# Equilibrium Class 11 Physics | Notes

## Equilibrium:

An object is said to be in equilibrium if the sum of all translational forces and rotational forces acting on it is zero.

i.e. for an object to be in equilibrium,

$\Sigma$F +$\Sigma \tau$ = 0 where $\tau$ is torque.

### Translational equilibrium:

An object is said to be in translational equilibrium if the sum of all translational forces acting on it is zero. i.e.

For an object to be in translational equilibrium

$\Sigma$F  = 0

### Rotational equilibrium:

An object is said to be in rotational equilibrium if the sum of all rotational forces acting on it is zero. i.e.

For an object to be in rotational equilibrium

$\Sigma \tau$= 0

## Moment of force:

Moment of force is defined as the product of force acting on an object and its perpendicular distance from the axis of rotation.

From figure,

Moment of force = F × OA

[ where the axis of rotation passes through O.]

Or, Moment of force = F × OB sin$\theta$ [$\because$ sin$\theta$ = $\frac{OA}{OB}$ $\therefore$ OA = OB sin$\theta$]

$\therefore$ Moment of force = F r sin$\theta$

### Principle of Moment:

Principle of moment states that for an object to be in rotational equilibrium, the sum of all anti-clockwise moments is equal to the sum of all clockwise moments.

Let us consider a rod pivoted at point ‘O’ is in rotational equilibrium under the action of different forces F1, F2, F3, F4 and F5 at points A, B, C, D and E respectively. The forces F1, F2, F3 try to rotate the rod in anti-clockwise direction whereas forces F4 and F5 try to rotate the rod in clockwise direction.

According to the sign convention, the anti-clockwise moment is taken as positive whereas the clockwise moment is taken as negative.

For the given rod to be in rotational equilibrium, we have

F1×OA +F2×OB+ F3×OC –F4×OD – F5 × OE = 0

F1×OA +F2×OB+ F3×OC = F4×OD + F5 × OE

$\therefore$ Sum of anticlockwise moment = Sum of clockwise moment

i.e. for an object to be in rotational equilibrium, the sum of moments of all forces must be zero.

### Torque ($\tau$):

Torque is the turning effect of force.

Mathematically, it can be defined as the product of force and its perpendicular distance from the axis of rotation.

In vector notation, above result can be expressed as

$\overrightarrow{\tau }$ = $\overrightarrow{r}$ × $\overrightarrow{F}$

Special Cases:

1. When $\theta$ = 0o

Torque, $\tau$ = F r sin0o = 0 (Minimum)

1. When $\theta$ = 90o,

Torque, $\tau$ = F r sin90o = F.r (Maximum)

### Parallel Forces:

Two forces are said to be parallel if their lines of action are parallel to each other.

### Couple:

Two equal and unlike parallel forces acting on an object at different points is known as couple.

### Torque due to couple:

Let us consider a uniform rod AB is pivoted at point O. Let two equal and unlike parallel forces are acting at point A and B.

Moment of force due to force at A = F× OA

Moment of force due to force at B = F× OB

Now,

Total moment of force = F× OA + F× OB

= F (OA + OB)

= F×AB

Since total moment of force = Torque

i.e. Torque due to couple = F×AB

Mathematically, the torque due to a couple can also be defined as the product of one of the forces of the couple and perpendicular distance between their lines of action.

## Centre of Gravity:

The centre of gravity of a rigid object is defined as the point at which the whole weight of the body is supposed to be acted.

### Centre of mass and it’s expression:

The centre of mass of a rigid body is defined as the point where the whole mass of the body is supposed to be concentrated.

Let us consider a rigid body of mass (M) consisting of large no. of particles of mass m1, m2, m3, … mn. Let the position coordinates of constituent particles m1, m2, m3, … mn are (x1, y1), (x2, y2), (x3, y3), … (xn, yn) respectively. Also consider the position coordinates of it’s centre of mass is (x, y).

Let F1, F2, F3, … Fn are the forces acting on particles m1, m2, m3, … mn. Then total force acting on the object can be expressed as

F = F1 + F2 + F3 +… + Fn

M×a = m1a1 + m2a2 + m3a3 + … + mnan   ……(i)

We know, acceleration can be expressed as

a = $\frac{{{d}^{2}}x}{d{{t}^{2}}}$

Taking x-component, of equation (i) we get,

M × $\frac{{{d}^{2}}X}{d{{t}^{2}}}$ = m1$\frac{{{d}^{2}}{{x}_{1}}}{d{{t}^{2}}}$ + m2 $\frac{{{d}^{2}}{{x}_{2}}}{d{{t}^{2}}}$  + m3 $\frac{{{d}^{2}}{{x}_{3}}}{d{{t}^{2}}}$ …………..+ mn  $\frac{{{d}^{2}}{{x}_{n}}}{d{{t}^{2}}}$

Integrating above equation twice, we get

M.X = m1 x1 + m2 x2 + m3 x3 ………….+ mn xn

Or, X = $\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}\ldots \ldots \ldots \ldots .+{{m}_{n}}{{x}_{n}}}{M}$

Here, M = $\Sigma$m = m1 + m2+ m3 ………….+ mn

$\therefore$ X = $\frac{\Sigma mx}{M}$…………….(ii)

Similarly, taking Y-component of equation (i), we get,

$\therefore$ Y = $\frac{\Sigma my}{M}$…………….(iii)

$\therefore$ The position of co-ordinates of centre of mass is

(X,Y) = $\left( \frac{\Sigma mx}{M},\frac{\Sigma my}{M} \right)$

### Centre of mass of two bodies system:

Let us consider a system of two bodies of unequal masses m1 and m2 such that m1>m2 is connected by a uniform rod as shown in fig above. Let ‘P’ be the position of the centre of mass which is at a distance of x1 from m1 and x2 from m2.

If the system is in rotational equilibrium about it’s centre of mass (P) then we can write,

Sum of anticlockwise moment = Sum of clockwise moment

m1g. x1 = m2g. x2

m1x1 = m2x2

Or, x1 = $\frac{{{m}_{2}}}{m{}_{1}}$ x2 …..(i)

Since, $\frac{{{m}_{2}}}{{{m}_{1}}}$< 1 $\therefore$ x1 < x2

Similarly,

Or, x2 = $\frac{{{m}_{1}}}{{{m}_{2}}}$ x1 …..(i)

Since, $\frac{{{m}_{1}}}{{{m}_{2}}}$ > 1 $\therefore$ x2 > x1

Equations (i) and (ii) shows that the centre of mass of two body systems of unequal masses shifts towards the larger mass.

## State of equilibrium

### Stable Equilibrium:

An object is said to be in stable equilibrium if the object regains its original equilibrium position when it is slightly disturbed.

### Unstable Equilibrium:

An object is said to be in unstable equilibrium if the object does not regain its original equilibrium position when it is slightly disturbed.

### Neutral Equilibrium:

An object is said to be in neutral equilibrium if the object always seems to be in equilibrium. In this case the height of the centre of gravity of the object remains the same.

### Conditions for stable equilibrium:

The conditions for stable equilibrium are;

1. The centre of gravity of an object should lie as low as possible.

2. The base area of an object should be as maximum as possible.

3. The downward perpendicular line drawn from the centre of gravity should pass through the centre of base.

The object in the figure (c) is in the most stable equilibrium state.

### Equilibrium of concurrent forces:

Two or more forces acting on an object are said to be concurrent if their lines of action pass through the same point. If the sum of all concurrent forces is zero then the object is said to be in equilibrium.

## Numerical Problems

Q. 1. Two forces of 1.5 N and 2.0 N act vertically at the two ends of a metre scale. Where and in what direction should a force be applied so that the scale remains horizontally stable?

Solution:

First force, F1 = 1.5 N

Second force, F2 = 2.0 N

Length of metre scale, l = 1 m

For the scale to be horizontally stable,

Sum of translational forces must be zero.

$\Sigma$F = 0

Or, F = F1 + F

Or, F = 1.5 + 2 = 3.5 N

The force 3.5 N must be applied in the opposite direction.

For rotational equilibrium,

Sum of anticlockwise moment = Sum of clockwise moment

F1 . x = F2 . (1–x)

Or, 1.5x = 2-2x

Or, 2x + 1.5x = 2

Or, 3.5x = 2

Or, x = $\frac{2}{3.5}$

$\therefore$ x = 0.57 m from F1 = 1.5 N

Q. 2. Two people are carrying a uniform wooden board that is 3 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift?

Here,

Length, l = 3m

Weight, W = 160 N

X = ?

For translational equilibrium, sum of all translational forces must be zero.

$\Sigma$F = 0

W  = F1 + F2

F2 =  W – F

F2 =  160 – 60 = 100 N

For rotational equilibrium,

Sum of anti-clockwise moment = Sum of clockwise moment

F1 ×1.5 = F2 ×

Or, x = $\frac{{{F}_{1}}\times 1.5}{{{F}_{2}}}$

Or, x = $\frac{60\times 1.5}{100}$

$\therefore$ x = 0.9 m from centre of gravity

OR 1.5 + 0.9 = 2.4 m from F1.

Q. 3. A roller whose diameter is 1.0 m weighs 360 N. What horizontal force is necessary to pull the roller over the brick 0.1 m high when the force is applied at the centre?

Solution,

Diameter of roller, (d) = 1 m

Radius, r = 0.5 m

Weight of roller, W = 360 N

Height of brick, h = 0.1 m

To pull the roller over a brick,

Clockwise moment about B $\ge$ anti-clockwise moment about B

Moment of F about B $\ge$  moment of W about B

Or, F ×BC $\ge$ W × AB

Or, F ×0.4 $\ge$ 360 × $\sqrt{O{{B}^{2}}-O{{A}^{2}}}$

Or, F ×0.4 $\ge$ 360 × $\sqrt{{{(0.5)}^{2}}-{{(0.4)}^{2}}}$

$\therefore$ F  $\ge$ 270 N

Also Read: Circular Motion Notes