Moment of Analytic Statics

CH

AP

TE

R 1Moment of Analytic Statics

1.1 Definition-I

Moment of a Force

Let F be a force and p be the ⊥ distance of its line of action from the fixed point O,

then moment of F about O = (F)(p).

Figure 1.1

Note that:

The moment vanishes if either F = 0 or p = 0 i.e. if the point about which the

1

2 CHAPTER 1: MOMENT OF ANALYTIC STATICS

moment is being taken lies on the line of action of the force.

Sign of Moment

The moment of a force F about a point O is said to be +ve or −ve according as the

force rotates the body in an anticlockwise or clockwise direction respectively.

1.2 Definition-II

Moment of a Force

Let A be any point on the line of action of F.

Figure 1.2

CHAPTER 1: MOMENT OF ANALYTIC STATICS 3

AO = γ and ∠OAN = θ then the moment of F about O

= (F)(p)

= (F)(ON)

= (F)(γsinθ)

= (γ)(Fsinθ)

= γFsinθ

= AO (Resolved part of F perpendicular to AO)

(MF)O = γ× F.

Varignon’s Theorem of Moments

The algebraic sum of moments of two forces about any point in their plane is equal

to the moment of their resultant about the same point, i.e

(MP)0 + (MQ)0 = (MR)0.

Note. Resultant of Two ‖ Forces The resultant of two like parallel forces P and Q acting

Figure 1.3

at A&B is equivalent to a force P + Q acting in the same direction at C such that

AC× P = CB×Q

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Proof. Case I: When the Forces are Concurrent Consider a particle A acted upon

Figure 1.4

by two coplanar forces P and Q i.e. contained in the same plane. Such forces are

said to be concurrent.

Let γ be the position vector of A relative to O. Then we may write

γ× (P + Q) = γ× P + γ×Q

γ× R = γ× P + γ×Q

(MR)O = (MP)O + (MQ)O

Note. Coplanar f orces Forces all of whose lines of action lie in one and the same plane.

Case II: When the Forces are ‖

Let P and Q be two like ‖ forces acting at A and B. Let R be the resultant of P and

Q acting at C. Then

AC× P = CB×Q. (1.2.1)


Figure 1.5

The sum of the moments of P and Q about O

= OA× P + OB×Q

= (OC + CA)× P + (OC + CB)×Q

= OC× P + CA× P + OC×Q + CB×Q

= OC× (P + Q) + CA× P + CB×Q

= OC× R− AC× P + CB×Q (∵ CA = −AC) (1.2.2)

Now

|AC× P| = ACPsinθ

|CB×Q| = CBQsinθ

From equation (1.2.1) we have

|AC× P| = |CB×Q|

=⇒ −AC× P + CB×Q = 0 (1.2.3)

Now equation (1.2.2) becomes

= OC× R + 0 = OC× R


=⇒ (MP)O + (MQ)O = OC× R

= (MR)O

Case III: When P and Q are unlike ‖ forces.

Figure 1.6

(MP)O + (MQ)O = OA× P + OB×Q

= (OC + CA)× P + (OC + CB)×Q

= OC× (P + Q) + CA× P + CB×Q

Put |AC× P| = |CB×Q|

∴− AC× P + CB×Q = 0

= OC× R− AC× P + CB×Q

= OC× R

Hence (MP)O + (MQ)O = (MR)O


1.3 Generalized form of Varignon’s Theorem

The algebraic sum of moments of any number of coplanar forces about any point in

their plane is equal to the moment of their resultant about the same point.

Example 1.3.1. A force F = Fx i + Fy j acts at a point A(x1, y1). Find the moment of F

about a second point B(x2, y2).

Solution.

Figure 1.7

γ = BA

= (x1 − x2)i + (y1 − y2) j

∴ M = γ× F

=

[(x1 − x2)i + (y1 − y2) j

]×[

Fx i + Fy j]

=

[(x1 − x2)Fy − (y1 − y2)Fx

]k (∵ i× j = k and j× i = −k)

M = (x1 − x2)Fy − (y1 − y2)Fx.

Example 1.3.2. A force F = Fx i + Fy j acts at a point of coordinates x and y. Find the ⊥

distance from the line of action of F to the origin O of the system of coordinates.


Solution. We know that the moment of F about a point (0, 0) is given by

Fy(x− 0)− Fx(y− 0) = M (1.3.1)

Also,

M = |F|OL

OL =M|F|

=Fyx− Fxy√

F2x + F2

y

Example 1.3.3. A 10 kg force acts at an ∠ of 60o on the corner C of a 12 by 1–cm plate

ABCD. Determine the moment of force about A.

Figure 1.8


Solution. Resolving the given force at C acting along AD and ⊥ of AD at A we get

F = 10 cos60o i + 10 sin60o j

= 10× 12

i + 10×√

32

j

= 5i + 5√

3 j

= Fx i + Fy j

Now (MF)A = (x1 − x2)Fy + (y1 − y2)Fx

= −(1− 0)5 + (12− 0)5√

3

= −5× 1 + 5√

3× 12

= 5(12√

3− 1)

Example 1.3.4. The point (5, 3) is on the line of action of the force P making an ∠tan−1(34)

with the horizontal. Find the moment of force about the origin of coordinates.

Figure 1.9

Solution. Resolving the given forces at A acting along AB and AC we get

F = P cosθ i + P sinθ j

(MF)O = (5− 0)P sinθ − (3− 0)P cosθ

= 5P sinθ − 3P cosθ

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= 5P(35)− 3P(

45)

= 3P− 12P5

=35

P

The moment of the force about the origin of coordinates (MF)O = 35 P

Example 1.3.5. Three forces P, Q and R act along the sides BC, CA, AB of a triangle

ABC. Show that if their resultant passes through the ortho-centre (Intersection of altitudes),

then

P sec A + Q sec B + R sec C = 0

Figure 1.10

Solution. Taking moment of force about O we have

POD + QOE + ROF = 0 (1.3.2)

Now, in4ODCODOC

= cos B (1.3.3)

in4OECOEOC

= cos A (1.3.4)

from equation (1.3.3) and (1.3.4) we have

=⇒ OEOD

=cos Acos B

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OE = ODcos Acos B

(1.3.5)

Again in4ODBODOB

= cos C (1.3.6)

in4OFBOFOB

= cos A (1.3.7)

from equation (1.3.6) and (1.3.7) we haveOFOD

=cos Acos C

OF = ODcos Acos C

(1.3.8)

From equations (1.3.3), (1.3.5) and (1.3.8) we get

POD + QODcos Acos B

+ RODcos Acos C

= 0

P + Qcos Acos B

+ Rcos Acos C

= 0 (∵ OD 6= 0)

=⇒ PcosA

+Q

cos B+

Rcos C

= 0

=⇒ P sec A + Q sec B + R sec C = 0

Example 1.3.6. Three forces P, Q and R act along the sides BC, CA and AB of a triangle

ABC. Show that if their resultant passes through the incentre (Intersection of the internal

bisectors of the angles of the triangle) and the circumcentre( Intersection of the bisectors of

the sides of the triangle), thenP

cos B− cos C=

Qcos C− cos A

=R

cos A− cos B

Solution. Let O be the incentre of 4ABC. Let OD, OE, OF be the ⊥ to the sides BC,

CA and AB respectively. If r is the inradius then

OD = OE = OF = r

Taking moments about O we have

P ·OD + Q ·OE + R ·OF = 0 (1.3.9)

⇒ P · r + Q · r + R · r = 0 (1.3.10)

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Figure 1.11

⇒ P + Q + R = 0 (∵ r 6= 0) (1.3.11)

⇒ R = −(P + Q) (1.3.12)

If O be the circumcentre

Figure 1.12

∴ ∠BOC = 2A

∠AOC = 2B

∠AOB = 2C

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Taking moment of force about O

POD + QOE + ROF = 0

ODR1

= cosA

OER1

= cosB

OFR1

= cosC

PR1cosA + QR1cosB + RR1cosC = O

PcosA + QcosB + RcosC = O (1.3.13)

from equation (1.3.12) we have

PcosA + QcosB− (P + Q)cosC = 0

PcosA + QcosB− PcosC−QcosC = 0

Q(cosB− cosC)− P(cosC− cosA) = 0

QcosC− cosA

=P

cosB− cosC. (1.3.14)

Also from equation (1.3.12)

P = −(Q + R) (1.3.15)

−(Q + R)cosA + QcosB + RcosC = 0

Q(cosB− cosA) + R(cosC− cosA) = 0

QcosC− cosA

=R

cosA− cosB. (1.3.16)

From equation (1.3.14),(1.3.16) we get

PcosB− cosC

=Q

cosC− cosA=

RcosA− cosB

.

Example 1.3.7. Forces of magnitudes 1, 3, 2, 5 act along the sides of a square taken in order.

Find the magnitude and line of action of their resultant.

Solution. Resolving the forces along AB and AD

Rx = R cosθ = (1− 2) = −1 (1.3.17)


Figure 1.13

Ry = R sinθ = (3− 5) = −2 (1.3.18)

Squaring and adding (1.3.17),(1.3.18) we get

R2 = 1 + 4 = 5 =⇒ R =√

5

Let the resultant meet AB in E.

Taking moment of the system at the forces about E we get

− 3BE + 2BC + 5AE = 0

− 3(AE− AB) + 2AB + 5AE = 0 (∵ ABCD is square ∴ AB = BC)

2AE + 5AB = 0

AE =−52

AB

|AE| = 52|AB|.

Example 1.3.8. Three forces P, 2P and 3P act along the sides AB, BC and CA of an

equilateral triangle ABC. Find the magnitude and direction of their resultant and find also

the point in which its line of action meet the side BC.

Solution. Let R be the resultant that meets BC in D. Resolving the forces along BX


Figure 1.14

R cosθ = 2P− 3P cos60o − P cos60o

= 2P− 3P2− P

2

=4P− 4P

2= 0

=⇒ R cosθ = 0 (1.3.19)

Resolving along BY

R sinθ = 3P sin60o − P sin60o

= 3P√

32− P√

32

(1.3.20)

R sinθ =√

3P (1.3.21)

Squaring and adding (1.3.19) and (1.3.21) we get

R2 cos2θ + R2 sin2θ = 3P2 =⇒ R =√

3P.


Dividing (1.3.21) and (1.3.19) we get

R sinθ

R cosθ=

√3P0

= ∞

tanθ = tan90o =⇒ θ = 90o

Taking moment about B

2P× 0 + 3P× BE + P× 0 = R× BL

3PBC sin60o =√

3PBD sin90o (∵ θ = 90o)

BD =32

BC

Example 1.3.9. Forces P, 3P, 2P, 5P act along the sides AB, BC, CD and DA of a square

ABCD. Find the magnitude and direction of their resultant and prove that if it meets AD

produced at a point E such that AE : ED = 5 : 4.

Figure 1.15

Solution.

R cosθ = −5P + 3P = −2P (1.3.22)

R sinθ = P− 2P = −P (1.3.23)


R2 cos2θ + R2 sin2θ = 5P2 =⇒ R =√

5P.


From equation (1.3.22)

R cosθ = −2P

=⇒√

5Pcosθ = −2P

=⇒ cosθ =−2√

5From equation (1.3.23)

R sinθ = −P

=⇒ sinθ =−P√

5P

=⇒ sinθ =−1√

5=⇒ sinθ and cosθ both are negative ∴ θ lies in the third quadrant.

θ = π + tan−1(12)

Let a be the side of the square. Taking moments about A

RAE sinθ = −2Pa− 3Pa = −5Pa

AE =−5PaR sinθ

AE =−5Pa−P

AE = 5a

AE = 5(AE− DE)

AE = 5AE− 5DE

=⇒ 5DE = 4AE

∴ AE : ED = 5 : 4

Example 1.3.10. Forces 3, 4, 5, 6 kg act along the sides of a square taken in order. Find the

magnitude, direction and line of action of resultant the said of square being a.

Solution.

R cosθ = 5− 3 = 2 (1.3.24)

R sinθ = 6− 4 = 2 (1.3.25)


Figure 1.16


R2 cos2θ + R2 sin2θ = 4 + 4 =⇒ R = 2√

2 kg

tanθ = 1 =⇒ θ = 45o

From equation (1.3.24)

R cosθ = 2

cosθ =2

2√

2From equation (1.3.25)

R sinθ = 2

sinθ =2

2√

2

=1√2

=⇒ sinθ and cosθ both are positive ∴ θ lies in first quadrant. Taking moments about C

R.CK sinθ = 6× CD + 3× CB

2√

2CK sin45o = 6a + 3a

2√

2CK√

2= 9a

CK =9a2


or CD + DK =9a2

a + DK =9a2

DK =9a2− a

=⇒ DK =7a2

.


1.4 Couple

Figure 1.17

Two forces F and −F having the same magnitude, parallel lines of action and

opposite sense are said to form couple.

Moment of the Couple

Let F and −F form couple. Let O be the center of moment.

Figure 1.18


Now

Moment of the couple = γA × F + γB × (−F)

= (γA − γB)× F

= γ× F were γ = (γA − γB)

M = |γ× F|

= γF sinθ

= Fγ sinθ

= Fd

where d is the ⊥ distance between lines of action of F and − F (called arm of the

couple).

Notation

Couple F and − F with the ⊥ distance between their lines of action will be denoted

by (F, d).

Resultant of a Couple

Theorem 1.4.1. The resultant of any number of couples acting in the same plane on a rigid

body is a couple.

Note. If the resultant of the forces is equal to zero then system reduces to a couple

R = RXi + RY j

RX = 0

RY = 0

=⇒ forms a couple.

Example 1.4.1. ABCD is a rectangle with AB = 4cm, BC = 3cm. Forces 2, 7, 6, 10

and 5 kg weight act along AB, BC, CD, AD, and AC respectively. Show that they are

equivalent to a couple whose moment is equal to 46kg.cm.

Solution. Let ∠BAC = θ then sinθ = 35 and cosθ = 4

5 .


Figure 1.19

Resolving the forces along AB and 5AD cosθ we get

RX = 2− 6 + 5 cosθ

= −4 + 5× 45= 0

RY = 7− 10 + 5 sinθ

= −3 + 5× 35= 0

=⇒ RX = 0 and RY = 0. The system reduces to a couple.

Now taking the moment of the forces about A we get

M = 7× 4 + 6× 3 = 46kg.cm.

Example 1.4.2. Forces 1, 2, 3, 5, P, Q act along AB, BC, CD, DA, AC and BD

respectively and ABCD is a square of side a. Find the values of P and Q for the

system to reduce to a couple. Find also the moment of the couple.


Figure 1.20

Solution. Resolving the forces along AB and AD we get

RX = 1− 3 + P cos45o −Q cos45o

= −2 +P√2− Q√

2

= −2 + (P−Q)1√2

RY = 2− 5 + P sin45o + Q sin45o

= −3 +P√2+

Q√2

= −3 + (P + Q)1√2

.

If the system reduces to a couple =⇒ RX = 0 and RY = 0

=⇒ −2 + (P−Q)1√2= 0

=⇒ (P−Q) = 2√

2

− 3 + (P + Q)1√2= 0

=⇒ (P + Q) = 3√

2

=⇒ P =5√2

and Q =1√2


Let a be the side of square and O its center. Then the distance of O from each side = a2

The moment of couple

M = 1a2+ 2

a2+ 3

a2+ 5

a2=

11a2

.


Example 1.4.3. In a regular hexagon, forces P, 2P, 3P, 5P and 6P act along

AB, BC, DC, EF and AF respectively. Show that a force can be determined to act along

ED, so that the six forces are equivalent to a couple and find the moment of the couple.

Figure 1.21

Solution. Resolving the forces along AB and ⊥ to AB we get

RX = P + 2P cos60o + 3P cos60o + Q− 5P cos60o − 6P cos60o

The systm redused to a couple if RX = 0 and RY = 0

=⇒ RX = P +2P2

+3P2

+ Q− 5P2− 6P

2= 0

= −P− P + Q = 0

=⇒ Q = 2P.

Now let O be the center of hexagon. Let distance of O from each side =√

32 a.

Moment of couple about O.

M =

√3

2aP +

√3

2a(2P)−

√3

2a(3P)−

√3

2a(2P) +

√3

2a(5P)−

√3

2a(6P)

=

√3

2aP(1 + 2− 3− 2 + 5− 6)

=⇒ M = −3

√3

2aP.


Example 1.4.4. ABCD is a rhombus of side ` and ∠ α. A square CEFD is constructed

on CD so as to lie on the side of CD away from AB. Forces P each act along the lines

AB, BC, DA, DF, FE, and EC respectively and a force 2P acts along CD. Show that the

system reduces to a couple, the magnitude of whose moment is 2`P(1− sinα).

Figure 1.22

Solution. The ⊥ distance between the sides AB and DC of the rhombus = AD sinα =

` sinα.

Similarly, the ⊥ distance between the sides AD and BC of the rhombus = ` sinα.

The force 2P along CD be replaced by two equal forces P and P along CD.

The force P along FE and a force P along CD form a couple of moment= (−P`).

(clockwise)

Similarly, the force P along DF and force P along EC from a couple of moment= −P`.

The force P along AB and force P along CD from a couple of moment= P` sinα.

(anticlockwise)

Then, the force P along DA and force P along BC from a couple of moment= P` sinα.


The four couples are equivalent to a single couple of moment

= −P`− P`+ P` sinα + P` sinα

= −2P`+ 2P` sinα + P`

= −2P`(1− sinα)

∴ The magnitude of the moment of the couple =2P`(1− sinα).


1.5 Equilibrium of a rigid body

Theorem 1.5.1. A coplanar force system acting on a rigid body is in equilibrium. Then

necessary and sufficient conditions are

F = 0 and M = 0

where F is the resultant of the forces and M is the moment of the system about any arbitrary

point in the plane.

Ist Equivalent Condition

A Coplanar forces system acting an a rigid blog is in equilibrium iff

∑ X = 0, ∑ Y = 0, M = 0

where ∑ X and ∑ Y are the algebraic sums of the components of the forces along

and ⊥ axes respectively.

IInd Equivalent Condition

A Coplanar forces system acting an a rigid blog is in equilibrium iff

X = 0, MA = 0, MB = 0

where A and B be any two arbitrary points on the plane and X is the sum of resolved

parts along any line which is not ⊥ to AB.

IIIrd Equivalent Condition

A Coplanar force system acting an a rigid body is in equilibrium iff

MA = MB = MC = 0

where A, B and C are not collinear ( or non collinear).

Example 1.5.1. A ladder of weight W rests at an angle φ to the horizontal with its ends

resting on a smooth floor and against a smooth vertical wall the lower end being jointed

by a string to the junction of the wall and the floor. Show that the tension of the string is12Wcotφ . Also show that the tension of the string is 5

4Wcotφ when a man whose weight is

equal to that of ladder has ascended the ladder three quarters of its length.

Solution. Let AB be the ladder and C the junction of the wall of the floor


Figure 1.23

Case I: Resolving the forces along AC and BC we get

X = O =⇒ R2 − T = 0 =⇒ R2 = T

Y = O =⇒ R1 −W = 0 =⇒ R1 = W

Taking moment about A we get MA = 0

−W(AF) + R2(BC) = 0

R2(BC) = W(AF)

R2(AB) sinφ = W(AE cosφ)

R2(AB) sinφ = W(AB2

) cosφ

R2 =W2

cotφ

but R2 = T

=⇒T =W2

cotφ


Case II: The man of weight w stands at D where AD = 34 AB (Given)

Let T1 be the tension in this case.

Resolving the forces along AC and CB we get

Figure 1.24

X = O =⇒ R1 = W + W = 2W

R2 = T1

Taking moment about A we get

MA = R2(BC)−W(AF) + W(AM) = 0

= R2(AB sinφ)−W(AE cosφ)−W(ADcosφ) = 0

= R2(AB sinφ)−W(AB2

) cosφ−W(34

ABcosφ) = 0

R2 = (W2

+3W4

)cosφ

sin φ

=5w4

cotφ

but R2 = T1

=⇒ T1 =5w4

cotφ

Example 1.5.2. A door of weight W height 2a and width 2b is hinged at the top and bottom.

If the reaction at the upper hinge has no vertical component. Find the components of reaction

at both hinges. (Assume that the weight of the door acts at its centre)


Solution. Let R be the horizontal component of the reaction at B and R2 and R3 be the

vertical and horizontal component of the reaction at A respectively. Resolving the system

Figure 1.25

along AD and ⊥to AD we get

RX = R1 + R3 = 0 (1.5.1)

RY = R2 −W = 0 (1.5.2)

Taking moment about A we get MA = −W(AF) + R1(AB) = 0

= −W(b) + R1(2a) = 0

R1 =Wb2a

(1.5.3)

from equation (1.5.1) we get

R3 = −R1 = −Wb2a

Example 1.5.3. A reactangular gate of l00 kg, 10 m wide and 6 m high is placed on two

symmetrically placed hinges 4 m apart. The lower hinge can exert only a horizontal reaction.


Find the reaction at both the hinge if a man of wt 50 kg is sitting on the gate at distance 8m

from its outer end.

Figure 1.26

Solution. Let E and F be the hinges having R1, R2 and R3 as the components of the

reactions along AD and ⊥ to AD respectively.

RX = −R1 − R2 = 0 =⇒ R1 + R2 = 0 (1.5.4)

RY = R3 − 50− 100 = 0 =⇒ R3 = 150 (1.5.5)

Taking the moment about E we get

ME = −100(AH)− 50(AK) + R2(EF) = 0

= −100(5)− 50(2) + R2(4) = 0

= −500− 100 + R2(4) = 0

4R2 = 600

R2 = 150

from equation (1.5.4) we got

R1 = −R2 = −150


Example 1.5.4. A uniform beam AB of weight W rests on a smooth floor at A and on a

smooth inclined plane of inclination α at B. The beam is maintained in a position inclined

at θ to the horizontal by a string fastened to B which passes over smooth pulling at the top

of the plane and supports a weight P hanging freely. Show that the beam will rest in all

positions if 2P = Wsinα.

Figure 1.27

Solution. Resolving the forces along AC and ⊥ to AC we get

R2cos(90o − α)− P sin(90o − α) = 0

R2sinα = P cosα(1.5.6)

R1 + R2sin(90o − α) + P cos(90o − α)−W = 0

R1 + R2cosα + P sinα = W(1.5.7)


Taking moment about B we get

MB = −R1(AC) + W(EC) = 0

= −R1(AB cosθ) + W(AE) = 0

= −R1(AB cosθ) + W(ADcosθ) = 0

= −R1(AB cosθ) + W(AB2

)cosθ = 0

R1 =W2

(1.5.8)

From eqution (1.5.6) we get

R2 =Pcosα

sinα

Now putting the value in R2 in equation (1.5.8) we get

R1 +P cosα

sinαcosα + Psinα = W

R1 +Pcos2α + Psin2α

sinα= W

R1 +P

sinα= W

using equation (1.5.8)

W2

+P

sinα= W

Psinα

=W2

Wsinα = 2P

Example 1.5.5. A beam of weight W is divided by its centre of gravity G into two portions

AG and GB whose lengths are a and b respectively. The beam rests in a vertical plane on a

smooth floor AC against a smooth wall CB. A string is attached to a hook at C and to the

beam at a point D. If T be the tension of the string and α and β be the inclinations of the

beam and string respectively to the horizontal. Show that T = Wa cosα(a+b)sin(α−β)

.

Solution. Now resolving the forces horizontally, we have

R2 − Tcosβ = 0 =⇒ R2 = Tcosβ (1.5.9)

Taking moments about A we get

−W(AE)− (Tsinβ)(AC) + R2(CB) = 0


Figure 1.28

−W(AGcosα)− (Tsinβ)(ABcosα) + R2(ABsinα) = 0

−W(acosα)− (Tsinβ)(a + b)cosα) + R2(a + b)sinα = 0

R2(a + b)sinα = W(acosα) + Tsinβ(a + b)cosα

using equation(1.5.9) we have

R2 = Tcosβ

Tcosβ(a + b)sinα = W(acosα)− Tsinβ(a + b)cosα

T(a + b)[sinα cosβ− sinβcosα] = Wa cosα

T(a + b)sin(α− β) = Wa cosα

T =Wa cosα

(a + b)sin(α− β)


1.6 Equation of The Line of Action of The Resultant

Let P1, P2, P3, . . . be forces acting at points A1(x1, y1), A2(x2, y2) . . .

Figure 1.29

Let X1, Y1; X2, Y2; X3, Y3; . . . be their resolved parts along the axes. Let R be their

resultant acting at A(x, y), then

R cosθ = X = X1 + X2 + X3 + . . . (1.6.1)

R sinθ = Y = Y1 + Y2 + Y3 + . . . (1.6.2)

Squaring and adding equation (1.6.1),(1.6.2) we get

R2 = X2 + Y2

R =√

X2 + Y2

and

tanθ =YX


Moment of X1 about O = −Y1X1

Moment of Y1 about O = X1Y1

∴ Moment of P1 about O = algebraic sums of the moments of X1, Y1 about O

= X1Y1 −Y1X1

Similarly moment of P2, P3 about O are (X2Y2 −Y2X2), (X3Y3 −Y3X3) . . .

XY−YX = G

which is the equation of line of actions of the single resultant. When G is the

algebraic sum of the moment of forces about origin.

Example 1.6.1. Forces 3, 4 and 7kg act along the sides AB, CB and CA of an equilateral

4. Taking A as origin and CA as x− axis. Show that the resultant acts along the line

7x− 5√

3y + 4a = 0 a being the length of the side of the4.

Figure 1.30

Solution. Resolving the forces along CA and ⊥ to CA we get

X = Rx = 7− 3cos60o + 4cos60o

= 7− 32+

42

=152

Y = Ry = 4sin60o + 3cos60o

= 4

√3

2+ 3

√3

2

= 7

√3

2


Taking the moment M about A we get

M = −4AD = −4

√3

2a (clockwise)

= −2√

3a (∴ AD =

√3

2a)

The equation of the resultant

xY− yX = G

7√

32

x− 152

y = −2√

3a

72

x− 5√

32

y = −2a

7x− 5√

3y + 4a = 0

Example 1.6.2. Find the magnitude of the resultant and line of action of a single force of 8

units acting along the Y− axis (vertically upwards) and a couple in xy− plane of moment

24 units.

Solution. Couple in xy plane of moment = 24 units = 8× 3.

The equation of the line of action of the resultant

X = 3

CH

AP

TE

R 1Catenary

Cable

A rope or cable is defined as material lines which can be bent arbitrarily without

changing its length.

The Catenary

When a cable hangs freely, under the action of gravity between two points which

are not in the same vertical line, then the curve formed by the cable is said to be a

catenary .

Figure 1.1

Span

1

2 CHAPTER 1: CATENARY

The distance AB is called the span of the cable if points of suspension A and B are

in the same horizontal line. Therefore

Span = AB = 2x.

Sag

The vertical depth of the vertex below the span of the cable is called sag of the

cable. Therefore

Sag = CD = y− c.

Parameter

The distance CO = c, i.e. the depth of the X− axis below the vertex of the catenary

is called parameter of the catenary.

1.1 Intrinsic Equation of a Catenary

If a uniform cable hangs freely under its own weight per unit length, then find the

Intrinsic equation of the curve.

Proof. Consider a uniform cable ACB hanging freely under its own weight per unit

length. Let CP be a piece of cable of length s.Therefore

Total weight of CP = Ws,

which is acting vertically downwards. CP is in equilibrium under the following

three forces:

(1) The horizontal tension H at C.

(2) The tension T at P.

(3) Ws, weight of CP, acting through the C.G of the arc CP.

(Since CP is in equilibrium , the lines of action of Ws, T and H meet in one point).

CHAPTER 1: CATENARY 3

Figure 1.2

Resolving the forces horizontally and vertically we have

H = Tcosψ (1.1.1)

Ws = Tsinψ (1.1.2)

where ψ is the angle which the tangent makes with the X − axis. Dividing (1.1.2)

by (1.1.1) we have

tanψ =WsH

(1.1.3)

It is convenient to write

tanψ =s

H/Wor s = ctanψ, (1.1.4)

where c = HW is the parameter of the catenary.

Eqn (1.1.4) is called the Intrinsic equation of the Catenary.

Cartesian Equation of the Catenary

If a uniform cable hangs freely under its own weight per until length, then find


the Cartesian equation of the curve.

Solution. We know that Intrinsic equation of catenary is

s = ctanψ.

Again tanψ =dydx

.

∴ s = cdydx

.

Now putting p =dydx

,

we have

s = cp.

Differentiating above eqn with respect to x we have

dsdx

= cdpdx

,

∴ cdpdx

=

√1 + (

dydx

)2 (∵dsdx

=

√1 + (

dydx

)2)

cdpdx

=√

1 + p2

∴dp√

1 + p2=

dxc

.

On Integrating we get

sinh−1(p) =xc+ A, (1.1.5)

where A is the constant of integration.

At the lowest point C, x = 0 and ψ = 0. Therefore

p =dydx

= tan0 = 0.


On putting these values in equation (1.1.5) we get

0 = 0 + A

=⇒ A = 0.

sinh−1(p) =xc

p = sinh(xc)

dydx

= sinh(xc)

On Integrating we get

y = c coch(xc) + B

At the lowest point C, x = 0 and y = c =⇒ B = 0.

∴ y = c coshxc

.

Which is called the cartesian equation of the catenary.

1.2 Relation between x, y, T, s and ψ

Parametric Equation of Catenary:

We know that, Intrinsic equation of Catenary is

s = c tanψ. (1.2.1)

Differentiating equation (1.2.1) with respect to x we get,

dsdx

= c sec2 ψdψ

dx√1 + (

dydx

)2 = c sec2ψdψ

dx√1 + tan2ψ = c sec2ψ

dψ

dx

6 CHAPTER 1: CATENARY√sec2ψ = c sec2ψ

dψ

dxdxdψ

= c secψ

dx = c secψ dψ

Integrating above equation, we get,∫dx =

∫c secψ dψ

x = c log {secψ + tanψ}+ C1

At the lowest point c, x = 0, ψ = 0

0 = c log {sec0 + tan0}+ C1

=⇒ C1 = 0.

Therefore, x = c log {secψ + tanψ}.

Again differentiating equation (1.2.1) with respect to y, we get

dsdy

= c sec2ψdψ

dydsdx

dxdy

= c sec2ψdψ

dy√1 + tan2ψ(

1tanψ

) = c sec2 ψdψ

dy

secψ(1

tanψ) = c sec2ψ

dψ

dydydψ

= c secψtanψ

dy = c secψ tanψ dψ.

Integrating we get∫

dy =∫

c secψ tanψ dψ

y = c secψ + C2.

At the lowest point C, y = c, ψ = o

∴ c = c sec0 + C2

=⇒ C2 = 0

∴ y = c secψ.


Hence, Parametric equations of Catenary are

x = c log (secψ + tanψ),

y = c secψ.

1.3 Relation between T, y and s

(i) y2 = c2 + s2

Since, y = c secψ. Squaring on both the sides, we get

y2 = c2 sec2ψ

= c2(1 + tan2ψ)

y2 = c2 + c2tan2ψ.

But s = c tanψ

=⇒ y2 = c2 + s2.

(ii) T = Wy

Since Parameter of the Catenary is given by,

c =HW

(1.3.1)

and

H = T cosψ. (1.3.2)

From equation (1.3.1) and (1.3.2),we have

T cosψ = Wc

T =Wc

cosψ=⇒ T = Wc secψ

T = Wy (∵ y = c secψ).

Prove that the span of the cable of length 2` and sagh h is 2`(1-2h2

3`2

):


Solution. We know that

s = c sinh(xc)

If 2a is the span.Then, x = a and s = `

∴ ` = c sinh(ac)

ac= sinh−1 (

`

c)

= log{`

c+

√1 + (

`

c)2}

a = c log{`

c+

√1 +

`2

c2

}But c =

`2 − h2

2h

=`2 − h2

2hlog{

``2−h2

2`

+

√1 +

`2

( `2−h2

2h )2

}

=`2 − h2

2hlog{

2h``2 − h2 +

√1 +

4h2`2

(`2 − h2)

}

=`2 − h2

2hlog{

2h``2 − h2 +

√(`2 − h2)2 + 4h2`2

`2 − h2

}=

`2 − h2

2hlog{

2h``2 − h2 +

√`4 + h4 − 2`2h2 + 4h2`2

`2 − h2

}=

`2 − h2

2hlog{

2h``2 − h2 +

√`4 + h4 + 2`2h2

`2 − h2

}=

`2 − h2

2hlog{

2h``2 − h2 +

√(`2 + h2)2

`2 − h2

}=

`2 − h2

2hlog[

2h`+ `2 + h2

`2 − h2

]=

`2 − h2

2hlog[

(h + `)2

(`+ h)(`− h)

]=

`2 − h2

2hlog[

h + `

`− h

]=

`2 − h2

2hlog[`+ h`− h

]


=`2 − h2

2hlog[

1 + h`

1− h`

]We know that, log

[1 + x1− x

]= 2

(x +

x3

3+

x5

5+ . . .

)=

(1− h2

`2

2h`2

)2[

h`+

13(

h`)3 +

15(

h`)5 + . . .

]

=

(1− h2

`2

2h`2

)2h`

[1 +

h2

`2 (13) +

h4

`4 (15) + . . .

]= `

(1− h2

`2

) [1 +

h2

`2 (13) +

h4

`4 (15) + . . .

]= `

[1 +

h2

`2 (13)− h2

`2

](Neglecting other higher powers )

a = `

[1− 2

3h2

`2

]Thus, Span = 2a

= 2`[

1− 23

h2

`2

].

Example 1.3.1. A uniform chain of length `, which can just bear a tension of n times its

weight is stretched between two points in the same horizontal line. Show that the least

possible sag in the middle is `[

n−√

n2 − 14

]Solution. Let A be (x, y). The tension at A,

T = n(weight of the chain)

that is T = wy

wy = n(w`)

y = n`. (1.3.3)

For any point of the catenary, we have

y2 = c2 + s2


Figure 1.3

Therefore for the point A

n2`2 = c2 + (`

2

2) (∵ s =

`

2and y = n`)

or c2 = `2[

n2 − 14

]c = `

√n2 − 1

4.

Hence, the sag in the middle is

CD = y− c = n`− `

[√n2 − 1

4

]

= `

[n−

√n2 − 1

4

]∴ CD = `

[n−

√n2 − 1

4

].

Example 1.3.2. If α, β are the inclinations to the horizon of the tangents at the extremities

of a catenary and ` the length of the portion, show that the height of one extremities over the


other is`sin

(α+β

2

)cos

(α−β

2

) , the two extremities being on one side of the vertex of the catenary.

Solution. Consider the portion PQ of the catenary. let P = (x2, y2) and Q = (x1, y1)

Tangents at P and Q make angles α and β respectively to the horizontal. Also arc PQ =

`(given). Suppose that arc CQ = s1 and arc CP = s2

=⇒ s2 − s1 = `. (1.3.4)

Figure 1.4

Using the formula

s = c tanψ

we have s1 = c tanβ

and s2 = c tanα


Therefore from equation (1.3.4)) we have,

c tanα− c tanβ = `

or c =`

tanα− tanβ.

(1.3.5)

Vertical height of one extremity P above Q

= y2 − y1

= c secα− c secβ

= c [secα− secβ]

=`

tanα− tanβ[

1cosα

− 1cosβ

]

=`

sinαcosα −

sinβcosβ

[1

cosα− 1

cosβ]

= `cosβ− cosα

sinαcosβ− cosαsinβ

= `2 sin( α+β

2 )sin( α−β2 )

sin(α− β)

= `2 sin( α+β

2 )sin( α−β2 )

sin(2α−β2 )

= `2 sin( α+β

2 )sin( α−β2 )

2 sin( α−β2 )cos( α−β

2 )

= `sin( α+β

2 )

cos( α−β2 )

.

Example 1.3.3. A given length 2s of uniform chain has to be hung between two points at

the same level and the tension has not to exceed the weight of a length b of the chain. Show

that the greatest span is√

b2 − s2log( b+sb−s ).


Solution. Let A be (x, y)

T = Tension at A

= wy = wb(Given).

∴ y = b.

Figure 1.5

For any point of the catenary we have

y2 = c2 + s2

=⇒ b2 = c2 + s2

c =√

b2 − s2.


Hence, the greatest span

= AB

= 2x

= 2c log(secφ + tanφ) (∵ x = Clog(secφ + tanφ))

= 2c log(yc+

sc) (∵ y = c secφ and s = c tanφ)

= 2c log(y + s

c)

= 2√

b2 − s2 log{

b + s√b2 − s2

}(∵ y = b&c =

√b2 − s2)

= 2√

b2 − s2 log{√

b + s√

b + s√b + s

√b− s

}= 2

√b2 − s2 log

{√b + s√b− s

}.

The greatest span

=√

b2 − s2 log{

b + sb− s

}.

Example 1.3.4. A heavy uniform string of length `, is suspended from a fixed point A and

its other end is pulled horizontally by a force equal to the weight of a length a of the string.

Show that the horizontal and vertical distance between A and B are (a sinh−1( `a )) and√`2 + a2 − a respectively.

Solution. Now as

H = wa ( Given ). (1.3.6)

Also, we have

H = wc. (1.3.7)

Form equations (1.3.6) and (1.3.7) we get

wa = wc⇒ c = a.


Figure 1.6

Now from the relation at the point A we have,

y2A = c2 + s2 = a2 + `2 (∵ c = a and s = `)

yA =√

a2 + l2,

and yB = c

but c = a

⇒ yB = a.

=⇒ yA − yB =√`2 + a2 − a.

Again fromsc= sinh(

xc) at the point A, we have ,

`

a= sinh(

xA

a)

or sin−1(`

a) = (

xA

a)

⇒ xA = a sinh−1(`

a).

Therefore xA − xB = a sinh−1(`

a)− 0 (∵ xB = 0)

xA − xB = a sinh−1(`

a).

Mohammad Arif

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Mohammad Arif

Pencil

Mohammad Arif

Typewriter

y_A-y_B

Mohammad Arif

Arrow

Mohammad Arif

Typewriter

X_A-X_B


Example 1.3.5. A rope of length 2` m, is suspended between two points at the same level

and the lowest point of rope is b m in below the point of suspension. Show that the horizontal

component of tension is W2b (`

2 − b2), W being the weight of the rope per meter of its length.

Solution. Now as we know that

y2 = c2 + s2.

At the point A(x, y), we have s = `

and y = b + c.

Therefore, we have (b + c)2 = c2 + `2

b2 + c2 + 2bc = c2 + `2

c =`2 − b2

2b.

Hence, H = Wc = W(`2 − b2

2b).

Example 1.3.6. Prove that when c is large compared with the span the catenary

approximates to a parabola.

Solution. As we know that equation of catenary is

y = c cosh(xc)

y = c[1 +12!(

xc)2 +

14!(

xc)4 + . . . ]

(∵ cosh(x) = [1 +x2

2!+

x4

4!+

x6

6!+ . . . ]).

Since x is small, c is large ∴xc

is small we get

y = c +(x)2

2c

y− c =(x)2

2c

=⇒ x2 = 2c(y− c).

Which is a parabola whose latus rectum is 2c and vertex (0, c), the lowest point of the

catenary.


1.4 The Parabolic Cable

Let the cable carry a uniform load of W0 per unit length. For CB,

The load = V = W0x

anddydx

= tanψ =VH

.

Now, we havedydx

=W0x

H∫dy =

W0

H

∫xdx

y =W0x2

2H+ c1.

Figure 1.7

If we choose as x− axis the horizontal line passing through C(0, 0) then, c1 = 0.

Therefore y =W0x2

2H.

This is a equation of parabola with axis vertical and vertex at origin.


The tension in the parabolic cable is given by

T =√

H2 + (W0x)2.

Example 1.4.1. If a cable is loaded with a uniform horizontal load of W and has a span a

and sag h, show that H = Wa2

8h and that

T =Wa2

(1 +

a2

16h2

) 12

,

where H and T are the tensions at lowest and highest point of the cable,respectively.

Solution. We know that, the equation of parabolic cable is

y =Wx2

2H.

Here y = h and x =a2

.

Now, h =W( a

2)2

2H

h =Wa2

8H

H =Wa2

8h.

Figure 1.8


The tension in the parabolic cable is given by

T =√

H2 + (Wx)2

T =

√(

Wa2

8h)2 + (W

a2)2

T =

√W2a4

64h2 +W2a2

4

T =Wa2

√1 +

a2

16h2 .

Example 1.4.2. A length ` of a uniform chain has one end fixed at a heights h above a

rough table and rests in a vertical plane so that a portion of it lies in a straight line on the

table. Prove that if the chain is on the point of slipping, the length on the table is

`+ µh−√[(µ2 + 1)h2 + 2`hµ],

where µ is the coefficient of friction.

Solution. Resolving forces horizontally and vertically, we have

R = Wx (1.4.1)

µR = T0 = Wc. (1.4.2)

From (1.4.1) and (1.4.2), we have

µ(Wx) = Wc,

⇒ c = µx. (1.4.3)

Using the formula y2 = c2 + s2 from the point A we have (∵ y for A =

c + h and s for A = `− x)


Figure 1.9

y2 = c2 + s2

(C + h)2 = (µx)2 + (`− x)2

(µx + h)2 = (µx)2 + (`− x)2

(µx)2 + h2 = (µx)2 + `2 + x2 − 2`x + 2hµx

x2 − 2x (`+ µh) + (`2 − µ2) = 0

Therefore, x =2(`+ µh)±

√4(`+ µh)2 − 4(`2 − h2)

2

x = (`+ µh)±√`2 + µ2h2 + 2µh`− `2 + h2

x = (`+ µh)±√

h2(µ2 + 1) + 2µh`

Thus, x = `+ µh−√(µ2 + 1)h2 + 2µh`.

The positive sign before radical is neglected for otherwise x > `, which is not possible.

1.5 Tension in a Cable

We write,

dsdx

=

√1 + (

dydx

)2

=

√1 + sinh2(

xc) (∵

sc= tan ψ =

dydx

= sinh(xc))


Figure 1.10

dsdx

= cosh (xc). (1.5.1)

Also, H = T cosψ

=⇒ T = Hsec ψ.

For any curvedxds

= cosψ

=⇒ dsdx

= secψ.

=⇒ T = H cosh (xc) (from equation 1.5.1).

1.6 Maximum Tension in the Cable

To prove that the maximum tension in the cable of weight W per unit length, span

2a and length 2` is(

Wa√

a6(`−a)

), approximately.

Solution. We know that the maximum tension occurs at A and B. Using the formula

s = c sinh (xc), at the point A


Figure 1.11

s = ` and x = a

We have,

` = c sinh (ac)

Or, ` = c[

ac+ (

ac)3 1

3!+ (

ac)5 1

5!+ . . . ] (∵ sinh x = x +

x3

3!+

x5

5!+ . . . )

` = a +a3

6c2 ( Neglecting the higher power terms )

`− a =a3

6c2

c2 =a3

6(`− a)

c =

√a3

6(`− a).


We know that, Tension in the cable is given by

T = Hc osh(xc) = H cosh (

ac) (∵ x = a).

Ifac

is small or c is large. Then, Tmax = H = Wc (approximitely)

= W

√a3

6(`− a)

Tmax = Wa√

a6(`− a)

.

Example 1.6.1. A cable 200m long hangs between two points at the same height. The sag

is 20m and the tension at either point of suspension is 120 kg Wt. Find the total weight of

the cable.

Solution. Given that T = 120 kg Wt

Figure 1.12


and y− c = sag = 20m.

2s = 200m Given

s = 100m.

y− c = 20

y = 20 + c.

We know that, y2 = c2 + s2

(20 + c)2 = c2 + (100)2

400 + c2 + 40c = c2 + 10000

40c = 9600

c =960040

= 240

y = c + 20

= 240 + 20

y = 260.

As, T = Wy

120 = (W)(260)

W =120260

=6

13

Total weight = 200× 613

= W (2s)

=120013

.

Example 1.6.2. A uniform cable hangs across two smooth pegs at the same height, the ends

hanging down vertically. If the free ends are each 12m long and the tangent to the catenary

at each peg makes an angle of 60o with the horizontal, find the total length of the cable.


Solution. At the point B, we have

y = h + c = 12

and y = c secφ.

Now, y = c sec60o

y = 2c

12 = 2c

c = 6.

As, s = c tanψ

= 6 tan60o

s = 6 (√

3).

Figure 1.13

Total Length of the Cable = 2s + 2y

= 2× 6√

3 + 2× 12

= 12√

3 + 24

= 12(√

3 + 2).


Example 1.6.3. The extremities of a heavy string of length 2` and weight 2`ω are attached

to two small rings which can slide on a fixed horizontal wire. Each of these rings are acted

on by a horizontal force equal to `ω. Show that the distance apart of the rings is

2` log{

1 +√

2}

.

Solution.

H = ω`

But H = ωc

⇒ ω` = ωc

⇒ ` = c

We know that S = c tan ψ

` = ` tan ψ

tan ψ = 1

⇒ ψ = 45o

Span = AB = 2` log{sec ψ + tan ψ}

Span = 2` log{√

2 + 1}.

Example 1.6.4. If T be the tension at any point P of a catenary and T0 at the lowest point

C, prove that

T2 − T20 = W2,

W being the weight of the arc CP of the catenary.

Solution. We know that

T = wy

Mohammad Arif

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Mohammad Arif

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is

Mohammad Arif

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where,


and T0 = The tension at the lowest point C

= wc.

Now, T2 − T20 = w2y2 − w2c2

= w2(y2 − c2).

But y2 = c2 + S2

T2 − T20 = w2(y2 − c2)

= w2(c2 + s2 − c2)

= w2s2

= W2 (∵ W = ws).

Example 1.6.5. A uniform chain is hung up from two points at the same level and distant

2a apart. If z is the sag at the middle show that

z = c(cosh(ac)− 1).

If z is small compared to a show that

2zc = a2 (nearly).

Solution. We know that equation of the catenary is

y = c cosh(xc).

Since the span 2x = 2a, x = a for the point A Given that z is the sag in the middle.

∴ z = sag = DC = y− c

= c cosh(xc)− c

= c[cosh(xc)− 1] (∵ x = a at A)

= c[cosh(ac)− 1].


Figure 1.14

Again z is small compared with a, therefore c must be very large. Now

z = c[cosh(ac)− 1]

= c[1 +122 (

ac)2]

14!(

ac)4 + · · · − 1

=a2

2c

⇒ 2zc = a2 (nearly).

Example 1.6.6. A uniform chain of length ` is to be suspended from two points A and B

in the same horizontal line so that terminal tension is n times that at the lowest point. Show

that span AB must be`√

n2 − 1log{n +

√n2 − 1}.

Solution. Let the coordinate of the A be (x, y). Then we have

span = AB = 2c log(sec ψ + tan ϕ)

where ψ is the inclination of the tangent at A to x–axis. Given that


Figure 1.15

Tension at A = n( Tension at the lowest point c)

= nT0

⇒ T = nT0 = n(ωc) (∵ T0 = ωc)

⇒ ωy = nT0 = n(ωc) (∵ T = ωy)

⇒ y = nc

⇒ c sec ψ = nc (∵ y = c sec ψ)

⇒ n = sec ψ (1.6.1)

∴ tan ψ =√

sec2 ψ− 1

tan ψ =√

n2 − 1. (1.6.2)

Again,

the arc CA = half length of the chain

s =`

2,

we have,`

2= c tan ψ (∵ s = c tan ψ)

=`

2 tan ψ

=`

2√

sec2ψ− 1


c =`

2√

n2 − 1. (1.6.3)

Hence,

AB = 2c log{sec ψ + tan ψ}

=`√

n2 − 1log{n +

√n2 − 1}.

Example 1.6.7. A telegraph wire is made of a given material and such a length L is

stretched between two posts, distant d apart and of the same height as will produce the least

possible tension at the posts. Show that

L =dλ

sinh λ

where λ is given by the equation

λ tan λ = 1.

Solution. Let T be the tension at either post. Then the catenary has x = d2 and s = L

2 . We

know that the tension in the cable is given by

T = H cosh(xc)

= H cosh(d2c

)

= Wc cosh(d2c

)

For maximum or minimum of T, we have

dTdc

= 0

=⇒ ddc

(W cosh(

d2c

)

)= 0


Figure 1.16

=⇒W[

cosh(d2c

) + c sinh(d2c

)(−d2c2 )

]= 0

or cosh(d2c

)− d2c

sinh(d2c

) = 0

ord2c

tanh(d2c

) = 1.

But λ tanh λ = 1 ( Given)

⇒ d2c

tanh(d2c

) = λ tanh λ

⇒ λ =d2c

.

⇒ c =d

2λ.

As we know, s = c sinh(xc).

Here, s =L2

x =d2

and c =d

2λ

Now, (L2) =

d2λ

sinh(d2c

)

⇒ L =dλ

sinh λ.

Example 1.6.8. If the length of the uniform chain suspended between points at the same


level is adjusted so that the tension at the points of support is minimum for that particular

span 2d. Show that the equation to determine c is

coth(dc) =

dc

.

Solution. We know that tension cable is given by

T = H cosh(xc)

⇒ T = H cosh(dc) = Wc cosh(

dc).

For maximum or minimum of T,dTdc

= 0

=⇒ dTdc

= W[c sinh(dc)(−dc2 ) + cosh(

dc)] = 0

=⇒ = W[c sinh(dc)(−dc2 ) + cosh(

dc)] = 0

=⇒ cosh(dc) = (

dc)c sinh(

dc)

cot h(dc) = (

dc).

Moment of Analytic Statics

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