# Equilibrium of Torques (F.Sc – Physics – Chapter 2)

* Equilibrium of Torques – Second condition of equilibrium*: let two equal and opposite forces act on a rigid body as shown in Fig. 2.17. although the first condition of equilibrium is satisfied, yet it may rotate having clockwise turning effect. As discussed earlier, for angular acceleration to be zero, the net torque acting on the body should be zero. Thus for a body in equilibrium, the vector sum of all the torques acting on it about any arbitrary axis should be zero. This is known as second condition of equilibrium. Mathematically it is written as

By convention, the counter clockwise torques are taken as positive and clockwise torques as negative. An axis is chosen for calculating the torques. The position of the axis is quite arbitrary. Axis can be chosen anywhere which is convenient in applying the torque equation. A most helpful point of rotation is the one through which lines of action of several forces pass.

We are now in a position to state the complete requirements for a body to be in equilibrium, which are

(1)

When 1^{st} condition is satisfied, there is no linear acceleration and body will be in translational equilibrium. When 2^{nd} condition is satisfied, there is no angular acceleration and body will be in rotational equilibrium.

For a body to be in complete equilibrium, both conditions should be satisfied, i.e., both linear acceleration and angular acceleration should be zero.

If a body is at rest, it is said to be in static equilibrium but if the body is moving with uniform velocity, it is said to be in dynamic equilibrium.

We will restrict the applications of above mentioned conditions of equilibrium to situations in which all the forces lie in a common plane. Such forces are said to be coplanar. We will also assume that these forces lie in the xy-plane.

If there are more than one object in equilibrium in a given problem, one object is selected at a time to apply the conditions of equilibrium.

Can You Do?

with your nose touching the end of the door, put your feet astride the door and try to rise up on your toes.

**Example 2.8:** a uniform beam of 200N si supported horizontally as shown. If the breaking tension of the rope is 400N, how far can the man of weight 400 N walk from point A on the beam as shown in Fig. 2.18?

**Solution:** let breaking point be at a distance d from the pivot A. the force diagram of the situation is given in fig. 2.19. by applying 2^{nd} condition of equilibrium about point A

400 N x 6 m – 400 N x d – 200 N x 3 m = 0

Or 400 N x d = 2400 Nm – 600 Nm = 1800 Nm

**Example 2.9:** A boy weighing 300 N is standing at the edge of a uniform diving board 4.0m in length. The weight of the board is 200 N. (fig. 2.20 a). find the forces exerted by pedestals on the board.

**Solution:** we isolate the diving board which is in equilibrium under the action of forces shown in the force diagram (fig. 2.20 b). note that the weight 200 N of the uniform diving board is shown to act at point C, the centre of gravity which is taken as the mid-point of the board, R1 and R2 are the reaction forces exerted by the pedestals on the board. A little consideration will show that R1 is in the wrong direction, because the board must be actually pressed down in order to keep it in equilibrium. We shall see that this assumption will be automatically corrected by calculations.

Let us now apply conditions of equilibrium

Σ Fx = 0 (No x-directed forces)

Σ Fy = 0 R1 + R2 – 300 – 200 = 0

R1 + R2 = 500 N …(i)

Στ = 0 (pivot at point D)

– R1 x AD – 300 N x DB – 200 N x DC = 0

– R1 x 1m – 300 N x 3 m – 200 N x 1m = 0

Substituting the value of R1 in Eq. (i). we have

– 1100 + R2 = 500

The negative sign of R1 shows that it is directed downward. Thus the result has corrected the mistake of our initial assumption.