Class 11 Equilibrium 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,  

Σ F + Σ τ = 0 where τ 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 

Σ 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 

Σ τ = 0 

Moment of force: 

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

Principle of Moment: 

Principle of moment states that for an object to be in rotational equilibrium, 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 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 

∴ Sum of anti-clockwise moment = Sum of clockwise moment 

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

Torque (τ): 

Torque is the turning effect of force. 

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

In vector notation, above result can be expressed as  

^→τ = ^→r ×^→F 

Special Cases: 

(i) When θ = 0^o,

Torque, τ = F r sin0^o = 0 (Minimum) 

(ii) When θ = 90^o, 

Torque, τ = F r sin90^o = 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 couple can also be defined as the product one of the forces of 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) consists of large no. of particles of mass m₁, m₂, m₃, … m_n. Let the  position coordinates of constituent particles m₁, m₂, m₃, … m_n are (x₁, y₁), (x₂, y₂), (x₃, y₃), … (x_n, y_n)  respectively. Also consider the position coordinates of it’s centre of mass is (x, y). 

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

F = F₁ + F₂ + F₃ +… + Fn 

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

We know, acceleration can be expressed as  

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 uniform  rod as shown in fig above. Let ‘P’ be the position of 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 anti-clockwise moment = Sum of clockwise moment 

Equations (i) and (ii) shows that the centre of mass of two body system 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 it’s 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 it’s 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 height of centre of gravity of the object remains 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 centre of gravity should pass through centre of base.

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

Equilibrium of concurrent forces:  

Two or more forces acting on an object is said to be concurrent if their lines of action passes 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.29. 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. 

Σ F = 0  

Or, F = F₁ + F2 

Or, F = 1.5 + 2 = 3.5 N  

The force 3.5 N must be applied in opposite direction. 

For rotational equilibrium, 

Q. 30. Two people are carrying a uniform wooden board that is 3 m long and weights 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. 

Σ F = 0 

W = F₁ + F2 

F₂ = W – F1  

F₂ = 160 – 60 = 100 N 

For rotational equilibrium,  

Q. 31. A roller whose diameter is 1.0 m weights 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? Ans: 270 N 

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,