -

rantof the circle with radius OB ,in the surface.- Take the radii 6 A and OB as x andj> axes respectively. Then referred to A these axes the ' equation of the circle is *a+^a«>a^ ...(1) <Eig. 190)

Centre of Pressure 169

where a is the radius of the circle. Consider an elementary strip of thickness 8x at a depth x

below the surface. Let S.r=the area of the strip=yS*=v'(o* *a) $x and/>=

pressure at anV point at a depth x = g?x=g kx.x, since p=/cx (given that density varies as depth).

Also fhe coordinates of C. P. of this elementary strip are

Hence if (x, j>) be the coordinates of the required C. P. of the

\xp ds \ x kgxi.y/(ai -x1) dx quadrant, then s=~ == - |

J p ds kgx^^-x1) dx

J -Kit sin3 6 cos* 8 de

= °

- I " H r j j £ 3 2.1 32a

[prfj |" kgxs.^(az x»)dx

V(fla -x s) x V ( a s - * a ) dx

„/2 , putting x=a sin S sin8 0 cos2 0 dd

0

and

f r*v(< Jo

a5 ( '£• f ir

Jo

a2 - x2j dx

IT/I

sina 0 cos8 0 </0 ^7i , putting *=a sin 0

sin2 0 cosa 0 d&

a r$T2 2T3_ '6a D2" 2r\ rin^is*

Hence the coordinates of the C. P. are ( rr, r / V Ans. V157T ISrcf

Ex. 11. Find fhe C. P of a quadrant of an ellipse just immer­sed vertically in a homogeneous liquid with major axis in the surface.

{Lucknow 79) Solution OAB is the area of the quadrant of the ellipse with

semi-major axis OA in the free surface Referred to OA and OB as x and y-axes, the equation of fhe

ellipse, of which AB is arc, is where OA=a, OB = b. •

170 Hydrostatics

cc Consider an elementary strip

PQ of thicknes gy at a depth y below the free surface

Then 85=area of the strip =>xly

from <i) p«=pressure at any point at a

depth y^gpy. The coordinates of C. P. of

this elementary strip PQ are ftx, y). (Note) (Fig 191) Hence if (x, y) be the coordinates of the required C P. of the

quadrant, then we have

/ xp ds \b

oixg9y.{albW(b*-ya)dy

Jo

I* {alb V(fca-^s) ytfib1-?) dy j LS.

\"yVJ*-y') dy Jo

.. aV(b2-?') ' X b

(Note) 1T/8

sin 6 ccs3 $ dd

Jo sin 6 cosa 8 dd

, putting >••= b sin 6

>\a f i r 2 w 2f§ la X , as in Ex. 10 (a) Page 167

\P ds

2TI T£\ 8

g?y (albW{b* y*) dy

V yW(fi*~ya) dy b* \" sina 6 cos3 6 dd caj_o ^ Jo

j"'yf(P-y*) dy ba l*" sin 0 cos8 6 da '

putting j = 3 sin b

-* ijix,rTfz - fe-as iD Ex-10 (a) P a § e 167

Hence the required coordinates of C. P. are (for, i%Arc). 4 ns.

\ ^ \ Centre of Pressure 171

X

'fflEx. 12. A segment of a parabola, cut off by a double ordi­nate at a distance h from the vertex, is immersed with this ordinate in the surface of a liquid, find the resultant thrust and the position of the centre of pressure on the lamina, neglecting the atmospheric pressure.

{Agra 81; Lucknow 76 / Rajasthan 74 ; Rohilkhand 81, 78) Solution. Let AB be the double ordinate which cuts off a

segment AOB, such that AB is at a distance h from the vertex. Let OX, the axis of the parabola be taken as x-axis and the tangent at the vertex O as y-axjs. Then referred to these axes, tht equation of the parabola is

f**4ax ...<J) Consider an elemen­

tary strip of thickness Sx at

A

y\

Sx\ " ..

B

(.h-x) 1

ftx.y) ± / , o r

(Fig. 192) a depth (h—x) below the surface or at a height x above the vertex O.

Then 8,?=>area of the strip=2>' §x**2y/(4as) hx=4tf(ax) Sx and /?=pressure at any point at a depth (ft -x)=gP (h—x).

Also the centre of pressure lies on She axis of the parabola due to symmetry.

If 3 be the height of C. P. of the area OAB above the vertex O

then \xp ds \ x gp (h—x) 4^/(ax) dx

x^J h S ' * ." gp (ft-x) 4<s/(ax) dx

X (h X)y/X dx a

(h—xWx dx

(hx^-xW) ds

(hx^-s*'1) dx s: } 8

(fAx8'"-!*5/*)'1

Depth of C. P. below the free surfaced - f A«=fh.

gPlh—x) 4<f(ax)dx

h(\~3)J$h tf-S) 7

Also resultant thrust- \pds~\

~W g9 f (hx^-xW) dx~4<?o gP hhx'*- ?xfi',T

=>\W<igphsli- • •*«". Ex. 13. A parabolic area is held vertically under a fluid of

uniform density such that the latus rectum is in the surface Find the centre of pressure. •

V

172 \

Hydrostatics

Solution. Proceed exactly as in the last example. Here h=>a, since the distance between the vertex and the latus

rectum is a and the latus rectum is in the surface Answer Depth of C.P. below the free surface=fa, where 4a

is the latus rectum of the parabola Ex. 14. Prove that the depth of C. P. of parabolic lamina from

its vertex immersed vertically in a liquid with its vertex at a depth k below the water sarface, and bovnded by an ordinate at a distance h from the vertex is f [C7k-f5h)/(5k-f-3h)] h, when the vertex is above the ordinate. l (.Agra 79)

Solution. Choose the axis of the parabola as x-axis and the tangent at the vertex as j>-axis. Referred to these axes let the equation of the parabola be y2=4ax. ...(l)

Consider an element­ary strip of thickness Sx at depth x below the ver­tex O. Then 8j=»area of the strip="2j 8x z=i2y/(4ax) 5x °W(ax), from (1) and /?«= pressure at any

point at a depth x below the vertex 0=gP(k+x)

Also due to symme­try, the C.P of the pat a-bolic lamina lies on the axis of the parabola. (Fig. 193)

Then if g be the depth of the required C.P below O, then

ds J xg? (h+x) 4 f (ax)Jx

Ipds i* g?(h+x).4y/(ax)dx Jo

P (*x3/a+*5'a) dx JO

•Exg/15.

[V*1 , a+x3 ' iVx Jo

ifcxs 'a+f x'> i n

I k x8'a + f *8'2 \

= f [(7k+5h)h]K5k-\-3h). An area bounded by the carve ay3'

Hence proved. >x3, tbe x-axis and

the ordinate x«=a is immersed in water with the axis in the surface. Find the co-ordinates of the centre of pressure ' {Lucknow 77)

Solution The curve ay%=xi is symmetrical about x-axis. Also the curve passes through the origin and the tangents thereat arej>2=0 (obtained by equating the lowest degree term to zero), i.e. there are two real ano coircide"nt tangents at the origin. Hence there is cusp at the origin. ^

Centre of Pressure 173

Also the curve does not exist on the negative side of the origin.

Consider an elementary strip of thickness 5.* a< a distance x from the origin. Then

Ss=>area of the strip =y8x-=>lxsl*(V<*) 8*

and p=>pressure at the C. G. of the strip=J ygp (N ote)

Also the C. P. of the strip (rectangular in shape) is (x, $y).

Hence if («, j>) be the required coordinates of the C. P , then

B_fxpds x t—J— ipds

B

(Fig 194)

S O v r3/» 1 f« x*g9.*7-dx -4-1 *B/a*s/2<**

o 2 \fa y/a J0

s: x*dx

x9 dx

[« y „*3 '2 , \Ag9^dx *X <f xVKx*'* dx

since y=x3i*y/a

«nd , - ' ^ £ f ypds ipds

\:

H =fa

%y.\g9 <**

Jo-o 2 V a

,*i°_ J'8*8/2 i *

x3/3 dx

[" yxsi* dx

_2_ - A - ^ 3Va' £a« C '»fo«

\ ZppB&acc the C.P. of the given area is (fa, ffa). Ans. - \ /Ex . 16. Find the coordinates of the centre of pressure of the

area between the curve V x + V y =Va» and the axes, taking the axes to be rectangular and one of them in the surface. {Lucknow 7S)

Solution. The shape of the curve is as shown. The curve crosses the axes at A, (a, l) and 27(0, a) res­pectively.

Consider an elementary strip of thickness $x at a distance x from the origin. Then &y«=area of the strip

•=ySx=.tfa->i/x)a 8x, (Fig. 195)

0

6

Sx J

A

I r

(0,<l)

174 Hydrostatics

from the equation of the curve putting the value of y and p=pressure at the C.G of the s t r i p = ^ p .

Also the C. P. of the strip (rectangular in shape) is (x, ly), Hence if (x, y) be the required co-ordinates of the C.P., thea

xpds \yis?.Wa~</xfdX

I * Wa—^x)1 dx ^ , sincey~Wa-i/x)*

(v'fl-Vx)4 dx

P Jo

sin1 / cos8 r 4 sin8 / cos t dt '^ji —•, putting x=a sin' /

a3 I cos8 t.4 sin* t cos t dt Jo

S «/a T4r5

s i n ' t cos8 tdt lD T' 2 • i

2T9 3fl n - « a » _ _

nrs 28

, . I ypds and y—

t co&91 dt nr„

j°Jy % gP.Wa-^/x)*dx

(« fw/a

(V°—\/*)8 dx Naa 1 cos1* t.4 sin3 t cos t dt , i _o a Jo

» f a s j-u/a »

I (v 'a —v/*)* a* a ' I cos8 tA sin8 f cos / dt putting »=»a s in 41

[ *'a • • , ii ," * ^ 2 r ? „ I sin5 / cos" t dt _ -^-r- _ 2o_ Jj = 2 a 2F9 ^ 5 a 3 • ( * ' • • s , „, ., = 3T2r5 = M 1 sin8 f c»ss f a/ JO IT!

Hence the required C. P. is (aB8a a, -& a). Ans.

Ex. 17. A circular disc of radios a is completely immersed with its plane vertical in homogeneous fluid. If h is the depth of the centre below the free surface of the liquid, prove that the distance between the centres of pressure of the two semi-circles into which the

disc is divided by horizontal diameter is -" „:;„—., . .

Solution. B'OB is the horizontal diameter through the centre O.

Centre of Pressure 175

Ccnsider an elementary strip of thickness SJC at a depth x below the centre O.

Then g l a r e s of the strip =*PQ Sx=2.JV0 5x «=2V(aa-*a) S*

and p=»pressure at any point at a depth x below O

~gp (h+x) If * be the depth of C.P. of semi •

circle B'A'B below O, then S - f *P ds

fpds

" x.(h+x) gP.2<?(a2-*8) dx

(h+x) g?.2^/(aa— x') dx

(Fig. 196)

h V xy/(a2~xl) dx+ I" x V ' a ' - x 8 ) a1* '° Jo

h j " V(a s-*a) <**+ 1° JcV(fl2-*2) dx

S ir/i r«/a sin 0 cosa 0 d9+a*\ sin" 0 cos8 8

a Jo

cos* 0 o"0 + o81 sin 5 cos8 0 rffl

<&

putting x=a sin (9

flA (i) 4 a* (An) a (I6fi4-37ra)

tfaeC,

OP^

4 (37tA + 4a) '

If Pj be the C.P. of the semi-circle B'A'B, then a (I6fe4-37T0) 4 (.inh 4-4fl)

Also if P be the C.P. of the whole circle with centre O, then OP=.(a2/4A).

If X and Z, be the fluid thrusts on the whole cirole and the semi-circle B'A'B, then X«="gpz S"=»gpfi.7M»a.

^i"=gp [h+(4ai3n)UneP=i (3n/?+4a) aBgP .*. The distance of the C.P. of the semi-circle B'AB from O.

_XOP X,OF,

*" nashgp—J (37rA-f 4a) aagp

176 Hydrostatics 136/11

ina2—,h a i\6h+3m) a (3m-16h) nh—| (3izh-t-4aj 4 ( 3 ^ 4a) '

A Required distance between the centres of pressure _a (3rcq+16ft) aCna-\6h) _bna (4ha a*) = 4{3nh+4a) 4 (3nh 4a) =(97caaa-16aa)

Hence proved. •Ex. 18. Find the coordinates of the centre of pressure of a

square-lamina, held vertically in a Quid, whose density varies as depth with a corner in the surface and the diagonal vertical.

Solution. The comer A of the square ABCD is in the surface and the diagonal AC is vereical. Let a be the length of each side of the square. Then AC<=>aiJ2.

Depth of C. P. of A

square ABCD is the same as that of A ACD. Also A ACD can be taken as the difference of A OAC and A OAD.

Consider a horizontal strip PQ of thickness 8x at depth x below the surface! Also PQ=PC tan 45°

Then 8s=aiea. of the strip=.Pfi 8x^(ay/2-x) Bx and p=»pressure at any point at a depth x

•=g kx*x, siace p varies as the depth x / . IfXi be the depth of C.P. of &.OAC below the suiface

(Fig. 197)

*gPx

then Xi» xp ds

!

AC

\AC

x.kx2g (atf2—x) dx

j P as j

i kx*g (oy/2-x) dx

x3(atf2-x) dx (aV2x4 _x5 ."1* V 4 5 K ?*<!!

x*(as/2-x)dx I as/2x3 -H the

3 4 Again the depths of CP.'s of A OAD and A OND are

same below the free surface. Also ON= \OA<= ia\/2=a/^/2. :. If xa be the depth of C. P. of A OND below the surface

then replacing a\/2 by a{^2 in xu we get x2=f (a/V2)=3a/5tf2. Let P, and P2 be thrusts on A 04C and A 04D respectively.

5 ra-va •p ds= \ kx*g (a</2—x) dx

, \a</2 x3 x4 1»*« , . . ^[—3 TJ Q -*-•*»

136/12 ' Centre of Pressure 177

and Ps=thrust on A OAD<=>2 times thrust on &OND =2 [kk'g CJa)4], replacing a\/2 by ajy/2 in Px

*=-A ^fl4-J. If JC be the depth of C.P. of the square ABCD below the

free surface, then x •• P,x,-.Paxa_J kgak jay/l- •3^fc|oM3fl/5V2)

V »f f>\V

Pt-P* ^kga*-a {V2-n/gy2)} 9^/r

= 14 '

kga*

ii-i) £x 19. The asym­

ptote of a hyperbola lies in the surface of a Quid ; find the depth of the centre of pressure of the area included between the immersed asymptote, the curve and two given hori­zontal lines in the plane of the hvperbola.

Solution. The asy­mptote OB is in the free-surface and asymptote OA is in the liquid. Let CD and EF be two given (Fig. 198) horizontal lines at depths a aad 8 respectively. We are to find C P of the area CDEF.

Consider an elementary strip of thickness Sx at a depth x below the free surface.

Then 8.y=area of the strip =»Pg 8a> and/7=pressure at any point at aNdepth x below the surface.

•^gpx^gp.PN^gp.OP sin A AOB ;. The depth of C P. of the area CDEF

\xp ds I x gp.OP sin [_ AOB.PQ dx \ xkdx

k dx \ p ds V g9 OP sin L AOB.PQ dx V

since PQ.OP sin [__ AOB=consta.at=>k (say) by property of hyperbola. (Note)

**£x. 20; A semi-circular lamina is completely immersed in water with its plane vertical, so that the extremity A is is surface and the diameter makes with the surface an angle a. Prove that, if E be the centre of pressure, and $ the angle between AE and the

37C+16 tan a 16+15TU tan «* (/• A. S. 74)

diameter, then tan

178 Hydrostatics

Solution. AOC, the diameter of the semi circle ABC, is inclined to the free surface at an angle a and A is in the free surface.

•> Take A as pole and AC as the initial line Then referred to these the equation of the semi­circle is r«=2o cos 0 .. ( ) Take any point P (r, 0) In the lamina. Surrounding P consider an element of area r dd dr.

Also depth of P below the free surface=i4P sin (0+a)=/- sin (0+a)

.'. p ~ pressure at P—gp.r sin (0+a)

.". If (x, y) be the coordinates of E, the C. P. of the lamina

!xp ds \ I r cos 0 gp r sin (0+a).r dO dr

men .* = = J*=o'f=o _ men *= f =» - , ; i r

\ p ds \ \ gp.r sin <0+«).r d6 dr

(Fig 199)

and y-\yp ds \ \ r sin 0.gp r sin (0± a.) r d0 dr J _ Ja-alr-o

i — c a — • -

S i * ' • t ' pi/f gp.r sin (0+«).r rf$ *

J»=0jr_a

...(2)

(Note)

.-(3)

Then if <f> be the angle between AE and the diameter AC, then cot $<=xly

11T/2 [ f I t cos 0 gp r sin (0+a)./" ^0 </r

or 0010.= ^ , from (2) and (3) 1 I r sin 0.gp r sin (0+a) rdOdr Je=oJo

ra cos 0 sin (0+a) d0

[r-—\ sin 0sin(0+a)</0

»/2 cos6 6 sin (0+e) </0

o

cos

-, V r=2s cos 0 from (1) cos40sin0sin(0+a)</0

f"2 f al cosB 0 sm 6 J0+sin a I

cos a l sina 0 cos4 0 •L* </0+sinal •B/l

sin 0 cos5 0 </0

Centre of Pressure 179

„„c i n H ) , . J5 3 1 rcl

jarfi , . rn.ni cos a. x cos a-t--3

£g- w sin a 16 cos a-f-15rc sin a -£*g 7r cos a + 1 sin a 3n COS a -f 16 sin a 16-H57T tana . . . . . . . _ ,

=*—i i/L t » dividing each term by cos a 37t + lo tan a '

i 37t+16 tan q — 16 f 157T tan a" Hence proved.

Exercises on C. P. by Integration Ex. 1. Find the C. P. of a parabolic lamina immersed verti­

cally in a liquid with its vertex at a depth k below the surface and the lamina is bounded by an ordinate at a distance h from the vertex which is below the ordinate Ans — n~-.——

7 (5K -3/J) Ex, 2. Show that the ratio of the depth of the centre of

pressure of an ellipse completely immersed in a liquid with its centre at a depth h below the free surface and with its minor axis horizon­tal in one case and vertical n another case is

(W+ayiW+b*}. § 4*10. C P. of an area immersed in n liquids.

Let the densities of the liquids be Pw p2, P3,—, Pn (Pi<P8<—<Pn) and the areas in contact with them be Slt Sz, <S3, .... S„.

Now the thrust on the whole area remains the same if we consider that the liquid of density Pj is in contact with the whole area(S,4S,2-,-S3 + ...-f s») t n e liquid of density (p2—Pi) is in con­tact with 'he area (5 a i5,-(S'1+ ~-\-Sn), the liquid of density (P3—p2) is in contact with the area OSj-f-S -f . +Sa\ and so on.

FmalJy the liquid of density (p„—p„_i) is in contact with the area S„.

if-now we take Pu P2. P„, , t P„_be the thrusts on the areas (51+S,+- .+&\{Sx+Szi-...+Sa\\S3+Si+... + Sn), S„of the above fluids and zu zs, r3, , be the depths of their C.P. is below tiie surface, then z, the depth of the C. P. of the whole area, is given

P,z,+P2zi+ ..+ Pnzn b y Pi+Pn+.-i-Pn ' ^plved Examples on § 4 10.

\?y(<\rfx. 1. A rectangle is immersed in two liquids of densities p and 2p which do not mix. The top of the rectangle is in the surface of the first liquid and the area immersed in each is the same ; prove thai the centre of pressure divides the rectangle in the ratio 7 : 3.

Solution. ABCD is the given rectangle, such that AD=a and AB •' b. LM divides the rectangle into two equal parts. ^ Let area ALMD^S^a^a. LBCM^iab , -.(1)

A

t

liBfl

iH-:-q

Z.

CO

\£ i

i r i

•\.p

- D -

M

•i

•1

m

180 ' Hydrostatics 384-66

Now consider that the whole area of the rectangle ABCD viz. 2S is in contact with the liquid of density p and area FBCM or $S is in contact with a liquid of density (2p- p) or p.

Let P, and P2 be the thrusts on rectangles ABCD and LBCM when they are in contact with the above liquids and /,*. za be the depths of their C.P 's In above cases below the free surface. Then Pt«= thrust on rectangle ABCD in contact with liquid of density p

P2= thrust on rectangle LBCM j^--/^~-~r:.=j^;5VF~x~ when in contact with liquid of density P=gP Qb).iS. (Fig. 200)

[Note. \ Here the depth ot G.Q. has been measured from the common surface],

r^depfh of C.P. of rectangle ABCD in contact with liquid of density p=ffi.

" and 22*= depth of C.P. of rectangle LBCM in contact with liquid of density P below <4D=.$H-§ ( ^ ) = | 6 . If 2 be the required depth of GP , then

Pi+P* - tbg?S+$bgPS l + l 30 10 Also this C.P. lies on EF, the line of symmetry joining the mid­

points of AD and BC. Let this C. P. be P, then / . ' " EP-&6 or PF~ b- ft&-&6.

^/^•s/ •"• • EP : PF=1: 3. > Hence proved. ^^fiL. 2. Show-that the ceatre cf piessme *>f_a ihemfcus, immer­

sed in two liquids which tfo.ncitanix wi<b cne another, with a vertex in the upper surface and a diagonal in the common surface, divides the other diagonal' in the ratio 5 : 3, the density of the lower liquid being three times that of the upper.

Solution. Let ABCD be the rhombus, with vertex A In the upper surface and the diagoaal BD in the common surface. The diagonal AC is vertical.

Let AC=2h. Then AO<=OC=h Then depth of C. P. of rhombus ABCD

=the depth of C. P. of A ABC. Let p and 3p be the densities of the liquids. Ler2Sbe the area

,of the A ABC ; then area <3£C=5'=area ABO. _ f

Consider rfcat a liquid of density p is in contact with the AABC i.e. wilh area 25 and a .liquid of density (3p—p) Le. 2p is,in contact wilh the area BCO or *S.

\ Centre of Pressure 181

Let Pt and P2 be the thrusts due to fhe above liquids on areas ABC and BCO respectively and r„ ?2 be the depths of tbier C P.'s below the upper surface. Then />,=thrust on the area

ABC In contact wish the liquid of density P -*P h 25 ;

Pa<= thrust on the area5C<9 in contact with the liquid of density (3P- p) i e. 2p (Note) -=^

-g.2p.tf S; ~ p Zi«= depth of C P, of area

ABC below the upper surface

= 2(«+0-fy)

0+h2A-(2h)s4-0+h.2h+0

ifi^^ii^h

(Fig. 201)

7A 2 iO+A+2/?) 6 '

and ?JB= depth of C. P. of area £C0 below the upper surface

Then if 3 be the depth of required C.P of the triangle ABC,

Pi+P, ' .(2P*ftS)jA + (fcPAS)fA *A+M 5A

(^PASHlfepAS) 1 + i 4 -

.*. If P be the required C.P. of the rhombus ABCD, lying on AC, the line of symmetry, then

AP~lh or PC~2ft-fA—£A

^ * « ^ : jPC=5 : 3. Hence proved.

Vs. A fluid of depth 2a and uniform density p is super­imposed on a liquid of density 2p and depth greater than a. A circu­lar lamina of radius a, is placed with its plane vertical and its cen­tre in ihe surface common to Ihe two liquids. Deteimine the depth of ihe centre of prts&ute below ibe centie of the lamina, neglecting the atmospheric pressure.

Solution. Consider that a liquid of density p in contact with the whole circle AlCD of area 7S (say) and a liquid of density (2p- P) i.e p is in contact with the lower half of the circle viz. area ADB which is of area S.

Then if Pt and Ps be the tbiusis on the whole circle and the area ADB respectively ard zu za be the depths of their C.P.'s below the centre of the circle, then •

182 Hydrostatics

and zaz

Pi •=• thrust on the whole circle when contact with liquid of density p ~gp2a2S ,-4flSgP ;

P8«=thrusts on the semi­circle ADB when in contact with liquid of density P~gP (40/371)5

=(4fl/37c) S g9 ; Zi=*depth of C.P. of the

crcle below the cen*re of the circle

a 'T of the

*a ' -a

»depth 4 i 2 a r

of C. P semi-circle ABC below

the centre O (Fig. 202)

'-,V« (see Ex. 7 Page 165) If z be the depth of the required C.P.. then P,zi+P2zz _{4aSgP) t3+U4a/37c) S gp} -fata

Pi+P* (4 as £p)+{(4a/3jt) S gP} U<*4 ,V> 9fl7C

{1+C1/37U)} Ans. 16<3TC+I)-

*Ex. 4. A vessel contains three fluids of densities p, 29, 3p. A triangle of area A is supported with one side in the surface of the fluid of density p and the opposite vertex in the fluid of density 3p. Jf3h, 2h h are the depths of vertex below the upper surfaces of the

/three fiuids, prove that neglecting the atmospheric pressure, the thrust on each face of the triangle is f P A h. Also prove that the depth of the C. P. of the triangle from the free surface is (59/36) h

Solution. ABC is . K v f the given triangle of M £ = •—^-area A. The horizontal lines DE and LM divide this area in three parts, each one of which is an contact with one liquid. Let Su S2. S3 be the L > - : ^ i ^ - r ^ £ areas of the three parts as shown in the figure

Consider th.it area {Sl+S2+S3) is in con­tact with a ^liquid of J density P, area (S2+Sa) . (Fig. 203) is in contact with a liquid of density 2p—p i.e. P and area 5s«s in contact with la liquid of densfty 3p—2P f e. p.

Centre of Pressure 183

Now area of & ABC^A (given)=5 ,1+52+5'3; area(5a+53)

=»area of A BDE<=%A, since the altitude and ba-.e of A BDE&T& equal to frd ot those of &ABC. Similarly area .S^area of i^BLM *=-SA, since LM~$AC and BQ~\BK.

A If Pu P?, P3 be the thrusts on areas (Sj+S.+Ss), (S2 + S3) and Sa of the liquids as considered abcne and z„ z2, za be the depths of their C.P.'s then

Piathrust on dSj+S^+SJ i.e. A -45Cwhen in contact with liquid of density P

- " s M S T - g p A ^ ; P2=thrusti on (S2+Sa) i.e. A BDE when in contact with

liquid of density P ""gP \h (2A) f .4, here z is measured below DE =•£* hA g? ;

7*3=thrust on S3 i.e. A BLM when In contact with liquid of density p

ag? \h.\A, here z it measured below LW =-hg? Ah i

Z!=»depth of C.P of t\ABC in contact with liquid of density P "i Oh);

z2r='depth of C.P. of &BDE in contact with liquid of density p

below AC=h f | (2/i)=»2A' and z3=depth of C.P. of A BLM in contact with liquid of density

P below AC~2h+i (h)=%h. •. The required thrusts on each face of A ABC

°='Pi+Pz+Pa=g9fiA+-i-i-gPhA + -2Vgp/z,4

•=Sfgp^=fgp/^-and required depth of C. P. of the whole triangle ABC

_Piz^P9rt+Para_g9hA {jh\ frg<.hA {2K\ + 2\- e9h A tjh} Pt+P2+P3 gphA + ffgthA + sygpft/t

^3° " 1 1 + 28»-+BV) 36" Hence proved. ^V\ M^s. 5. A rectangle is immersed in fluids of densities p 2p,

3?,..., np ; the fop cf the ief tangle being in the surface of the first liquid and the area immersed in each liquid being the same Show that the depth rf the centre of pressure of the rectangle is \ [(3n+ 1 )/(2n-r 1)] h, »bf re h is the depth cf lower side.

Solution. ABCD is the given rectangle immersed in n liquids with side AD in the free surface.

Consider that the whole rectangle Is in contact with a liquid of densiiy p, the rectangle of height \h- (h/n)] is in contact with a liquidof density (2 p—p) f.e. p. the rectangle of height [h—tlh/n)] is in contact with a liqu'd of density (3p-2P) i.e. p and so on Lastly the lowest retcangle of height (h/n) is in contact with a liquid of density np —(«—l)p=p.

184 Hydrostatics

Thrusts on these rectangles are respectively gP i (k) AD:h ;

**(*-£M*-i!):

gpi * - ? M *-")•• efi\-\.AD.k- • A

or &P AD.(h)s ; \ gv.AD. [ ft-£j' ;

tgp.AD[h- 2-^J; ...; igp.AD Wnf ^ g

: • ? - • • :

1

B C cs\\Avy I

(Fig. 204) (Note)

i (z in each case has been measu­red below "the upper edge of the rectangle under consideration).

Also the heights of C. P. of these rectangles above the lowest

side BC are resepectively | , i \k~ £1 , | f A - ^ 1 _ , | [ J l

.*. The height of C.P. of the whole rectangle above^BC

\g?AD.(h)*. \+\g?.AD. [ h~J J [ft- ^]

J +-+feMP [^.igl

»f'+(«-i),4(«-|),+'.+(iV

i»+t'-ir+(H)+-+ii)' ft r»34-(H-l)3+fH-2)3-j-...+l3-l A_f2W»*|

a3« Ua+(«-l)a-h(«-2)a+... + laJ=3«L2;naJ Ji f in8 («+l)3- 1 h («+l)

'3nL |«(«+1)(2«+1) J==r2(2«-Hl>

»fV

. . . , n . , ft ( B + D (3«+?)

v < «

The required depth of G. P. of the whole rectangle ft

Hence proved. y. Exercises on § 410

\E£ 1. Prove that the C.P. of a rhombus immersed in two liquids which do not mix, with a vertex, in the upper surface arid a diagonal in the common surface divides the other diagonal in the ratio of 27 ; 11, if the density of the lower liauid fs twice that of the upper liquid. , (Jodhpur 76)

Centre of Pressure 18 5

Ex. 2. A rectangle is immeised in three liquids of densities p, 2p and 3p; the top of the rectangle being in the surface of the first liquid and the area immersed in each liquid being the- same. Show that the depth of the centre of pressure of the rectangle is (5/z/7), where h is the depth of the lower side. (Jodhfur 76 -

[Hint : See Ex. S Page 183. Here n=>3]. MISCELLANEOUS SOLVED EXAMPLES

Ex. 1. From a semi-circle whose diameter is in the surface of a liquid, a circle is cut oat, whose diameter is the ?ertical radius of the semi-circle Prove that the depth of the C.p. of the remainder is 9a7c/[8(16—in)], where a is the radius of the semi-circle. (Lucknow 77)

Solution. Let Pi and Pa be the thrusts on the semi-circle and circle respectively and zlt z2 be the depths of their C.P. below the free surface.

Then Px=»thrust on the semi-circle 4a

>"gP.zS"-=gp 3TC'

TOT , S/iS aagp

and z2

(Fig. 205) Pa=thrust on the circle of radius la=gp \a.K (ha)z=%naagp Zj=depth of C.P. of semi-circle below the free surface

=>-1Vra...sea. Ex. 7 P. 165. depth of C.P. of the circle below the free surface

*+£-m- tW a a +

5a

; . The required depth of C. P 2 T 8 8

free surface= PiZi-Pi*, (fa»gp) \A of the remainder below the

xsoiz - fg

p p «-9aw/[8 (16—3«)]

*Ex. 2 (a). A regoJar hexagon of side a is immersed in water with one side in the

.surface. Show that the depth of its 'centre of pressure is to that of its centre of gravity as 23s l8 . (V. P. P. C.S. 80)

Solution. ABCDEF is the regular hexagon of side a, whose iside AFv& in the surface Let G be the C. G. (also the centre) of the hexa­gon. Join all the vertices to G. Then the whole hexa­gon Is divided into six equi­lateral triangles of equal areas', and the depth of G below the surface be h.

— t&Tia3gp] is"

a ina—Arita (fa3 g?) - (Uas gp) r2-

la i f f ]

Hence proved.

(Fig, 206) Let the area of each triangle be S

iLet Plt P2, P3, Pif PB and P6 be the fluid thrusts and zl3 z2, z„ zt, zs and z6 be the depths of the C.P.'s of triangies AFG, ABG, BCG, CDG, DEG and EFG respectively.

186 ( Hydrostatics

Then P.-tbrust on A AFG—Ug& S"~gp ^.S .. (1) . Pa«»thrust on A ABG*=gp f A.S~ 2P2.. from (•) P3~thrust on A BCG~gP.ih+ih) S*=§h g^S^4P1 .. f.om (1) ' P4=>thirust on A CJDG^gP (A- f/t> S~ f A gPS=»5Pi .. from (1)

> Ps^Ps (by symmetry)=4P, (Proved abovej Pe^Pa (by-symmetry)^2P, (Proved above)

and z1=depthofC.P. of {±,AFG<=*\h. \ za= depth of C.P. of A ABG=\h, z8<= depth of C.P. of A BCG, whose three vertices are at

depths A, 1h and A below the free surface " v ' Aa+(2A)a-f A2+A.(2A)j {2h)h+h.h 11A " VVi+lh+h) *" 8 *

z4=» depth of 'C. P. of A CDG, whose three vet tices are at depths A, 2A, 2A below the free surfaee

h>+(2h)*+(2fi)a+h 2A+2A.2A+2A.A_m = 2(A+2A+A) 10 '

z6>=»za (by symmetry)=V"h and z6=za (by symmetry)** |A J. The depth of C.P. of the hexaeon below the free surface

1 SPl Pl+P a + P8+Pl+PB+P6 Pi ( W + 2 P , (f/»)+4P1 (Vtt+Sf, (^^?+^Pi (V-ftH-2P,tgA)

= P1+2P1+4P1+3P1+4P1+,'P1

" 2MPtJ23h D 18Pi D18 " |

& If 0 be the middle point of side AF and P be the C.P. of .the hexagon, then OP«=ff/z. Also CG=>A, where G js the C.G. of hexagon. /." OP : OG=>\%h : A=23 : 18. Hence proved.,

Ex. 2 (b). A regular hexagon is immersed in water with one side in the surface, find the C. P. of the upper ball (Mogadh 76)

Solution. Refer Fig. V06 oJ Ex 2 a) Page 185. Here we are to find the C.P. of the figure ABEFA i.e. the

upper half of the hexagon. Proceeding as in Ex. 2 Ca) above, the required depth of C.P. below the free surface

SPj P1+P2+Pe ' ' (Note) - f L i * + i ^ i * i B B i l l E x 2 ( a ) page 184,

^ y ^ / l p ^ " ' . - 2 * - f t see Ex 2 (a) P. 185.

^ ^Jr^C&UoG. C The C.P. divides OG in the ratio 7 : 3. Am. "*Ex. 4. A trape2inm AECD is immersed with the side AB

3o the surface of water and the side AD=a. BC«=b are vertical. Prove that the vertical line through tire centre of pressure divides AB irtthe ratio (aa-f 2ab+3bJ): (3a2+2ab-}-bs) and the depth of the

centre of pressure is I ( a + , + | £ + f f . ' '

Centre of Pressure 187

Solution Let/iiB=c. Take AD and AB as x and 7 axes respectively Join 'AC, then the trapezium ABCDi- .

*=AADC+&ABC. Now we know that the

C.P. of a triangle coincides with the centre of three parallel forces acting at the middle points of its sides and whose magnitudes are gp^Sx (depth (Fig. 207) of the middle point at which it !s acting), where S is the area of the triangle. (Note)

The coordinates of the mid-points9 of the sides of &ABC are R [it, c], E [%b, ie] and F [0, fr].

And the area of A ABC^^AB BC=>\cb. Now since the forces at these mid. points R, Eand Fare

igP-S, (depth of the po(nt). .'. The force at R=igp.lcb.ib**--&btc gP,

force at E=£gp \cb.\b=>'-}$b%c gp and forces at F=%gP.lcb.0=0. Again the area of (\ADC<=>\a.c and the coordinates of the mid.

points of the sides of A ADC are E lib, |e], N [|a, 0] and L [\ {a+b), is]

:. The force at E=igp ^.(depth of E)^\gp.\ac \b <**£$abc gp

force at iV=>$gp \ac (depth of N)<={gp.$ac \a=*-£i&c gpt and force at Z,«4$p.£ac (depth of L)=Jgp.4ae.J (a-j-6)

=3-^ (6+0) ac gp Hence the C.P. (x, y) of the given trapezium is at the centre of

these six parallel forces acting at mid. points whose coordinates are given above.

=depth of C. D - — X l X *= P.«a

SPX * ! . c

f f p * j . ^ • L n n i . ' a f i ' : g p * b.c^cgpa (a+^acgpa+b 10 « r O ~ n , M - i t w - U I p , - n ' T T " • n " 1 P i • o 12 s 2 ' 12 12 12 ' 2 12

IPfgP.tPc , n , * gp 1 «2c g? , (a+6) acgp 1 2 " t " i 2 * P + U i 12" ^ 1 2 * 12

&gc4-fc3^4-a6»c-l-a»c+(fl4-6)3gc 2^+ f l f e s +o 8 +a («+6)a__ 2l&aci-6ac-|-tt6c+a i !c+Ca-f-6Jac] ' = 2 [26a+fl6-|-aa+(a$.&)a] 2& 3+2f l6 2 f2a 84-2f l s i . b3+ab'+as+asb (b+a) (b*-{-a*)

2(26a-h2a6-r-2as)

And

A2c b2c ^ 1

2(6*+ab+c8) 2 (a2+fl»+b*)

abc r JgP.c+T2gP.- 2-+0f l T^.- 2- + l2-gP.0+ pj

(3+a) o^p ^_ 8 2

b2c piP-b-jgV-i

b2c.abcgp , a*c_^q+b)ac 12 7lrl2 +«n-gp#v

12 •gp*

188 Hydrostatics

c [26a+5H ah 4 a (a+b)] c 3b*+2ab+a*

.*. If P be the C. P. of the trapezium and PK be the vertical ,. 4. ' _ .. . „ . c(3624-2a&+o2) ' . line through P, then AK=y- 4 ( a2+ f l»+ g 6 ) -

.-. KB~AB-AK<=c-AK c(3A2+2a6+a') c (fea+3oi!+2fl»)

"=,C 4 (a* -t- 62+ad)" "" 4{.di+bt^ab) Hence .Iff: £B~(«8+2a6-f-3£a): (3a2+ts+2a&). Ex. 4. A system of coaxial circles is immersed in water with

the line of centres at a ghen depth. Pro^e that the centre of pressure of those circular areas which completely immersed, lie an a parabola.

Solution. Take the line of centres and the radial line as the axes of* and y respectively. Then the equation of any circle of this system of coaxial circles can be taken as

x*+y*-2gx+c=0, where g varies from circle to circle whereas c Is constant.

The centre Q of a

- ( I )

circle is (g, 0) and radius r Is given by

=g*-c

its

.(2) If d be the depth of

the line of centres below the free surface and P (x,y) the G. P. of this circle, then

x-=OC=g and y=CP="a*f4h"

>(r2/4rf)=fea •e)f(4d)t from (2) or y<**(x2—c)/(4d), since #=»*

P and repre-Hence proved,

a liquid is in the

or 4.y,d=xa—c, which is the equation of the locus of sents a parabola.

**Ex. 5. Show that if a lamina, totally immersed in is a quadrant of a circle of radius a, of which the centre surface, the locus of the C. P. lies in a straight line

x+y~|B86,a(TC-)-2). Solution OAB is the quad-'

rant of a circle of radius a, with OA, one of the bounding radii/ inclined at an angle a to the free surface and the centre O in the surface.

Referred to O as pole and OA as initial line, the equation of the circle is r=a. ...(1) (Fig. 209)

Take any point P (r, 6) in the lamina. Surrounding P consider an element of are^r dd dr. From P draw PL perpendicular to the

or

and

Centre of Pressure 189

free surface. Then the depth of P below the surface = P £ c = , r sin ( 0 + a ) .

Hence if (*, y) be the coordinates of this quadrant , f [J/J [a \xp ds I 1 r cos 6 gp,r sin ( 0 + a ) r dd dr

then * = . J » -lfl=0 J r - ° (Note) \p ds \ \ gpr sin ( 0 + a ) r dB dr i J 0-0 Jf=0

cos 0 sin ( 0 + a ) ( r - J ^ fl=n _ Z _ _ P _ _

sin (0+a) ^ - j </0 e=,o w /o

(ir/2 f * /S

« cos a cos 0 sin 0 d 0 + s i n a I cos2 0 dB -"= ~A f»/a '~

sin ( 0 + a ) <*0 Jo

1 w/2 f ir/2 sin 20 d9+sln a \ (1 + c o s 20) dB

__ _ _ o Jo -8 r \*12

-cos (0+a) multiplying numerator and denominator by 2

f T"/2 T '„ T / a

- cos a! - $ cos 20 I -fsina 0+$sIn20j / —cos ($jt+a)+cos a I

3aU cos a + re sin a) 16 ^cos a-)-sin a)

\yp ds \ \ r sin B gp.r sin (0+a) r dd dr

I /> </J I gp r sin (&-\-a) rdBdr i J 8 c O J « , = 0

| sin 0 sin (0+a) | ^ -1 ^0

, cos a 1 sin20 J0+sln al sin 0 cos 0 d B

sin(0-fa)rf0 7 Jo fn/S f«/»

3 cos a 1 (1-cos 20) d0+sin a I sin 20 dB

I - c o s (0+a)

multiplying numerator and denominator by 2

190 Hydrostatics

or

or

3o cos a | d— —-— I +s in os j '

cos 2e\-"a

2 o ' 8 [—cos ( i n + a j + c o s a] _ 3a (2 sin a+rc cos a)

x+j>

16 (cos a+hin a) 3a (2 cos a+w slfl a) , 3a (2 sin K+TT cos a)

"16" (cos oc+sin a) 16 (cos a -j-sin a)

[1 (cos a+sin a) +-t (sin a + cos a)] 16 (cos a+s in a;

= & a (2+*) . .'. The locus of C.P. (x, j ) is x ^ ^ - A - fl (2+w).

Hence proved. Ex. 6. A circular sector, centre O asd radius a, symmetrical

about the radius OP, is compietely immersed with F in the surface and O p vertical. Determine depth of its centre of pressure.

Solution OAP3 is the sector of the circle with OP -vertical and P to the sur­face Let /_AOB<=2K.

From symmetry the C.P. of the sec­tor will lies on OP. Referred to O as pole and OP as the initial line, the equa­tion of the circle is r—a.

Take any point P' (r, 6) fn the sector. Surrounding P' consider an element of area dS = r d0 dr

From P' draw PL' perpendicular to (Fig. 210) the surface and P'N' perpendicular to OP. Then depth of P' below the surface=-PX'=OP ON=a—r cos 6. ,

/ . pressare at P '=gP P'L'=gp ( a - r cos 8) If x be the distance of C P. of the sector from O on OP,

then %= I xp ds r r r cos 0.gP ( f l - r cos 0) r d8 dr

' P \ \ g? (a—r cos 6) rdS dr ' 1 8—ojr_# a\\ f" r2cos Bdddr- [° ["

J 8=-aJr_o J 4 - _ a J r r3 cos20 d0 dr

a\ \ rd& !#=_ojr=o

COS 0 fi?0 10

ra cos fl rffl dr H" i" Je„-ajr= cos2 6 d&

2 cos3 0 d§

a3 [a . a3 f°

Centre of Pressure 191

or 8 j sin 0 T — 3 \" (1+cos 20) dO

jfcfl t ^1 1=5 12 [ 0 T - 8 [sin 0 |°

o [ l 6 s i n a - 3 | e+ is ta2f i l ° 1

24a - J 6 sm a. a [8 sin a—3a—3 sin « cos «]

1=1 4 (3a—2 sin a) (Note) depth of C.P. of the sector below the surface=a—».

a (8 sin a—3a—3 sin a cos a) 4 (2a—2 sin a) '

a [15 a-4-3 s'n a cos a—16 sin a] 4 (3a—2 sin a) Ans.

Ex. 1. A plane quadrilateral ABCD is entirely immersed in water with tbe side AB ID tbe snrface. If the depths of C and D be­low the surface are 7 and 8 respectively, and that of the C. G. is h ;

prove thai the depfh of the C. P. is —-6 h

Solution. Join BD. Let the areas of A ABD and ABCD be Si and S2 respectively.

The depths o[ their cen­tres of gravity are $8 and i (y+S) respectively.

/ . The depth of C.G. of the quadrilateral

S -i-S or 3/z (Sx+s jUs^+f r+S) S2

or Si(3A-8)=(y+8-3A)S, or

(Fig. 211)

S» y+8-3 f t Sa

= (3A-&)~

- (1 )

...(2) Also the depths of G P. of A ABD and A BDC are JS and

" ( ^ H O " ° r 2(y+S) resPect^y-Also if Pj and P2 be thrusts on A ABD and A 5CZ>, then

Pi<=- thrust on A ABD=gp $8 S, and ,Pa~ thrust on A ^CZ?= gp § (y+ 8) 5a

J*. The depth of the quadrilateTal ABCD

l-Pi limp* a (s3+y3+ys)/(y+s)] ' P 4-P

„ | g p 8^ ( | 8 H i (7+8) gP s, ft (8Hya+y8)/(y+8)]

192 Hydrostatics 136/12

from (1) T 8'fS1/S,)-Hy'+8'-fr8) .. . . . ' fc , . '

=*• aa t ^ g j - n } — • d i v , d i n § e a c h t e n n b ? ,s« , 3a r(y4-8-3ft)/(3/z-8)I+(y*+8a+rS) '

~* - 3A [{(y +-S-3A)/(3/I-7S»+1] ttom(2)

§2 (y+S-M)+ph-*) fya+Sa+y8) ==*' 3/z(y + 8-3A+3A-8)

8ay+(8-3A) fSa y a -8 a -y8) " * ' 3/*y

, 8 H ( S - 3 A ) ( - r - g ) , - 7 8 + 3 / ; (7+8) "*• 3/* ~ * 3A _

= * (*+*)-* K r W Hence proved. EXERCISES ON CHAPTER IV

Ex. 1. An equilateral triangle, each of whose sides is 6\/3 feet long is immersed vertically in water with its side in the surface which is open to the air. If the water barometer stands 34 feet, find the depth of the centre of pressure of the triangle. Aas. 3-7

a4- feet.

Es. 2. A parallelogram is immersed vertically in a liquid with a pair of sides horizontal and at depths A, and h2 below the surface. Find the position of its centre of pressure.

[Hint. Take p as constant in Ex. 2 Page 160]. Ex. 3. A square is immersed with its diagonal vertical and its

lowest point as deep again,as its highest point. Find the depth of the centre of pressure. (Magadh 73)

Ex.4. The centre of pressure of a parallelogram completely immersed in water is in one of its diagonals. Prove that the other diagonal is horizontal. (Lucknow 83)

CHAPTER V

Floating Bodies **§ 501. Conditions of equilibrium of a body freely floating

in a liquid. (Agra 80, 77, 73 ; Avadti 83, 81 ; Bhopal 81, 80 ; Garhwai 81; Gorakhpur 80, 77, 75, 73 : Magadh 75 ; Mhhila 81 :

Rohilkhand81. 79 ; Vikram 83) Consider a body floating freely wholly or partly immersed in a

heavy liquid which is at rest. The forces acting on the body y^"""\

are i— / \ (i) The weight FF(say) of the / \

body acting vertically downwards ErOrOE^ G ^ ? ^ r ~ through G, the CO. of the body; and - EGr3-3' I w ^r j I^E

<il) The resultant fluid thrust ^p3 G\w' Ef^S t.e. the force of buoyancy W (say) E^LJ^^^-^~~~-> - ^ H T ~ which is equaJ to the weight of the —--^-^^-^FI^^^^^rLrzl liquid displaced, acting vertically upwards through G', the centre of (Fig. 212) buoyancy i.e. the C. G. of the liquid displaced.

•Now we know that when two forces acting on a body keep i« in equilibrium then the two forces must be equal in magnitude but opposite in direction and their lines of action should be the same.

Hence in our case, the conditions of equilibrium for a body floating freely in a heavy liquid are :—

(a) The weight of the body must be equal to the weight of the liquid [displaced by it.

(b) The centre of gravity of the body and the centre of buoyancy (i e. the centre of gravity of the liquid displaced) must lie on the same 4 vertical line.

§ S'02. Volume immersed. If a solid of volume V and mean densfty o is floating in a liquid of density p (> f), the volume immer­sed is (Vajp).

Let V be the volume immersed. Then the weight of the liquid displaced^?" gP.

Also the weight of the solid=Fag. Now for equilibrium, wdght of the liquid dispjaced=weight of the solid

or V Pg=Forg 'or F'-=(Fo)/p. Hence proved. Note 1. Since V carmot be greater than V, therefore from

the above result we conclude that a cannot be areater than p i.e. 136/13

194 Hydrostat'-tJ

a solid cannot floaf In a liquid whose density is less than that of It. 4 If the riersUy of the solid is greater than that of the liquid then the solid will sink. r

Note 2. In the above formula . .„ , . , . . . mass of the soMd

mean dens.ty of the solid- v o l H m e o f t h e s o l i d

Here we must remember the well known example of a ship made of iron floating in water whereas the density of iron is greater than that • of 'be water^ since the mean density of the ship is less than that of water.

Solved Examples on § 5*01 and § 5*02. Ex. 1. (a). What fraction of a piece of cork will be outside

water ia which it floats, the specific gravity of cork being 0*24 ? , \ Solution. Let V be the volume of the cork and V the volume

V of the water displaced by ft. Let the density of water be p then the / density of cork=0 24 p.

f The weight of the cork=the weight of water displaced by it.

A FX(0-24P) g^V'xg? or V'~?0%V~-£SV A Volume of the cork outside the water-=F-F'= V—.&V*= Jf V.

•n • J e *i» volume of the cork outside water Required fraction— y:

•=-it«=>0-76. Ana. Ex. 1. (b) What fraction of the volume of the boffy will remain

Immersed if the density of the liquid is three times the mean density of the body ? * >

Solution. Let V be the volume of the body and <V the volume of the liquid "displaced by it. Let the mean density of the body be P, then that of the liquid is 3P.

Since the weight of the body=>,the weight of liquid displaced by It.

^ J*. 'V*g~V"i?.z or V'*=\V :. Volume of the body outside the liquid =?F- F'«=» V— JF= \V

. T, . . , . „.. Vol. of body outside liquid A Required fraction= '—~z • —

- S F/F-f . ins . " Ex. 2. The specific gravity of sea water is a and of ice P (P>w).

What fraction of the volume of an iceberg floats out of water ? Solution Let Fand V be the volume of the iceberg and the

volume of the iceberg immersed in water respectively. V The weight of the iceberg

• - = weight of the water displaced by it " i Vvg^V'og * or V'~V9lo

Floating Bodies 195

& Volume of the iceberg outside the water=»K-F'

„ . . „ A. Volume of the Iceberg outside water A Required fraction- Volume of the iceberg -

„ F ( g - p ) / p _ g - P K «» • Ans.

Ex. 3, The specific gravity of ice is 0"92, that of the sea-water is 1'025. What depth of water will be required to float a cubical iceberg whose side is 100 feet ?

Solution. Let the iceberg sink In water to a depth h feet while floating in it.

Then the volume of the water displaced by the Iceberg =(100xl00)x /2cu . ft.

Also the volume of iceberg= 100 x 100 x 100 cu. ft. Let p be the density of the standard substance, then the densi­

ties of ice and sea-water are 0'92P and l'025p respectively. A The weight of the iceberg.

«=>the weight of the water displaced by it. i A <100XlOOxlOO)xO-92p=(100xlOOxft)xl055p

or 100X0-92-1-025A or A - 1 ^ ^ ™ ? - 9 * ™ = 8 9 - 7 5 6 f t .

X The least depth of water required to float the iceberg is 89-756 ft. Ans.

Ex. 4. A wooden solid cute of side one meter floats in water with fths of its volume immersed. Calculate the depth to which it will sink in a liquid of sp. gr. 0*8.

Solution. Let a and p be the densities of the wood end water respectively.

The volume of the cube=>l cu. meter and the volume of water displaced by i t = f cu. meter (given).

Then as the weight of the solid=»weight of water displaced by it _•. lXga=igp or a=|p _(1)

Let the cube sink to a depth h meter in the liquid of sp. gr. 0-8 or density 0'8P. Also the volume of liquid displaced=>l X1X A.

Then as the weight of the cube=> weight of the liquid displaced .*. lXgX0=(lXlX/j)X£XO-8p

or a=0'8p.h or tP=x% M, from (1) or | = f A or A = i f mefre=if X100 cm.=>93-75. cm. Ans.

Ex. 5. A piece of wood weighing 24 grammes floats in water with f rds of its volume immersed. Find the density and the volume of the wood. . *

Solution. We know that the mass of 1 cu. cm. of wafer «=•! gm.

m tfy&QStfcttaa 3S4-46

Let the density of wood be e and the volume of the piece of wood be F.<

Then the weight of the wood=F«^=»24 grammes wt. (given) Also the weight of the water displaced=»§7 l.g. V weight of the wood«=the weight of the water displaced £» Vagt=lVXg. or o«=>| Ans. Also Vog<=24 gms. wt. or Vot=*24 gms.

or F.§=»24 or P=24xf~36 cu.xm. ^ Ans. Ex 6. A uniform cylinder when floating with its axis vertical

in distilled water, sinks to a depth of 3'2", and when floating in alcohol sinks to a depth of 4s. Find the specific gravity of alcohol

Solution. Let W be the weight of the cylinder and A be the area of Its base. Let p be the density of the distilled water.

When the cylinder floats In water. the weight of the cylinder •=• weight of alcohol displaced by it

or >W={A0.2))g? , ...<D Let s be the sp. gr. of alcohol, then the density of alcohol=jp .*. When the cylinder floats in a Icoho!, we have the weight bf the cylinder^^eight of alcohol displaced by it

or W~(Ax<t)sgp . , ...(2) A From (1) and (2), {Ax 4) s?g~(Ax 3'2) pg

or -/**•

s = £<3°2)=0-8 Ans. Ex. 7 A thin rod of weight W is loaded at one end with a

weight P of insignificanrvyolume. If the position with (l/n)th of its length out

( n - l ) P = W . Mithila 82, 80 ; Patna 82

Solution. Let AB=2a (say) be the rod floating in water with its length BC ' outside the water and the weight P attached at A Let the rod" be inclined at an angle 6 to the horizontal.

Then BC<=(\jn)AB =2a}n

rod floats in an inclined of water, prove that (Jodhpur 78 ; Magadh 74 ; Ranchi 80, 75 ; Vikram 75)

L . - _ - V - - - i

lllJ^^^X --^"-i^—yi-i-f--

A***-£-2.(t!) (Fig. 213) Let G and Gt be the centres of gravity of the whole rod AB

and the volume^of the liquid displaced i.e. portion AC of the rod. Then G and G are the mid-points of AB and AC :. AG^ AC^\.2a {(n-l)/«}=a {(«-!)/«}

and AC=\ (2a)=a A GGx=AG-AGl'='a—a{(n—\)ln}'='aln

""" The forces acting on the rod are : x (1) the weight W of the rod AB, acting at G vertically down­

wards.

Floating Bodies 1*)7

($) The force of buoyancy V acting vertically upwards through G1 and (3) the weight P at A.

Taking moments of these' forces about Gt (in order to avoid force of buoyancy) we have PxAGt cos B=* Wx GGt cos 0 ' or Px.AG1^WxGG1 or P.a{(B-l)/n}=W(a/n) or P (n — l) => W. Hence proved.

_^Ex. 8. A small hole is drilled at one end of a thin uniform rod and it is filled with some much denser metal. It is observed that the .rod can float in water half immersed and inclined at an angle to the vertical. Show that the sp- gr. of the rod is J.

x Solution. Let AB be the rod with a small hole drilled at A and filled with some denser metal of. weight Wi say.

Let a be the area of the cross-section of the rod. Let AB=>2a, then if G be the C. G. we have AG*=>a. Let G\ be the C.Q. of the"water dis­placed, then Gi is the mid. point of AG as the rod is half immersed in water or AGt=\a.

The forces acting on the rod ABsue : — (1) its weight W<=>2aag?, where P is the density or the rod,

acting vertically downwards through G, (2) the force of buoyancy V<=*weight of the water displaced

•=>aa. l.g, acting vertically upwards through Gx (regarding density of water as unity), and

(3) the weight W\ of metal at A acting vertically down­wards.

Taking moments of all these forces about A, we have -V^AGy cos8=-WAG cos 0

or flag §a=2a:rgp a or P=J, which is the density of the rod. But we have taken the density of water as unity. Hence sp. gr. of the rod is P<=»J. . Hence proved.

**Ex. 9. A rod of small section aod density m, has a small piece of lead of weight (l/n)rh that of the rod attached to one extre­mity, prove that the rod will Moat at any inciiaation in a liquid of density m', if (n+1)* BQ=>n2m'. (Lucknow 81, 74; Rajasthan 76; Ranchi 74;

U.P.P.C.S.77) Solution. Let AB be the rod of length 2a floating at any

Inclination 6 to be horizontal and with a length AC==2b of it immersed in the liquid.

Let a be the area of cross-section of the rod. Let G and Gj be the centres of the gravity of the sod AB and

the liquid displaced respectively. Thea AG™* and AGX=\ AC<=>b.

in ftydrosfattes

The forces acting on the rod are :— (I) its weight W<=>2a%gm,

• acting vertically down­wards through G,

(ii) the force of buoyancy V =weight of-the liquid

displaced by the rod =2A« gm', acting verti­

cally upwards thro­ng0 *?i / '

(iii) the weight {Win) attached (Fig. 215) at A, acting vertically^downwards. Taking moments of these forces about centre of buoyancy Gu

we have (W/n) AG,, cos 6=W.GGt cos 0

or (l/«) AG^GGj, or (l/«) b<=>{a-b) ...(1)

*(J+l)-« or 6 -^j (2)

Also for equilibrium of the rod under the action of ihree para­

llel forces we^have V~W-^- or V~wft~\ or >2ba.grri <=>2a«.gm\(n+\)lri[ or bm'^am [(n+l)/n] or, [annn+l)]m'='am[(njrl)l"], from (2) or »*«'=(«,+• l ) 2 « . Hence proved.

{Gorakhpur 81)

Ex. 10. A thin uniform rod of weight W has a particle of weight w attached to one £nd. It is floating in aa inclined position in water, witn this end immersed. Prove that the length of the rod above water is w/(w-t-W) times its whole.length and that the speci­fic gravity of the rod is W2/(w-f-Wj.

Solation. Let AB be the rod of length 2a floating at the incli­nation 0 to the horizontal and with a length AC=2b of it immersed in the liquid. v

Let a be the area.of the cross-section of the rod. Let G and Gt be the centres of gravity of the rod and the water displaced by it respectively.

Then AG=a, AG^AC^b, GG^AG-AG^a-^b. , (Fig. 216)

Let s be the sp. gr. of the rod and P the density of the water. - Then the density of the rod=sp. The forces acting on the rod AB are 1— (i) its weight W=>2aas?g, acting vertically downwards

through Gt

(li) The force of buoyancy K= weight of the liquid displaced m2k&gt acting vertically upwards through G»; and

Floating Bodies $9

fri) the weight w attached at the end A, acting vertically downwards.

Taking moments of all these forces about Gu we get w.AGk cos 6= W.GG% cos 8 qr w.AGx=>W.GGx

or wb<=*W{a-b) or (w + fV) b<=*Wa or d^aJP/OH-JF) ...(I) A. length of the rod outside water=J5C=^B—AC

e=2a-2b=2a-{2aW/(wi-W)} „.from (1) w w

.2a=- X (length of the rod). w±W w+W

Also for equilibrium of the rod under the action of three parallel forces we have V^W+w.

V W+w y a w~, dividing both sides by W 2bx?g^ W+w ^ p m t i n g t h e v a l u e s o f K a n d w

or

or

or

2a« jpg W b_=W.+w

as W or s=-W

WA •w

( W \a

I F^j •from <!> Hence proved.

- Ex 11. A bady is composed of a hemisphere of radios r, with a cone, of heigh 2r, on the same plane bise. In a liquid of density P1} it floats with the whole body just immersed, and in a liqaid of density P, with the hemisphere just immersed; prove that Ps ""

4wa.2r= §w3 Solution. Volume of the cone and volume of hemisphere=Sw8.

Let p be the density of the material of the cone and the hemisphere.

V weight of the body =weight of the liquid displaced

4 So we have, in the liquid of densitg px weight of the whole body

=weight of the liquid displaced by the whole body.

Ue. (fnr84-iwr8) gp~ (Sw! ± fnr8)p,g P~Pi

And in the liquid of density p, weight of whole body=weight of the

or (Fig. 217)

...0)

Ue. (f7rrs+5jrr8) eg=(f-tr5j P2g or 2p=pa or p=fP2

t. From (i) and (ii) we get p1=|p a ' •Ex. 12. A hollow conical vessel floats in water with its vertex

downwards and certain depth of its axis immersed j when, water is poured into it upto the level originally immersed, it sinks till Ha

liquid displaced by the hemisphere

(Note) «.(li)

or f a ^ p i .

3ob Stydrosfafe

moath is on a level with the surface of the water. the axis was originally immersed.

Solution. Let V be the vertex and a the semi-vertlcal angle of the cone. Let h be its height and let it be immersed up to a height h', i,e. VC~h! originally.

Let W be the weight of the vessel, v The weight of the liquid displaced originally=f reft'3 tan1 agp, where p is the density of the water.

jS. The weight of the body=»weight of the liquid displaced.

/ . W=^7rA'3tana«fP

What portion of (Rohilkhand 79)

(Fig. 218) •d)

i.e.

or or \Or

Let water be poured in the vessel upto a height h'. Then the weight of the vessel wiih water ^

=awelgnt of the water displaced in this case weight of the cone •+-weight of the water contained

=weight of the water displaced W+lTih'* tan3 ag?=*faffl tana a g?

JJTA'3 tan3 a.g?+$Tzh'3 tan2 a gp=4wA8 tana a gp, 2/z'3*=A3 or

from (1) h'={iyt»h. "" Ans.

Ex. 13. A solid homogeneous cone of -sp gr. o and height h floats in a given liquid of sp. gr. p. Find the position of equilibrium when (i) the vertex is down and base up,

(ii) the base is down and vertex up. Solution. Let V-be the vertex

and « the semi-vertical angle of the cone. *

Case I. The vertex is «?OWE and base up,

Let ti be the length of axis immersed in the liquid.

For equilibrium,' weight of the cone<= weight of the

liquid displaced. JTCA3 tan8 a go=fah'* tan8 « gp

L^J&'Ss

or OS

or

=A'3p (Fig. 2i9)

Ans. haoc

A'»«Aa (or/p) or h'=h (O/P)1'*.

Case II. The base is down and vertex up. Let h' be the length of the axis outside the liquid.

the volume Immersed in the Hquld«=>volume of v the Then

rluating Bodies 20,

whole cone—volume of the smaller cone outside the liquid

=£ nh3 tan3 a—%nh's tan* oe —JTB (h3-h'3) tan2 a

For equilibrium, weight of the cone=>we!ght of the liquid displaced by it or \nh3 tan2 « go

*=\niji3-h'3) tan» agp or h3a=.(h3-h'3) P or A'8p=/j3(P-a) ~ ^ ~ ~ • or h'3=.[(9-o)l?] h3 - - - " or fi'=[(?-°)lp]lia h (Pig. 220)

.*. The length of the axis immersed in the liquid = A - / i ' = A-{fp-a)/p}V3^=/j[J_{(p_ ( r) /p}l/3j i A n s #

v-^Ex. 14. A frustum of a right circular cone, cut off by a plane bisecting the axis perpendicularly, floats with smaller end in water and its axis just half immersed. Prove that the sp. gr. of the cone is 19/56.

(Bhopal 8 s ; Rohilkhand 78)

Solution. ABCD fs the frustum of the cone with vertex V and base AD such that the height of the frustum is half of the height of the cone. Let h be the height df the cone and a its semi-vertical angle.

The volume of the frustum *=voiume of the con2(K, AD) —volume of the cone {V, BC) —jwA" tan2 a - | R {\hf tan2 a «=>I7/24.TTA3 tan8 a - - ( l )

Let the frustum be immer­sed upto the level of EF, such (Fig. 2.21) that EF bisects this portion of the axis between At) and BC.

Then the-distance of EF from the vertex K= JA+J ( J & H P .'. The volume of the liquid displaoed

•=volume of the frustum BCFE •=vol. of the cone {V, EF)—\o\. of the cone {V, BC) - t i c {\hf tan2 «—ire {\hf tan* • - £ * * «A8 tan2«. ...(2)

For equilibrium, weight of the frustum ABCD=*weight of the liquid displaoed or -2\-jrA

8 tan2 a i^g^-^s JT/J3 tan2 a p#, from (1) and (2) where s Is the sp. gr. of the cone and p the density of water. or &«=-i.Va- or •$=»!!. Hence proved.

*Ex. 15. A cubical box of one foot external dimension is made" of a material of thickness one inch and Boats in water immersed to a

^3^3==^£tf-:K-^£=£=~===3

Hydrostatics • --

depth of 3 | inches. Not taking air into consideration, determine the weight of tbe box and the amount of water that must be poured in so

• that water inside and outside may stand at the same level. [Solution. The depth immer- l2"

sed In w a t w - r - 4 5 ^ = * ft-I. volume of the water

^displaced=lxlxi5a"=-A-cu ft. .'. weight of the water E~±-.

displaced =>-A gp, ...(1) ==Y=i_, where p is the density of water ^f^€k^^r^=,-='=^===^:A

«=i%x62i lbs., since "->« "" ~~~ -1 cu. ft. of water weighs 62J lbs. (Fig. 222) = is

8-XA§fi~¥/="l9-5 lbs. nearly For equilibrium, weight of the box

=weight of the water displaced=»19"3 lbs. Ans. Now let water be poured in the box to a depth of x ft. then in

order that the water inside and outside the box should bs in the same level, the box will be immersed to a depth (JC+IV) ft. in the water (since the thickness of the box is 1" i.e. -& ft.)

Also the inner dimensions of the box are 10" x 10". I. The volume of water inside the box

=JSxJtx*=ff xcu. ft. The volume of the water displaced by the box=> 1 x 1X (x +-1

1a-)

=(*+i \ ' ) cu. ft. • j Also as proved above, the weight of the box«=i66-xgP, from (1) '

where P is the density of water. For equilibrium in this case, weight of the box+w'eight of the water inside the box

^weight of the water displaced by the box or \%gp+Sf*-gp=(^+«)gP or ft+if»-*+ft or < l - H ) * - f t - A or ifrx^iMr or * - 4 f V x f M f t . - o »

0 Also the amount of water inside the box «= 10" x 10" x 9"=900 cu. in. Ans.

Ex. 16. A m&n whose weight is equal to 160 lbs. and whose sp. gr. is l 'l .j can just float in water witn his head above the surface by the aid of a piece of cork which is wholly immersed. Having given tnat the volume of his head is one sixteenth of his whole volume aod that the sp. gr. of the cork is 024, find the volume of the cork, the weight of a cubic foot of water being 6251 lbs.

Solution. Let the volume of the man be V cubic feet. Then the weight of the man =>7xl-lx62 5

,*« T / U V6 2 * , v 160x4*128

or WWXfiX-jg «* P - n x S T ^ - * * -W

floating lodfes 263

Let the volume of the cork be V' cubic feet. \ Tne weight of the cork=K' X 0-24 x 62-5.

Volume of the head of the man=-1V V (given) £, Volume of the man immersed in water= V—£g V**\%V,

The volume of the water displaced =»H V+volume of the cork=if V+V

% Also the weight of the man-f-the weight of the cork =the weight of liquid displaced

or 160+P'X0'24x62-5=(£fF+P")x62'5 or 160-UVx 62-5=7' [625-0-24 x 625] or 160-tf X f f x W - K x W (1-0-24), from (1) or V'x^xO-ie^m-^^W3-

T/, 260 2X100 104 or K - j p X ,-25^76=209C U ' " ' Ans.

'Ex. 17. A steamer is going from salt water into fresh water was observed to sink 2", bat after burning 50 tons of coal to rise J". Supposing the densities of salt and fresh water to be as 65 : 64 ; find the displacement of the steamer in toas. (1 ton =-2248 lbs.)

Solution Assume that the area of cross-section of the tteamer near the level of water is uniform and equal %o A, say.

Let the steamer sink a inches in the salt water. Let the densi­ties of the salt and fresh water be 65/cP and 64/cp respectively.

Then in the salt water, the weight W of the steamer=»weight of salt water displaced

or W-(A.-&).65 kpg. _ ( l ) In the fresh water the steamer sinks 2" more i.e. the steamer

sinks i1,- (a+2) ft. in fresh water, so in fresh water the weight W of the steamer

=» weight of the fresh water displaced or W~A ft (a-t-2).64 keg ...(2)

From (1) and (2), we gee A.&a 65 k?g=A.-& (o+2>.64 k9g or 65a=64(a+2) or a=128. ...<3)

Now it is given that after burning 50 tons of coal, the steamer rises 1' i.e. whsn the weight of the steamer is (W—50x2240), the steamer sinks (e-f-2—O i.e. (a-J-l) inches In fresh water.

,\ From the principle, the weight of the body =• the weight of the liquid displaced,

we have (0r-5Ox224O)<=»4.i,if (a+1) 64 k?g =A iV (a +2) 64 kfg-^hA 64 k?g (Note) =W— **A.i*4 kPg from l2)

or 50X 2240~iV4 64 k?g or j j A k9g=- A' (30 X2240)= 1750 .*. From (I), W•=*{;& A kpg)x65, rearranging the terms

=17s0x 63 x 128, V a-128 from (3) 1750X65X128, < « « • « -

"" ~—2240 |ons«ao500 toas.

204 Hy&ortfi&sft

.*. The displacement of the steamer «the weight of the steamer=6500 tons. Ass.

Ex. 18. A ship sailing from the sea into a river sinks a inches and on discharging x tons of her cargo rises b inches. If the sea water be one fortieth beaveir than river water, prove that the weight of the ship is (41ax/b) tons. Assume the sides of the ship to be vertical at water level. M«« 79; U. P. P C. S. 79)

Solution. Assume that the area of cross-section of the ship near the level of water be A. Let the ship sink h inches in the sea water. Let the density of the river .water be p, men the density of sea water is f J p, since the sea water is one fortieth heavier than river water.

Let W lbs. be the weight of the ship. Then in sea water, the weight of the ship=the weight of sea water displaced, te. W=(A.-A-h)4b gp. ...(I) In the river water, the ship sinks a" more i e. the ship sinks

^2 (h + a) feet in river water, hence in river water,, the weight of the ship=the weight of the river water displaced

^or , W=>A.ft(h + e)pg , TO(2j A From (1) and (2) we get A ft h. H pg*=*A ft (h+a) pg.

or Uft^h+a or A=40 a „.(3) Now it is given that after discharging x tons of her cargo, the

ship rises h inches i.e. when the weight of the ship is (W— a; X 2240). the ship sinks ft (h+a—b) feet in river water.

I. From the principle, the weight of the body«= the weight of the liquid displaced

--we have {W—x x 2240)=-^ ft (k+a—b) pg *=A.ft (Zt+e) Pg~ ft Ah Pg =» W~ ft Ahgp ^ from (2)

or ».2240»il# Ab gP or (xfb) 2240«ft A gP Substituting this value of ft A gP in ^1), we have the weight

of the ship~ JP=»A f J (x/ft) 2240 lbs. 0 =»40fl ih.(xjb) tons, V /j=>4Ca from (?)

=•{41 ax/b) tons. Hence proved. **Ex. 19 A ship sailing out of the sea Into a river sinks

through a distance b feet, on unloading a cargo of weight P, the ship rises through c feet: show that the weight of ship afier unloading is

I { - p t ~ * **» w a e r e ° aB0" P a r e %e SP- §r- °^ s e a w a t w a nd river water respectively. The cross-section of the ship near the level of water may be assumed to be uniform. (Avadh 83 ; Gorakhpur 74)

Solution -Suppose that the area, of cross-section of the ship near the level of waier is A. Let the ship sink h feet in tfee sea water. Let the weight per cubic foot of sea and river water be ow and Pw respectively? Let W be ffae weight of the ship.

Floating Bodies 20$

Then In sea water, the weight of the ship=the weight of the sea water displaced

or W&Ah o w ...(1) In the river water, the ship sinks b feet more I.e. the ship sinks

(h+b) feet in the river water. Hence as the weight of the ship=the weight of the river water displaced, so we have

W~A(h + b)pw ...[2) *\ From (1) and (2), we get A (h+b) pw=Ah aw

or h(a~P)=ftp or A=»6p/(«r—p) ...(3) Now it is given the after unloading a cargo of weight P, the

ship rises c feet i e when the weight of the ship is (W— P), the ship sinks (h+b—c) feet in the river waer.

From the principle, the weight of the body=the weight the liquid displaced

we have W-P^A (h+b-c) pw =A [h+b) pw Acpw^W—Acpw . ...fiom (2)

or P=Acfw or Aw=P/cp. Substituting this value of Aw in (1), we have

the weight of the ship^H^Aa Aw**{ho P/cP). ,*. The weight of the ship after unloading a cargo of weight P

oW-PJjL-P=(t - j) P> whQTeh-J!L cp up ; O-P

4 - 4 f — v - l 1 f 4 / " v - l l l * . Hence proved. Lcp(er p) ] lc(o~p) ]

**Ez. 20. A solid displaces 1 'a, 1/b, 1/c of its volume respec­tively when it floats in three different fluids. Find the volume it displaces when it floats in a mixtare formed of (1) equal volumes of the fluids and (2) equal weights of the fluids

(Agra 83 f Rajasthan 74 ; Ranchi 76) Solution. Let V be the volume of the solid and a its density.

Let pJt p2. pa be th'e dersities of the three flu'ds. \ Then as the weight of the body=weight of liquid displaced, so

we have Vogs=>(V/a) pjg- or p1'=aa ...(1) Simiiary Vag^(Vlb) p2e or p2=»6a „.(2)

and Vog=.(V/c)pag or p3=ca ...(3) Case I. When eqnal volumes of fluids are taken. Let p be the densi'y of the mixture and Vt be the volume dis­

placed by the solid in this mixture/1

T h e r | ^ P d l | + P 3 = ( a ± * + ^ j f t o m (1>> (2} a n d ( 3 )

Also as the weight of the body«=>weight of the liquid displaced

'... F c f - r l K - K , ( 2 ± | ± 2 ) ^ or V-V^-^*)

or V^ZVKa+b+c). Ans.

208 Hydrbifaflcs

Case II. When equal weights of fluids are taken. '=

Let p' be the density of the mixture and Vt be the volume displaced by the solid in this mixture.

Also as the weight of the body^welght of the liquid displaced

OF

or

3K, 1/fl+l/ft + l/C ( or ^ ± + ^ . + 1.)

*W V(ab±bc+cg) abc ADS.

Ex. 21. A hollow cylinder of external and internal radii a and b is open at the top and the thickness of its base is c. The cylinder floats in water with its axis vertical and a depth h immersed. If the cylirder develops a small'leak, show that it will never sink if its height is greater lhan (aah-bsc)/(a2—b2).

Solution. Let H be the height of the the cylinder and p the density of its material.

H—

(Fig. 223)

Then the volume of the cylinder =>(volume of external cylinder of height if and radius a) —(vol of internal cylinder of height H—c and radius b) (Note) va-K&H- nb* (H-c)~n (aa-6a) H+nb*o. Then for equilibrium, when there is no leak,

the weight of the cylinder=»weight of the liquid displaced by the • cylnder with a depth & immersed

[nta*-b*) H+nb*c] Pg=(ita*k)Xg PJ=flM(fia=fi«L2±63cL ^.(1)

Le.

Floating Bodiea 207

le.

le. le. le. U.

i.e.

Now If the cylinder develops a leak, it will not sink If tha weight of the cylinder < upward fluid thrust on the cylinder when

it is full of water if the weight of the cylinder < weight ofwater displaced b y . .

ihe whole cylinder if [w (a8-£») H+*b2c] 9g < [re (a'-b*) H+ub*c].l.g if 9<l if a*hl[(a*- b*) H±b»c]<l -from (1) if aaA < (aa-b*) H+b*c le. tfa'h-b*c < (a*-*8) H

" n > (a» _ g3) Hence proved. **Ex. 22. A triangular lamina ABC of density p floats in a

liquid of density o with its plane vertical, the angle B being in the surface of the liquid, and the angle A not immersed ; show thai

p: o=sin A cos C : sin B. (Bhopal 80; Gorakhpur 78, 75 ; Lucknow 83

Solution. ABC is the given triangle with the portion ABE out­side the liquid. Let D be the mid­point of the side BC. Join AD and ED. Let G and G1 be the centre of gravity of triangle ABC and the centre of bouyancy (i.e. the C.G. of A BGE) respectively. Then DG=>\AD and DGt=i DE Then for equilibrium GGt must be vertical. Also in triangles DGG1

, _ , „ DG . DGt • and DAE, we get ^ = s = j ^ f •

Hence GGi. is parallel to AE. le. AC is also vertical.

Also for equilibrium. weight of the body=>welght of the liquid displaced

ot A ABC gp= A BCE go . 9 A BCE i BE EC EC p BC cos C

Or — — •— -~

(Fig 234) But GGx is vertical, hence AE

(Note)

&ABC \AC.BE AC°T <**~ AC But we know that for any triangle ABC,

a sin A BC_sm A y ^ s i n ' 5 °r AC~sin B

sin A.cos C from (1), we get T-'

/ sin B

(1)

Hence proved.

JThis result can be expressed in another way as

BC „ 'AC C0S C= ,C0S C~ b \ 2ab J W— J

m mm *Ex. 23. A uniform prism whose cross-section is an Isosceles

triangle of vertical angle 2«, floats freely in a liquid with its base just Immersed, one edge being in ihe surface ; show that he ratio.of_its density to that of the liquid is 2 S*2 «. « » « ' 8 1 • l n d o r e 72>

Solution. Refer Figure 224 Page 207- Ar MA ABC is the given isosceles triangle, where AB=AC and

LBAC-U or LDAC-* and £ 2>G4-90°-«. As in the last example, we can prove that AC Is vertical.

Let p and a be the densities of the prism and the liquid respec­tively. Let h be the length of prism.

For equilibrium, wt. of the prlsm=weight of liquid displaced (MBCxh) g9=*(&BCEXh) go 9 ahBCE^mEqaEC a AABC \BEAC AC « W

In A BCE, EC/BE^cot C=cot (90°-«)=tan a, EC=*BE tan a=AB sin 2a tan a,

V in A -<4B£, BE<=*AB sin 2a. £C=>24B sin a cos « tan a=»2^S sin3 «, ,

2.4B sin2 a

or

or

or

or

from (1), =2 sin2 a, V AB=>AC. o AB

Hence proved. **Ex. 24. A hollow closed cone of semi-vertical angle sin-1 &

of metal whose specific weight is w is made of such uniform thickness that it will float in all positions wholly submerged in a liquid of specific weight w. Show that the thickness must be

Jh {l-d-w/w') 1 ' 3}, where h is the external height of the cone.

Solution. Let VAB be external cone and V'A'B' be the internal cone. Let k be the thickness of the metal and a the 'semi-vertical angle of the cone.

Then sin a=»| (given) ...(1) Let O be the centre of the base

of the external cone. Then VO=h. In A VNV',NV'~k. VV'-=>k cosec a.<=kx3, from (1) Height of the internal cone

*=>V'0'<=VO-VV~0'0 «=A—3k—k^fi-M.

.*. volume of the metal «=(yol. of the external cone)—(vol. =JTT/I3 tan2 a—ft* (h - 4k)3 tana a} = 4 * {/j8-(/z-4fc)3} tan2 a.

For equilibrium (when the cone is wholly submerged) the weight of the cone=weight' of the liquid displaced when the

cone is wholly submerged

(Fig. 225)

of the internal cone)

136/24 Floating Bodies 209

or ^ lh*-(h-4fc)8] tan" «.w'«= JnA8 tana a.W or [^-(A-4fc)8]v>--A*w

or h*-th-4ky~h\ 4 or A3-^=(A-4Jfc>8

or A3 [l-(w/vt>')]~(A-4fc>8 o r ft [l-(w/»v')]1"'=(A-4fc)

or k~lh[l-{l-(w/w')y3]. Hence proved. **Ex. 25. A ithin hollow cone with a base floats completely

immersed in water wherever it is placed ; show that the vertical angle is 2 sin-1 £.

(Agra 79; Bhopal 82; Lucknow 79,', Rajasthan 77 ; Vikram 83, 77,75) Solution. We should first ofall determine the C. G. of the

whole cone with a base. Let h be tne height, r the radius of the base and / the slant side of the cone.

Then the area of the curved surface»7c/7 and the area of the

The 0. G. of the base is Its centre and the C. G. of the curved surface is at a distance %h from the centre of the base on the axis of the cone.

J, The distance of G. G. of the hollow cone with base from

the centre of the base= — ^ - ^ — ^ - ^ /.T , . nlr+nr2 (Note)

-ft/13 </+r)] Also in the case of the cone completely immersed in water the

displaced liquid is in the shape of a solid cone, hence the C.G. of the liquid displaced is at a distance of lh from the centre of the base of the cone.

Now the cone floats completely immersed in water wherever it is placed, therefore C. G. of the cone with base must coincide with the C. G. of the liquid displaced. I.e. ft/p (/+r)]- |A. or 4/=3 (l+r) or J=3r or r/ /=J or sin a=4, where a is the semi-vertical angle of the cone or • a>= sin-1 J.

.'. The vertical angle of the cone=>2<x-=»2 sin"1 $. Hence proved.

Exercises on § 501—5 02 *Ex. 1. The specific gravity of sea water is 1*028 and of Ice

is 0*918. What fraction of the volume of an iceberg floats out of water ? Ans. 55/514

Ex. 2. How much water will overflow from the edges of a cup just full of water when a cork 2 cubic inches in volume is gently placed in it so as to float ? (The specific gravity of cork is 0*24). • Ans. 0'48 cubic in.

EX. 3. A spherical shell of copper (sp. gr. 8) floats In water with half of its surface Immeised. Find its internal radius if the outer one be 10 cms. . Ans. 10 (15; 16)1-* cm.

m Hydrostatic*

- *Ex. 4. A .solid displaces $, J and J of its volume respee- * tively when it floats In three different fluids, find the volume it displaces when it floats in a mixture formed, first of equal volumes of the fluids, second of equal weights of the fluids. ;

- - (Gorakhpur79; Fanchi79) Ans. i and t l of i*s volume.

Ex. 5. A block of wood floats in a liquid with | th of its volume immersed. In another liquid it-floats with frd of its volume immersed- If the liquid be mixed together in equal quantities by weight, what fraction of the volume of the wood would be now immersed. Ans. -2:/24.

"Ex. 6. A hollow conical vessel floats in water with its axis vertical and (1/n) of the axis immersed, t i t d to what depth must the vessel be filled with water so that it may just sink till its mouth is on the level with the surface of water outside. (^*ra 8S)

t 5*03. Bodies floating In more than one liquid. A body is floating partly immersed in one liquid and partly In

another, to find the conditions of equilibrium. iGorakhpur 74 ; Jodhpur 76 ; Magadh 76)

Solution. Let a body be floating in two liquids of densities ?t and P,.

Let the volumes of the two liquids displaced be V^ and Fa.

Thea the weights of the two liquids displaced are Vtfi g and PaPa g acting at Gx and G2 respec­tively.

Let (Vtfig+VM Le. the resultant of these two forces

on Vt9ig and V%?2g act at G lying theiineGiGa.

Hencclhe body is in equili­brium under the action of two

==.-_™l <5\

. : ; . . * . • . : ; : ' . ; * ,

(Fig. 226) forces viz the weight of the body acting vertically downwards at G, the O G. of the body and the resultant fluid thrust (F1pJg+ Va?2g)

acting vertically upwards at G. Hence these two forces must be equal in magniiude but opposite in direction and also their lines of

action should be in the same line i e. G and G, the centres of gravity of the body and the liquid displaced, should be in the same vertical line, since one of the forces viz. weight of the body must act in a vertical line.

Hence the required conditions of equilibrium are ;— (i) Tne weight of the body must be equal to the sum of the

weights of two liquids displaced and (ii) The C. G. of the body and that of the liquids displaced

must he in the same vertical line. N

- » Note.. the above * theorem Includes the case of the body float-

Floating Bodied 211

ing partly immersed in liquid and partly in air. / Solved Examples on § 5-03

v— *%t. 1. A cylinder of sp. gr. o floats with its axis vertical partly in liquid of sp. gr. o1 and partly in another of sp. gr. «ra.

Show that the common surface divides the axis in the ratio. a-a2 : ffj— a - tAvadh 82 ; Rajasthan 76)

Solution. Let hi and h2 be the lengths of cylinder in contact with the liquids of density o and 0j respectively.

Let r be the radius of the base of the cylinder. v

Then if P be the density of the standard substancee the weight of the cylinder

Also the sum of the weights of fhe liquids displaoed=7tr2Ai<r1Pg f •nr^hio^'g-

Z. For equilibrium, the weight of tbe body«=sum of the weights of liquids displaced

(Fig. 227)

or or

or

nr* (Aj-f A„) t>Pg**nr2h1<r1Pg+nrzhzoapg (^i + ht) a =» hiP!+Aaera or ht (a - cr,)=Aa (<"a - a)

h .°2-°^a-a2 h2 o—ot a1~o' Hence proved.

Ex. 2. A mass composed partly of gold (sp. gr- 19.25) and partly of silver (sp. gr 10 5) floats with if th of its volume immersed in mercnry (sp. gr. 136) and the remainder ia water. Compare the weights of gold and silver in the mass.

Solution. Let the volumes of gold and silver be Vx and Fa respectively. Let p be the density of water.

J5, Weight of rhe mass=Fi x 19*25 Pg-f F2 X 10*5 Pg. Also volume of mercury displaced=Jf (F,—Vt)

and volume of water displaced =»*11« (Fj+Fa)

.*. weight of mercury displaced=^f (Vx-{-V2) 13*6 pg and weight of water displaced«=>iV (Vx+ V2) Pg.

For equilibrium, weight of the n>ass=sum of the weights of liquids displaced

or (Fix 19-25P^+F 2 X lQSpg)-it (Fi+Fi) l?6Pg+& (Vt+VJ pg or

or

or "I

Fix 19-25 + F2X105 1925__2j)5] J"205_I051 100 16] " L'16 loj

Fx/F2=37/103 weight of gold _ FtX 19-25 pg

or -•El 37x19*25

weight ot silver FaX"10-5 pg 103 X 10*5 407

'618* c*i o* ADS*

m Hydrostatics 3&4-€6

Ex. 3. v An alloy of gold (sp. gr. 19) and silver (sp. gr. 10*5) floats with 11 of its volume in mercury (sp. gr. 13 6) and its remain­der in a liquid of sp. gr. 6'8. Find the proportion of gold and silver in the alloy.

, Solution. Let the volumes of gold and silver in the alloy be Vx and V2 respectively. Let p be the density of water.

Then the weights of the a l loy-^X 19pg+F2X 10'5 Pg. Also volume of mercury displaced«=f f {Vx+V2)

and volume of liquid of sp. gr. 0 8 displaced «il8- (V\+ V2)

* weight of mercury displaced"!| (Fi+Fa) X 13*6 pg and weight of liquid of density 0'8p displaced •=& (V1+Va)X G-8pg

For equilibrium, weight of the alloy=sum of the weights of liquids displaced

(VxXWgxV^XlOSpg) * ^n (Vi+rjxi3-6pg+-h (vt+v2)xo-tog

(19FH-10-5 V2)^(Vi+Va) [4IX13-6+AX0-8]

-(K,+n, [«?^l»]=v<w,) Yx [57-40]= V2140-31-5] or YlV^l^V^

Fi 5 J V ^ : 17=1 :2 . Ans.

or

or

or or

Ex 4. A cylinder of wood (sp. gr f) of length h, floats with its axis vertical in water and oil (sp. gr. | ) , the length of the solid in contact with oil being a(<3&h). Find how much of the wood is above liqaids ; also find to what additional depth mnst oil be added so «s to cover the cylinder.

Solution Let * be the length of the cylinder above liqu.ds. Then the length In water is h-(a^-x), where a is the length in oil.

Let r"be the radius of the base of the cylinder and p the density of water.

Then the weight of the oylinder=(nr2A) \$g.

The weight of oil displaced =7i;raa \Pg. and the weight of water diplaced rarer

For equilibrium, the weight of the cylinder=the sum of the wts. of liquids displaced or v:rahlpg=i:r2a^g+7tr

9%(h-a-x)gp

or ' ih~ia±(h -a-x) or x^{\h-\a) Ans. Here x is j - ive as a < \h (given)

^ Let oil be added to an additional depth h\

' t •rt. ••'

t

^_~^=

•a (k-a-

CZL_r_

rT "!~ ^

CH *

i

\

(Fig. 228) -x).?g. ^

T X J,

, . C * 1*"

Mm 'S-CLrs/Oz?

/

floating Bodies 21*

Then the length of the cylinder In oil=a+fi' and in water

A In this position, for equillbruim as before we have 7rrsA.|P^=7rr2 (a-t-h') tfg+nr* (h-a~h') ?g

or l>i<=>\(<i+h')+{h-a-h')~h-la-W or i f t ' - iA- lo or h'^Qh-a) Ans,

••Ex. 5. A right circular cylinder of sp. gr. o floats in water with its axis vertical; one third being above the water. If p be the sp. gr. of air, prove that

3<r=»Z+p. Solution. Let A be the

area of the cross section of the cylinder and 3/i be Its height.

Let Pi be the density of water.

Then the weight of the cylinder»^.3AaPjg

Then weight of air dis- " '—• placed=^Ppx g, since the length of cylinder In air is \ (J/j) i.e. k.

The weight of water displaced =» ,2APag. For equilibrium,

the weight of the cylinder =.the sum of the wts. of liquids displaced or A3/jffP1g=/4ftpp1? + ^2ftP1 or 3<r«=P+2. Hence proved.

Ex. 6. A cylinder of sp. gr. 2P floats with axis vertical bet­ween two liquids of sp. gr. P and 3P respectively, the height of the cylinder being equal to the depth of the upper liquid. Prove that the pressures on its ends are as 1 j 5.

Solution. Let A be the area of the cross-section of {he cylinder

iK (Fig. 229)

(Kg. ZM)

m Hydrostatics

and h its height. Then the depth of the upper liquid is h. Let the cylinder be immersed to depths hx and (h—hj In

liquids of densities p and 3 P respectively. Let cr be the density of the water.

Then for equilibrium, the wt. of the cylinderasum of the wts. of liquids displaced.

{Ah 29fg)<=Ah^og-\-A {h —hj 3fog 2A=M-3 {h-hj or 2h^h or hx=lh ...(1)

.*. Pressure at <?i the centre of the upper end of the cylinder «=>gPor (/j - heifer (h ili)>=*lh g?a /

Pressure on the upper end=^ x \hg9a ...(2) Also pressure at 02, the centre of the lower end of the cylinder

»[APw + (A-*i) 3Pa] g~[h$3 {h-ih)] Pag, from(l)

or or

01

•*M<fg A Pressure on the lower eod=Ax f l.o?g A " From (2) and (3), we havs Pressure on upper end A \hgjo

.(3)

. 1 • 6 -Pressure on lower end A.%hgpa 6" Hence proved

Ex. 7. Two liquids which do not mix are placed in the same vessel; the density of the lower liquid is p aad that of the upper Is mp ; a cylinder floats in them with its axis vertical and is completely submerged ; its density being np, fiad the condition that it may be half in the lower liquid. " (Osmania 69)

-Solution. Let h be the "height and r the radius of the base of the cylinder. The cylinder is immersed to depths %h and \h in the two liquids

Now for equilibrium, the weight of the cylinder =the sum of the weights of liquids displaced

i.e. rcra hn?g =-Ur*.$h) m?g f(wa.JIi) pg or 2n=m+l is the requited condition.

Ex. 8. A cylinder of density P, floats with its axis vertical in two liquids, the density of the lighter being p,, (hat of the heavier P2 The length of the cylinder is n times the depth of the lighter liquid. Prove that the depth of the upper end of the cylinder is,

P <3 Pa and > P 8 - ( -^JJ-^-1

Solution. The height of the cylinder is h (given). Let A be the area of the cross-section of the cylinder.

.'. The depth of \hs lighter liquid is h)n.

Then weight of the cyliud««»(4 k) ?g. (Fig, 23U

Floating Bodies 213

Let a length x of the axis of the cylinder be immersed in lighter liquid of density Px. Ihen the length of the axis immersed in heavier liquid of density p3 is (/z - *).

For equilibrium, the wt. of the c>Under =wt. of liquid of density Pj displaced

-j-wc. of liquid of density pa displaced (4 h) Pg"(A.x) Pig+A.ih-x) pag

Ap=PiX-j>(A-je)Pa or A (p-Pa)«=* (Pi-Pa)

x = hJh_J)

= (P2-P1) . . .(1)

J". Required depth of Ou the centre of the upper end h h h (P.—p) n " (P2-P1)

Also p < pa otherwise the cylinder will sink entirely in the liquid of the density p2 and no part of it will be in the lighter liquid of density 9V

Also if z is positive, then •P)

or or

or

h h (Pa-_ « (Pa-Pi)

or (P2-Pi) >n<P a -P)

or , «P > np2—Pa+Pi or p > pa-[(Pa-Pi)/«] Hence proved. L-^Ex. 9. A right circular cone of density p, floats just immersed

with its vertex downwards in a vessel containing two liquids of den­sities "i and <72 respectively, show that the plane of separation of the two liquids cuts off from the axis of the cone a fraction [(p-ffjiVfo-CTa)]1'3 of its length. (Jodhpur 76)

Solution. VAB is the given cone. Let its height bs h and semi-vertical anglg a. Let tue base AB of the cone be m the surface. CD is the surface of separation of two liquids O and 0 ' are the centres of the base AB and surface of sepa­ration CD respectively.

Let VO'<=z then 00 '= / ! - z

.*. The weigbt of the cone =i(vol. of the cone) pg *=*\nh3 tan3« Pg.

k Volume of liquid (of density ox) displaced •^volumes of eone (V, Cp^ins? iao8 a

(Fig. 233)

216 ' - Hydrostatics

and volume of liquid of density oz displaced ' «=»volume of the=fcustum 4&DC

~ ( M 3 tan2 a) - faz3 tan2 a). .*. For equilibrium,,

weight of the cone=(weight of liquid of density ^ displaoed) +(weight of liquid of density aa displaced)

or Irih* tana aPg=^jcz3 $ana oca^-f^re (A3 -zz) tan2 a.o2g . or . Wp^z^+Qi3—z3) a2

. or h* lP-a2)=23 (or,- a2) or z=ft [ ( P - 0/(»i-aa)l».».. . . Hence proved.

•Ex. 10. A solid sphere floats just immersed in heterogeneous liquid composed of three liquids which do not mix and whose densi* ties are as 1: 2 : 3 . If thickness of the two upper liquids be each one-third of the diameter of the sphere, show that the. density of "the liquid in the middle is equal to the density of the sphere.

(Agra81;Garhwal79) Solution. Let a be the radius of the sphere and a its density. Let the densities of three liquids be P, 2P and 3p. £. The weight of the sphere=f ;ra3org. The volume of the upper most liquid displaced-

. ' ' =the volume of the lowermost liquid displaced, by symmetry. - f JTC O W J ) [3ra~ W+dJAdflY*- - see list Pf formulae

[3a»-(£a2+£as+fl2)], ' _ _ - = _ _ _ _ ' since the distances of the \:::::r.::;::.::::*^ j N ::.V.::.vjs.v!:V:r

two parallel faces from the centre :;zy;^,:,7 ..'- J \.:::;i :":'~ are a and a/3 in this case JIjIlllilillllF • "~ ! ~~\|||I|I||IIIIHIII

- W M I ^ - W ] 111 [ • <€ IIM . «ffi7«J8. ===^k—i—jgigij, ,V Tne volume of the middle -=£15=^.' I j&^fp^-

liquid (of density 2P) displaced = £ ^ ^ ^ ^ J = ^ 5 S ^ S 4 i l | £ -«avolume of the whole sphere " , =--~^^r=-==£bz-==s . -*2 (volume of the uppermost ' . :^~ liquid"displaced)' ^ (Fig 234) -af7,;a8-2 (ffrtfl8)=|fjra3.£ ; .

.•.' The weight of liquids of density p disp]aced=§fjwi8P& the weight of liquidiof density 2P displaced^! f jro3.2pg

and the weigh^oriiqddjDf density 3P displaced=»ff;ra8.3Pi£. Forequilibrlumv .'•:•-

the weight'of the sphere=the sum pf the wts. of liquids displaced or iwaVs^gfTra'Pg+ffrcaS^ps+ffTttjS.Sps or 108a=28P+104p+84P=216P or <r=2P. Hence proved.

Ex. 11. A solid sphere of radius a is just immersed in three liquids whose densities are 1 : 3 : 5 ; the two surfaces of separation A of the liquids are. at depths fa and fa from the top, JProye tiiat J

Floating Bodies 2fy

the density of the sphere to that of the topmost liquid are in the ratio 71 :27.

Solution. Let the density of the sphere be a and those of the liquids from the top be P, 3p and 5 p.

Let EF and PQ be the surfaces of separa­tion of the liquids, mee­ting the vertical diame­ter AD of the sphere in B and C. Alsa we are given that AB<

Then OA=>a, OB=a- la=\a and OD=a. v

.'. The weight of the sphere» And the volume of uppermost liquid of density p displaced

-Jw (a - *a) [3a»-(iaa+a.Ja+fla)] =$f Tea3, on simplification

Similarly the volume of lowermost liquid of density 5p displaced = J K (a-fa) [?a«-{(fa)*+|a»+a2}] =**.*« [3««-Jfl« (4+6+9)]=^- Tra3

The volume of the middle liquids of density 3P displaced a (volume of the sphere)—(sum of the volumes of the

uppermost and lowermost liquids)

(Fig. 235) •faand i4C«fo. s OC=AC-AO'=*y-a'=>la

%Ka*og

I.

4 a /28310s, 8x0*1 72 . 8

+ J" 9*a°

or or or Henoe proved.

\ 81 T 81 / 81 X The weights of the liquids of densities p. 3p and 5P dis­

placed by the sphere are l\naz Pg, fno8 3pg, and fita?^9g For equilibrium, the weight of the spheres the sum of the weights of liquids

displaced by 1( * f jcflsffg<=Sf7rflsPg + §TCas.3pg+-8

s1-Taa.5Pg.

108«r =28P+216p+40P«284P < 7 / P , = , 1 0 B = 2 7 -

Exercises on § 5*03 Ex. 1. A body is floating partly immersed In one liquid and

the rest in another. Find the conditions of equilibrium. (Bhopal82) Ex. 2. A vertical cylinder of density 7P/4 floats completely

Immersed in two liquids. The density of the upper liquid Is p, and that of lower is 2P, find the position of equilibrium.

Ans, I of the cylinder is in upper liquid. *§ 504. Tension in the string supporting a body. A body is entirely Immersed tn a liquid and supported by a strings

to/tad the tension In the string.. • («**»*« 78)

2ife Hydrost^es

Let T be the tension ia the string at A (say) supporting the body. Let W be the weight of the body and W the weight of the liquid displaced by it.

The body is in equilibrium under the action of the following farces:— Jv (1) The weight W of the body J ^ictiQg vertically downwards through ^^j^==^T{^^JF~= G, the C. G. of the body. B^E-^^E^^^^^E

(2) the force of buoyancy "E^~ZEIE=f ^S^f^lFJri which is equal to W, tne weight of r^r£H>f^r F ^ ~ ^ r the liquid displaced acting vertically EsE^rz-jf t F>~-^r2E upwards through the centre of buoy- ErEEEEp' AW '£EEE^G ancy (since the body is entirely ^^f 1 % £ ^ F J ^ = immersed so the centre of buoyancy —~:/ ]& \^szr~-will also be G in this case) and, EEE3 \ w \ji=£=

(3) the tension T of the string, = E E \ j ^ - - - - ^ acJing at A vertically upwards if the E\EE=!h^ _ ^ ^ J r E I density of the body > density of ^-p^-z^^^^r^^ls-^f^z liquid (otherwise if the density of the ' ~ body < density of the liquid, then (Fig. 236) this tension T acts vertically downwards). (Note)

A, For equilibrium, we have the resultant of the forces acting vertically upwards=the resultant of the forces acting vertically downwards or T+W'^W or T=W~W or tension in the stringent of the body—wt. of the liquid displaced

In case the density of the body < density of the liquid, then tension T of the string acts vertically downwards and for equili­brium, we have W'=T+W or T^W'-W or Tension in the string «wt. of liquid displaced—wt. of the body, if the density of the body < density of the liquid. /

§5-05, Weighing in a liquid. If a body of weight W immer­sed in a liquid is weighed, then the tension T in the string suppor­

t ing the b >dy will be equal to (tie.true weight of the b3dy)—(the weight W of the liquid displaced by it)

This tension T is called the apparent weight cf the body. From the' previous article, we have T<=>W - W or W=T-\- W

Of True weight of a body ^apparent weigh t+weight of the liquid displaced by it.

§ 506. Weighing in air. Let a body of weight (true weight) W be weighed in air. Let

W be the weight of the air displaced by it and JV0 be the weight of the "weights" that are used to weigh the.body.

Then true weight of the body «=»appareat welghtH-weight of {ho air displaced by If.

6, . w^w^w\

floating Bodies 219

To obtain a psrfcctly accurate weight of the body, it should be weighed in vacuum, winch being not practicable, a correction is applied for weighing in air (as given in the next article) If a grea­ter accuracy is required.

**§ 5 07. A body of density p is weighed in air by means of "weights" whose density is p'. If a be the densit v of the air,- to find the weight of the body when its apparent weight is W0. •

(Avadh 79 ; Gorakhpur 76 ; Magadh 74 ; Patna 82 ; Ranchi 79. 74)

Let W ba the true weight of the body, whose apparent weight is W0 which is nothing but the sum of the " weights" used.

Since the body on one side balances the /'weights" on; the other 6ide.

.*. the tensions in the stricg supporting the scale pan contai­ning the body=tt-e tension in the string supporting the scale-pan containing the "weights". or (weight of the body—weight of the air displaced by the body)

=(the weight of the "weights") —(weig \t of the air displaced by the ''weights")

(See § 3-04 Page 217) or W—(volume of the body) a#= ^—(volume of the ''weights") og

"-Q>t-w'-{%)"

o, w.w. (.-;)/(.- i)-r.(.- ;)(.-!)-or „-„.(!-^)[l+f+(f)"+...].

^ - i expanding by Binomial Theorem

Hence W<=»Wa ( l ~ pT + "p")» neglecting powers of a higher •

than the first, o- being small ia comparison to p and P'. Solved Examples OH § 5*04 to § 5*07. *'Es. 1. If W and w be the weights of a body in vscno and

water respectively, prove that tae weight ia air of sp gr a will be W -a (W—W). (Agra 82 : Garh »al 81 : Magadh 77 ;

Sajasthan 75 ; Rohilkhand 80 ; Vikram 78) Solution. The true weight of the body=»weight of the body in

vacuo *=-W. The weight of the body in water

eafiue weight—weight of displaced wajes or v**W-vgt,

• 220 ' Hydrostatic

, where v is the volume of the body and p the density of water. or vg9=*W— w ...(1)

Also weight of the body in air =»true weight—weight of air displaced =• W— vgop=>W— o (W— w), from (lj. Hence proved.

* Ex. 2. A piece of silver and a piece of copper fastened to the ends of a string passing over a pulley, hang in equilibrium when suspended in a liquid of density 1/15. Determine the relative vol­ume of the masses, the densities of silver and copper being 10'47 and 8*89 respectively. \

Solution. Let v and v' be the volumes of the piece of silver and copper respectively.

V The tension in the string supporting piece of silver =*the tension in the string supporting piece of copper i. Weight of the sliver—wt. of the liquid displaced by silver => weight of the copper—wt. of the liquid displaced by copper,

or vg (IO-47)-vs (l-15i=v'g (,&'69)~v'g (I'l3) or v (10-47-l-15)=v' (8 89 -M5) or v (9-32)==v'(7-74) or v/v'- iK—H*. Ans.

Ex., 3. A body immersed in a liquid is balanced by a weight P to which it is attached by thread passing over a fixed polity, and when half Immersed, is balanced in the same manner by a weight 2P. Prove that the densities of the body and liquid are 31 2.

Solution. Let v be the volume of the body and ot p be the densities of the body and liquid respectively.

When the body is completely immersed, tension in the string=weight of the body—weight of liquid displaced or P=>vg<J-vg9 or .P=»vgO—p) ...(1)

Similarly when the body is half immersed, (Note) 2P**vg<J—(v/2) gp

%OC • v 2P<=vg(o-l?) ^(2) From (1) and (2), 2vg (o-p)=w» (w—£P)

Or ff-=2P—JP=|p or <y/P=3/2. Hence proved. Ex. 4. A body is weighed in tv?o liquids, the first of sp. gr. 0 8

and the other of sp. gr. 1*2. In the two cases its apparent weights are 18 and 12 grammes respectively. Find its true weight and specific gravity.

Solution. Let v c.c. be the volume and o the sp. gr. of the body. Let w be the weight per c.c. of the standard substance.

Kow we know that apparent weight of a body =true weight—weight of the liquid displaced

£ In the first liquid, we have U=vow- v (0'8) w<=v (<y—0'8) w ...ft)

In the second liquid, we get t2^yQW-Y{VZ)\»^(a-VZ1w „,({$

Floating Bodies 221

Dividing (i) by (ii) we get ^ - ~ |

or 3 (a—1'2)=2 (o—0#8) or a=3'6—1"6«=2 A sp. gr. of the body=2 Ans. .'. From (i), 18=v (2-0*8) w or vw=18/(l'2)«=15 .*. True weight of the body=vorw=l3x2«=30 gms. Ans.« Ex. 5. The apparent weight of a sinker weighed in water is

fonr times the weight in vacuum of a piece of material, whose sp. gr. is required, (hat of the sicker and piece together is three times that weight. Show that the sp. gr. of the meterial is 0'5.

Solution. Let V and Vx be the volumes of the piece of material and sinker respectively. Let s and sx be the sp. gr. of the material and the sinker respectively. Let p be the density of water.

Then the true we'ght of the piece of meterla] «=its weight in vacuum =* Vgsp.

Also apparent weight of a body —true weight—weight of liquid displaced.

.*. The apparent weight of the sinker =>True weight of the sinker—weight of water diiplaced by it =Vxgxs9-VigP, But apparent weight of the sinker n given to be 4 times the

weight of the material in vacuum (given). .-. 4Vgsp=V1gs1p-V1gP or 4V1s=V1s1-V1 ...(1) Also apparent weight of the sinker and piece of material =(true weight of sinker and piece together;

—(weight of water displaced by them) ~ ( J W + Vgsfy-fy^+VgP). But apparent weight of the sinker and piece together = 3 times the weight of the material in vacuum (given). .-. 3Vgsp~(V1gs1P+ Vgsp)-(VlgP+ VgP)

or 3Vs=Wvh+Vs)-(V, + V) ...(2) Subtracting (2) from (1), we get •

Vs<=*—Vs-\V or 2sV=*V or s=>i. Hence proved. Ex. 6. A solid whose sp gr. is 1 85 is weighed into a mixture

of alcohol of sp gr. 0-82 and water. It weighs 28 8 grammes in vaccam and 141 grammes in the mixtnre; And the proportion of the

- alcohol present. Solution. Let F e e . be the volume of the solid and s the

sp. gr. of the mixtire of alcohol and water. Now true weight of the solid=weight of the solid In vaccum

or • F.g(l-85)=28-8g (Note) „ 28-8 '2880 576 F=T85 = T85 = ~W ° °-

Also the weights of the solid in mixture wits true weight—weight of mixture displaces! by it

223 Hydrostatic*

or I4'lg~(28'8)g-Pgs or tfy«=(28'8-14-l)~14-7 14-7 147x37

or ^ i r - T ^ f f r ...0)

Let Vx and Va be the volumes of the alcohol and water in the

mixture, then the sp. gr.-of the mixture<= • y ,y

or * v1+v* L e t J V f T W or V2~V'-Vt

Vx (0'82)+(V'-V1) 147x37 V-Vi AH8) & 5 r a _ o r j ^ j g - F ,

or 147K37F'=>S760F'-F,x576xl'8 or Vt (576X 1-8)=F' (5760-5439)=32IF'

V 321x10 535 . . . , 31 0 1 P " 576X18 " T728 " 3 1 n^X^\m or F ^ i W " or volume of alcohol>=>31% of the whole volume.

Ex. 7. A body is found to sink in each of liquids whose sp. gr. are 0*7 and 1*3 and its apparent weight in the two liquid* is found to be 14 end 8 grammes respectively. Find the sp.gr. of the body and show that it will (float half immersed, when placed in a liquid of sp. gr 4*2.

Solution. Let V be the volume and s the sp. gr. of the body. Let p be the denity of the standard substance. Then the

densities of the two liquids are 0-7p and l'3p respectively. True weight of the body—Vgs?. Now apparent weight of the body =struo weight of the body—weight of displaced liquid. .'. we have U*=V gs9-V (Q-lf) g=Vg9 (s-0-7)

and 8=V gs?-V(l-lP)~VgP(s-V3) or 14=F£P ( J - 0 - 7 ) and %=>VPg(s-l-3).

Dividing, we g e t ^ - - ' — ^ or 7 (j—1-3)=4 (s-0'7)

or 3s=(9 ' l -2 ,8)=6'3 or s=2*l. \ Ans. Also let Vx be the volume immersed In the liquid of density

4-2P. Then as the wt. of the body=>wt. of liquid displaced. & VgspcaVxg 4'2P, where 5=»2-l (proved above)

T. Vs T(2-l) V r Vl 4-2 4-2 2

.*. The body will float half immersed in liquid of sp. gr. 4*2. Hence proved.

Es. 8. A solid of weight W is weighed in air by a spring bal­ance and Us weight is found to be W, prove that W = W {P/(p—a)}

~ where a and p are the specific gravities of the air and the solid respectively. • .

Floating Bodies 223 •

Solution. V The apparent weight of the body In alt| «=»its true weight-the weight of the air displaced

.*. W^W-Vgo?! ...(1) where Vis the volume of the solid and Pj is the density of standard substance

.'. weight of the body=>FgPPi or W=V gP9x or Vpg^W/p •

A F r o m ( l ) , P r = F F - - c = ^ ( l - ^ ~ ^ ( ^ )

or W<=>W [p/(P- a)] Fence proved. ^Ex, 9. A body is floating partly immersed in a liquid and the

air in contact with it is suddenly removed. State giving reasons whether tf>e body wi8! rise or sink.

1 Solution. ' Let the volumes V1 and V2 of ihe body be. immersed in the l'qu'd and air respectively. Let p and o be the densities of liquid and air respectively.

If Wbe the weight of the body, then from the principal, the weight of the body=tfee sum of the weights of the fluid * displaced, we have W-^V^g-^-V^og •• -0)

Let V be the volume of the body immersed in the liquid when air is removed, then from the above principle, we have

W=V?g ~(2) A From (1) and (2), V 9g=* Vtfg+ V&g

or ( K - ^ p - K ^ V V& is positive. .*. (V- Vx) P is also .positive

or V—V1 is positive or V- Vx > 0 or V > Vt

I. e. when air is removed, the volume immersed > volume immersed when air is not removed. i e. when air is removed, the body will sink. Ans.

**Ex. 10. A body of specific gravity a when weighed against a weight of sp. gr. p in water (the whole balance being immersed) appears to have a weight W. Show ihat its true weight is *

o — l p (Lucknow/82, 74; Muzaffarpur 80; C ^ v 7 Vikram8-j)

Solution. True weight of the "weights"=fF V Volume of the "weights"^ W/g A Apparent weight of the "weights" in water=true weight

of the "weight^" — weight of the liquid displaced by 'weights"

Let a) be the true weight of the body of specific gravity a. Then volume of the body=w'g» .\ Apparent weight of the body in water

«=the weighs of the body—weight of the liquid displaced by it. -v-i<Slgo)A.g**a>[l-{lIo)\

/

224 Hydrostatics 136—14

A When the whole balance is Immersed in the liquid, the apparent weight ol body=apparent weight of "weights"

or » [ l - ( l / " ) ] = ' n i - 0 / P ) ]

« _„(i- i ) / ( i - l )_„(-fj)( fc!) Hmcoproyed

• **Ex. 11. If a body of mass mx and density ^ is balanced by a mass m8 and density ^ w n e n placed on the pans of a common

balance show that m1»='ma •, ;f3 \\ , where ^ is the density of air. Ya VVi Wa)

(Rajasthan 74) Solution. The volume of the body of mass mx and density, tyt

& The true weight of the body of mass m^nijg and weight of the air displaced by body of mass mi^QmJtyJ 4*agr •

«!.«JThe t#8& weight of the body of mass wx ' vyv =its true weight- weight of air displaced

Similarly apparent weight of the body of mass m% '

V The two bodies balance each other when placed on the pans of a common balance.

.*. apparent wt. of the body of mass m^apparent wt of the body of mass ma

* 2*a (vi-va) ' Hence proved. *Ex. 12. If the sp. gr. of air be s and W and W be the weights

of a body in air and water respectively, prove that its weight in vacuo is W-f Js / (1- s)] ( W - W>. Ulgra 8 )

• Solution. Lew a be the weight of the body In vacuo and a be its sp. gr.

A The volume of the body =><n\og. Apparent weight of the body in air

«=(ifs true weight)—(weight of air displaced by it)

or W=ca—— .sg=*to[l ) ... °S \ <* ' •. (1)

Also apparent weSght of the body in water (its true weight)—(weight of water displaced by it)

or W'~a>-—. lg=a>(l--\ •(2)

From (1), — =«>- W and from (2), - ~ c o - W\ (X m Q

136/15 Floating Bodies 225 \

Dividing, we get -j-= =r, or s (o> — W'^co—W

or at(g-l)—sW-W or a>(l-s)=*W-sW'<=>(l-s) W+ TVs-sW (Note) or CJ=W+[S{W W')l(\-s)\. Hence proved.•

*Ex. 13. A body floats in a liquid of sp. gr. s with as much of its volume out of the liquid as would be immersed in a second fluid of sp gr. s', if it floated in that fluid. Prove that the sp gr. of the body Is ss'/(e+s'). - (Agra 81,78)

Solution. Let a be thesp. gr. of the body. Let Vx and VQ be the vojumes of the body immersed in liquids of sp. gr. s and s' res­pectively.

Then the volume of the whole body=(F1 + F1), since the volume out of the first liquid=>voluoie immersed in second liquid.

Also as the weight of the body=weight of the liquid displaced so we hwefVi+VJag^ViSg or V^i=Vx{s—a) ...(I) and (Vi+VJ ag-VJg or V2 (s'-o)=Vla ...(2)

Dividing (1) by (2), we get -r^-*"-—

or a*=(s'—o) (s—o)=ss'—(s+s') (r+«r* or (si-s')a=-ss' or tf=ss'l(s+s'). Henoe proved.

**Ex 14. Ifw„ w2, w8 be the apparent weights of a given body in fluids whose sp. gravities are s1; s, and sa respectively, then show that w, (sa-s,)£Wj, (sa—s^+w, (Sr^s^O.

(Gorakhpur 76 ; Lucknotf78% 76J Magadh 75 ; Patna 82 ; v _ SRajasthan 75I Vikrcan 74)

Solution. Let V be the volume of the body and TV its true weight Let P be the density of the standard substance.

V The appatent weight of a body in a fluid. <=>(its true weight)—(weight of the fluid displaced-by It)

w^W-Vsfg ' - . „.<1) Similarly. wa~*W-V ss9g —(2) and ws=-W—V.s,pg ...<3) Subtracting (2) from (3), (3) from (1) and (1) from (2), we have

H'j, -w8«=F (st -st) g? ...(A) Wt-w^VfSi-sJgp ...(B)

and w»-wi=V (Si—sJ gp ...(C) Multiplying (A), (B) and (C) by wu wa and w3, and adding we

have wu (wa—wa)+w» fa —wj+w, (w»—Wi) •=*Vg? [wt (f,-st)+w% (Si-sJ+Wa fa-^)]

or WJC ,—s,)-{!-W)i (*!—sJ+Wf (5-j—J2)="0. Hence proved. **Ex. 15. Two solids are each weighed in succession in three

homogeneous liquids of different densities, if the weights of one are w1( wa and wg and those of the other are Wlf W9 and W„ prove that

" ' w, (W.'-W^+w, (W,-W,)-!-w, (W, -Wg)=0. <** (Agra 79, 74 ; LucknowfSll; Mithtla 81;

Muzzafarpur 82 ; RohilkhandVJTtf!) I Vikram 80)

226 Hydrostatics

Solution. Let the volumes of the solids be v and V and their true weights be w and W respectively.

Let Pi, p2 and P3 be the densities of three givtn liquids. .!. The apparent weight of a body in a .liquid

=»(its true weight ) - (weight ot the liqu'd displaced) £ For the fiist solid we have

Wi=>w—vPig, wz=w—vp2g and wa=M'—vpag Whence as in last example, we have

w1—wa=vg (P2—Pi) •• C1)) wa-wa~vg(?3 p2) > ...(2) wa-w1<='Vg (P1-P3) , - ( 3 )

Similarly for the second solid, we have Wt-W^Vg Vt-Pj .-(4) Wz-W^Vg (Pa-?2) - .-(5) Wf-W^Vg fa-9j - ( 6 )

Dividing (4)1by (1), (5)<by (2) and (6) by (3), we have

vhence fPi-fF2«=X (wi-Wa), —(A)

and JFS—H^W (w3-wi) .. (Q ' Multiplying (A), (B) and (C) by ws, wx end w2 and adding we get wa (Wi-WQ+Wj, (W2~ JF3H ws (JFa'- WJ*=0. He»ce proved.

Ex. 26. A body crnsisJs of an alloy of two m«tais of sp gr st asid s2 respectively, its, weight in vaiuo is w and in vtaier w'. bbcw that the pioportion of two nelals by lolnme is

_ s2w'- (sa—1) w : (sx— 1) w- Sjw'. Solution. Let F, and F2 be the volumes of two netals in the

alloy and p the density of the water. ' > •• Then^ota l weight of the alloy in vacuo is given by

W^F^pg+F^apg • ^ .. (1) ' Also the weight o! the body in water

=(true wt. of the body)—(wt of the water displaced) or M , 'Q M ,_( j /1 +^) p^ ,

•=(?&•+V&J gt-iVj.+ VJ & * r ..'.torn 0 ) or w'~Vx ( j j -1) gp+ F2 (s2-l) g? - (2)

Now from (1) and (2) we get w . w' {

and p = , ' iVlSx + VJt) g """ v> i V1 ( ^ - ' 1 ) + F2 ( j , 1)} g

' .We. have; — '(Fjji-h F2s2) F , C J 1 - D + V* ( J 2 - 1 ) or - TP {Fi [sx-1) f F2 ( 4 - l)}=u>' {Pi*•+ F2ja}

Floating Bodies 227

or Yl ~s*w' fa ' *> w

V2 t^ -1) w -Sjw' ' Hence proved. **Ex. 1? A body floating ip water has volumes vlf v2, v3

above the surface when the densities of the surrounding air are respec­tively Plf Pw p3) prove that P « - ^ ! + ^ f ^ f * = 0 .

^J^~xZera 83 / Avadh 80 / Bhopal 81 ; Garhwal 79 ; Lucknov{82, 79, 75\Rancht 75 ; Rohilkhand82 ; Vikram 78) <

Solution. Let v be tTTe-voiume and PFthe weight of the body. Let p be the density of water. Then the volumes immersed in water in the three cases are (v—v^, (v -v2} and (v —v8).

Then as the weight of the body <=> weight of water displaced +weight of air displaced

.*. When the surrounding air is of density ?l we have W=(V - V,) Pg+V1gp1 Or W— VgP=rV! (Pi -p) g

Similarly in the other two cases we have W-vgP~v2 (P2—P) g and W-vgP<=>va (Pa-p) g From these three results, we have

Vi (pi PJg~Vt (Pi-P) 8=Va (P3-p) g=*W-vgp or Vl (P1-P)=v, (P, -P)=»V, (Pa-P)-X (say), whence P2 - p =» (x/va) ...(1)'

P2-P«(A/v2) ...(2) and P3-P-(X/v8) ...(3)

Multiplying (1), (2) and (3) by (P2-P8), (P,-ft) and (Pl-P2) respectively and adding, we have

. (Pi-P) (P2-Pa)+(Pa-P) (P j -P i )+(P 8~P) (Pi-P»)

L vi v2 • v„ J

o r 0=X r« tZ&^ p J=& + 5i=&l or '£=b*&=h+*3=h'-Qm L Vj Va Va J VX V2 Va

Hence proved. •Ex. 18. Prove that if volumes v and V of two different subs­

tances balance in vaccam and volumes v', V balance when weighed in a liquid, the densities of the substances and the liquid are as

v ' - V v ' - V /v' _ V \ - T~ : V ! I v V /'

Solution. Let Pj and Pa be the densities of the two substances and P that of the liquid.

Then as the volumes v and V of the two substances balance in vaccum. .*. vgP,=»FgP2 or vp^Fpa ...(1)

Also the volumes v' and V of the two substances balance when weighed in liquid of density p.

.*. apparent weight of volume v' in liquid of density p «=apparent weight of volume Pin liquid of density P

or v'p1g-v'pg=F'p2g-rp£ (Note)

228

or

or

or

v'P,.

*[?-

-v'P=

v'

,.}

i

= F ' P a -

(?) = p ( v ' -

p2

"' H

Hydrostatics

•V'p or v ' P j - F ' P s - P t v ' - F ' )

-V'?2*=?(v'-V'), f rom(l)

T,A pa V - r fv'-

p

"c-ci

384-

..

-V)

• «>

-66

.(2)

-(3)

or

or

Again substituting the value of P, from (1) in (2), we have

v' h-V ( ^ ) - P ( v ' - n or Pl (v«-^rV]=.p(v'-n

*L V-r) V'~V' P ~ I , V'vf /v ' V \

Pl p

m (T-T) Hence from (3) and (4), we have

o /v ' - r \ . iv'-r \ (v' V'\

Hence proved. *Ex. 19. Jf a tody floats in a liquid with volumes v„ ?a and y„

above the surface when the baionetric heights are hM h, and b8, prove that b,v, (v a -v s H bav2 (v,- v,)-f bava <vx- v8)=.0. (K P. P. C. s. 81)

Solution. The densities of the surrounding air are proportional to barometric heights. Let the densities of surrounding air be plt Ps and Ps when the barometric heights are hu h» and AB.

Then P^AA^ Pa=AA2 and P8=AA8 . .(1) Let the volume of the body be v and its density be o. Let the

density of the liquid be P. Then \olumes immersed in liquid in three cases are,

(v—Vi), (v—v2) and (v—v3) respectively . Now for equilibrium of the body floating in liquid,

wt. of the body«=Wf. of ihe liquid displaced-)-wt. of air displaced. .". In the first case (when barcmetric height is Ai), we have

VCTg=.(v- V!> Pg-l-VjPjg

or ya=(v—Vl) P+ViPi or vx (P- Pj)=v (P— o)

or vz=>-^+— (p— a), dividing each term by p P P

Vi*Ai . v . N

substituting the value of P, from (I)

Floating Bodies 229

Similarly we have

v . - ^ ' + y <P-°) .-(3) and v a - ^ » + y (P—) ...(4)

From (4) and (?), vie have v8—Vs=»(X/p) (vA v3fc3), by subtraction

or AiVi(vs—v3)=(A/p) AxV], (va/z2-vsA3) Similarly />ava(v3—v,)="<A/P) V 2 ( v A - v A )

and A3Vj (vx -va^=(A/P) V i (Vi—vA> Adding these results we get the required result

V i (v a- v,)+^ava 0-3-Vi) - fisr, (v!-v2}=0. Hence pioved.

'Ex. 20. A given mass weighed in air en a spring balance indicates a weight W, it is ihm compressed to i]/n)tb of its foimer volHme and uppeais to migh W . |io^e ibat its weight in vacuo Is(nW'-W)/(n-l) .

Solution. Let w be the weight of the given mass in vacuo and v its volume.

Then as the weight of body in air =(its true weight)—(weight of the air displaced)

'. We have W^w—vog, where 0 is the density of the air. or • v<fg-=w—W .. (.)

Also when the body is compressed, its volume becomes (v/«) and from the above principle, we have*

W-w-[(vog)l«] or [lyog)ln\*=>w-W ...(2) Dividing (1) by (2) we have

n>=-(w-W)/(w-W') or n (w-W')=w-W nW'—W

or w (n—l)=nW'—W or w= — — j — . Hence proved.

Ex. 21. A substance whose density is P} is weighed by means of weights, the density of which is p' : if a be the density of the air find what is the true weight corresponding to any apparent weight. If the density of the air increases turn a to a' ; piove that the sppaient weight of the body is less than its foimer weight by fiaction

IP— a) (?'—°'{ 0 ? t h e ,&tter> p' be ,D8 greater than p.

Solution. Let Fbe the volume sr,d W \he trut weight of the substance. Then V=(W/gP). ...(1)

Let K0bethe volume and W0 the true weight of the "weights" when (he substance is weighed in air of density a.

Then V0~(W/g?') ...(2) Now apparent weight of the substance

=apparent weight of the "weights"

230 Hydrostatics

(using the result apparent wt. in air , . =true"weight—weight of the air displaced)

or " W[l-(<rfp)}-W0 [1-(*/?')]. , ...(3) Similarly if W'0 be true weight of the "weights" when the

substance is weighed in air of density a', then we have W[l-(o'/9)] = W'0[l-(°'in\ - (4)

Or

or

(?-oi _ W0(?'-o) (P-°'> W'nP-o') or 'WB \9'-a')\p-a )

Wa (P ' -«0(P-°) 77, on simplifying the numerator

or W0~W\-

( P -a ) (P ' -a ' ) :

(p'-P)(ff '_or) WA. (P ' -a ' ) (p-a) ""' Hence proved.

§ S"08. A body ts floating partly immersed In liquid and with a string attached to a point of it. To find the condition of equilibrium when

(I) the sp. gr. of the body > the sp. gr. of tfie liquid and {II) the sp. gr. of the body < the sp. gr. of the liquid.

Case I. sp gr. of the body > sp. gr. of the liquid. Let W be the weight of the

body acting vertically downwards through G, the C. G. of the body. v LetW'bsthe force of buoy­ancy, which is equal to the weigbt of the liquid displaced and acts.verti­cally upwards through Glt the cen­tre of buoyancy or the C. G. of the liquid displaced.

V sp. gr. of the body > sp. gr. of the liquid and volume of liquid displaced is also less than the volume of the body

.'. W>W Hence for equilibrium-the tension T of the-string should be

acting vertically upwards. Let the string be attached tothe^body at P. , , . , ,

Let the verlical lines through P, G and Gx meet the surface of the liquid in A, B and C respectively.

V The vertical forces acting cm the body are T, W, and W and the body is in equilibrium,

(Fig. 237)

Floating Bodies 231-

(Fig. 238)

A the algebraic sum of moments of all the forces about A is zero.

ie. W.AB<=W'.AC Hence the conditions of equilibrium in this case are (1) W~T+W and (2) WAB-W'.AC Case II. sp. gr of the body < sp. gr. of the liquid. As in case I, let W and W be the

weights of the body and the liqufd displaced.

Also in this case sp. gr. of the body < sp. gr. of the liquid.

«*. The force of buoyancy W will be greater than the weight of the body and the body will ascend.

.'. In order to prevent the body from ascending a string will be attached at P and the body wii) be pulled down­wards.

Hence tension T in the string will be acting vertically down­wards.

Hence as in the last case, the conditions of equilibrium are (1) W'=T+W and (2) W.AB=> W'.AC. (Note. In case I the line of action of W will be between these

orcesTandtf" while in case II the line of action of IF'will be between T and W\.

§ 5 09. A body floating under constraint. Case I. When one point of-the body is fixed.

{Bihar 75, Mlthila 82, 80; Muzaffarpur 81 / Ranchi 82; Rohilkhand 82) Let the point P of the body be

fixed i.e. the body is capable of mov­ing freeJy about this point. / L

The forces acting on the body ====^ r aie (1) the weight W of the body , f^-§.-^0^^ acting vertically downwards through (7, the C. G. of the body '

(2) The force of buoyancy W which Is equal to the weight of liquid displaced acting upwards to rough Gv the C.G. of the liquid displaced and

(3) the reaction R at.the,point P. (Fig. 239) Since two of these forces viz. W and W are acting vertically

and body is in equilibrium A, the reaction R should also act in the vertical direction.

232 Hydrostatics

(Miihilf> 82)

(Fig. 240) these moments will be

.'. For equilibrium, W=W'+R, tfW<W and W'~W+R if W < W.

Alsa taking moments of the forces about P, we have W.PA^W PB

• (Note. If W > W\ R is acting vertically upwards and if W < W, R is acting vertically downwards)

Case II. Woen two points of tbe body are fixed The two points A and B of the body

are fixed I.e. the body is capable of moving about the line AB. There are reactions at A and B.

This is a three dimensional ease. ^ Here we-are to take moments *f :

the forces about this Jine AB m order to 5 avoid the reactions at A and B. Hence : we have to find the moments of two forces viz the weight of the body and the force of buoyancy about the line AB and equal as befcre Solved Examples on § 508 and § 5'09.

*Ex. I. A balloon of volume V contains a gas whose density is to that of the air at the earth's surface as 1: 15. If the envelope of the balloon be of weight w but of negligible volume, find the accele­ration with which it will begin to ascend.

Solution, Let a be the density of the gas, then that of the air _>-is 15f.

Then the weight of the balloon. == weight of the gas+• weight of the envelope= Vga+w.

If/be the required acceleration of the balloon acting verti­cally upwards then from "mass x acceleration=forces acting in the

' sense of acceleration", we get [(Vga+w)(g]Xf->force of buoyancy—wt. of the Balloon with gas

•=K.15o£—{Vvg+w) or - ' f~>[(.l4Vg<r - w)l(Vg<r+w)] g. Ans.

Ex. 2. The mass of a balloon and the gas it contains is 6500 lbs. If tbe balloon displaces 48000 ca. ft. ol' air and the mass of a ca. ft. of air be 1*25 ozs., find the acceleration with which the ballooa commences to ascend.

Solution. Mass of the balloon=3500 lbs. F J . . . . 48000x1-25 Mass of displaced air=»

16 lbs. =.3750 lbs.

I f /be the required acceleration of the ballon, then as in the last example

3500 X/^Yorce of buoyancy-weight of the balloon

Floating Bodies 233

or 3500 X/= (weight of air displaced)-(3*00 g) •=3750g-35f%~25Og-

- f- I I -w-2 ***-* «•'-'• *Ex. 3. Two cubic feet of cork of sp gr 0*24 is kept below

water by a rope tfassteaet? to the bottom. Prove that the tension of the rope is 95 pounds. (Vikram7$

Solution. Toe weight of the cork=2x0"24w=0'43w, where w is the weight per cu. ft. of water.

Force of buoyancy <=> weight of the water displaced<=2K>, acting vartically upwards

.*. For equilibrium, force of buoyancy=weight of the cork +tension in the rope ...(see § 5*08 Case II Page 231)

or 2w=0,48+tension in the rope or tension in the rope=»2n>—048w=l'52w

= i § n X <>25 'lbs. wt.=35 lbs. wt. Hence proved. **Ex. 4. A uniform rod capable of taming about one end which

Is out of the water, rests inclined to tha vertical with one-third of its length in some water. Prove that its sp. gr. is 5/9

{Agra 82 ; Garhwal 78 ; Lucknow 77 • Rohilkhand 83, 77)

Solution. AB is ths rod of length _ ^ 2a say, with end B fixed and length ! ] ~e^ " AC<=*\ (2a) immersed in water. | ^ /

Let « be the area of cross-section \y *~ of the rod and o its sp gr. Let p be the t_C density of water. r^3ZHL|J^L

The rod AB is in equilibrium T-^^^GfslMM^fMfi^^L under , the action of the following ? 5 ^ ^ < = = f f ; ^ ^ = ^ f ^ 3 forces:

(i) The weight 2aaa?g acting (Fig. 241) vertically downwards through Gt the C.G. of the rod%

(ii) The force of buoyancy V or the weight of the watae displaced =>AC.aPg=% (2a) apg, aoting vertically upwards through Gu the centre of buoyancy or C.G. of the water displaced.

(iii) The reaction at the end B, which is fixed. If 8 be the inclination of the rod with the horizontal, then

taking moment of these forces about B, we have 2fla<rpg BG cos 0=4 (2a) apg.BGt cos 6

or BCra 1 RG M , 3 BC±\iAQ_ § (2fl)+j (fa) or BGo=\BG1 or o - g ^ g - • (2f l ) = ^ or cr^S/9. Hence proved.

Ex. 5 (a). A uniform rod 6 ft. long can move about a fulcrum one foot above the surface of the water. In the position of eqnili-

234 • v ' - ~ Hydrostatics L

brinm fonr feet of the rod is immersed. Prove that the sp. gr. of the rod is 8/9.

Solution. Given that the rod i AB~6 ft. _ jy c

The immersed length AC-=>4 ft. =™E=3{er^\_-|_ _ =-------.. the length outiide the \^^£=£^^g-^^_^._

water=£C=6—4=2'. EE£^£=r-3-^iH:-='-=-3r^rE£=€5 Let the rod make an .angle 6

with the veitical. (Fig. 242) The forces acting on the rod are (i) the weight W of the rod=6xaoP£, acting through G, the

"C.G. of the rod, where a, o and p are the area of cross-section of toe rod, sp .gr. of the rod and density of water respectively.

,<!i) the force of buoyancy K—4XaPg, scting -.vertically up­wards thiough Gu tbe C G. of the water displaced and

(iii) the reaction at the fulcrum at B. Taking moments of these forces about B, we have

W.BG sin dcaV.BGx sine or " 6«agp.BG=4<tpg.BG1 or 2a BG^IBG^ nr 2.BG{ 2(BC+hAC) 2 (2+2) * „ , °r° a~TBG—mAB) '~10r~9- Henceproved.

Ex. 5 (b). A uniform rod 3 metres long is free to turn about one of its eeds which is out t>f water and is in equilibrium being inclined fo the veitical. If its specific gravity is 5/9, find the length immersed. (Bihar 76)

Solution. Refer Fig. 242 above. Given that the rod AB=3m. Let this rod be free to turn

about, its end C and AC be the portion immersed. Let AC=x x metres. So the length outside the water=BC*=(3—x) metres. Let

the rod make an angle 6 with the vertical. The rod. Is in equilibrium under the action of the. following

^forces;— (i) the weight W. of therod=3x«CTPg, acting through G,

the C G. of the rod, where «, a and p are the aiea of the cress-section of the rod,4p. gr. of the metal of the rod and density of water respectively (Here c=>5ftt given);

(ii) the force of buoyancy V^ACXapg i e. V<=>xa.?g, acting vertically upwards tirough Glf the C. G. of the water displaced

and . . " (iii) tfie reaction of the fulcrum at J5. Taking moment of these forces about B, we have

W.BG sin 0= V.BG, sin 6 or 3aapg BG<=>Xoipg BGU as sin .9^0

Floating Bodies 235

or 3o (iAB)=.x(BC + CG1), where CG^IAC . or 3(5/9)4 (3)=.X[(3-JC +\ {x)] or S=* [2{3—x)+x] or S~>* (6-x) or xs—6x+5=0 or x = l , 5

Butx=5i.e > 3 is impossible. Fence *=1 metre. Ans. Ex 6 A uiiform rod rests in position inclined to the vertical,

with half its length immersed in water, and can turn about a point in it at a distance equal t» ith of the length of the rod from the extre­mity below the water. Prove that the sp. gr. of the rod is 0'125.

Solution. AB - 2a is the rod whose middle point 0 is in the surface and the point C on the rod, such that AC=>% tAB)<=°\a, is fixed and the rod can turn about C.

' Let a hi the area of cross-section of the rod and a be its sp. gr. Let p be the density of the water.

The forces acting on the rod are : (i) its weight W'-=>2aa.opg, acting

downwards through G, the C. G. of the rod ; (Fig. 243)

(ii) the force of buoyancy V=ax?g, acting vertically upwards through Gl5 the G G. of the water displaced and

(iii) the reaction at C. Let 6 be the inclination of the rod to the vertical. Taking moment of the forces about C, we have

W.CG sin 6=V.CG1 sin d W.CG=V.CG1 or 2axaPg{AG-AC)=*atfg (AGt-AC) or

or 2o (a—\a)=%a—\a or a (fa)=£a or »• =0123 Hence proved.

•*Ex. 7. A uniform rod of length 2a, can turn freely about one end which is flxed at a height h (<2a) above the surface of a liquid; if the densities of the rod and liquid be p and a, show that the rod can rest either in a vertical position or inclined at an angle 6 to the vertical such that cos 0=>(h/2a).\7"[<»/(a—p)].

(Agra 82, 78. 75 ; Avadh 81; Gorakhpw 77; Lucknow78i. Rajasthan 78, 75 ; Ranchi 76; U. P. P. C. S778)

Solution. AB=>2a is the rod of which a length AC is immersed in the liquid and B is fixed.

Let the rod be inclined at an angle 8 to the vertical in any posi- * EPA* tion. . —**--

Height of B above the surface ^^^^'^WrJIE} «=»fflV<=A (given). ~ ~ ,~ /_ BC-Asectf . (Fig. 244)

236 Hydrostatics'

pi ^4C=(2ti-ftsec 0) Let ss be the area or cross-setion of the rod.

^ The forces actirjg on the rod are :— (i; its weight W^2awg, acting vertically downwards through

G, the C.G. of the rod,\such that BG^a; * „(ii) the force of buoyancy V=AC a.<jg<=(2a—h sec 0) a, ag, acting vertically upwards through Gl5 the C G. of the liquid dis­placed, where

CGX=\AC~ \ (2fl- h sec 0) or BC^BC^ CGX= h se-c 0 f CGj' or BG-^h sec 0-fi {2a—h sec 0)=(a+i/i sec 0) and (iii) the reaction at B

Taking moment of all the forces about B, we have \ W.BG sin 0 - V.BGX sin 0=0 or sin 6 [2anpg a- (2a- h sec 6) aag {a-\- (\h) sec 0}1=O,

INote. sin 0 is not to be cancelled as we are to find the values of0].

or sin 0 [2osP- {a (2a- h sec 0) (2o+A sec 0)]=O ox - $ sin 0 [4a2p- <* (4a2- A8 sec8 0)]=>O or ' i sin 0 [4aa (p- a)+ aAa sec8 0]=O

.!. Either sin 0 =0 i e. the rod is vertical or . 4aa(P oH<7/jaseca0>=.O

1 // g ) h*o -* — 2a,sjU-9V Hence proved.

Ex 8. A heavy uniform rod of small cros<5-sectioB and of length / ts movable io a vertical plane about a hinge at one of its extremities situated at a depth h (</) below the surface <*s a liquid of density p. If ibe,density of tbe rod be CT (<P), show the inciina-tioiB of tbe rod to the vertical in equilibrium is cos"1 [(h/0V(P/ff)]

Solution. rB -1 is the rod with a length AC immersed in the liquid

rand the end A is fixed. Let 0 be the inclination of tbe

rod to the vertical in any position. Depth of A below surface

=AN=>h (given). A AC-=h 4ec 0 Hence if Gj be (Fig. 245) ,

the mid. pt of AC then AGx=\h sec 0. Let « be the area of cross-section, of the rod.<4#

The forces acting on tie rod AB are : (i) its weight W—hPg, acting vertically downwards through

G, the C. G. of the rod AB, where AG<= \JiB - \l; (ii)' the force of buoyancy V^ACapg^h sec 0.<xpg-, acting

vejtically upwards through Glt the C. G. of the liquid displaced such that AG^lfrsec 8 ; and . ,

4a*(o~p) or sec20=' n or cos 0 .

Floating Bodies 237

(iii) the reaction at A. Taking moment of all these forces about A, we have WAG sin 0-=*V.AG, sin (9 or WAG=VAGlf ~ sin 0^0

or / og.\l=h sec 0 aP? \h sec 5 or l'ac=,h2P sec3 5 or cos3 9*=(ha?il2o) or cos 6=(hllW(P/cr) ot 8-'cos"1 [ih/l)y/(?lo)]. Hence proved.

*Ex. 9 A rod o£ length 2/ and sp. gr. P is hinged at one end at a height fa above the surface of a liquid of sp. gr. np (n > i) The rod remains in equilibrium in an inclined position partly immersed la t5ie liquid. Prove that the inclination «f the rod to the vertical is cos-1 [ h/2/)v'{n/(n -1)}] and the lecgth immersed is' /

2> [1_^/{1_ (1/B)}]. (Lucknow SOyRohilkhand 78)

Solution. Proceed as in Ex. 7, Page 235. "**^ Here "2^"=»2/s '<T'=«P.

= ( A / 2 / ^ { B / ( J I - 1 ) } ...<)) Also the length immersed=^C=',2a—A.sec 0'

= 2/- h (2//A)V[.«-!)/«], from (1) =2 / [ l -V r {(» - l ) /« ) ] -2 / [ l - ^{ l - ( l / i i ) } ] .

Hence proved. Ex. 10. A uniform rod of length 2a floats partly immersed in a

liquid, being supported by a string to one of its ends. If the density of the liquid is f times that of the rod, prove that the rod will rest with half of its length out of the liquid. Find also the tension of the string. {Ranchi 81)

Solution. Let AB=2a be j the rod with a length AC immer- \ sed in the hquid aad supported JQ by a string at the end B. Let a J/ be the density of the rod, then | . C^& the density of the liquid will be H E I I 1 ! ^ ! ^ ^ — ^

Let AC~2x. Then ifGt Yg^^^^ff§^f^^ and G be the centres of gravity ~^^^.^^^§^^^^^^===z of the liquid displaced and the I . € # ^ € ^ ^ ^ € = _ f 5 ^ ^ i ^ ^ f 7 rod respectively, then CG^x and BG^a. (Fig. 246) ;

«. BG1=BC+CG1=(2a-2x)+X'=(2a-x) The forces acting on the rod are : (i) its weight W^2a a.ag\ acting vertically downwards through

G, the C. Q. of the rod and a is the area of cross-section of the rod ; (ii) the force of buoyancy V=2x«, fag, acting vertically up­

wards through <?„ the C. G. of th£ liquid displaced and

238 ~ ' Hydrostatics

(iii) the jtension T at B. Let the rod AB be inclined at an angle 6 to the horizon, then

taking moment of the forces about B, we have " V.BGi cos &<=*W.BG cos 0 or (2x«.fo-g) (2c—x)=(2aa.ag) a

4x(2a—x)*=3a2 or x=ia or- 2x=>a AC=a=\ AB i e. C is the mid-point of AB.

„•. The rod is half immersed in the liquid. Also from the figure it is evident that T+ V—W

T^W— P=2a.K ag—2x«.%og, where 2x<=a (proved).

or or

or

Ex. a 11.

-2aa.ag— 'A thin

faaag. aajag^ weight of the xod. Ans.

gr. % and length 4 rod «f

sp gr. % SBU leugm <* metres floats partly immersed in water, being supported by a string fastened to one end of the rod ; find how much of the rod is immersed ?

If the upper end of the rod be 1 m above the surface of water, prove that the rod is inclined at an angle 30° to =~~~^^=^~£^^^£L==~=± horizontal.

Solution. Let .45= 4m. ' (Fig. 247) be the rod with a length AO=>2x immersed in the water and the end B fastened to a string. Let p be the density of water then the density of the rod is |P. Let a be the area of cross-section of the rod.

The rod is in equilibrium under the action of the following forces :—

(i) the weight of the rod W=tfa..\?g, acting vertically down­wards at G, the middle point of ths rod /

(ii) the force of buoyancy V=2x aPg, . such that AGi<=>x or 5(71=(4—x)

and (iii) the tension T at B, Taking moment of all the forces about A, we have

4a.fP.g- 2 cos 0=2a:apg.<4—JC) cos '0 3=»JC(4-X) or xs~4x+3=0 or *=»3 or 1 2x=6 or 2. But 2x=6 m. Is impossible since AC <3 AB

.*. .4C=2*=»2 m. i.e. the rod is half immerad. A\so in A BCL, sin 0=£Z,/5C<=4 or 6=30°. Hence proved.

"Ex. 12. A aeriform rod of length 2a and density P is movable in a vertical plane about one end which, is fixed in a liqaid of density P' at a depth 3h below the surface. A liquid of smaller density a is added on the top of the first liquid ; if in the oblique position of equilibrium the rod is just covered by the liquid, prove that the

or or

l

Floating Bodies 239

[Jj / p' a 1/2 1 ,

~\?ZI7} ' (Agra 80, 77) Solution. AD=2a is

the given rod with end A fixed as a depth 2/z below the surface ot separation of two liquids. Let G, G1 and G2 be the middle points of AB, AC and CB respecti­vely. Let a be the area of cross-section of the rod. If 6 be the inclination of the rod to the vertical, then

AC<=-2h sec 8 (Fig. 248J and BC=?a—2h sec 0.

The rod AB is in equilibrium under the action of the forces j (i) The weight W=2a«pg acting vertically downwards through

G, the C. G. of the rod ; (ii) The force of buoyancy Vx-=* ACa.9'g ^>2h sec 8 ap' g. acting

vertically upwards through Gx due to the liquid of density P': (iii) The force of buoyancy V2=BCa.og>=:(2a~2h sec d) nog,

acting vertically upwards through G, due to the liquid of density a and

(iv) the reaction at the end A. Taking moments of all these forces about A, we have

W.AG sin e^V^AG-L sin 6+ V2.AG2 sin 9 2aaPg.a=(2h sec 6) aP'g iAC+(2a~2h sec 6) aag AG2>

aaP=/! sec 0 p'.f/j sec B)+{a-h sec 5) <* (AC+CG2) a*P<= P'AS sec20i(a-h sec 5) cr (2/z seo 0+£ #C)

•=p'A2 sec2 e+(a—h sec 5) cr {2/z sec 0+(a | h.sec 6)} •=• 9'h* see3 6 -f or (a-ft sec (?) (a+A sec 5) '

P'A8 sec8 0-fa (a2-A2 sec2 0)~oaa+h* (P'-<x) sec2 (9

or or Of

or

or

a2(P-a)=A2(p'-or)sec20

cos -afeT - --^f*(fe?n Ex. 13. A semi-circular cylinder floats in water with its axis

fixed in the surface of the wafer. If this cylinder be movable about the feed axis and if its density be half that of water, show that it will be in equilibrium in any position.

Solntion. Let the semi-eiicle ABCA be the cross-section of the given cylinder through O, the middle point of the axis of the cylinder. ACDOA is the sector immersed in the water of density p (say) such that [_ AOD=>2B.

240 Hydrostatics

Let OC be perpendicular to AB' then G the C!G. of the cylinder lies on OC, such that OG^4al3n, where a is the radius of cylinder (Note)

Let P be the middle point of arc ACD. Join OP, then Gt the C G. of the liquid displaced will be on OP,

136/15 /

such that OGt> a sin 0 6

If p be the

density of water, then the density of the material of the cylinder =4P (Fig. 249) (given). Let h be the length of cylinder. The forces acting on the cylinder arc ;— '

(i) its weight W=>(%na2h) \9g, acting vertically downwards through G, .

(ii) the forc6 of buoyancy F=(area of the sector ACDOA) x/iPg={$a2 (26)} hPg acting vertically upwards through (7»,the C.G. of the liquid displaced and ^

(iii) the reaction of the axis of rotation acting at O. Taking moment of these forces, about the axis of rotation

through O, we have the sum of moments of these forces = W.OG cos Z COD - V.OG1 cos L POD ...(1)

Now [_ COD^L AOD-L -40C=>20-9OO

and L POD-\.L AOD=9. 1. From (1) the sum of the moments

=W.p cos (20-90°)- F.§ a-^~ cos 0 5K 6

4a >lxa*hpg.Z- COs (9Oo-20)-flW£.5

S7C

a sin 0 cos 0

=4a8Apg [cos (9O°-20)-sln 2fl]=JaB/jpg [sin 20-sin 20] =0, for all values of 6. Hence there is equilibrium for all values of 6. Hence proved. *J5K. 14. A semi-

circnlar lamina has one of the tends, of its dia­meter smoothly hinged to a fixed point above the surface of a liquid and floats with its plane vertical and its diameter half immersed. If the incli­nation of diameter to the horizon is \IT, prove that the ratio of the density of the liquid to that of the lamina is

4(3rc-4) : 9 n - 8 . (Fig.-250)

136/16 Floating Bodies 241

Solution. ABC A Is the given semi-circular lanjlna of radius a (say) and centre O with the end B of the diameter AB fixed above the surface of ths liquid and the sector ACDOA immersed in the liquid. OC is the radius of semicircle perpendicular to the diameter AB Then Gr the C.G. of the semi-circle lies on OC, such that OG =*(4a/3jz). Let a and P be the densities of lamina and the liquid respectively. *

Let P be the mid point of the arc ACD. Join OP. Then Glt the

O G. of the liquid displaced lie on OP such that OGV=f.a s i° 6 f

: 20 is the angle AOD. Here J_ AOD=

Hence 0Gi«=f.

•K—£»

d sin (3:

or 26' <3JT/8)

«/8)

=5* 16a

or 5= 4 3TC

-g»

«

The forces acting on the lamina keeping it in equilibrium are f (i) its weight W={lna2) <rg, acting verticlly downwards through

G; (ii) the force of buoyancy F=(area of the sector ACDOA) Qg

•=C}a9.3:r/4) p^«=(3jc/8) a*fg, acting \eitically upwards through Gj and (iii) the reaction at B.

Also ZCOZ)=90°-Z_^OD=90°- 45°-Jn and l_POD=l ( 20 )~0= |TC (proved) _\ The horizontal distance of B from 0=>ML—OL—OM

-OA cos" -OO cos LCOD-a. * - <? . J - - a <3"74> V2 3ff v 2 ' 3 v 2 t

The horizontal distance of B from G1=NL=OL—ON

*=0A cos jj — OC?! cos /_PODr=a cos^ —g— sin l -^j cos I -^ J

1 8fl/_ . 3re 3rc\ a 8a . 3*

_fl 8 a [ J L \ ° (9w-8) D V 2 M V 2 /"" V2re "

Taking moments of the forces acting on the lamina about B, we have W.LM^V.LN

, . a (3re-4) 3ff 9 a (9a—8) 01 ****• • "37^r - T a9g- -w^r or (3:r-4) a=J (9«-8) P or ?- - f ^ " ? / _

v w (9n - 8) Hence proved. *Ex. 15 A solid hemisphere floats in a liquid, completely

immer&ed with a point of the rfm joined to a fixed point by means of a string. Find the inclination of the base to the vertical and toe tension of the string; p, a being the densities of the solid and the liquid. . *

m Solution. 1M the base of the hemisphere be Inclined at an

1 angle 0 the vertical and lot the point B of the rim be joined to some fixed point by means of a strisjg. Let a be the radius of the hemisphere.

The hemisphere Is In equili-^k-lum under the action of the following forces :

(i> its weight W- IncPgp, act­ing vertically downwards through

> G, the C. G. of the hemisphere, such that OG=3?o/8.

(ii) The forces of buoyancy V>=> incFog, acting vertically up­wards through G, which is also the C. G. of the liquid displaced as the hemisphere is completely imm­ersed and

(Fig. 251)

(iii) the tension Tin the string attached at B. Since two of these three forces viz. W and V are acting vertically I.e. their lines of action are parallel, hence third force viz. tension T is also acting vertically and its line of action also will pass through G„

QG (3a/8) 3 " " 8 * Hence in A BOG, tan 0«

OB a or fl^tan""11 which gives the required angle.

A!so resolving the forces vertically, we get T+ r«= or T= W- V=> fTtfl3. ?g - j j»f lV-1 naB (P-o)g.

W

Ans. Ex. 16. i n equilateral triangular lamina suspended freely from

A rests with tSje side AB vertical «r.d tjse side AC bisected by the surface of a heavy Quid. Prove that ike density of f he lamina is to that of the fluid as 15 : 16.

(lucknow 76J, Solution. ABC is tbe given

lamina suspended from A and side AB is verti 2al. I

'Let F be thft mid noint of AB and D that of AC. DE lies on the free surface and area DEBCD immersed in the liquid. Let CF=A, then DE=\h.

Also AE^\4JP

(Fig. 252)

Floating Bodies 243

Area of AADE<**\X DEx AE<*>\.\h.\AF*=*\.h.AF^>S (say) Area of A ABC^l area of AACF^2x\.CF.AF *

^2xlh AF<=*h.AF=*%S. ;. Area of the figure DEBCD*=%S-$=.7S. Also the distance of the C.G. of AABC of area 85 from AB

=4CF=4A and the distance of the C.G. of A ADE of area S from AB

- i (DE)^i (A/2)=A/6. .•. The distance of Glt the C.G. of area DEBCD from AB

_iS.jh-S.jh (f-i)A 7 A*

The forces which keep the lamina in equilibrium are $ (i) its weight Jf= 8Sag, acting vertically downwards through

G, the C.G. of AABC and a is the density of the lamina. (ii) the force of buoyancy V<=>1S ?g. acting vertically upwards

through G, the C.G. of figure DFBCD. Here p being the density of the fluid and (Hi) the tension in the string at A.

Taking moments of all these forces about At we have

%Sog.lh**7S9g. ~ or °- « !- | or a i p-15 i 16 14 P 16

Hence proved. *Ex. 17. An equilateral triangle ABC of weight W and sp. gr.

B is movable abont a hinge at A and is In eqnilibrinm when the angle C is immersed in water and the side AB is horizontal and above water. It is then furred abont A In its own plane until the whole of the side BC is in water and horizontal; prove that the

pressure on the hinge in (his position io to2 ( ^ S l W. \ Vs /

Solution. In the given triangle ABC, let each side=*2fl. Let D be the middle point of AB. Join CD. Then G, the C. G. of the AABC and Gx that of the Immersed portion A CFQ will both lie on CD.

Also CD-=>BC s?n 60°«*2a sin 60°=

(Fig, 233)

244 , ' Hydrostatic* 484-66

And PQ*=>2.eo cot 60°='22 (1/^3), where OC^z (say). Let P be the density of the liquid! The forces acting on the triangle are ; (i) Its weight W

=(area of AABC) sPg^d.AB C£).Jpg«=4.20.ffV3 s?g \ *=a2\/3sPg acting vertically downwards through G ;

(ii) the force of buoyancy F=(area of & CPQ) Pg=.|?Q.COpg

,-_-. *' ^ 3 zvg </3 ' acting vertically upwards through G2 aad

(lii) the reaction of the hinge at A. I. Tak'ng moments of these forces about A, we have

aPtfl sPg.AD=>Z^§-. AD or 3ass-=>z*

or z«=o ^ ( 3 J ) •••(!) In the second position when 5C is wholly immersed in liquid

and horIzontal,--the depth of BC below the free surface^z and height of A above the free surface i

=2a sin 60°—z*=>a*J3 —z=a<tf3— a-\/(3s), from (1) = 6 ^ 3 (1—-\/j)=z' (say) Also action at the hinge=wt. of AABC—v/t. of displaced liquid «= IF-(area of the figure CBQ'P') ?g . .= JF-(areaof hABC area of AAP'Q') Pg ~W-(\.2a.a</l-\P'Q!.z') ?g « = W - ( a V * - i 2 z ' cot 60°.z') Pg, where z'^ay/3 ( l - \As)

B IF - jaV3-3fl" (1-V^)2- 4 g - ] P#

= J P - [a V 3 ffc- o V 3 Pg (1 - VJ ) 2 ]

-^'--i+(o+.-H-,p(2-7r)—^('vf-')-' negative sign shows that the direction of action at the binge is vertically downwards. ' v Hence proved.

Ex 18. A triangular lamiBa ABC, of which tbe sides AB, AC. are equal, floats In water with BC vertical, and three quarters of its lesgth immersed, being &ept in equilibrium in this position by means of a string fastened So A and the bottom of the vessel. Find the sp gr. of lamina and show that the tension of the string is -a\th of the weig»htofttbe lamina,

Floating Bodies 245

Solution. ADC Is the given triangle with side AB vertical and a string attached at A with tension T acting downwards. Let D be the mid-point of BC. Join AD which is horizontal.

Let PQC4P be the area im­mersed, with PQ in the surface. Since f of BC is immersed in the liquid so Q is the middle point of BD. Hence P is the mid point of AB, as PQ is parallel to AD, Let AD~h, thenPfi=4/j. . (Fig 254)

.'. Area of A BPQ=1 x PQ X B(2=4 X }/iX \BD = %hBD=-S (say)

Area of AABC-=*%XADxBC=kxhx2xBD=*h.BD=*&S .*. Area of figure PQCAP=V>S-S=>1S Also the horizontal distance of G, the C. G. of A ^5C of

area 8S from BC=4A And the horizontal distance of C. O. of A 5PQ from BC

=4 P0=JA .*. the horizontal distance of Gu the C O . of area PQCAP

Let p be the density of water and s the sp. gr. of the lamiaa. The forces acting on the lamina are:

(1) its weight W=>%S s?g, acting vertically downwards through G, the C.G. of A ABC.

(ii) the force of buoyancy V^IS pg, acting vertically up­wards through G, the C.G. of the displaced water and

(iii) the tension T in the string at A, which also acts in the vertical direction as the other two forces are acting vertically.

And the horizontal distance of G and Gj from A are §ft and (ft—Aft) l.e 1ft and f4-ft.

Taking moments of all the forces acting on the triangle aboufi A, we have hS s?g lh<=*7Sgp Aft or

or

¥*• or J = s l , which gives the sp. gr. of the lamina. Also from the figure it is evident that T+W<=V

T~>V-W~7Spg-$S spg=il-8s) S?g = ( 7 - ¥ ) Spg~lS9g

Also W-=-8S spg~8S.U Pg-*£Spg. m

i From (1), r = i Sogt=-hW. Hence proved.

Ex. 19. ABC is an isosceles triangle right aogled at A composed of,,two heavy rods AB, AC hinged together at A and light string BC ; it floats Tvith angle A imsaersed ire \?ater„ Show that the tension of

SA5 * , Matrices \

the string is ^W (a—b)/a, where ,2a is the length of a rod, 2h \\i& iesgth immersed and W the weight of each rod.

Solution. In A ABC, AB and AC are rods and BC is a string. Let AP and AQ be the portions of the iod immersed in water.

Then we are given that AB=AC^2a and AP=AQ=*2b. LetC?!, G?8, G'x, andG 'gbe the middle points of AB, AC, AP

and -4g respectively. Then AGy*=-AGtf=a and AG\=AG'2-=.b. Let J-K be the weight of eac1* rod. Ktheforcoof buoyancy,-

(acting vertically upwards) on each rod and Tthe tensioa in the

(Fig. 255) spring. Then resolving vertically the forces acting on the system as ] 1 a whole we have 2K«=2W or V=-W " ' ...(1)

The rod AC is in equilibrium under the action of the following . forces i—

(0 its weight W, acting vertically downwards through G2, theC.G. of the rod AC, ' \

(ii) the force of buoyancy V, acting vertically upwards through G'a, the Q.Q. of the liquid displaced by the rod AC.

(iil) The tension T at C and (iv)' The reaction of the hinge at A. Taking moment of these forces about A, we" get T.ACxos \l_BAC+V.AG'% sin \i_ BAC~>W.AG% sin \ U BAG,

where £ , BAC=9\)° (given) I. T.la. cos 45°+ V.b sin 45°= W.a sin 45°

- or T.2a+W.b~Wa, v ftom(l)V=W or T<=*W(fl-6)/(2a). Hence proved.

v Exercises on § 508 - § S'G9

Es. ; 1. A uniform rod, eight metres Iongs cau move about a fulcrum which is above the surface of water. In the position of ' equilibrium six metres' of the rod are immersed. Find its specific gravity.; AES (15/16)

E x . 2 . As uniform rod of specific gravity 2 is suspended by vertical strings attached to. it§ axtremifjes. such | h a | {J3/4J ths 6f,

Floating Bodies '247 •

it is immersed in a water. Find the ratio of jhe tensions in the strings. *

HARDER SOLVED EXAMPLES Ex. 1. A solid hemisphere of radius a and weight W is floating

in liquid, and at a point on ihe base at a distaacc c from the'centre rests a weight w, show that the tangent of the inclination of the a^is of the hemisphere to she vertical for the corresponding position of equilibrium, sssuming the base of the hemisphere to be entirely out of the fluid, is f (cw)/(a.W).

Solution. The weight w is placed * at P, such that OP==-c (given). 4 ^

The fluid thrust at each point / \ \ \ of the(curved surface being BOrmal to / \ \ \ ' the surface passes through the centre \ N \ \ O of the base. > \ • G X T \ \ '

Hence the resultant flaid thrust \ Fon the curved surface in contact « \-c_ with the fluid will be passing throuah =^=^^^v

The hemisphere is in equili- E = € € ^ C -

brium under the aciion of the follow­ing forces :— * (Fig. 256)

(i) Its weight W acting vertically downwards through G, the C G. of the hemisphere, where OG=3a(8.

(ii) The force of buoyancy V, acting vertically upwards thro­ugh O, and

(Hi) The weight w placed at P, acting vertically downwards. Let 0 be the inclination of the axis OC of the hemisphere to

Jhe vertical. Taking moment of these forces about O, we get

W.OG sin &*=w.OP sin (9Q°-0) OK W f a sin Q*=>w.c. cos 0 or tan 0 m 8 cw/QaW^

*Ex 2- * A thin uniform wooden rod AB is in ecgnilibriusi \a dfi inclined position with one end A iomersed in a bowl of water, one point D supported on the edge of the bowl, show that the sp. gr. of the

. AC 2 A D - A C rod is A B 2 A D _ A B •

Also prove that the greatest fraction of the length of the rot! which can remain immersed is 1—<?(\~o), where a is the sp. gr. of rod.

Solution. AB is the rod with a point D of it in contact with She edge of the bowl and portion AC immersed.

Let « be the area of\;ross-sectiion and o the sp. gr. of the rod. The forces which keep the rod AB In equilibrium are 5— (i) the weight of the rod W*=>AB os erpg, acting vertically down«

W&rdi through G, the C.G. of the rod AB (P Is th« density of water);

24B Hydros atieS^

(Fig. 257)

(ii) the forces of buoyancy V =* AC a. pg, acting vertically upwards through G„ the C. G. of the liquid displaced by AC such that

AG^iAC *or DG^AD -AG,,

*=>AD-\AC, and (iii) the reaction of the bowl at D.

Taking moment of these forces about D, we have W.DG cos 0= V.DG1 cos 8,

where 6 is the inclination of the rod to the horizontal. . W.(AD-AG}=V(AD~IAC)

ABx°9g {AD -lAB)=>ACa.?g (AD-IAC) o AB (2AD-AB)=>AC (2AD-AC)

AC HAD-AC) °~TB

or or or

or - C ) (2AD-AB) " * ' Hence proved Again the weight of the portion DB of the rod has a teadency

to lift the rod out of water, hence when BD Is least the immersed portion of the rod is greatest and the least value of BD is zero. Hence when B and D coincide7 i.e. when AD^AB,. the length immersed is greatest.

Let AClAB*=k (say), _. , „ . kAB(2AB~kAB) fc(2-fc) Then from (1), a . AB{2AB m — —

or l-<j=»l-2fc+fcs=C&~fc)a

or ^(1—w)-=l— k or k<=\—^{l <*). Hence proved. Ex. 3. A uniform FO(B, loaded at one end so as to float upright,

lioats in a liqaid of density P, with a length a aoimmersed, in a liquid of density p2*wi4h a length b uDimmersed. Show tbat the length that

' is unimmersed when the rod Moats in a liquid of ddnsfty p is , [«Pj (P - Pa). *Pa ( P - P J ) ] / { P (Pi - Pa)}

Solid ton. Let / be the length of the whole rod, a its area of cross-section and wits weight.

_•. When length a is unimmersed, length (I - a) is immersed and so on.

Since weight of the body=weight of the liquid displaced. „\ In liquid of density pu w=-(l—a) apg .. (1) In liquid of density pa, w •=(/-&) op2g ...(2)

and in liquid of density p, \v=(l -<) ctpg, ...(3) where c is the length unimmersed in liquid of density p.

From (I) and (2), we get Q-a) Wig*=>{l—b) «P„g

Floating Bodies &#

or /(Pi-P^«-aP»-6P, or lJ^-^- • ,,.

From (1) and (3), (l-a) rtg^Q-e) aPg or (l-a) Pi=(/—c) P or cP=/ (p— P,)+flPa

or fp= P ln

< ? =^- (p-Pi)+aPi, from (4) Pi — *2

^ (aP.-ftP,) (P-PQf aft fa-P,) Pi-Pa

or c=[aP1(P~p3!!-iP2(p-P1)]/{p ( P I ~ P 2 ) } . Hence proved. Ex 4. A rod of density a and length / is freely movable about

one end fixed at & depth h below the surface of a liquid of density p ; show that the rod may rest in a position inclined to the vertical provided that 1 < (P/cr) < (P/ha). (Lucknow 75)

Solution. AB is the . "~^/3 given rod of length /, with ^<****^ portion AC immersed and ^^"'^'^ ' the end A fixed at a depth __^ l?-"'^__ A below the surface. Let d_ JX'>f^r^rz^__-Z_ -^=^=^L^^ a be the area of cross- ^%QJ0^^^L-^J^^^^^J^^Z^J^ section of the rod. Z^^^Lrj^^^^^^^kM^'J^.

Let G and Gx be the ^ mid-points of AB and AC (Fig. 258) respectively, then AG^ih sec 5 and AG=\l.

The forces acting on the rod are : (i) its weight W=>faog, acting vertically downwards through (?„ (ii) the force of buoyancy F=(A sec S) afi.g, acting vertically

upwards through Glt and (iii) the reaction at A. Taking moments of these forces about A, we have

{k sec 6) x.pg AGt sin Q-=la,ag AG sin d or (h sec 6) P (JA sec 0)«/or J/ or AaP see3 0=>/sa or cos0 e=>ha9ll2o

If the rod is at rest In a position inclined to the vertical, $hen 8^0 and therefore cosa d < 1 or A2P//V < 1 or P/a < /»/A" „.<!>

Also If the rod rests partly immersed in the liquid then a < P or 1 <3 (p/o) ...(ii)

Hence from (i) and {ii), 1 < (P/») < (/a/As). Hence proved. Ex § An equilateral triangular lamina ABC of sp. gr.

na (< 1) is moveable about.a Ssed hircge at A and is in eqaiiibriuni when the corner C is immersed in water and AB horizontal and above the water. If the triangle be turned about A in its vertical place and kept in position with the side BC immersed and horizontal, prove thai

•

SiO Hydrostatics

the action of ^ie hinge ia this position is (2/a) (1—n) W, where vV> is the weight of the lamina.

Solution. Proceed exactly as in Ex. 17 Page 243. Here s=na, • rEx. 6. A rod floats upright partially immersed in a hofro-

geneous liquid. Prove that a small increase of atmospheric density will produce a small rise of the rod proportional to the square of the length of the unsniisersed portion.

Solution. Let / be the length of the rod and a length x of it be immersed in the" liquid. Then the portion out of the liquid • = ( / - * ) .

Let cr, p and p' be the densities of the rod, liquid and air respectively. Let a be toe area of cross-section of the rod. Also for the equilibrium of the rod, weight of the rod=weight of air displaced+weight of liquid displace! 'or fao i=(l— x) a Pg-fxa.pg or h—(/—x) P'«=JEP

/fr-p') o r i X=I / ( « - P ' ) _ / f t - q ) (P-P') (P -P ' J ( P - P ' J „.(i)

If P' changes i.e. if atmospheric density changes then x chan­ges hence x is a Inaction of P'.

A From (i) oiSerentiating both sides with'respect to P', we get

or x=-

dx H?~o) dx I (P-q) or

(/-*)" * ' "(P-P')2 "rfp' (P-P ' ) J U P ^ J '

f r o m ( i ) ; As the rod is floating partially immersed, therefore a <3 P.

, Hence tfx/tfp'=• negative, which shows that x decreases as p' increases i e. the rod rises as atmospheric density increases.

Also dxjd?' is proportional to (/—x)3 i.e. the square of the 'length of the unlmmersed portion. Hence proved.

**Ex. 7. A cone af given weigh* ., aad volume floats in a given flaiei with its vertex downwards, show that the surface of the cone in contact with ihe fflaid is leas'? when the vertical angle of the cone is 2 tan-1 (l/V-2).

{Rohilkhand 77) Solution. Let the semi-vertical

angle of the cone be « and W be its weight. Let the cone be immersed to a depth z>

Then for equilibrium, Weight of the liquid displaced

' ^-weight of the cons or J7rr3.tana a.cD=>W, where w is the weight per unit volume of the liquid. or zs=(3$7wu) cot2 a or z^tfWlno))1!3 cot2?3 a D,.(i)

*% If 5 be the surface of the cone in conlao? with the liquid "fern S« rc p? Ian «) (a sec a^rcsft tan QS sec <&t

(Fig. 25PJ

Floatfag Bodies lM.

\>i S=n Q\Vlnu))W col*'-* a tan a, sec a, from (i) -»A.cot"3 a sec a, where X'KSW/TWO)2'3. * "

,S=>X sec3'3 a cosec1/' a If 5 be minimum, then rfS/rf«=0

X [—sec2'3 a.& (cosec a)-*'3 cosec a cot a -hf (sec a)_:l/a.sec a tan a cosec1'3 a]=»0

\\ sec2'3 a cosec1-3 a [-cot a-t-2 tan <x]=Q 2 tan a—cot os=0 or tan8 «=>£ o r tan a«=(l/\/2)

Now we can prove that d2S/dx2 is positive for tan a«=(l/V2) Hence 5 is minimum if tan «>=1/V2 or a=>tan-1 (1/V2) the vertical angle of the cone=>2«=2 tan -1 (1/V2).

Hence proved. *Ex. 8. Show that a right circalar cone of density P and ^semi-

angle a can float vertex do innards in a liquid of density a with gene­rator vertical and the base just clear of the liquid is

or

or

or or

or

P«=<r (cos 2K)°'a. Let h be the height

(/. A. S. 73) Solution.

of the cone and O the centre of its base. Let BC denote the surface of floatation. From O draw 00' perpendicular to the surface of floatation, meeting BC in O'. Then V is the mid-pome of BC as OO' is parallel to AC, both being vertical. weight of the cone^lnh* tan2 « gP. B=W=^£f)t

...(f) = - - - - - - - * • =

Now the section of the cone (Fig. 260) by the surface of floatation is an ellipse with centre O' and BC ai major axis and let 2a, 2d the lengths of its axis.

Then 2a=BC= VB sin 2a, from A VBQ =>h see « sin 2a=2/i sin «

Through O' and C draw EF and CD parallel to AB. * Also from geometry b^^EO'xO'F (Note)

or tf"=\ABx\CD „. V In A ABC, O' la themid-pt. of BC and EO' is parallel to AB. Similarly in A BCD, O'F^i.CD

- 1 (2ft tan *).\CD. CD VC

Now in similar A-s VCD and VAB, -^ *=—. AB VA v

or

OS

CD* VC

AB= VC ,2/j tan «=>2FCsines

VA'" Ji sec e' =>2.F5cos 2« sin «... Y in A VBC, Va^VBcos 2a =2iA sec « cos 2« sin ««=»2A tan « COS 2«

from (ii)» 6a«=i {2k tan «).$ 2ft taa « cos 2««fia tan8« cos 2® d«a>6 jac.®,^' (cos 2a) *

: ^ 2 Hydrostatics

-.'. area of tbe ellipse with O' as centre and BC as major axis, «="7ra6"«=7r \h sin al h tan K.V(COS 2«) =7rfta sin a tan a A/(cos 2a)

.*. Volume of ths coaevimmersed in liqu'd «=»£ area of the bass x height of the vertex V abave the base ^krJi* sin « tan a \/(cos 2a) FC ra^Tuft9 sin a tan a\/(eos 2a) h sec «'cos 2a -

... V VC<=*h sec a cos 2a •=47r/;3 tan3 a (cos 2a)s/a

, For equilibrium, weight of the cone=weight of displaced liquid

or \ii)f tan2 a. gp^^nh? tan2 a (cos 2a)3'2 go or P=a (cos 2a)3/a. Hence proved.

*Ex. 9. The vertical angle of a solid right circular cone" is 60°: prove that it can float in a liqaid wiih its vertex above the surface and one point of its base in the surface, if the densities of the cone and the liquid are in the ratio of (2^2 -1) : 2^/2.

' Solation. T et h be the height of the cone and O the centre of -its base*. Let BC denote the surface of flotation. Let O' be the middle point of BC.

Now G, the C. G. of the whole cone will lie on VOp such that OGs=*\h, where YO=h.

Also let G1 and Ga be the Q.G.'s of the portions outside and inside the liquid respectively, Then G, lies on VO' such that O'G^l.VO*. , (Note) ' ,

•and C8 will lie on tbe Has {Fig 261) « ! G produced. In similar triangles VOO' and VGG& we have

OG~iVO and O'G^IVO' OG VO

le. Q~Q~=YQ' o r °0' is parallel to GGX

y But 00' is parallel to CA O and O' being mid-points of AB and CB

U GGX or GG% is parallel to CA. Now for equilibrium there are only two forces viz* the weight

of the,body acting vertically downwards through G and the foroeTor' buoyancy acting vertically upwards through G2. Hence these are equal in magaitude but opposite in direction and their points of application viz. G and G he in $be same vertical ime *,<?, G(?a is Vertical. Heace CA is vertical.

Floating "BodSea 233

Now as In last example, the section of the cone by the surface of flotation is an ellipse whose area

«=7rf!a sin a tan «-\/(cos 2a) =5tAa sin 30° tan 30V(cos 60°), since here a«=30° =nh\(l[2) (1/tf 3) (l[</2)=nh*l2tf6

& Volume of the cone outside the liquid •=4 (area of this ellipse) X VC *' *~1.(KFI2/2\/6) h sec 30° cos 60°,

• • yc*=> VB cos 60°«="/i sec 30° cos 60° 7ths

lS%/2' And the volume of the whole cone

- H i " tan* a=ijcfi3.(tan 30°}"—S«*B . A Volume of the cone immersed in the liquid

nh9 nil* 042-1) 18^r:

7tA3. 9 18V2

Also for equilibrium, the weight of the cone«=»weight of the displaced liquid.

or 7th*

rT aSz (2K/2-1)

9 ~° 18V2 the cone and the liquid respectively,

nh3gp, where a and P are the densities of

or 2^2 a**(2\/2—1) p or a : p« »(2^2- l ) :2 \ /2 . Hence proved

*»Ex. 10. A uniform isosceles triangular lamina (sp. gr. a) floats in water with its plaae vertical, its vertical angle (which is equal to 2a) immersed and the base wholly above the water. Prove <hzt in the position of equilibrium in which base is not horizoBtal the sum of the lengths of the immer­sed portions of (he two sides is 2'» cos2 a, where a is one of the equal sides.

Also prove that a < cos4 a as well as cos %%. (Fig 262)

Solution. ABC is the given triangle with portion APQ immer­sed in the liquid. AB=AC==>a and [_ BAC-=2O.. Let D and O be the mid-points of /JCand /"^-respectively Then G and Gt tne C.G.'s of A ABC and A APQ will lie on AD and AO respectively, such that AQ^iADaadAG^AO.

For equilibrium,, since there are only two foiyes viz. the weight

of the triangle acting through G and the force of buoyancy acting 1 through Glf keeptrg the triangle ABC In equilibrium, so G and Gt must be1 in the same vertical line. Also in A ADO, since

AG_ alAp aAD c AGt " f ^O "^O*

A Z)0 Is parallel to GGt, and hence vertical. • Hence in A PDQ, the base PQ is horizontal and median DO

is vertical. A DP=DQ ...(I) Also in A APD, DP2~AP*+AD9-2AP AD cos a And fn A AQD, DQ}*=>AQ*-\-AD*-2AQ.AD cos a But from (1), DP=>DQ or DP*=DQZ

X AP*+AD*—7AP.AD cos «=AQ3+AD2—2AQ.AD oos a or (AP*-AQ')-2AD cos a WP- /J2)=0 or. ' (AP-AQ) [AP+AQ-2AD cos «1=>0. ~ ' But AP^AQ, since then BC becomes horizontal which,is against hypothesis.

v > Hence AP+AQ—2AD cos a=>0 or AP-\-AQ*=*2AD cos a=2 AB cos a cos a, t AD*=*AB cos a ' or s ,4P-M0=>2acosaa ~.(2)

Hence first result is proved. Again for equilibrium, Weight of the lamina"=>we}ght of the liquid displaced,

or (area of A ABC) opg—(area of A APQ).Pg, where P Is the density of water.

,Ot $ AB.AC sin 2a.og*=\ AP.AQ.sin 2a g, V p—1 or ^ AB AC ov-AP.AQ or a2 a=AP.AQ, as AB^a^AC

AP.AQ '

Now we know that arithmetic mean > geometric mean

- A-' ^ ± 4 ? < ViAP.AQ) er AP,AQ < (^0)a

- :. f r o m k . < f ^ ^ % . , . < ( ^ ^ ; f r o m ( 2 )

or o < cos* a. Hence second result. Again as o increases, the area of the immersed portion increases

and the maximum portion of the triargle will be immersed (keeping BC out of the water) when C is on the surface of water i.e. when AQ^AC^a. v '

In this case from (2) we get AP-=*2a cos" a—AQ=*2a cos2 a—a •~ ,~\ i r (2a cos8 a—a) a A From (3), max. value of a=> g —

«=(2 cos" a—l)«=cos 2a. -Hence a < coa 2« ~ Hence'the third result. *Ex. 11. A triangular lamina ABC of specific gravity s floats

with its plane vertical in water. A being outside the liquid and BC

JPSoat!»g Bodies - , 255

not horizontal. The angle A Is a right angle and AB*»AC, If 0 be the inclination of A B to the vertical, prove that •

sin 20~[(2-2s)/(2s-l)], given s > f.

Solution. ABC is the fi given triangle with portion /k-PCBQPimmersed in water of / | u density p (say). Let the side / j \\© \ . r> AB be inclined to the vertical . ___ff i g \\)Q ^ ^ H at an angle. 0 then the side AC f i l ^ y L

r\ will be inclined to the hod- =§=E5' } _ zontal at an angle 8, since \^^^^^=^^^-^%'§'=^.-Z. CAB=>90° (given). I f i N ^ ^ J i f ===W^¥¥B~iP

Let i? and 0 be the mid- ^ e ^ - ^ - - - - - - = — - — -_ -points of CB and P<2 respec­tively. Then G and G2 the (Fig. 263) C.G.'s of A ABC and A APQ will lie on AE and AO respectively such that AG>=>1 AE and AG^=lAO.

AG AE „ &. In triangles AGG2 and AEO we have ^ - •= -JQ- Hence

GG2 is parallel to OE. Since A ABC [floats freely, Jk G and Ga

must be in the same vertical line. Hence OE is also vertical. Also weight of the lamina=wejght of the liquid displaced.

or A ABC jp^=(area PCBQP) Pg or A ABC H A ABC- A APQ) or (1 - j ) A ABC^L APQ —0)

From ^ draw AL perpendicular to the free surface. Let AL^z, then AP=z cosec B and AQ=z sec 5.

J. Area of A APQ=*l AP.AQ<=>\ z cosec 0.z sec 6 •=2a/sin 20

Also as O is the mid-point of PQ and /_ PAQ<=-90°. :. the circle with centre O and radius OP or OQ will pass

through O.' Hence* OA=> OP=>\ PQ^i AP sec 0 =\.z cosec 0 sec 0«=2/sin 20

Also /IE being the bisector of [_ BAC, /_ /^J?=45° .*. AE=>AB cos 45°«=(1/^2).^J5 . r /,% /i s A APQ fz2/sin 20)_ 2zl

2za

o r ^"^""^B^BinM ...(2), since AB*=>AC Now Z. LAE=L BAL-Z. BAE~6-lit. A Z. OEA~L EAL'=(0-\n) ' Also In AAOO,L AQO~9(f-6 and 0^=OQ A, L OAQ~L OQA~90°-6

25$ Hydrostatics 136/16

or

or

or

Exterior angle ^OZ,«=(i80?-2»)+S0o«=(270o-2fl) In A AOE — \=?{TkJ- A0E sin (270°-2e) cos 20

/ * ' ^O sin LOEA^ sin,(0-irc) "sin ( ivt- l ) M / ^ 2 ) ^ g _ \ cos 20 (s/sin 26) "^{(i/VZ) cos d-ilj^2) sin 0}

y4BsJri20 .^2 cos 20 . ' " 2z cos 20 , — or AB*

AB>

z*/2 - (cos 0—sJn 0) 2r(cos 0-f-s!n.e)

sin 20

sin 20 (cos 0—sin fl)

, Y cos 2 0>= cos" 0 - sin* 0

from (ii), (i~s)~~ 2zs slri 20 • sin 26. 2 (1+sin 20)

or or-

b(i-4

4z* (cos 0 + sin 0)a

2<1— s>+2 (l-s)sin20=»sin20 . . i sin20,=,2(l—s)l(2s— 1). Hence proved.

. »Ex. 12. A cylindrical backet with water in it balances a mass M over a "pulley. A piece of cork, of mass m and sp. gr. or, is then tied to the bottom of the bucket so as to be totally immersed. Piove that the tension of the spring will be

2Mmg (2m+m)

Solution. Since mass M over the pulley balances'the bucket with/water, hence the mass of bucket with water is M. which becomes (M+m) when a cork of mass m is tied to the bottom of the bucket. Let in this case the,

v acceleration of the system be/, then for the bucket with cork we have the equation of notion

{M+m)f~(M+m)g-T . ; ' .. (i) v And for the mass M.Mf=T-Mg. „.(ii)

r- where T is the tension in the string round the pulley. •

r\

Mc f

Adding (i) and (ii) , /~- mg »-(A) (Fig. 264) (/«-f-2M)' The forces acting on the cork are : - -(i) its weight mg acting vertically downwards,'. " (ii) the.force of buoyancy, actirg vertically upwards,

and (iii) Ihe tension T in the stiiug tied to'the cork which also acts downwards in the vertical direction as the oiher two forces are vsrtical.

Now' the mass of the cork«=/w and its sp. gr.=er. Let p be the density of water, when the mass of* the liquid displaced by the

c6rk«=»(its volume) x p«=> — P=»—. . per . o

136/1? Floating. Bodies 257

Bui <hh ttass fs descending with acceleration / and if P be the upward force on this mass we have t

(mJa),f^(rn/a)g-P or P^(mfo)(g-f) :. The upward thtust on the cork=>(m/o) (g-f) .". For the motion of tbe coik in space, we have

mxfc=mg+T'-(mla)(g f)

or T'= •m

-m

(4 ->)•

(Note)

- (B)

m is

mg m+2M!

2mMz 'a{m+2M)\ a 1}' Henceproved.

Ex. 13. Two bockets containing water, the. mass of each bucket with tbe cODtaised water being M, balance each other over a smooth pulley. Two pieces of wood of masses m and in' and specific gravity a, a'are then tied to the bottoms of the buckets so as to be wholly immersed prove that tbe tension of the string attached to miss

2m(M-fm')g M \ l2M4mjm') \ a 1J' Solution. Whea wood of masses m

and m' ere tied to the bottoms of buckets given, tbenlhe equations of motion of the two buckets are

{M-\-m)f=(M+m) g—T, where/is the acceleration of the system and T the tension in the string, and (Mim')f=T~(Mim')g

(Here we have supposed that the bucket with wood of mass m is moving downwards).

Adding the*above two we have

j€B(lM-tm + m') ...(I) Now we can proceed for the wood of mass m as in the last

example and prove that T', tension in the string attached to mass m

is given by T'=m

(Fig. 265)

>m

( 1) (S f) ...See result (B) of last example

\a L)\* (2M+m + m')\ 2m(M+m')g (2M-\-m-tm' Hence proved.

tit •%<3?os»4tte

*§ 510. Pressure flerftaiive in terms of force. Let a fl,uid be in equilibrium under the action of a given

system of eAternal forces. Take two points P and Q situated along the given direction

and in the fluid, where PQ-= Is and is small. Join PQ and with PQ as axis construct a cylinder, whose cross-section is small and a closed curve of any type whose aiea is a (say). Ifp and p-\ 8/> be the pressures at P and Q, then the fluid contained within the cylin­der is in equilibrium under the action of the following forces :—

(i) the thrusts pa. and (p+$p) a on the bases acting in opposite directions at right angles to them as shown in the adjoining Fig 266.

(ii) the thrust acting at each point on the curved surface in directions pespendicular to the line PQ and

(lil) the force F (pa8s) acting In the direction PQ, where .Fis the component in the direction PQ of the given force per unit mass of the fluid and P is the density of the fluid. (Fig. 2(6)

Hence resolving all the forces in the sense of the line PQ, for equilibrium of the fluid contained within the cylinder, we get

pa+F(i?a.Bs)~{p-\-8p) a«=0 (see figure 266 above) or 8/>=P F Sj

' Now as SS-*0, p becomes the density at P and F becomes the component at P of the given foice per onit mass is the direction PQ and we have

S 7 m p P- - 0 ) (Remember)

i.e. the rate of increase of pressure in any direction is eqttolto the product'of the density and the resolved pert of the free in that direction.

COR. lix, y, z be the cartesian co-oidiiates of the point P in the fluid and X, Y, Z be the component foices per unit mass in the increasing directions ofx, ^ and z respectively, then from (1)

above we have / c p j l = P y l = P z , dx ' dy dz \ .. (2)

where/)1 is the pressure at P (x, y, z) and P is the density of the fluid. Also we have dp*=* ~ dx+^~ dy+~ dz

= PX dx±9Y dy-£?Z dz, from (2) os dp** P (X dx+ Y dy+Z dz) ^ ^

Floating Bodies 259

This, is known as equation of pressor©, which determines the pressure p at a point. {Rohilkhand 78)

Note. If the liquid is homogeneous I.e. p«=»constaHt, then from (3) we get

XJx V Ydy-\ Zdz=> — dp*=>d (^\ 9 P \ 9 ) ...(4)

i e. Xdx-\-YdyiZJz is a perfect differential i e. the forces X, Y, Z belong to a conservative system. (See Author's Dynan/ics)

Hence a homogeneous liquid under the action of a system of forces is in equilibrium only if the system of forces belong to a conservative system.

A We may write Xdx+Ydy+Zdz^—dV, where V is poten­tial function. * or d(pjP)~-dV, from (4)

Integrating we have p/p^ — V+ c, where c is constant of inte­gration, or (pl?)+V~e ...(5)

•*§ 5-11. Condition for the fluid under the action of forces to maintain equilibrium.

To obtain the necessary and sufficient condition that must he yrjlsfiedby a given distribution of forces X, Y. Z (say) in order that the fluid may maintain equilibrium. (Agra 82)

Necessary condition. If p be the pressure at any point P(x,y, z) in the fluid and P be the density of the liquid, then we

Now as/> is a function of the independent variables x, y, z so ay dap J ^ ^ J V . _9V_=_a^_

we have &y fa^^y 8z dx^dx dz* dx dy^dy dx ' ...(2)-/ « \ &P 8*P 9 idP \ d4dP \

From (2) we have ^-^WTy o r - ( - J - ^ - j

or | - (PZ) -^ - (PY), from (1) o r 9oJ+Zo^ PdT+Ydi °lP\dz- dy)Zdy Ydz ...(3)

Similarly from other two results of (2) we can get pidZ_8X\ 8±_ dP_ . p[dx d~z r x dz zdx -.(4)

a « d P W dx~) Y dx Xdy . -<5J

aw> Hydrostatics 4 8 4 - 6%

Multiplying (3) by X, (4) by Y, (5) by Z and adding we get

which is the required condition. Sufficient condition. Let us suppose that

(8Y_dZ\ (8Z__8X\ ldX_8Y\

Then X dx-\ Y dyiZ dz is exact differential. (See Differential Equations)

I e P (X dx'S-Ydy-l-Zdi)^ an exact differential =<//>, (say)

Equating coefficients of <&, 4>, dz on both sides, we get

JL^vX, . - " P r , / - 3 PZ, which we know are the equations dx oy dz 1

of equilibrium. (See § 5'1Q COR. Page 258) Hence the,condition (6) is also sufficient. Note 1. Ifshe iiquid is homogeneous, Le. P=> constant, then

from iesults (3), (4) and (5) above we get dY eZ 8Z dX ' dX=dY x

dz " By' 8x^8z dy dx ...(7) Note 2. If the liquid is heterogeneoas, then P is a function of

K, y, z and then from (2) above we can-get

h {9Z)~h™ h w-k {:z)>h ( p 7 J = l m ••(8) *3'S"12. Surfsccs ef Equal Pressure. (Vikram 81) We know (see § 510 COR. Page 258) that for a liquid at rest,

"the equa'ion of pressure at any point (x, y, z) is given by dp~? (X dx+ Y dy+Z dz) .. (J)

This on integration gives p=tf> (x, y, z) .. (2) If />=cons tan t=c (say), then from (2) we hase

<P (v, y .z j^e, ..(3) which is the equation of a surface at every point of which the' pressure is constant and equal to c.

A surface of this type is known as surface ofeqaal pressure As c t-ikes different (constant) vajues, lben from (3) we obtain

different surfaces which are all such that at eveiy point of tbem the pressure is the same and thus we get from (3) the smfaie of equal pressures for different ^alues of c

The equation of the free surface (or the external surface) can beoblained b'y putlir/gp equal to^he external pressure to the fluid. If

Floafjag Bodies - 2o\

there Is no external pressure, thcap^Q gives the equation of the free surface as j> (x, y, z)*=Q. #

s§ 5 13. Lines of Force. {Vikram 81) Definition. The line of forct is defined as the curve (or line)

which is such tnat the tangent io it at every point is in the direction of tise resultant lores at thdt pomt.

Equations. Let X, Y, Z be she components of the resultant force acting on ti e iluia at any point (x, y, z) parallel to coordinate axes. Then the direction cosines of trie line or actioa or this res­ultant force arepioportional to X, Y, Z.

Also, from geometry, we know thai the direction cosines of the tangent to a cur\e at any point (x, y, z) oa It are proportional to dx/ds, dy[ds, dz/ds.

From the definition given above, these two directions are the dxlds dylds dzfds .

same ana so we have —y - = —— •=• —j-

c r ^ = ^ = 5- , which are the differential equations ol the line A. X Z

of force. An Important property. The lines of force and the surface of equal pressure are ortho~

gonal. (Rolulkhand 79; Vikram 81)

A surface of equal pressure is given by p<=j> (xty, 2)=constant, ...(1) (See § 5'12 Page 260)

The direction cosines of the norma] at any point {x, y, z) to

this surface are proportional to—- , — , ~- / A w . dx' op* ds (Note)

I e. proportional to g , | , g , v from ( 0 g - g et*

i.e. proportional to PX, pF, pZ, uliere X, Y, Z are «he com­ponent forces per unit mass acting as {x, y, z) m t"rse increasiog, directions of x, y and z and P is the density of the fluid and

|£«PAT etc. (See § 5 i0 COR. Page 258). dx

l.e. the direction cosines of the normal at any point (x, y, z) to the surface (1) are proportional to X,Y,Z, I.e. proportional to the direction cosines of the lines of force.

Hence iines offeree are parallel to the normals to the surfaces of equal pressure i.e. the lines of force and surfaces of equal pressure are orthogonal, § 5'14. Equations of the curves of Equal Pressure and Density.

p {Rolulkhand 83, 77) Let the fluid be iacornpresiib'.e and let its density be a function

|<52 hydrostatics

Then the surfaces of equal pressure aie given by ^=>constani or 4»=»0 or XVx+Y dy+Zdz^O (see 5" 10 COK-. Page 258) - (i)

And the surfaces of equal density are given by p=constant

or </e=o <*£*+£*+£*-°- ~(«i) The required curves of equal pressure and density are the

curves of intersection of (i) and (ii). Solving (i) and (ii) simultaneously, we get

dx dy dz V8P _ 7 9P a

7 » _ y d ? " x 3P _ y 8 P dz dy dx dz dy. dx ...(iii)

Also from the conditions of equilibrium (see § 5*11 Page 259; we have .

_ 9 P VSP /BY 3 Z | . vd? dp !?Z dX\ zTy-YTz=9\df~dJl' J fe-Zax~ p \"^/

V3P YSPaJdX dY\ a n d Ydx~XTyapW~dx)

Substituting these in (iii) we get dx dy dz

dY_dZ VZ_dX dX_d_Y dz dy dx dz dy dx ...(iv)

which are the required differential equations of the curves of equal pressure and density. Solved Examples'on § 5 10—§ 514 .

*Es. 1 (a). If X=*y (y+z), Y~z (s+x), Z&y (y-x), show that surfaces of equal pressures are-given by y (x-$z)=>c(y±z) aud the curves of equal pressure and density are given by

j'(x+z)=constant,y+2=constant (Agra 8 ,77) Solution. We know that for the surfaces of efiual pressure,—""

" p -constant or dp=>0 or X dx+Y dy+Zdz^O or y (y+z) dx+z (z+x) dy\-y (v—x) <fe=>0

dx , zdy (y—x)dz A ,. . ,. . U+x)+7iffW+ <yw*+x)**> dlvldlng cach

term by y (y f z) (z+x).

ot dx 4- xdy A(y+z)-(z+x)ldz n

(z+x) "*" y(y+z) "*" (y+z) (z+x) u« (Note) writting y—x as (y+z).—(z+x) -dx , z dy dz dz

— 4 . ^ - - r r ~Q *+# y{y+zj z^x' y+z j ({fete)

or

or

or

or

Moating Bodies 263

Iz+x^z+x] ' L>' (y+z) y(y+z,j (Note) 'dx+dzl r (z+y) dy-jy dy + y dz)~\ ^z+~x~\ + L y(y-fz) J = (Note)

rf (.v+z) , dp _ <*Q+z) (x+z) + ^ (y+z)

Integrating, log (x+z)+log j>-]og Q;-|-z)=>log c, where c is constant of integration. or log [(x+z)yHy+z)]=logc or y (x+z)<=c {y+z), which is the required equation of surface of equal presrure.

Again the curves of equal pressure and density are given by dx dy d

dz 'dy dx ~ dz dy dx (see§ 5-i 4 Page 261)

dx. dy dz ° r (2zfx)-(2j>-*) ~{-y)-{y) "(ly+zi-'z)'

sustituting values of X, Y, Z from the given example. dx dy dz

or —— D •—.. = — z+x-y -y y „.(i)

From last two fractions we have dy+dz<=0 or j + z=constant Also using the multipliers y, x+z and y from (i) we get

ydx+(x + z) dy +y dz=0 (Note) (y dx+x dy)+{z dy+y dz)<=0 or d (xy) + d (yz)«0

Integrating, xy+yz*=-constant or y (JC4-z)•= constant. Hence the curves of equal pressure and density are given by

/-t-z=constant, y (x+z)<= constant. Hence proved. Ex. 1 (b). . A mass of fluid is at rest under" the forces

X=(y+z?-x*, Y^(z+x)2-y\Z^(.x+y^-z\ Find the snrfaces of equal pressure and curves of equal pressure and density. {Agra SO)

Solution. We know that for the surfaces of equal pressure we have =p constant ' " or dp = Q or X dx-[ Ydy{Zdz~Q

[(y+zf-x2] dx+[(z+xf-y2\ dy + [(x f ;)«-4»] di**Q [(y \-z-x) (y+z+x)] dx+[(z+x-y) (z+x-\-y)] d$

+ [(x+y-z) (x+y u)] & = 0 or iy+z—x) dx+{z+X-y) dy+(x+y-z) dz<~0,

V x+y+zjLQ Or (p dx+x dy) +{z dx+x dz) + {z dy + y dz)

*=xdx+y dy'+z dz (Note)

or

or or

BY dz

dx

dZ ' i dy

264 Hydrostatics

or d(xy)^d (zx)-\-d (yz)~>x d$-\-ydy\-z dz Integrating, we havs

xy+zx+yz-l (xa4-J'4-fz2)-l c\ where c is constant of integeation.

or x*+y*+z2-2xy-2yz-2zx=c*, , which is the required equation of the surface of equal pressure.

Again the curves of equil pressure and density are given by dx dy dz

~"dZ_dX ra9£3iZ dx dz ~dy dx

dy rfz 0 r 2 T ^ * ) - 2 ( * - f W ='2(xTy)-2{y+z) ^ 2{y+z)-2 {z+xf substituting vajues of X, Y, Z f f o m the given example

dx dy dz o r = 3 • * = " — — * . ,

z-y x~z y—x ...u) From (i) we have dx-\-dy -}-^-=0 (Note) Integrating, v/e have *+7-r-z=oonstant —00 Also using the multipliers x. y, z from (i), we have

xdx+y dy+zdz=0 Integrat ing, x a + j ? a + z 2 = c o n s t a n t _— <iii> Hence from (ii) and (iii), the curves^ of equipressure and

equidensity are given hy AJ+.y-f z=cons t an t , x2-\-y% f z2=>Constant. Ans. Eak 1 (c). UX~y%+yn-z; Y ~z2-t-zx-\-x\Z~xs+xy-{-y\

Had curves of equipressure and cqai-density. (Rohilkhand77) Solution. We know that the curves of equi-pressurc and equi­

density are given by dx dv dz

, -m cz dy dx dz dy ~ dx dx , dy dz

or &z+x)-(x+sy) {.2x±y)~{y+2z) ( y+z)—(z+2x) » substituting values of X, Y, Z from the givsn example.

dx dy dz , o r —•— • = = '

z-y x-z y~x „.(i) From (i), we have dx+dy+dz*=>0. (Note) Integrating we get x+y \ z=-constant. ...(ii) Again from (i) we get

X dt+y dy . dz x{z -y)+y (x - z) a y~^x (Note)

X dx+y dy dz x dx+y dy dz Off y— •— *3» •*—— Oi" ~Z-^~ -—

Hosting Bodies 265

'or x dx+y dy+z dz^O. • Integrating we get xz+y2+z2<= constant ...(iii) Hence from (u) and (iiH the curves of equ!-pressure and cqui-

density are given by xa-\-yij-za'='ComU, x+j+z=const . Ans. Ex. 1. (d). A fluid rests in equilibrium In afield of forces

where X=y3-\-z*~xyr-xz ; Y=za+:xa -zy-xy ; Z^r5-^ '2—zx- vz*. Show tbat the curves of equal pressure and density are a set of circles.

(Rohilkhand S3) Solution. We know that the curves of equi-pressure and

equi-dcnsity are given by dx __ dy_ <h_

WjbZ™ dZ_dX~~ dX_dY dz dy dx dz • oy dx dx dy = dz

o r (iz~y)- {.y-zr&x -V) ~{2z-x) (2y-x)~Ux-y)' substituting values of X, Y, Z from the given example.

dx dy dz or

or

*{z-y) 3 ( w ) 3(y-x) -(1) From (i) we have dx+dy+dz<=>Q (Note) Integrating, we get x+y+z<=*constant ~ (H) Again from (i) we get

x dx+y dy dz x (z-y) i-y (x-zfy-x (Note) x dx ±y dy dz x dx+y dy ^ dz

o r z{x-y) ~-(x-y) ° r z " - I x dx+y dy+z dz^O

Integrating we get x*+ya -j-2a=constant. ...(iii) Hence from (ii) and till), the curves of equi pressure and

cqui-dens'ty are given by *2-hv2+.za=constant and x+j-f z°cons-tant, which represent a set of circles. Hence projpdk

' *Ex. 2. if the forces per unit of mass at the point (x, y, z) narallel to the coordinate axes are y (a—z), x (a—z), xy ; find the surfaces of equal pressure and also the curves of equal pressure and density. *

Solution. Here we are given X--y {a -z), Y~x (a-z), Z^xy, ...(i)

The condition of equilibrium Is

which here reduces to yifl-t) [ -* -* ]+*<«-* ) ]y-{-y)\+xy [ (« -z ) -<«-*) ] -0

o r - - 2 ^ ( a - 2 ) + 2 ^ ( a « 2 ) - | - ^ ( 0 ) » 0 , which being $rue the fluid is at rest.

BY cZ dz By

-'dz dz

dx d\ —x—x

d* _ —X

y-i--dy~

y •

dX dX dz By

1

-?) dz 0

( a -

37 dx

dz Z)-(d- -z)

266 Hydrcstatks

Again the surfaces of equal pressure are given by /^constant or dp=0 or Xdx+Ydy+Z dz~0 or y (d~z) dx+x (a -z) dy+xy dz=-0, from (i)

or —+—-f-, =0, dividing each term by xy (a—z). x y (a—z) a J J v '

Integrating, log x-f log y—log (a—z)>=>Iog c, where c is cons­tant of-integration, or log [xj>/(a-z)]=>Iog c or xy=c (a—z), which is the equation of surfaces of equal pressure. 'Ans.

Again the curves of equal pressure and density are given b^ dx __ dy dz

or

or From the last fraction we get dz*=0 or z=constant. From the first two fractions we get

' " y dx+x dy=0 or d (xy)^=0 or x^=constant

A The curves of equal pressure and density are given by z>=constant, x^=constant. Ans.

Ex. 3. If the components parallel to the axes of the forces acting on the element of fluid at (x, y, z) be proportional to j>a-f2W2+za, 2a+2fi,;x+*2, x*+2vxy + y*\ then show that if equilibrium be possible we must have 2A=2ft=2v=l. (agra 79)

Solution.' HefeAr<=ce(,j'aH-2\pz-f z2), F = K (za-f 2pzx+x2) and £=>a (x* -\-2vxy j-ya), where a is constant of proportionality.

• * The condition of equilibrium is .

* lif-^i£-iFM£-£ J-or * (j>8+2tyz j zaJ [a (2z+2/*x)- « (2vx+2;>)] + a (2B+2J»JX+JC«)

X [<* px+2vy)-* (2),y+2z)] {- a ( A 8 + 2 V ^ f >S) [a (2j>-f-2Xz) —a (2^+2*)] =0

or C,+2A^z + 2a) [z+nx-vx-y] + (z*+2pzx+x*) [x+vy-ty-z] + (KZ \-lvxy \- ysf[y-\-\z-fi.z-x]=*Q. _<:,)

If 2x=2j*=»2v=l, then (i) reduces to ^ (^+2+z 2 ) ( z -^ ) + (2a + zx+x2)(x^-z) + (x*+xy+y*) (y- x)=0

or (z8-j>8) + (x8-z8.)+0'3—x8}-0, which is evidently satisfied. Hence the#condition of equilibrium is satisfied i.e. the equili­

brium is possible"if 2,A=2p =»2»>=1. . Hence proved.

Floating bodies \tfl

Ex. 4. A liquid of given volume V is at res<funder the forces X=- — px/as, r=>—/Liv/fia, Z=«-fiz/ca; And the pressnre at any point of the liquid and the equation of tbe surface of equal pressures. (Agra78)

Solution. We know the pressure/? at any point (x, y, z) in the liquid Is given by dp=9 {X dx+Y dy+Z dz) (see § 510 Page 258)/

or dp•= - P ^ ^ ~ dx f £ dy+J- <fej

Integrating, we ge tp—Jp/* j^ - - f^ -+ | j - J+C,

where Cis constant of integration. The free surface (i.e. surface of zero pressure) is given by /?=9

_ ^ g + > l + ! ! / ] + c = o X1 V2 22 1C

or -y-f ra"+-T"= —, which is an ellipsoid whose volume is given O D C r r

t o b e F

.'. From (i) we get

P—^[Xi+£+%]+^ (4 |L)2/3 ' „ (i0 which gives the pressure at any point.

Surfaces of equal pressure are given by/?=constant. te. dp=*Q or X dx •{- Y dy-fZ dz=Q

or -*• [jf d*+£ *y+^ ^ ] = ° o r £ *<+£ AH-C*» ^ ° o ' ^2

Integrating we get,— f „- + -a =*constaDt, which is the requ­

ired equation of the surface pressure. Exercise on § S-10-§ S14

Ex. Show that the forces represented by X<=(i (ys+yz+z2), y^fi (za+zx+x*)t Z=>fA (x8 f xy+y2) will keep a mass of liquid at rest, if the density varies Inversely as (distance)3 from the plane «4-y+z~0. Also show that the curves of equal pressure and density will be circles. (I. A. S. 72)

(Hint i See Ex. 1 (c) Page 264)

7, < „V*

' 2(78 Hydrostatics

, EXERCISES ON CHAPTER V

Ex. 1. A clo_se cubical vessel with walls one inch in thickness is to be made of ,metal whose sp. gr. is 27/19. Show that io order that the vessel may float in water iw internal volume must be at least 64 ou. inches. <Ga hwal 78)

Ex. 2, A piece of metal weighiug 32 lbs. in water is attached to a piece of wood whose weight is 30 lbs and then the compound body is found to weigh 12 lbs. in water. Prove that the sp. gr. of the wood is 0'6.

Ex 3, A hemisphere of weight Wis floating in a liquid and a weight w is placed on the rim. Show that the base will be inclined to the horizontal at an angle tan"1 (HWJ3W).

Ex 4. Two uniform straight rods leagth 2a, 2b and sp. gr. P, a respectively*, are joined together to form one straight rod If it-floats freely in water in an inclined position, the rod of length 2a and part of the other being immersed, prove ihat

a*? (p~J)+2a<3<> ( P - l )+6 2 a (a - 1 )^0 . Ex. 5 A rectangle, movable about an angular point which is

fixed below the surface ofja liquid, fbats with it? sides equally inclined to fee vertical and with half its area immersed in the liquid. If the lengths of the sides be a and b and one of the sides of length b entirely immeised in the liquid, show t&at the ratio of the density of the body to that of tae liquid is (a -b) : 4a.

Ex." 6. A square lamina is floating in a liquid of density twice its own with its plane vertical. - Find the positions of equili­brium. ' [Vikrain 82, 81)

Ex. 7 A uniform hemisphere of weight W with a small >- weight w placed on its rim floats in a liquid with its entire rim above

the surface of the liquid. Show that the plane of {he rim is inclined to the horizontal at an angle tan'1 ifrwIZW). (U. P. p. c. $. 79)

[Hiat : Use c=a in Ex. 1. Page 247],

CH\P?ER VI

'Atmospheric Pressure

§ 601. Gases and liquids.

In § l'Ol, Pege 1 we have already mentioned that gases and liquids have many common properties and both aie commonly termed as fluids. Gases are very compressible as they can be made to ocouoy a small or large space by Increasing or decreasing the pressure e.g., air in toy-balloons. §6-02. Air.

(a) Air exerfs pressure. This can be proved by the following experiments :—

(i) If two hollow hemispheres are so fitted that they are air-t'ght and if air is drawn out, then a great force will be required to «rparare them since the air from oatsids will press the hemispheres from all directions and keep the hemispheres tightly pressed against each other.

(li) If a rubber balloon is inflated, it expands due to the pressure of the air within it on its surface.

(bl Air has weight. It is found by experiment that a flask full of air weighs a little more than wfien it is empty (,/ e , the air of the flask being exhausted by means of exhaust-pump). This proves that air has weight.

*§ 6 03. Torricelli's Experiment. In 1643 Evangelista Torri-cellJ, a pupil of Galileo, performed a simple experiment to show that air has weight.

A glass tube AB about J feet in length, closed at one end and open at the other, is filled with mercury, Ths tifbe is thenjcm-porarily closed and inverted and reopened with its lower endIB below the surface of mercury in a basin.

The mercury then sinks in the tu^e leaving a vacuum AC as shown in the figure on Page 270. The height of C above the level D of the mercury in the basin is about 29 inches The space AC in the tubs which contains no air is generally called Torricellian Vacuum.

Now the pressure inside the tube at level D is the same as the pressure outside the tube on the suriaee of mercury in the basin. The pressure inside tbe tube is due to a depth CD of the mercury i e, gv.CD, where a is toe density of mercury. Hence the re muse be an atmospheric pressure of tbis amount, on the surface of the mercury in the basin and this pressure, is due to the weight of the atmosphere.

m Hydrostatics

c

D

'B

•Such an apparatus for measuring atmospheric pressure 5s the simplest form o% Barometer.

Note. Liquids other than merdry might also be used but mercury is generally used for its high specific gravity i.e, 13 6* nearly. If water wereyused, we would have required a tube of 13*6 times leDgth of the tube required for mercury ie., a tube of nearly 35 ft. in length.

§ 6'04. Pascal's Experiment Pascal suggested that the mercury column would decrease as we go up since in that case the atmospheric pressure decreases

§ 6*05. Homogeneous Atmosphere. We know that the density of the atmosphere gradually decreases as we rise above the surface ! . of the earth. If we replace the atmosphere by a column of homogeneous air of the same (Fig. 257) density as at the surface of the earth and of such a height as would give the same pressure at the surface of the earth as the actual atmosphere, then this height is known as the height of the homo­geneous atmosphere.

If A be the height of the homogeneous atmosphere, then ftXsp. gr of air=height of mercury barometerxsp. gr. of mercury.

^sp. gr of mercury X height of mercury barometer sp. gr of air

13-596x(30/'2) = 0-0013 "

height of mercury barometer=30") =»5 miles approximately.

§ 6*06 Relation between pressure, density and temperature. The relation between the pressure, density and temperature of

a given mass gf gas are expressed by the following two Jaws which a/.e*onfirmed by experiment and found to be holding good In a wide range of circumstances.

**(a) Boyle s La w. If the temperature of given mass of gas is constant the pressure varies inversely as its volurr e.

I.e. p oc l/v or pv=constanr, where p denotes the pressure andv the vohime

(Magadh 74 • Mithila 82) Note, (i) This law was discovered by Robert Boyle in 1662. (ii) The equation />i>=consiant is known as the isothermal

equation of the state. (iii) For gases (e.g air, hydrogen, Qxygen etc ) which are not

easily liqueflable, the law is very nea'rly true, whereas for gases (e g. carbon dioxide, water-vapour etc.) which are easily liqueflable this law is not quite trQe since in such cases volume decreases more

i.e.

{7 sp. gr. of air=0 0013 and

Atmospheric Treasure 271

rapidly then tbe pressure increases. The gas which obeys Boyle's Law accurately is known as Perfect gas. t

(b) Cbatle's Law. Jf the pressure of given mass of gas is consfanf, the volume increases by a definite fraction of the volume at 0°C,fJr every degree centigrade bji which the temperature is raised.

(Bhopal 82 / Magadh74 ; Mithila 82) Hence if v0 and v be the volumes at 0°C and t°C respectively.

then v-v0*=veat or v=v0 (l-f-at), where a is a definite fraction.

Note, fi) a is called the coefficient of expansion and is equal to 2} 3 for air and most of the gases.

. (ii) Th-h law also does not apply to vapours. § 6"07. Deductions from above laws ! (i) Relation between Pressnre and Density.

(Bhojwl 83 ; Renchi 80) Let/? v and p denote respectively that pressure, volume and

density of a given mass o r a gas. Let v' and p' denote the corres­ponding volume and density when fbe pressure is charged fiom p top', temperature remaining consfant.

Then from Boyle's Law we have pv=p'v' ...(j) But as the given mass of the pas remains unaltered so we have

vp=v'P' ...(ii) A From (i) and (ii), on dividing we get pl?-=p'/P' Hence the quintity fj?/p) is constant for a given mass. Let plP*=k, a constant, or p=kp -.(iii) i.e. the pressure of ages varies as its density, provided the

temperature remains constant. (Ranchi 80) The equation (Hi) is another form of stating Boyle's Law. (ii) Relation between dtmsities at different temperatures. Let v and v0 be the vo'umes : p and pn be the corresponding

densities at the temperatures t°C and 0°C respectively. Then as mass remains unaltered, we have vp=v0P0«

o r P5L<=,JL=voJleh^-, from Charle's Law [See § 6'C6 (b) above] P v0 v0

or P o = P ( l + aO, ...(Jv) (iii) Relation between Pressure, Densify BBd Temperature. Let/? be the pressure and p0 the density of a given mass of gas

at the temperature ..°C-Then we know p=kpa. ...(v)

where k is a constant depending upon the nature of the gas. Let P be the density of fbe gas when the temperature is raised

to t°C, the pressurep "remaining constaot. Then as proved above P0=>p (1-(-«/) -.(vi) , \ From (v) and (vi) we get. p<=»kp (1-f at) ..,(vu)

272 Hydrostatics 136/17

§6'08. Absolute Temperature. (M!thila82) If the pressure of a given mass of gas of constant volume

vanishes, then from p=k9 (14 at) we "get 14 af=0 . or r=-(l /a)=.-273°C.

This temperature —27S°C Is called the absolute zero and the tfmperatue measured fiom this point are called absol&te teropc-

•rature. The absolute temperature is generally denoted by T. So that T=>i\/a)+t<=>271 + t ...(i) le. the absolute temperature T corresponding to t"C is

27341 degree centigrade. § 6 09 Relation between Pressure, Voleme and Absolute Tempe­rature. {Bhopal82>

We know p=kp (1 + o.t) ...See § 6'07 (vii) Page 271) =*P« [Cl/a)+/]=*p« T, from § 6*08 (I) above.

& If v be the volume of a certain quantity of perfect gas,

then p-= •=> g v =ka (vp)=-ka (mass of the gas)

=constant, since mass of the gass is constant. or pv=*R.T, where R Is a constant depending on the nature of the gas.

Hence the prcduct of the volume of a certain quantity of perfect gas and its pressure varies a? its absolute temperature.

Note. If the pressure, volume ard absolute temperature of a given mass of a gas change from p, v, T top', v', V respectively,

then * L - ^ 1 T T'

§ 6'10. Mixture of gases. If tw o vessels of different volumes contain two different gases at the same pressure and temperature and a communication be opened between the vessels, the gases form a mixture whose pressure and temperature are the same as

,.laifore provided no chemical action takes place. The experimental fact enables us to deduce that the pressure

of the mixture of two gases occvpring a given volume V would be the sum of the pressures that each gas would have when occupying sepa­rately the same volume V at the same temperature.

Let/7| and/72 be the pressures of the two gases when volume of each is V. Let the volume of first be altered before mixing so that its pressure changes from/?! top2.

Then by Boyle's Law p-y^p^v, where Vx is its new volume, or Vi=>py,pz. The two gases are now both at the same pressure Pi and have a total volume (V+Vi) i e..V{l+(-px!p2)}.

Now if the gases are mixed, the volume of the mixture is V{\ + {p-i!p^} and by ihe experimental law stated above its pressure isp%. If the volume of the mixtire is now reduced to Vt

136/18 Atmospheric Pressure 273

then the required pressure p Is given by ' t

u pV~Pi.V {l+lpJpJl-toV+PiV or P=Pi+Pa> <•

This law is known as Dalton's'Law for the'pressure of mixture of gases and a similar result holds good for a mixture of number of -gases at the same temperature. •

**§ 6*11. t Pressure of a1 mixture of two gases of, different " volumes. '

{Bihar 76, 74; Lucknow 77; Manadh 73 ; Mithila 81 ; Muzaffarpur 80 ; Patna 82 ; Ranch! 79 ; Rohiikhand 83, 81)

If volumes Vt and V2 of gases at pressurespx and pt are mixed together without chemical changes to form a mixture of volume V, to find the pressure of the mixture, the temperature being constant.

Let us suppose that before mixing, the volume of each gas Is changed to V, then by Boyle's Law their pressures are Plyr- and P-^~ respectively. \

\ When these are mixed, the required pressure p of the mixture when its volume is V is given by

P° ?Y1+P-p. by § 6-10 Page 272

-tPiVt+frVJIV § 6 12. Pressure of a mixture of n gases of different volumes.

Ifanumber of gases of volumes Vi,va, va, .... v» and pressures Pi'Pa-Pa •••Pa respectively be mixed together to form a mixture, of volume V, show that the pressure of the mixture, assuming the tempe~ rature to be the^same in each case, is

' (PiV1-f/>2va+...+/>nv„)/K. {Rohiikhand 82,80) Let us suppose that before mixing, the volume of each gas is

changed to V, then by Boyle's Law their pressures are (PivJV), (P2VsIV),...,(pnv.lV).

When these .are mixed, the required pressure p of the mixtuw > by Dalton's Law (See § 6 1 0 Page 272) is given by

P y ~~y t'"'^"~y •^(PiVi+PsVa+.-.+p^jV. Hence proved.

Solved Examples on § 6 0 1 to § 6 1 2 . Ex. 1. (a). Find the height of the homogeneous atmosphere

corresponding to a barometric height of 760 m.m. of mercury, taking the sp. gr. of air as 0 0013 and that of meroury 13 596.

Solution. If h be the height of the homogeneous atmosphere, ' then the pressure on one square centimetre is O'0013/i. Hence we have

0-0013A-»^oaX 13-596

Ex. 1 (b). When the water barometer is standing at 33 dm. a / bubble at & depth of 10 dm. from the surface of water has* a volume Of 3 cc. - At what depth will its Volume be 2 cc. ? (Bhopal 83)

* ^oluiion. The pressure at a depth of 10 dm. from the surface of water is due to a column (33+10) dm of water.

- Also 2 c.o =.2/00 X10 X 10)=.2/1000 o. dm. and 3 c.c.^3/1000 c. dm. , >

I. From Boyles law we have , . , . ' ' (33+10)Xt3/1000)c=(fi+33)X(2/1000).

where h dm. is the required^ depth or ' '43x3^(7i+33)x2 or 2ft=129-66'=63 v J or 'k-='(6i)l2':>3Vl*dm. Ans.

Kx 1 *c). When the true barometric height is 30", the mercury stands at 29'8" in the barometer with a defective vacuum. What fraction of the space above the mercury would the air till if it were compressed to atmospheric pressure ?

Solution. Pressure due to air-=>30ff—29'8"«=0-2* Let air column^/" and area of cross-section of the barometer

-»fl square inches. ;." volume of air in the tnbe«=>/a cu. in.- ' -

' \ \ From Boyle's Law we have 0'?x/a=30.x.d -0J2/ _2_/ l i . '

or * * D 30 "aoo^iso or x//-1/150. ( Ans.

Ex. 1 (d). A litre «f air at 0°C and under atmospheric pressure rwefchi 12 grammes. Find the mass of the air required to produce • at !8°C a pressure of 3 atmospheres in a volume of 75 cu. cms.

Solofion. Let J^cubic centimeters be the volume of the required air at 0°C and atmospheric pressure.

Also from Note on § 6*09 Page 272 we know that if the pressure, volume and absolute tempeYalure of a given mass of a gas change from p, v, T to p', v'. T respectively, then

pv/T=p' v'lT' • ( / ...(i) s

Also from § 6'08 Page 272 we know that the absolute tempe- ' rature T corresponding to t°C is 273+? degree centigrades^ ...(ii)

Z, In this problem, from (i) and (ii) we have I x F 3x75 273 "273 + 18 (Note)

T, 3X75X273 61425 >. o r » K"=» —— ==3

o r 291 291 ,". The^required mas? of the ^ r°?91xl0Q0 ° ^ ' 5 3 3 ?m8#

Atmospheric Pressure 273 *

•Ex. 2. A barometer tube rises to a height of 33'above the external mercury surface, and contains as much air1 as would, at atmospheric pressure, occupy 1" of the tube. , Ihe tube barometric height is known to be 29*6 inches ; find the length of tube actually occupied by the air.

Solution. Let o be the area of the cross- • section.of the tube. Let x be the length of the f - f ^ t , , tube actually occupied by the air. y

Then the apparent reading"^' (33—*). w ,, 29-6 296 37 '*| True reading- — - j ^ - ^ *»| A True atmospheric pressures f \ gP,

where P is the density of the atmosphere. From Boyle's Law we have/?v=constant

or " PV<=H gp.a xV^constant • ...(i) "(Fig; 268) Now pressure at any point on the surface of the mercury in

the basin^-A- (29 6) g?=-x\' (33-*) gP+p, ' „.<ii)' where p is the pressure of the imprisoned air.

Also from Boyle's Law we get />v*=»constan|

or p. J2 •0CCTconstant ' '

_•. From (i) and (iii) we get 37 1 x 37 Q , T5g9'*12~P'l2'« 0 r ' - I T * * '

Substituting this value of p In (ii) we get -29'6 . /33-X), ,37 37 3 3 - * , 37

•+TT.. 29'6 . / 3 3 - x i ,37 37 33-a

: "72 g 9 ° (-12 J gP+lT* gP ° r VT-~12 ' 15* or 4x37*=5 (33-*)*+(4x37) or 5*a—17*—148 = 0 or Sx"~37*+20*-148=0 or *(5*-37)+4(5*-37)=0 or *=>37/5, —4. But —4 is Impossible, so *=»37/5=>7? inches. Ans.<

*JEx. 3 (a). A hollow cylinder, of height h and open at the top, is inverted and partially immersed so that a length k of it is under water. Prove that the air inside it occupies a length x given by the equation x2+x (H+k—h)=Hh, where H is the. height of water barometer. (Lucknow 75 ; Ranchi 75 / V. P. P. C. S. 78)

Solution. Atmospheric pressure=II=gp# ,1((i) When we immerse the cylinder into water, it contains air at

atmospheric pressure due to which the level of water inside the cylinder will be lower than that of outside.

AB and CD are "the level of water inside and outside the cylinder.

A Pressure at any point on the level AB

m $$$<**§§

where x is ttie length occupied by the ale inside the cylinder.

Now aa' Boyle's Law />v=»oonstant holds good, ' . ;- v *

:. Equating this constant finally and Initially we have ,

P.A.x^ll.A h, wh?re A is the area of oross-seotlpn of the cylinder

„ lift gpH.h .) /jv e 7 p«=>—«=>2l , from (i)

C

h ..

0,

= 6

-^r

(Fig. 269) Ax From .(H). the pressure at any ' , point on {he level ABt we get (gp H h)/x<=gp fi+gp (k-ti+x) -o r . Hh=-xH+x<Jc-h^x)

or ' xM-x (.ff+fc-A)-#ft=>0, where the true value of x is the . positive root of this equation. Ans.

Ex. 3 (b) rk hollow cylinder of certain height closed at theN

bottom nd open at the top is inverted and poshed down vertically in water until water rises 2 feet in it. If'water barometer be 34 feet high; find the height of the cylinder. / . (Bihar 75)

Solution. Refer Fig. 269 above. , Let' h be the height of the cylirider. When the cylinder is.

Immersed in water,, it contains air at atmospheric pressure, due to, which the level AB of water inside the cylinder will be lower than the level CD of water outside the cylinder.

-Now according to the problem, the height of AB above the top (shown the lowest in the figure, as the cylinder is inverted)-is

" 2fcfct.; - ; - . '. -Also II, the atmospheric pressure is glyeri by

%\ «.(i) , II«=gPX34 . .!. Pressure at any point oh tbe level A3

•=*U+gpx, see figure 269 above •=gpX34+gp(ft-2) (Note) =gP.(ft+32)-P (say), „._(ii)

Now as Boyle's Law (viz. /jv=»constant) holds good, therefore equating this constant finally and initially we have

P A.x<=-ll. A. h, where A is the area of the cross-section ofthe' cylinder. „ • - ' . , ,6r gp (/z+32). (ft-2)=.(gp.34).ft, from (i), (ii) ' or \ • t A2+30/j-64==>34A or. /j2.-4A~64=0 'or ' 'ft=H4±V(i6+256)]=.2>2^(17),as/jisnotlesS'than2' i

Hence ft~2 [1+^(17)] ft '. . \ Ans.. •* Ex. 3.(c). A hollow cylinder ofSleight b opfn at top, is inver­ted, and partly immersed in waftr with its length a in the air. Prove

Atmospheric Pressure lH

that the difference x between the water levels otstside 8nd inside the Cflifflder is the positive root of x2+(a+/r) x~{b—a)n—Qt h being the height of water barometer,

Solution. Since h is the . height of the water barometer, therefore, atmospheric pressure

>=>g?li=Il •• ( i )

When the cylinder is Immersed in water, it contains air at atmos­pheric pressure due to which the/ level of water inside the cylinder will be lower than that of outside.

AB and CD are the levels of water inside and outside the cylinder.

A Pressure at any point on the level AB=>g?!i-b-gpx<~P(sa.y), ...(ii) where x is the difference between water the cylmdsr

Now as Boyle's Law (viz. pv=>constant) holds good, therefore, equating this constant finally and initially we have

P.A. (a+x)'='ll.A.b, where A Is the area of cross-section of the cylinder.

_ lib gphb . ...

or g?h +g9X=gphbl(a+x), from (ii) or (h+x) (a+x)=*kb or xt+la+h) x—{b-a) ft=0. Hence proved.

Ex.4. A thin closed cylindrical ve«sel, of height a, contains air at atmospheric pressure and floats in water with axis vertjjgal aad^ length b immersed. -If a smaH bole is made in the bottom of the vessel, show that water leaks in until there is a depth ab/(h f-b) inside, h being the height of the water barometer.

' (Fig. 270) levels outside vand inside

Solution.*• Wnen a and due to the weight of the water on the base of the vessel, the vessel will go down.

Lety.be the depth of the base frorn the * free surface in this posi­tion and let x be the height of water in the

small hole is made, the water will rush in

<Fsg,27l)

M , liycfrostalicl

The pressure at any point on the surface of water in the cyliA-dei=hgP+(.y~x)gP='(,li-ty-x)g?. #

The volume of the air in the second position will be (a —x) A, where .4 Is the area of the cross-'section of the cylinder. ,

Then from Boyle's Law (pv=> constant), equating the value of 'pv' initially and finally we have

A a.gPk=A (a—x) (h-\-y-x) gP •or ' (h-\-y-x)(a-x)<=-ah .,.(i)

Also we know that weight of a body=weight of the fluid displaced. And weight of the gas in the vessel will remain the same in both the cases.

J. Weight ot the gas=%=Q> -x) g$ (Glvenrfhat the weight of the vessel is negligible)

or y-x<=>b £ From (1) we get (.h+b) (a~x)<=>ah or " * (h$b)=ab

or x=>abl'Ji i-b). Hence proved. **Ex. 5. A closed air tight cylinder, of height * \

2a, is half full of water and half full of air at atmos- " " ~—~ pheric pressure, which is equal to that of a column, of height h, of the water Water is introduced without letting toe air escape so as to fill an additional height k of the cylinder, and the .pressure-of the base is thereby doubled: Prove that k=>a4rh—y(ah -j-ha). " <?•

Solution. Given II«="gp/j, whe.re P is the density \ of water. - >

Let p' be the pressure of the imprisioned air (at (Fig. 272) any point) when compressed.

Then equating the value of pv ^constant, initially and finally we have Il.Aa=>p'A {a—k), where A is the area of cross-section, or gph a=p' (a—k), V UpgpA (given)

rfr • p'ragPhl(a-k) * : ...(1) Originally the pressure at any point of the "base

=II+gPa=>gPh+gpa' ^ x " , Finally, the pressure at any point on the*' base '

•=/+gP ia+k)~[ag9hHfl-k)]-\-gP (a+/fc), from (i) ,

A By hypothesis, 2 (gPA+gPa)- -^^+gp(o+*) (Note)

or 2 (fe-ho) {a-k)^ah+0+k) (a-k) or fc2-2fc (fl+A)+CaA+o*)=0 , or &=»(a+/0± <f (h2 \-ah), solving the quadratio equation. or k=(fl+h)—^(W+ah.), other value being impossible as k<at

_ Hence proved,

Ex. 6.' *At a depth of 10 ft. in a pond the volunae of an stir babble Is O'QQOl of a cubic idch ; tfud approximately what it will be

Atmospheric Pressure • 2'7"$

when It reaches the surface, if the height of the barome/er is 30' and the specific gravity of mercury is 13 5.

Solution. Given the atrnosphejic pressure II«=»gl3"5P — •=-—-gP, 12. 4

where P is the density of the water. *"* The pressure at a depth of 10 ft. below the surface of water is "

due to a column of water=II+gP.i0=«l'Llil + 10 gP^^l*- gp.

If V be the required volume of the water bubble, then from ., , , . 01J5 V 175 0-0001 " Boyle's Law we get — « P . 1 2 x l i x l 2 - — * P . 1 2 X l 2 x , U o r F a0-0001 X175 m f J . C 0 0 i 3 c u b i c i n c h n e a r l Am

1 133" **Ex. 7. A barometer stands at 30 inches, and the space occu­

pied by the torricellian vaccum is then 2 inches ; if now a bubble of air which would at atmospheric pressure occupy half an inch of the tube be introduced into it, show that the surface of toe mercury in the tube will be lowered 2 inches. Show also that the height of a correct barometer, when the incorrect one stands at x inches, is x+{15/(32—x)} inches. -

Solution. Atmosphericpressure"=ffgP, where pis the density of mercury.v

When bubble is introduced, it will go up, the pressure of air would increase and the level of mercury wi.il go down.

£. From the Boyle's Law (i.e. i?p=>cons5ant) we have i8*PtiX&.«)=constiintt - , _x ' - . (1 )

where a is the is the area of cross-section of the tube. Let z be the height occupied by air in the tube whose total -

height is 30+2~32» Then the height occupied by mercury=(32—z) inches; Let p

be pressure of the air. . # , Then from Boyle's Law we have p\ •£$z os=cons(ant ~ ^ f i ^ .'. From (i) and (ii), we ge$ , z 30 1 , 30£P 5 „

Atmospheric pressure (on the surface of mercury in the basin) =?fgp—h (32-z) g?+p' (Note) . 30 „ J 2 - Z . . 5 n . ......

or i2gp==~T2~ 8 " l ~ 4 F f f P , f r o i n ( m ^ or 30~(32-z)+(15/z) or z a - 2 z - 1 5 ~ 0 or, z « 5 . - 3 or z=>J" (since —3 is impossible)

Hence the level* of mercury is lowered by (5—2) or 3". Part II.

Let//" be the pressure of the impmioned air. - • Then from Boyle's Law (/^constant), WJ haVo

Hydrostatic!

bf •J, 3QgP P '24(32--*) N ...(iv)

Let the true reading be (y+h) Inches. Then equating the atmosphgric pressurs

S>n the surface of mercury in the basin in the two cases, we get

from (iv)

<3»

32-X :=5

x

12 s r 2 4 (32-x) '

* 3T

15

Or y-r-»=x+[15/(32-x)]

; . Required true reading=ij;+/i=x+-22Z^"

**Ex. 8.* The height of a Torricellian vacuum in a barometer is a inches and the instrument indicates a pressure b inches of mer­cury when the true reading is c incises. The

- faulty readings are doe to an imperfect'vaccum. Prove that the troe reading corresponding to an apparent reading of d inches is

d+{a(c-b)/(a4b-d)}. Solution. True reading=c" A True atmospheric pressure

=»& <&P Lei pressure of the imprisioned air be Pi

then A-^P=P4-4„-ftPo aS 'u

(Fig. 273)

. Hence proved.

9/

IT

(Fig. 274) W«P-P+ftfttf> m •••<0>

Let a be the area of the cross-section of the tube.

J5, Volume of the-Torricellian vaccum<= -&a*. '

In the.second case, apparent reading=dff and let Jrue reading ^5& (x+d) inches. ^ .'

A The volume of. the Torricellian ' j vacuum^}1,- (a+b—d)x and let P' -be the «,+£-< pressure of imprisioned air in this case.

Then from Boyle's Law (/>v=constant) we have

-f P.-A-floc^P'.i^ta-ffi-^oc ...(ii) / ; (Note)

Also equating atmospheric pressure on the surface of the mercury in tne basin we get

lf t (x+<^-A«Pfr- * ...(Hi) From (i) and (ii) we get

\xscg^\1tbgP+P'{{ai.b-d)la} „.(iv) (Fig, 275)

d* \

Atmospheric Pressure 281

Prom (iii) we get P'<=>-?$xgp .'. From (iv) and (v) we get , ^ ,.,(y) -A cgP~?i bgP+& xg? {(a+b-d)/a}

or ca=ba+x (a+b—d) or x*=a(c-b)l(a+b-d) •

/ . True readifag="X+^=rf+{a (c—b)/(a+b-d)} ~ Hence proved:

Ex. 9. A hollow closed conical vessel of height h floats par­tially immersed in water with vertex downwards and axis vertical. A hole is then made very near the vertex and water allowed to come' into the vessel so that no air escapes from within. If the vertex was originally at a depth b and H is the height of the water baro­meter, prove that the new depth c of the vertex is given by

b 8 = J c _ H ( c « - b 8 ) ] 8

D I h3~(c8-b3jj '

Solution. Let a be the semi vertical angle of the cone and W its weight.

Also as the welght,of a body=weight of water displaced by It Theref re W-=- £TC&8 tan8 a gP, ..(j)

where p is the density ofjhe water. Now when a hoJe is made In the vessel, let x be be the depth of

water inside the vessel, (

/ . weight of the wafer inside=%nx3 tan2 a.gP And^in this case, weight of the cone+weight of water inside

»weight of the water displaced. i.e. Inb* tan8 a gp-f | T « 3 tan* oc.gp^Jnc3 tan2 cc.gP from (i)

or ' ' » 8~c 3 -4 3 ^ U ) Again the air originally occupying the/whele volume of the

cone at the atmospheric pressure^ gp H now occupies a volume \K (A8-*8) tan8 a under a pressure gP (H+c-xj which is the pre­ssure at the level of water inside the vessel.

A By Boyle's Law. we get JTTA3 tan2« gW-^fa (h3-x3) tan8 « {H-\ c-x) gP

or pH~(P-xs) (H+c-x) • x*H H (c*~b*) .

or ^ « - « ~ ffZ?-«- h»-{c*-b°) f r o m <lfa)

Hence from (ii) we get c8-A8- * 3 - { c - MS^^l V

Hence proved. *Ex. 10 (*). A thin conical surface of weight W just sinks to

the surface of a fluid, when'immersed with its open end downwards but when immersed with its vertex downwards, a weight equal to mW must be placed within It to make it sink to the same depth as before. If i i be the length, of the axis, h the Height of the baro-

1*1 - Hydrostatics

metric coiuma o | the flaid, show that H=hm (t + m)V3. •-, ' , {bucknow 80, 79, 76 ; Rohiikhand 80)

Sohttion. Let H be the height of the . cone and « its semi-vertical angle. • , "** Originally the cone is full of air at

atmospheric pressure«II*=>gph, where p is ihe density of the fluid.

-.Let * be' the depth of the surface of 'the liquid inside the cone below the free • surface. Let P be the pressure at any point of the imprisioned air, thenv

. P=>gph-\-gpx , ' ' ...(i) Frorh Boyle's Law G?v=.const^ht)

on equatingjhe initial and final values of */>v' we have

(Fig. 276}

gp/ifaH3 tan2 «=JP.|5rx3tan2 a

•00 . o r gph.H*=T.xs-=(gph+gPx) xa, from (i)

or. ' fl3A~*"A+**,' which gives the value of x. .

Also for equilibrium, • ' . weightof the cone-f weight of imprisioned air.. •'•

" ' i'r ; .=» weight of liquid displaced. or W^\nx3tan2 «.$P. ' (Note) - (iJI>

xwhere Wis the weight of the cone and the weight .of air inside the cone is negltgible.in comparison to the weight of the cone..

In the /case when the >' vertex is downwards, the forces acting OH the cone are (i) total weight force (i.e. W+mW-\-weight of air Inside the cone) acting vertically downwards, and (ii) the fluid thrust (which is equal to the weightof the liquid displaced) acting vertically .

jjjj^ards. • (

_ A For equilibriuum, we have (Fig. 277) ' y + m ^ l ^ t a ^ a g p / n e g l e c t i n g the weight of the air (l+m) fax3 tan2 «.gP =IJCH* tan*« gP, from (iii) .

H

t j — - • —

/ ! \ :

Of

or *s=*#s/(l+»0 or •x=*Bf(l+m)ita Substituting/this value of x in (ii) we have -

Jtlh rJP_ --. % •'•- ". <l+m)" r( l+m) ,( l + »0V".

~<iv)

IPh*

Of

or

•H or

H V+m) ' (l+m)W m (l+m^h [X~ (db)l

H-mHl+mW.- ' . Hence proved! - a • u 1 0 i ( b i A s s a m i n g t h e h e i 8 h t o f w a f e r barometer to be H find to what deptn-a small inverted conical glass must be lowered so'

• <Mwm 751 mnctti 79).

Ex.

'that the water may rise half up it.

Atmospheric Pressure 2t%

-*-* - 1

t / •^h /

iA~

—

T — « -• 1

"~*-fc-.

dotation. Let A be the height of the cone and a its semi- • Vertical angle.

Initially the cone is full of air at atmos­pheric pressure =>g?Ht where p iS the density of water.

Let the vertex of the cone be at a depth x below the free surface,' when the water rises half up the cone. Let P be the pressure at any point of the imprisioned air in this condition, then P*=gpH+gP (*+P) ...(i)

From Boyle's Law (/>v=const) on equating the initial and final, values of pv we have ' - -'

gpH (JTTA8 tana a)«=.P.$rc (ihf tan2 a (Note) # (Fig 278) =-a\Tr/j3 [gP (H t-je+JAJjW a

or 8#=- H+x+lh or x-=*!H-\h. Ans„ *Ex. I i . If the volume of a certain quantity of air at rempe~

rature of 10°C, be 300 cu. cm., what wili be its volume when the temperature is 23°C ?

Solution. We have by Charle's Law p*=>kp(l+<rt) or pv=*k vP (1-foef)

or pv=*km (ITJ-OGO, when m is the mass of the air A Here we have p.30Q=km (1+«.10) „.(!) and p.V=km (l+«.20), N .. (ii)

where V is the required volume. Dividing (ii) by (i) we get J ^ ^ T O w h e r e «~l/273 where a=

=•300 (273>20) 300 l«flO«x*

or V <1+ 10a)~300 (l+20a) or V (273+10)= or 7=(30O x 293)/283=>310 f f §• cubic cms. Ans.

*Ex. 12. If p l t Pi, tx ; p2t pa, t 2 ; p3, Pg, ts be the corresponding values of the pressure, density and temperature of the same gas,

(Magadh 76 ; Ranch! 80, 79) Solution. From Charle's Law "p=kp (1+otf)" we have

Pi^k?! (l+atx) I.e. pJPt^k+akti Similarly pJV^k+akh •••<") and/»8/pa« Subtracting (iii) from (ii) we get

Pi

k+aJcta

»<x& ( f j , - / , )

Similarly f5- -^-=«* (/,-/,)

and

$et

PI PI

...CI) •(ill)

.(iv)

...(v)

(vi)

--«k (tt-tt)

Multiplying (Iv) by tu (v) by ta, and (vi) by /s and adding we

11U pJ+,*U h)+r ,U pJ

\ -

* <=«A*[0]=0. ' , Hence proved. *Ex. 13. A bubble of, air having a volume of one cubic inch

N at a pressure of 30" of mercury escapes up a barometer tube whose crgss-section is one square inch and whose vacuum is 1" long. How much will the mercury descend ? - (Lucknow 77)

Solution. Let the mercury depend x inches. Then the pressure of the air wh'ch occupies a volume of (1+x) cubic inches of mercury is that due to x inches of mercury. '

** •"• By Boyle's Law we have lX30«=(l+*).x or x a ix—30=0

or (x+6) (x—5)=>0 or *«=5 inches, neglecting the negative value. ADS.

*Ex. 14. Masses m, m' of two gases in which the ratio of the pressure to the flessity are respectively k and k' are mixed at the same temperature. Prdve that the ratio of the pressure to the density

J n the compound is (mk+m'k')/(m+m'). ' (Muzoffarpur 81 >; Ranchi 82, 76)

Solution. Let v v'; P, P' and p, p' be the volumes, densities and pressures of masses m, iri respectively.

Vhen v=m/P, v'^m'jp'* Also p=A:Pandj/='ftrp' -Let P and o be the pressure and the density respectively of the

compound. v

T h e n ' P = ^ ^ — , vhere Fls the volume of the compound

(See §611 Page 273)

. ' {(m+iw')M ' \ i n ; v or' - Pjo^(mk^m'k')/(m^m^) Hence proved.

**Ex. 15. Prov^that If volumes vx and v2 of atmospheric air aj^^rced into* vessels of volumes \ 1 and V2 and a communication is opened between them, a mass of air of volume (ViVa-iVs^/^+V.) at atmospheric pressure will pass from one vessel to the other.

Solution. Let a mass of air of volume Fpass from vessel of volume V2 to the other of volume Vlt at atmospheric pressure II.

Then in the vessel of volume Vit there are two gases one of - volume V-i at pressure p r (say) and the other of volume V at pressure

II. ,*. the pressure of the mixture in the vessel of volume Vx

^yji-vn ( B y D a I t 0 Q, s L a w j

And in the other vessel of volume V2, a mass of gas of volume V at' pressure II is taken out from the gas of volume F» at pressure

Atmospheric Prsssu** 28S

The pressure of the gass left In the vessel of volume Pa

V9 , / (Note) Now the pressure of gases In the two vessels will be equjj, in

the position of equilibrium, so we have

(0 vt v2 Also from Boyle's Law (i.e. j?v=constant), we have

vJl^Vrfi and Va'I^J^Ai A Pi^ViliiVi and ^a^Vall/Fa .

/ . From (i) we have F , n [ ^ ' ] - ^ - ^

or V ( 7 X + V ^ v ^ - v ^ or V^V^V^V^ VJ i Hence proved.

Ex. 16. A cylinder contains two gases which are separated from each other by a movable piston. The gases are both at 0°C and" the volume of one ga« is double that of the other. If the temperature of the first be raised -t°, prove that the piston will move through space 2/at/(9-f 6at>, where / is the length of the cylinder, and a is the coefficient of expansion per 1°C. /

Solution. Let p and P' be the densities of the lower gas in the initial and final positions.

Then mass of the gas=P &l) 4«=»p' (f/ + x) A, where x is the distance through which the piston moves when the

1 2b UJ

(Fig. 279) • temperature of the lower gas Is raised t° or p'=2/P/(2/+3x) . ...(1)

Let/» and/?' be the pressures of air in'the lower portion in the initial and final positions. ,

Then we have p=k9, from Boyle's Law and

or

p'=kp' (l+at), from Charle's Law p kP ' p (2/+3x) ,. n ,...

7 = ^TTT«T) = 2ipTT?^y f r o n i ( l )

p_ 2l+3x p'^llil-^t) • m

g|6 Hydrostatic*

In the final and Initial positions, the piston Is in equilibrium. • A. The pressure upwards oh the piston Is equal to pressure

downwards on it. . ' " • Hence the initial and final pressures of the upper gas also will

bg^and/j'. ,,. i. - Now from />v=>constant, equating the initial and final values^

of 'pv' for the upper.gas we have p.^LA^p' (£/—x) A p V-x l^-3x ' " ' .

' * * ' - .: P'. " 1 / / ; -( i i i ) . , Equating thei values of p/p' from (ii) and (Hi) we have

or x (6a?7)-9)==2/af or »=»2/ar/(9+'far). Hence proved. 1 . **Ex. 17 fa) Two volumes vx, v2 of different gases at pressures Pi, Pi and absolute temperatures 7\, T* are mixed together so that the

. volume of the mixture is Fand the absolute temperature T. Find the pressure of mixture. ' ,{Lucknow74;Mogadh77;Ronchi75)

.. Solution: Let the volumes and absolute temperatures of both the gases be changed to V and T, then if their corresponding'pres-sures are p' and p", we have for the first gas . ,

- * ' • PJLirjT • " • - ' ' " - Tj. T . (See Note on § 6-09 Page 272)

Similarly, ifor the second gas, we have p"<=* - ^ ? >m

Now when the temperature remains T, letthe gases be mixed in snch a way that the volume of the mixture remains V, then the pressure of the mixture .

«*sum of the pressures of both the gases

' ~ / + ^ ^ + | ^ , from (I) and (H)

w, T [Pi*- +Pivt 1 V \, Ti Tz J Hence" proved.

*Ex. 17 (b) If volumes v,, vai , vn of gases at pressures Pi> Pas •• -i Pn and absolute temperatures t1( ta, ..,, t„ aie mixed with­out chemical change and V, P, T denote the volume, pressure and absolute temperature of the mixture, then prove thai

Solution. Changel the volumes and absolute temperatures of all the-gases to V and T', then their corresponding pressures ase.

AtttMpltato Vnum ' &1

f ViPi W VtP% & VnP*

Now when the temperature remains T, let the gases be mixed In a ves'el of volume V. Then we have

Pt= Sum of the pressures of all the gases ^»

L h h t„ J" Hence proved. **Ex. 18. A barometer with an imperfect vacuum stands at

29*8 and 29'4 inches when a correct barometer indicates 30*4 and* 29 8 inches respectively. When the faulty barometer stands at 29", what will be thcreading of the correct barometer ?

Solution. Let the length of the tube be /*. Then by Boyle's Law we h3.veplyi=pavt

or </-29-8) (30'4 -29-8)=.(/-29'4J "(29'8-29"4). or (Z-29'8) (0-6)~>(7 -29'4) (0;4) or / (0"6-0-4)=(29 8) (0-6)-<29 4) <0'4) ^

"or /(0-2>=.17-8S-ll'V6«612 n . 6-12 612 v 10 306 ; 01 '-or-ioo^"^-306 . ...(o

Again if H be the required reading of the correct barometer then we have (/-29'8) (30'4-29-8)-=(/-29) ( # -29 ) or (30-6-29 8) (0'6)=(30 6-29-0) (H 29), from (1) or 0-8xO'6«(l'6)(H-29) or 0"48=(l-6) (ff-29>

„ 0Q 0-48 48 10 3 _., '

or #=»29-f-0-3=>29-3 inches. Ans. * Exercises on § 6 01—§ 6 12

Ex. 1. Given the sp. gr. of air at the earth's surface at a given place is 0-0013 and of mercury is 13596, when the barometer is at 30", find height of the homogeneous atmosphere. .

L • Ans. 5 miles nearly" Ex. 2. A piston without weight fits iato a vertical cylinder

closed at its base filled with air, is initially at the top of the cylinder. If water be slowly poured on the top of the piston, show that the upper surface of the water will be lowest when the depth of the water Is [<</(ah)~ h], where h is the height of the water barometer and a is the height of the cylinder. (JLucknow 82)

Ex. 3. A barometer whose cross-sectional area is one sq. cm. has a liUle air in the space above the mercury. If it is found to read 71 8 cms, when the true height is V7-l cms, determine-the , volume of the air present, in the -tube measured under normal conditions. * Ans. * 0*0686 of the original volume.

Ex. 4. A barometer wh:ch has a little air in it reads 29 "6 inches, the encl of the tube being 6" above, the top ofthe^mercury^

m Hydrostatics 1&/I*

when the true, pressore, of the atmosphere Is 30*. What Is the reading of the barometer'when the true pressure is 29" ? {Mithila 81)

**§ 6'S3. Theorem. If the atmosphere be at rest snd if the force due to gravity and Its temperature be constant, then if points •te taken whose heights above the earth are in A. P., the common

'difference being small, the densities at these points are in G.P. Proof :' I et there be a column of air of small hori­

zontal section. Let O, Alt A2 A3, . . A„,... be a series of points, as shewn in the adjoining figure 280 such that OAi'=*A1A2<=A2Ai*='...='A„-iA„*=h (say), where h is small.

I.e. Alt A2, AB...ate a series of points in A.P. If n be very large i e. hbe very small, then the den­

sity of each layer OAls A^A$,...is practically constant throughout. • y -

Let the densities of the layers OAx, AiA%, A2As,...be Pi. " ?2» Pa»-

pressures are the pressures at the points

respectively. Their corresponding i

which ruay be taken as O, Ax, Ag,„.s

The difference of pressures on the faces at O and A, is equal to the weight of aiT contained in the layer OAv

Hence k?i~ fcpa=gp1A Similary, for the layers A^, A^A^ we have

Apa—fcPa gPa h, k9a-k%i~g?s h,

k9n-l~ kP^gPn-! h :. From.(i) we get p2=Pj (l-(ghlk)] Similarly from (ii), Pa*3 Pa [» ~ (ghlk)) >

•=Pi[ l - (^ /W.from(i i i ) In a similar way, we get

'A* A»-f

<?

A, o

(Fig. 280)

«.<ii)

-.(HI)

p4-=ps [\-{ghim=h ii-(ghik)]a

n-a P.-P,-! [l-<ghlk)]~9i tt-(ghlk)]*-Hence as the altitudes increase in A.P., the densites, consequ­

ently the corresponding pressures decrease in G.P. Hence»proved. **§6*M. Density of air at a given altitude.

(Asm 83 ; Lucknow 83, 81,79; Ranchi 82 ; V. P. P. C. S. 79) If in the last article p be the density just above A„, then

P - P. [1 - (gh(k)]•= Px [1 -(gA/fc)]" ^ Let nh=H. Then p is the density at a height H .above the

ground and we have

•*(•-• i r = 4 i -vZ

gHzlh

<b-m where z«

-gHIk

•hklsH

136/W Atmospheric Pressnri

,\ when n + eo, H remaining constant, z-+oo,

1 1 -+e

A, we get p=P1e_»H'*, which gives, a relation between the demlty

at, a height H and that at the surface of the earth. Alitcr i Let p be the pressure and'P the

density at a height z. Then />*= ftp , —(0 Let/>-f 8/> be the pressure at a height z+Sz. Consider the equilibrium of the element $z

of the thin column. The forces acting on it are (as shown in the Fig. 28!) the pressures p and p+Bp acting in the dircections as shown in the figure and the weight gP$z of this elementary column * acting vertically downwards.

Resolving these.forces vertically, we have ~ . x p~(p+Sp)+gP$z (Note)

A In the limit, dp/dz^—gp i Also differentiating (i) we get dpfdz=k.dp/dz,

— gP^k dP/dz, from (ii) dP g

>**/>

'm

ip

(Fig. 281) .(ii)

or

or ^ ~ - 4 < f e . P A; Integrating, log p== - (glk) z+ C, where C is the costant of inte­

gration. On the surfaee of the earth z-=>0 and p^Pj C«=Iog P] °

Hence log p=.-(^/fc)z+l'og Pi or log (P/P1)'=—(glk) z. v

or P<= p^"*/* (#• P. i\ C. 5. 7P) ' Cor. If/? and Pi be the pressures corresponding to the densi-

-gz/k ties P and P^ then p^pj e Ranchi 78, 76 / Magadh 76;

; J W J 82)' the same

/» the

(Bhopal 81 M'thila82 ; M*zxofarpur 88

§6*15. The height of the homogeneous atmosphere is at all places at the same temperature.

Let h be the height of the homogeneous atmosphere, pressure and p the density of the air at that point.

Then/?=£P/j or /*•=-£-=>-?, •/ n=>&p gp gP y

or /j=fc.£=constant. .. ' Hence proved. **§ 6.16. To determine heights by Barometer.

(Lucknow 74 ; Magadh 75 ; Mithila 80 ; Pxitna 82 ; Ranchi 74 ; Rohilkhand 80)

Case I. When the temperature ramains constant.' , ^ Let * be the difference.in the altitudes ot two stations and let

h and h' be the heights of the barometer at the upper and lower stations respectively. Also let/?, p' be the pressures and p, p' be the densities at these stations. > v Then if (he temperature remains constant throughout, we h»v e

' h p p r - * - - (See § 6'14 Page 2SS)

jghicb gives the height of one station above the other by means of barometric observations at ?hese stations.

Case II. When the temperature is not constat^. We- know by Charie's Law, the relation betv/een p and P is

j?=fcp(l+<Xf) Also (see cor. § 6*14 Page 288) we know that if p a n d / be the

iressures corresponding to the densities P and p' then p/p'^e''"'11

or log (Plp')~-(gxlk), ...(i) where x is the difference in the altitudes of the two stations and g is taken to be^constant. '' " <

Now if we suppose that tB is the temperature at the earth's surface and ^ at a height z, taea the temperature of the atmosphere is their mean i e. t =ȣ Ua+h)- H ki corresponds to this mean ^ejmperature, xhen we have

ft1=fc[l+*.J('o+01 '•••(") (Note)' Hence from (i) we have

log (p/po)0—(gzlki),' where p0 is the pressure on the surface of the earth. >

or z<*>—- log ( — J, where kt is given by (ii).

Solved Examples on § 6*13 to § 6*16. **Ex 1. A box is filled with a heavy gas at aniform tempera-

tee. Prove that if a-is the altitude of the iiighest point above 'the lowest anil p and p' are the pressures at these points, the ratio of the pressareto the density at any point Is equal to ag/log (p'/p).

(Osmania 69 ; Lucknow 80, 76 ; Ranchi 78 ; U. P. P. C. S. 80) Solutiou. We knowp=>/Jie-"*/* (See'§ 6*14 cor. Page ^89) «*., He,re p&p'e~Ba'i> _ ,

"ST #»l*=p'lp or ^ . - j o g ^ j '" , -CD Also p=»fcp or fc=>jp/p

& from (i), — caioj; {£- 1 or —«=•,—°f ,, -r- Hence proved. P \P f P logjp/p)

*Ex. 2. Assuming that a change from 30" to IT in the height of the barometer corre>p^nds to a rise in the altitude of 2700 ft. ," fiBd the altitude corresponding to the height 21 "-87" of the barometer.

(.Mog dh 74 ; Ranchi 81,74 ; Biter 76, 74) Solution. We know p pt e~""ft

Let h be the height of the first station and pi the pressure at the paiut from which tne height is measured

" Tflen fr»m (i) at the two stations we have iO=i7ie-

B*/* ...(ii) and 27~/>Ie-» (*«><»+*>/» ...<iii> ,* *

Atmospheric Pressure £9\ *> f

Also 21 •87=»/?1 «-».(»+»»>/», „.(ivy where V is the required altitude.

Dividing <ii) by (iil) we get §ft—e»«>w» ,..(vj Dividing (li) by (iv) we get ^L=e0W*

•" *»/ ~ ( v l ) ^ From(v), 2700s/fc~log (30/27)~log (10/3a)=log 10-2 log 3

= 1 - 2 log 3 _(vi) From (vl), gA7&~log(30/2*.87)=>log (10/7-29)nIog(1000/729)

~log (10736)=3 log 10 -6 log 3 = 3 - 6 log 3 ...<vlii)

Dividing (viii) by <vil) wc get ^ " ^ l o l f " 3

or A'~8100 feet. Ans. *Ex. 3. A gaseous atmosphere in equilibrium 4a such that

p=kPv=«RpT, where p, P, T are pressure, density1 and temperature at a height z, and k, 7, R are constants. Prove that

*E -JL y~1

dz = R' 7 ' U.A.S. 75) Solution Given p<=>kPv*=>RPT „.(i) From p-=kpv we have P~(/>//e)I'y ...(ij)

Also from § 6*14 result (ii) Pa ge 289 we ha.ve-~±=>—g(>

01 %-°~ S<W*),/v.from(ii) 0 r pw ~Waz ...(iii) s

Also from (i) we get p^RpT or P=R wltyc.T, from (ii)

- '""";- £ r • (Not.) Differentiating both sides with respect to z, we get •

V y)p dz W dz \y J/»V JfcW d i

o r ^ - ( 3 ) Jza^>^,from(lii)

dTa_g_ 7-1 dz , i?' 7 " Hence proved. Ex. 4 A hollow gas-tight sphere containing hydrogen requires

a force mg to prevent it from rising when the lowest point touches the ground ; the total mass of sphere and hydrogen is M. Showlhat the sphere can float in equilibrium with its lowest point at a height h

above the ground, where h= - log I—«*•-)» and k is the ratio of

the pressure of the atmosphere to its density. (Rohitkfumd79)

Z%X "• Hy«roa?&tl6s 484- 186 • • S c *

, ! Solution. Given pulling force«=rog and total welght«=»Af£. .'. Actual pulling foice^mg+Mg. Let V be the volume and P0 the density at surface. •"' .*. Total air displaced=«^p0g On the surface of the earth equating the forces acting upwards

to the forces acting dowDwards, we have mg+Mg<= Fp„g or i»+Af= V p„ ...(i)

• At height h, the forces acting are weight and surrounding thrust, so we have Mg^ V(g, where P is the density cf air at height h. or , M^VP ...(5i)

Dividing (i) by (11) we g c t ^ ± - - ^ - _ (H1)

„- Also WJB are givenlt**p\Q 6r />«=>£? ...(iv) / e. tempe-ature remains constant. .'. On the surface of the earth/>0==frp0 ...(v) Also we know /"=jV02 ' s

« Here 7>«=./v>-<w* or p0/p=em^ or Po/p=eBWfc. from (iv,s and (v) " A From (Hi) we h a v e ^ = ^ / * or ^ ~ l o g ( 5 ± ^ )

or A=-Io lm*M\ g ' M ' Hence proved.

Ex 5 A hollow gas-tight balloon, containing helium, weighs W units When its lowest point tourh'es the ground, it requires a force of w us its to prevent it from rising- Show that it can Scat in equilibrium at a height H log 11+ =r ] , where H is a constant and

the atmosphere is assumed isothermal. ('• A-£- 72) i Solution. Proceeding exactly as in Ex. 4 above we can find

that the required height h is given by / > . . k . lmA-M\ k. / - , m \

-fflog(l+g) |Whiwff-J

•=>H log 114- SF )> "•" u»=mg and W<=*Mg (given).

Hence proved. **Ex 6 Taking >nto account the variations of gra'vity with

height and aborning that the temperature of the air is constant at all heights, prove that at a height x the prsssure p of the air given by

n ff s x * log —=— . , " y where a is the earth's radius, k=>p0/P0 aod Pn- Po> So a r e the values of pjessure density aad gravity at the earth's surface (Agra 75 ; Lucknov> 83)

Solution. From Dyn amies* we knpw that acceleration due to gravity at a distance r from the centre of the earth is given by jt/ra.

Atmospherio Pressure j ^

or

A On the surface of the earth, go"3"-? or fi<=fla^

.*. gfit a height x from the surface of the earth is given by

ga (a+x,* (a-txf „.(i) (Note),

Also we know j-=-- gp, See § 6'14 (ii) Page 289

<=—g(plk) ... V />=/cP 'dp gp dp g ,

rfx k • p k

Integrating, log/>~ ^ °^+c ,

where c is constant of integration. Initially x=0, ^ P o

or' - l 0 g , - l 0 g ^ ^ [ ^ - l ] = - ^ | _ or log (p ]p0J •=» —goaxlk (a +x). Hence proved.

**£x. 7. Tlia height of a balloon is calculated from the baro­metric pressure rcadiag (p) on the assumption that, the pressure of this air varies as toe density. Show that if the .pressure actually varies as the'nth power of the density, there will be an error

in the calculated height, where h„ is the height of homogeneous atmosphere. • • *

Solution- If the pressure varies as deasijy, then p=>kP ,..(i) .\ From (l) oa the earth, p0=>kP0 , ...(ii) Actually pressure varies asihe »ih power of the density

i.e. p*=>k\9», ...(jii) whence on the surface of tne earth, we get p0 =>k'PQ

n _.(iv) Let P be the press are at a height z above the surface of the

earth, then from «V/?/rfz=>~gP" we get

£=-*(£)'", from cm, ^ 0T phn {k'fl"

2 ^ Hydrostatic^

On the surface of the earth p-=pQ and z>=0

» b H - 1 / ; ^ (/e'),/n + n - l ^ Aceording,tothis law suppose z comes out to be zt

Then — «<-«/ •„ _ - I _ *,+ liPoHnOl

Also from (iv) />0~&'V=£P<A)i/Where ft0 is the,height of homogeneous atmosphere or Po=Polgho and Pli

n=pjk'

Substituting this value of &' in (v); we got

(.Note) n - 1 £/io n - 1

or *.-£&»r [B-ri {& tn-1)/n' <ri«-**}]

D ik_r 1 _/^ \ ( n l , / n l («-i)L ' A / J »(vi)

Also from i'p=p0e~z"1" we have p=poe~"l"l0 (Note)

or ZiWfo log (/;0/p) _ ...(vii) J". The required error=za—zx

' -*°[«"=11*" (^) (nl)'n}~108^]' fr°m (Vl> and C v i ° ' Hence proved.

Exercises on § 6 13 to § 6*16 K Ex 1. If a change from 30" to 27* in the barometric height

correspoilds to a nse in the altitude of 2290 ft, find the altitude 'which corresponds to the barometric height of,24ff. (log 2=»0'J0J0 J log 3«0'477i)."

Ex 2. A volume of air of aay magnitude, free from the action of force, and of variable temperature, is at rest 3 ifth^s temperature , at a series of points be in A. P., prove that the densities at these points are in H. P. (RohilkhandSO) § 6 17. Thermal Capacity and Specific Heat. Thermal capacity.

-, Definition The thermal capacity of a body is defined as the quantity of heat required to rise its temperature by one degtee. Specific Heat.

Definition 1. The specific heat of a gas is defined as the quantity of heat required to laise the tempei attire of a unit mass of the gas by one degree.

Atmospheric Pressure; S>$

Defiaition 2. The specific heat of a body i§ defined as the thermal capacity of a unit mass of the body.

DefThition 3. < The specific heat of a body JS defined as the ratio of the amount of heat required to increase the temperature of-

the body by one degree to the amount of heat required to increase* the temperature of an equal weight of water by one degree. _.

If an amount dQ of heat raises the temperature of a- unit mass

of gas by a small amount dt, then specific heat=-p .

Also the temperature of gas can be altered either by ke?plng volume constant or by keeping pressure constant. In the former case (/ e. when volume is kept constant) tiie specific heat is denoted by Co and in the latter (/ e. when pressure is kept constant) the specific heat is denoted by cp. %

It is evident that cv > c„, because when the pressure is kept constant the heat is required not only to raise the temperature but also to expand the gas. *§ 6 18. Internal energy of a gas.

The amount of energy contained by a given mass of gas de­pends upon the configuration and motion of its molecuies. If is observed that the difference in the amounts of energy in two given states of the same mass is due to these states and not due to the mode of passage from one to the other. If E be the difference in the amounts of internal energy in a given state and in some standard state, then dE is an exact differential of a'function determined by {he state of gas.

Also • we should remember that the state of a mass of gas de­pends on Its volume, pressure end temperature and these are not Independent but connected with each other.

The first law of thermodynamics for a gas can be expressed as dQ^dErpdv, ...«) where dQ Is the heat imparted, dE is the Increase in internal energy &ad pdv is the work done In expansion, where h is the pressure aad« dv is the increase fn volume.

Now let us consider perfect gas only. The perfect gas is aa ideal substance for which pv -RTholdsj where R is a constant and T dependsfm the absolute temperature. For such a gas £ is a func* tion of T alone as proved experimentally.

First let us suppose that while a quantity dQ of heat! is impar­ted the volume Is kept constant. Then from (i) putting dv—Q we have dQ=*dE ,#K . ,jjj

Also from § 617 Page 294 in this case we have dQldT>=*Cn or dQ=>c0dT

I. From (ii) we get cv-dT-^dE ...flil) Now here we should note that Els& function of T alone

(property of perfect gas) and it is an experimental fact that for permanent gases c» is Independent of T at all temperatures except &{ very high and now, so c, i> a QQ»s$ajaf,

. ' 2 9 6 « Bydrostafilea • • , . .^ ' * - * v ,

* From (i) and (iii) we get dQ**c, dT+p dv »...(Iv)' Also for a*perfect gas % py=RT ' S"6 that p dv fv tfp^iJ «/r or p <ft"=»i? ^7—v </p , ,'. From (iv) we get dQ=c0dT+R dT-ydp l ~.<v)

^ Now suppose that while a quantity dQ of heat is imparted"1 the pressure Is kepc cda'stant. Also trom & (>'17 Page 294 we know that in this case

d&ldT<=*cp or dQ=cP dT ...(vi) Also in this case dp =1), so from (v) and (vi) we have

t cpdT=>ctdTi-RdT or cp-r0=>R, ...(vii) which shows that cp > c, and cB isalso constant fora perfect gas. ••§ 619. Adiabatic changes.

Definition. • When a change of state takes place in a gas in such a way that no heat is either gained or lost, then it is called an adia-batio change. •

As no heat is gained or lost In this case, so dQ^=Q .". From 'result (iv) of § 648 above we get

Q~cvdT-\-pdv ' ^ ...(i) Also for a perfect gas we knojv

pv=*RT and R=>cp^-c„ (see § 6'18 above) A pv=(cp—c0) T ^ So that, pdv\-vdp^(cP-cu)dT ( _ „.(ii) Eliminating dT ftom (I) and (u) we get

, pdv\-vdp=.(cD-cv)^^)T-{^-^pdv

or , * — S V * * « ' * • + (&) £-.0 CD P \C0 / V

o r • * + y * L r a 6 , where yJ^ > . ' , p V Co Integrating, logp 4-y log v=log k, where k is constant

' or i pvy^constant ...(iii) * y denotes the constant ratio of spe;ific heats i.e. cP'cv. Its

jiumerical varue is about 1'4 but ic varies slightly for differ.nt gases. ' • The relation (iii) gives the law of sfdubatic expansion This

law is assumed to no id good in all changes which take place, so rapidly that there is no time for heat transfer. **$ 6'20. Work done in compressing a gas.

Le( the volume of the gaswnenits pressure is p. bs v. And let If be an external atmospheric pressure on the surface of tne containing vessel.

Consider an element dS of this surface, then pre-sure on this e\ement="(p-II) dS. Now let us suppose that dn is the inward normal jdisplacement of this element dS, then the external work necessary to overcome the opposing pressure is\p ~ll) dS dn. x

Summing for all elements of the surface which undergo dis­placement, the total work required=(p—II) 2dS dn. , ...(t)

But 2 dS </««=<deciease in volume due to the displacement of ^ho surface -»-dp • * (Note)

> 1.* I

Atmospheric Pressure &1

i . From (i), the total work required to overcome the opposing pressure <=—(/> —If) dv •,

A 0 If the volume is to be reduced from V to V, the required

S V (p—ll) dv • ^ -.j.^"

Now following two cases arise *'"* Case I. When the change takes place isothermally. " ' In this case;;i"=constant=C,isay)

l.e. p<=*CJv A, From (li), the required work done v

- C log (VIV')-ll (V- V) (Note) <=PVlog(VIV')-il(V~V), (Cil i>

where P is the pressure when V is the volume / e. PV<^ C Case II. When the change takes place adiabatically.

In this case />vv'=constant=C (say) < Then P=>CJvv

A From (ii), the requited work done

~]v (£~n) *— [c=l+IJ^~IIv]r " y-£ Leo*-1' ~ 7*=*- J -11 (K_F'J „.(iv>

Let P and P' be the pressures when volumes are V and F ' respectively, then py*=*C=>P (F')y ...(v)

"A From (iv), the required work done

y - l U n " P j U V K r ' (Note)

m J _ [P'y'-PVl-II (V~V). from (v) ( v j )

And if T and I" be the absolute temperatures when the volumes PV P'V

are V and V, then -^^ mjr^K (say)f see § 609 Page 272 # _

Then from {vi), the required work done

**§ 6 21. Kelvin's hypothesis of Coavective Equilibrium. If is observed that temperature of the atmosphere diminishes

slowly as the height above the earth's surface increases. In static atmosphere theiefore the conduction will try to equalise the tenv perature but it would take sufficiently long time and in actual atmosphere we find that wind and convection currents contioually change the state.. Lord Kelvin's hypothesis of convectlve equili­brium is a better hypothesis than an isothermal state and by con-vective equilibrium a state is implied in which if equal masses of air aj; any two stations were interchanged each would assume the

•A .

$!& &ydros£a$cl

pressure, temperature and' density of the other and this change Would be adiabatlc. '•• Now the pressure at a height z is given by • ' '-1 •

•» dp^-gPdz, ...(i) where /) and P denote the pressure'and density at a height z.

*"*• Also on adf&batic hypothesis, . ^-p*±kl,v • °«»(ii) • A Frora(i)and(li)wehaveft7P''-1:rfp=>—gPiz .."'.' ,or i Ayp v : a d?*=*—gdz

Integrating, we have -^— py-};-=*C—gz

or - ^ r ^ C - g z , since i=>/>/pr from (11) / ^ P

( • - ' • . y • - . • • "

or —-. RT=C-gz, •y~l . -

as / p=*R?T ...(in) . If T0 Is the absolute temperature when z^n, we have

- r y—1 -»(iv) Eliminating C fiom (in) and (iv) we have <

. ; JLRiT~T0)-~gz or T-T^~y^~J-~

T y-\ gz ,_ Ta" 7 ~RT*L' . ...(v)

dividing eabh term by T0. " ' , . • ' • . . . * If/fdenofes the height of the homogeneous atmosphere, then

'- ftoT^Po^gPoH i.e. RT0=gH * , -:.' F r o m ^ w e h a v e ^ l - ^ . p ^

Correction for variation of gravity. , If we. consider the variation in gravity at different heights and

use gr'ftr 4- zf instead of g in (i), and where r is the radius of earth, > then we havo • •"' • , . •> ' • • • . . ,c. " • " • • ' • a " * * "

dp^^^L^ds . . . , , , - ;

or fc V 1 # - - -J^<fc, using p«=«&p*. I'."*" Z)

or : hyp'* de*^,—[gr*Kr+z)*] dz

... Integrating, we have — - p^»«C+rsL^ ,

";% . y ^ i | ^ C + ^ ) ' s t a c e / c ^ / P y

or

It T^ta when s?**0. wc have r~~. r ^ C f #> M« (viii|

Atmospheric Pressure 2 '$^

Eliminating C from (vii) and (viii) we get'

" If # be the height of atmesphere, then as before ^ RT0<=gH

7— 1 srz From(ix) we have T-T0~- y - ^ q ^ )

0 1 - fo = l - ^ ^ ~ , dividing each term by rfl

" ' fr^^^HO'^^W - ...<xi) Solved Examples on § 617 to § 621. Ex. 1. A gas sraturated with vapour is at pressure II. It is

compressed without change of temperature to (l)n)th of its former volume and the pressure then ob&erved to he II ' . Show that the pressure of the vapour is (nil— II')/(n—l) snd (hat the pressure QX. the air in the original volume without its vapour is equal to

Solution. If* be the original pressure of the gas then the pressure of the vapour=II—x, which remains unaltered during compression because the pressure of a saturated vapour at constant temperature does not depend on Its volume. \

But the pressure of the gas will change due to compression and if p be the pressure when i£s volume is (1/n) of Its former volume v (say), then from pv= constant, we have

P (V//J)=>*V or p=nx «.(1) Therefore according to given problem we have ir~/>+(II-x)~BX+(II-») , from (i)

or (n-1) x = H ' - I I or se-(H'-II) / («- l ) Also pressure of the vapour=>H-x=-II—[(II'—II)/(n-l)l

' - = ( B I I - I P ) / ( B - 1 ) m Hence proved. . *Ex. 2. In a vertical column of perfect gas the . pressure sad

absolute temperature at any height z are p and T. Prove that _ Po \Po T dp

&nTu)p p ' . z where pM p„, T0 are the pressure, density and absolute temperatore at the bottom^

Solution. We know for a perfect gas p=RpT „.(i) At the bottom, p0-=>Rp0T0 ...(ii) Now the pressure at a height z is given by

'dp= -.pg dz ~-{PIRT)gdz, r P~pl(RT) from (i)

or dz^-{RTIg).(\lp) dp (Note) —-(toW, T£ {Up) dp, from (ii)

• - • ( ,

306 JaydWfatieb

IntegTat&g between the proper limits we have • At \ [P ' T dp p0 tp0 T dp

,gPo T0 f )p0 p g?a T0 )p p„ ' % „ 'Hence proved.

^ - t o ^ * * E s . 3. Assuming the temperature of the air to diminish anjforaily witi) the height, prove that the difference of lerel between

two stations is H ^ ~ - T^^T^T, where H is the height of the

•homogeneous atmosphere at 0°C. T0 and Tx are the absolate tem­peratures at the stations and h0, hx the barometric heights redaced to ,o°C (consider gravity, as constant), rrova also that, if as an approximation the temperature were taken constant and equal to £(T„ tTj) , (be calculated height would be too great by a fraction of its true value equal roughly to £ [(T0 -Xi)J/(,T0 + T;)]a.

Solution. At a l?eigbiz above the lower station, let the abso­lute temperature T bs given by T=T0(L-hz) r ...$i)v

— Also we know from § 6-08 Page 272 that the relation between pressure, density and temperature a"t this level is given by

p-fcpT/273 _ „.(ii) Also we ino N that the pressure at a height^is given by

dpea—gPdzi:* r~- gdz, from (ii)

dp 273 g , c ... / p k T0 (1 —Az)

Integrating, log p -C-H273£/fcr0\l log ( I - \ z ) , ...(Hi) s lip0 be the pressure at the lower station, then from (lif) we get

log/>a=C +(273 g//croA) log (l) or C=»log/>0 C From(iii)wegetlo^^/p0)=T(273<?/<fcT0X)]og(l— hz) ^(lv) ,£* If Pi be the pressure at the upper station at a height zx

above the lower one,, then T=T0 (1 — U J i ...(v) and from (iv), log Oi//>0)=i273s/fcr0\) log (1-AZi) or lag(pJp0)-~i273glkTal) log (TJTJ, from (v) - „.(vl) " *"• •Also at 0°C we have gf>QH=>pa-=?k9Q

v©r^ k=*gH Then from (v) we have zx* T9-Tx

KT0 (Note)

Again if temperature=4 (r0-|-Tj-) aad assumed to be constant, then from (ii) we have p<=ȣ k? [TB +7'1)/273-=/c'P, say .J(viii)

And the pressure equ i tion dp =• gPdmz redu«es to - ''dp**-(gplk'tdz or i,llp)dp=*~(glk')d&

Integrating, logp=*C'—(glk') z •> At the lower statioa/?«=>#}, so we get log pae*Q

0

..(xi)

(Note)

on expanding

Atmospheric Kressajpe J ^

If zi' be the calculated height on this hypothesis, then u »

beforerfrom (\) we get z / ~ - - log (Pl \ •= ~ log ( ^ \

M *1'" ^ T ^ " log (§• )• w ^ f r * - ^ -.(*) From (vli) and (x) we have

?L=&-^-i4Js±z»L log(a\i_i *i * I 2<rB-rt)

iq8lrJJ But log l r o ^ i e g f i ± i ^ Z ^ i ± M -

.*. From (xl) we have 1 ' ~fr ^° M , on expanding *i Wot J n s _

\ , Hence proved. Exercise on § 617- § 6 21

Ex. Prove that if toe temperature in the atmosphere falls uniformly with the height ascended, the height of a station above sea level is given by z*=a {l—(h//i0)

m}, where A'and //<, are the reading of the barometer at the station and at sea level respec­tively, a, m are constants t (Agra 79)

MISCELLANEOUS SOLVED EXAMPLES **Ex. 1. The reading of a faulty barometer, the tube of which

contains a quantity of ajr of length 3&', is 28'; when the reading of a true barometer is 30'. What will be the reading of a faulty baro­meter when the true barometer reads IV ?

Solution. Let the cross-section of the tube be a arid z be the required reading when tbe barometer reads 29*.

At 30* atmospheric pressure the volume of the contained air=>3$« and the pressure of the contained (or impxisioned) air is 30"-28"/.e 2\ 4 . .

A *p.v'-2x3$«~*,r\x ....0) Also the total length of barometric tube=28"+3£"~31 J* Again volume of imprisioned air at 29" atmospheric pressure

«=(31 J*z).a and the pressure of this air is (29-z) .-. «/>v'=(31!-z)a.(29~z) - ...(it*) Also from Boyle's Jaw pv= constant, so fiom. (i) and (ii) we get

- (2>;~z).(3!i-z)a=.s3

fla or (29-z)(94-?z)=20 or 3za- 181z 1-2706=0 or * - 3 3 , Vi.

But z=-33" is impossible, as total length of barome(rictube<33* z=27£". ABS.

**Ex. 2- The readings of a perfect, mercurial barcmefer are-« and p when fBe torresponding readings of a faulty barometer are a an<f b. Prove that the ccrject"on to be applied to any reading c

. . . . . . . . fa-a) f@~bl (a h) of the faulty barometer is ( a _ CJ ^ ^ y ^ ^ - •

(lucknow 8J ; U. P. P. C, S, 71)

N§2 ttj^rostatkS

,. Solution. $et z be the required correction and h be the length of the tube of the false barometer.

& From Boyle's Law (I e. />v=»constant) we h#ve * •

or

mJO'

(h-a) (a-flH(A-A) (fr-b)^h-c) z . From {h-a) (a -o)=>(A-ft) fli-ft) we have

h («-a-p+3)=a (B-a)-b ($-b) . , ft~{a («-*)-£ ( p - ^ ^ a - ^ - a + A )

Agam from (i) we have (k—c) z^(h~b) (S-&) z~{h-b) ®-b)l{h-c)

...<i)

- (H)

(«—a} (a—c)-

Also from (H), h-b~aJ*=*=*^

and ft-<:~ ° <«-*>-» » - *> -^(^rgl ig «—p-a+o

X from (lfl) we have r= patting the value of (ft - *).

"" **Ex. 3. A closed straight tube of fine uniform bore contains mercury in the middle and air at each end. When

.tube is vertical the portions occupied by air are of lengths a and b respec­tively, which becomes a', b' wben the tube is inserted. Prove that .tbe lengths When the tube is horizontal are

aa' (b-f b') bb' (a+aV ab+a'b' • ab+a'b' *

, Solution. As temperature dees not vary,

* «?. A be the the tube.

&

—<iif> (cc-a) (a-b)

(«-a)-({i-6) tf)-(3-ft)(6-g)

(as-a)—tp-ft) («-a) (a-b) (3-6)

-(p-6) 0-c) ' > Hence proved.

Boyle's Law holds good. Let area of the cross-section of

(Fig. 282) . . From '/n^constant', equating the value of 'pvy for 'the

j portions of air whose height changes from a to a' as the tube is '• inverted in the vertical position, we get PxaA^Pba'A ...Ci)

' Similarly, for the second portion of air, PsbAfPt b'.A ...(ii) , / Also as the volume of mercury remains constant and the

'length of ljube js also used, so we have a+b^a'+b' § > — < i i i ) < Now when tbe tube is in horizontal position, the pressure on

> both sides of mercury will be equal (say P) due to equilibrium. :. P5~P,~P Let x mdy be the lengths

occupied by air, which were pre­viously occupying distances 6 and a respectively. '

From Boyle's law 'pv^cons-tant', equating the values of 'pv' we have P,Aa'=»Pi.A.b''=>Pi.Ab

/*/* • k.-

1 MERCURY

*-y<-*

A/A

%

s J C - *

(Fig. 284) ••J5eo(i}]

t

r Atmospheric Prcssort , ^3©^

and P.AJ**P*AM,—PJU ...[Seefli)] W xP~b'Pt<=b Pa 1 and 0 yP=a'P8c=fl Pj J «.(iv)

\ Also considering the equilibrium of mercury in first (wo cases, 'wehave Px~>Pa~— geh and P^-Pt=-~gph

A ^ ^~/Wa or - j v fl,, from(iv)

«

or -Also as in (Hi) we have a+b^d'+b'^x+y —(vi)

From (v) we have,- ( * ± r ) = * ( * £ r )

or | c+a' vra / 6-fb'\~ a+a ' , A+b' ) yw) ~^"v ••• \ ad ) \ bb' } aa' " bb'

(xA-y) bb' (q+a') . (x+y) aa'{b+b') aa' {b+b')+W (a+a')'y ad (b+b'Hbb' {a + a')

(a+b)bb' (a+a') (a+b) aa' jb-\-b') 0 r Xaab (a'+b'Ha'b' (a+b) ',y~ab (a'+b')+a'b' (a+b)

...'.' x+y=°a+b, from(vi) bb' (a+a') aa' (b+b')

since a-f 6—a'+6' from (\i). Hence proved. •Ex. 4. If the 'absolute temperature T at a height yis a given

function f (z) of the height, show that the ratio of the pressures at two heights Zj and z2 is given by log j^2J=.— f V-fj-r, k being the

' P i / " K }Z11 {%) constant in the equation p^=kpT.

Solution. Given p=kQF or 9<*=>plkT ,..(i> Also we know "dp/dz^—gP"

i. From (i) we g e t g — ^ or f — jL* ~

Integrating, fi>gp-=*—°\~=,dz4-A, where T<=>f{z), given *

> Let p^and pa be the pressures at heights zt and za, Theafrom (ii), ldg/>,~- | ( | ' ^ + ^

Subfaethg^e g.t !og (ft) _ _ J J ^ * . ^ ^

**Ejf 5. A heavy gas at constant temperature is confined in a vertical cylinder of height h. If p0 be the density at the base prove that the mean density is (kn0/gh) (l-a~«"ft).

* ytgm 811 litckww 82* 77, 74 ; Rohilkhand 83)

S&4,\ IjfydroKatles mm

Solntioq. Let P be the density at a height z and a be the radios of the cylinder. 1 *" Then the total mass of gas In the cylinder f \

•»( P»oa<fe ' _ A. J.-o (Note)

{*

(See § 6-14 Page 288) f e~ow in

* mr« - j - -i* mass of the gas A Mean density =»—, , . ~g*i.—. ' volume of the gas *=kp0[l— e'owygh Hence proved.

• EXERCISES ON CHAPTER VI •Ex. 1. A cylindrical we]] of depth h and section A is main­

tained at constant temperature. If p0 and p, are densities of the air at the top and bottom, show that the total amount of air contained is Ah (P i~ P0)/Pog P i - l o g P„]. iAgra f2, 78; I. A. S. 73 ; Zucknbw 75)

Ex.2 . If (the pressure of the air at any height varied as the /wthpowerof its density, show that the height of the atmosphere would be mHI(tn—l), where H is t ie height of the homogeneous atmosphere.

Ex. 3. If a change from 30' to 28' in the height of barometer corresponds to 15a change of altitude of 18C0 ft., find tbe change of altitude' corresponding to change of barometric height from 30" to 264'. N log 2«=-0-3010 ; log 3=0'4771 ; log 7-~0'84Sl ! log 11«=»1'04141

J£x 4. The readings of a faulty barometer In which there is some air are 30 and 32 inches when the corresponding true read-ings^are 30 2 and 32'3 inches. What will be the true reading when the reading in, the barometer is 33 inches ? (Magadh 73)

Ex. 5. If the pressure of the air varied as [I+(l/m)] th power of the density, show that neglecting variations cf temperature and gravity, the heJght of < atmosphere wcjild be equal to (w-J-1) times

'the height of the homogeneous atmosphere. (Agra 83,80) Ex. 6. Show that in an atmosphere at rest and at uniform

temperature, the density p at height h is given by p<= p0 e"«», where a and P0 are positive constants. Interpret these constants in physical terms. * (U.f.p. c.S.79)

IHinti See § 6-14 Page 288]. Ex. 7. The, readings of a faulty barometerJn which there is

some air, are a and b when the true readings are,« and p. Find the true reading when the faulty barometer reads c. (RmicMS^)

[Hint i See Ex 2 Page 301]. v * ^ _„» , Ex. 8. The same quantities o£ atmo'spheric air are contained

in two .hollow spheres, the internal radii bflng r and r' and the temperatures t and t', respectively, compare the whole pressures on the surfaces. i " (Rohilkfy>nd82) ,

sr

3**ft*^taT vfk^ *n$ Wr

ACC No

Call No

Author -*

So/63 62X'L C 33?//

Cha./?OJL£ . ftAJ \ \\\JS • • « • * • * • • # • • • • • • • • * • * • • •»• •»* • * • • • a • • • ««• • • • • • • • • • • • •

Date of Return

.*

Borrowere No,

• •• ••• •••

"&

Date of Return

Borrowars No.

•