Lecture Note on Unit-I of Mechanics (DSE 3T) by - Narajole ...

Lecture Note on Unit-I of Mechanics (DSE 3T) by

Dr. Shreyasi Jana, Assistant Professor, Dept. of Mathematics

Narajole Raj College, WB, India

DSE3T(SEM VI), Unit-I, Coplanar Forces, Equilibrium of Coplanar Forces

Syllabus

Mechanics (DSE 3T)

Unit-I

Co-planar forces. Astatic equilibrium. Friction. Equilibrium of a particle on a rough

curve. Virtual work. Forces in three dimensions. General conditions of equilibrium.

Centre of gravity for different bodies. Stable and unstable equilibrium.

Coplanar Forces: Coplanar forces means the forces in a plane. When several forces act

on a body, then they are called a force system or a system of forces. In a system in which

all the forces lie in the same plane, it is known as a coplanar force system.

If the forces are having a common line of action, then they are known as collinear

whereas if the forces intersect at a common point, then they are known as concurrent.

Example: If two people stand at the ends of a rope and pull on it.





Composition and equilibrium of coplanar forces : In this part we show how several

forces whose lines of action lie on a plane and pass through one point, can be balanced

by a single force with line of action passing through the same point. The method is to

find the one force, called the resultant, which is equal to the sum of the original forces,

and then to balance this resultant with an equal and opposite force, called the

equilibrant. The resultant of the original forces is found by the method of vector

addition.

When a number of forces, F1, F2, F3, for example, is acting on an object and lines of

actions of all the forces pass through one point O as shown in Figure below, the

resultant of the forces, R, can be found by arranging the vectors (forces) by forming the

sides of a polygon as shown in Figure below. The vector that connects the tail of the first

vector with the tip of the last one is the resultant of the vectors being added.

F3 F2 F3

F1

F2

O R

F1

O

Rectangular resolution and equilibrium of coplanar forces: In this part we show how

a single force may be resolved into two mutually perpendicular components which

together may be regarded as equivalent to the given force. As shown in Figure below, a

force vector F can be resolved along mutually perpendicular x- and y-axes.





Y

Fy

F

O X

Fx

By drawing perpendicular lines from the tip of the force vector F to the axes, the

projections Fx and Fy along the x- and y-axes are the rectangular components of the

force F. The combined effect of Fx along x-axis and Fy along y-axis is the same as that of

F. If there is more than one force acting on an object, all the forces can be resolved in

rectangular components. The sum of all the x components will be the x-component of

the resultant force and the sum of the y-components will be the y-component of the

resultant.

Condition of Equilibrium:

Theorem: Triangle of Forces. If three forces acting at a point, be represented in

magnitude and direction by the sides taken in order, of a triangle, then the forces will

be in equilibrium.





Theorem: Polygon of Forces. If any number of forces acting at a point be represented in

magnitude and direction by the sides of a closed polygon taken in order, then the forces

will be in equilibrium. The converse of the polygon forces is not true.

Reduction of Coplanar forces:

Theorem I: Any force acting at a point of a rigid body is equivalent to an equal and

parallel force acting at any other arbitrarily chosen point of the body, together with a

couple whose moment is equal to the moment of the given force about the chosen point.

Proof: Refer to textbook.

Theorem II: Reduction of a system of Coplanar forces:

Any system of coplanar forces acting on a rigid body can be reduced ultimately to

either a single force or a single couple unless it is in equilibrium.


Theorem III: Any system of coplanar forces acting on a rigid body can be reduced to a

single resultant force acting at any arbitrarily chosen point in the plane , together with a

resultant single couple whose moment is equal to the algebraic sum of the moments of

the given forces about the chosen point. The chosen point is a point on the rigid body in

the plane of the forces.






Ex.1 If a system of forces in one plane reduces to a couple whose moment is G and

when each force is turned round its point of application through a right-angle it

reduces to a couple H; prove that when each force is turned through an angle ; the

system is equivalent to a couple whose moment is 𝑮 𝐜𝐨𝐬 𝜶 + 𝑯 𝐬𝐢𝐧 𝜶.

Sol. Let the forces be given by Fr acting through the point (𝑥𝑟 , 𝑦𝑟) inclined at 𝛼𝑟 to axis

of x (where r = 1, 2, 3, ….), then

𝑅𝑥 = ∑𝐹𝑟 cos 𝛼𝑟 = 0; … … … … … … (1)

𝑅𝑦 = ∑𝐹𝑟 sin 𝛼𝑟 = 0; … … … … … … (2)

𝐺 = ∑(𝑥𝑟 sin 𝛼𝑟 − yrcos 𝛼𝑟) 𝐹𝑟; … … … … … … (3)

at the origin.

If each force be rotated through a right angle in the positive sense, say, resultant is R’

having components 𝑅′𝑥 and 𝑅′𝑦, then

𝑅′𝑥 = ∑𝐹𝑟 cos( 𝛼𝑟 + 𝜋

2) = −∑𝐹𝑟 sin 𝛼𝑟 = 0; 𝑏𝑦 (2).

𝑅′𝑦 = ∑𝐹𝑟 sin( 𝛼𝑟 + 𝜋

2) = ∑𝐹𝑟 cos 𝛼𝑟 = 0; 𝑏𝑦 (1).

and the resultant moment of the forces in their positions about O is given by

𝐻 = ∑{𝑥𝑟 sin(𝛼𝑟 + π/2 ) − yrcos(𝛼𝑟 + 𝜋/2)} 𝐹𝑟 =

∑(𝑥𝑟 sin 𝛼𝑟 + yrcos 𝛼𝑟) 𝐹𝑟 … (4)

Now if the directions are changed by α in the same sense and the resultant is R’’ with

components 𝑅′′𝑥 and 𝑅′′𝑦, then

𝑅′′𝑥 = ∑𝐹𝑟 cos(𝛼𝑟 + 𝛼) = ∑𝐹𝑟 (cos 𝛼𝑟 cos 𝛼 − sin 𝛼𝑟 sin 𝛼)

= cos 𝛼∑𝐹𝑟 cos 𝛼𝑟 − sin 𝛼∑𝐹𝑟 sin 𝛼𝑟 = 0; 𝑏𝑦 (1) 𝑎𝑛𝑑 (2);

and

𝑅′′𝑦 = ∑𝐹𝑟 sin(𝛼𝑟 + 𝛼) = ∑𝐹𝑟 (sin 𝛼𝑟 cos 𝛼 + cos 𝛼𝑟 sin 𝛼)

= cos 𝛼∑𝐹𝑟 sin 𝛼𝑟 + sin 𝛼∑𝐹𝑟 sin 𝛼𝑟 = 0; 𝑏𝑦 (1) 𝑎𝑛𝑑 (2).

And the resultant moment of the forces in their final positions is given by

𝐺′′ = ∑{𝑥𝑟 sin(𝛼𝑟 + 𝛼) − yrcos(𝛼𝑟 + 𝛼)}𝐹𝑟

= ∑{cos 𝛼 (𝑥𝑟 sin 𝛼𝑟 − yrcos 𝛼𝑟) + sin 𝛼 (𝑥𝑟 sin 𝛼𝑟 + yrcos 𝛼𝑟)}𝐹𝑟

= 𝐺 cos 𝛼 + 𝐻 sin 𝛼. (proved)





Equilibrium of a Rigid Body acted on by three forces:

Theorem: If three forces, acting in one plane upon a rigid body, keep it in

equilibrium, they must either meet in a point or be parallel.

Proof: To prove this, suppose that the given forces P, Q and R are not all parallel. Then

at least two of them, say, P and Q must meet in a point O. R must balance the resultant

of P and Q as the three forces P, Q and R are in equilibrium. Since the resultant of P and

Q passes through O, R must also pass through O, i.e. the three forces are concurrent.

But if two of them are parallel, their resultant is parallel to them, and therefore so is the

force, which is to balance them, i.e. the third force is also parallel to them.

Exercise: 1. Forces equal to 3P, 7P and 5P act along the sides AB, BC, and CA of an

equilateral triangle ABC; find the magnitude, direction, and line of action of the

resultant.

2. A system of forces in one plane is equivalent to a couple of moment G. If the line

of action of each force is turned about its point of application in the same direction

through a right angle, prove that the new system is also equivalent to a couple. Also

prove that if the moment of this couple be H and if the lines of action of the original

forces be each turned through 2tan-1 [H/G], they would still be equivalent to a couple

of moment G.

P

Q

R

O

P Q

R





General Condition of Equilibrium: Analytical Method:

Theorem1: The resultant R and the couple G must separately vanish.

Proof: We know that the system of forces acting at different points may be reduced to a

force R through an arbitrary point O and a couple G. Since a finite force R cannot

balance a couple G, it is necessary for equilibrium that the resultant force R and the

couple G should separately vanish. The vanishing of R involves the condition that Rx =

0 and Ry = 0.

Note: Now we arrive at the following necessary and sufficient conditions of

equilibrium. A system of forces in a plane will be in equilibrium if the algebraic sums of

their resolved part in any two perpendicular directions vanish, and if the algebraic sum

of their moments about any point also vanishes.

Theorem 2: A system of forces in a plane will be in equilibrium if the algebraic sum

of the moments of all the forces with respect to each of three non-collinear points is

zero.

Proof: To proving this theorem, consider the origin at one of the three points and let the

coordinates of the other points be (x1, y1) and (x2, y2).

If we denotes the algebraic sum of the moments of the forces about the above points by

G, G’, G’’

then by G = 0,

G’ = G – x1Ry + y1Rx = 0,

and G’’ = G – x2Ry + y2Rx = 0.

These reduce to – x1Ry + y1Rx = 0 and

– x2Ry + y2Rx = 0

Since the three points are not collinear, y1 /x1 ≠ y2 /x2.

and the above conditions reduce to

Rx = 0, Ry = 0 and G = 0, a set of conditions involved in Theorem 1.





Theorem 3: A system of forces in a plane will be in equilibrium if the algebraic sum

of the moments about each of any two different points is zero and the algebraic sum

of the resolved parts of the forces in any given direction not perpendicular to the line

joining the given points is zero.

Proof: Let the two different points be A and B.

As before we take origin at A and x-axis along the given direction.

Let the coordinates of B be (x1, y1).

Then the conditions are

G = 0,

G’ = G – x1Ry + y1Rx = 0,

Rx = 0,

which lead to Rx = 0, Ry = 0 and G = 0 provided that x1 is not zero, a set of conditions

involved in Theorem 1.

Example. Two equal uniform rods, AB, AC, each of weight W, are freely joined at A

and rest with the extremities B and C on the inside of a smooth circular hoop, whose

radius is greater than the length of either rod, the whole being in a vertical plane, and

the middle points of the rods being joined by a height string; show that, if the string

is stretched, its tension is W(tanα – 2tanβ), where 2α is the angle between the rods,

and the angle either rod subtends at the centre.

Sol. Let O be the centre of the circular hoop O BFC, AB, AC the equal rods and DE the

string joining their middle points.

O

β

A

D α E

B C

F





If R be the pressure of the hoop on either rod, then resolving vertically for the system

consisting of the two rods

𝑊 = R cos 𝛽 … … … (1)

Taking moments about A for either rod, suppose of length 2l,

we have,

𝑅. 2𝑙 sin(𝛼 − 𝛽) = 𝑊. 𝑙 sin 𝛼 + 𝑇. 𝑙 cos 𝛼

𝑇 = 2𝑅 sin(𝛼−𝛽)−𝑊 sin 𝛼

cos 𝛼

= 2𝑊 sin(𝛼−𝛽)−𝑊 sin 𝛼

cos 𝛽 cos 𝛼

= 𝑊(tan 𝛼 − 2 tan 𝛽). (proved)

Exercise:

1. Three equal uniform rods, each of weight W, are smoothly jointed so as to form an

equilateral triangle. If the system be supported at the middle point of one of the rods,

√𝟑W/6, and that at each of the others is show that the action at the lowest angle is

W√(𝟏𝟑/12).

2. An elliptic lamina is acted upon at the extremities of pairs of conjugate diameters

by forces in its own plane tending outwards and normal to its edge; show that there

will be equilibrium if the force at the end of each diameter is proportional to the

conjugate diameter.




DSE3T(SEM VI), Unit-I, Astatic Equilibrium, Friction

Astatic Equilibrium: Let a given set of force be in equilibrium. Now if each force be

turned about its point of application in the plane of forces through the same angle in the

same sense and after rotation the given system be still in equilibrium whatever be the

common angle of rotation. Then it is called astatic equilibrium.

Astatic Centre: Again if a system of coplanar forces acting at different points of a body

have a single resultant force and if each force be turned into the plane about its point of

application through the same angle in the same sense, there resultant will always passes

through a fixed point in the plane. That fixed point is known as astatic centre.

Friction: Friction is the force resisting the relative motion of solid surfaces, fluid layers,

and material elements sliding against each other. There are several types of friction.

Static friction is friction between two or more solid objects that are not moving relative

to each other. For example, static friction can prevent an object from sliding down a

sloped surface.

https://en.wikipedia.org/wiki/Force

https://en.wikipedia.org/wiki/Sliding_(motion)

Lecture Note on Unit-I of Mechanics (DSE 3T) by - Narajole ...

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