Figure 12-23 shows a horizontal block that is suspended by two wires, Aand B, which are identical except for their original lengths. The center of mass of the block is closer to wire Bthan to wire A.

(a) Measuring torques about the block’s center of mass, state whether the magnitude of the torque due to wire Ais greater than, less than, or equal to the magnitude of the torque due to wire B.

(b) Which wire exerts more force on the block?

(c) If the wires are now equal in length, which one was originally shorter (before the block was suspended)?

1. The block is in equilibrium and is held by two wires.
2. The center of mass of the block is closer to wire B than wire A.

We use the concept of static equilibrium, i.e., balanced torques and balanced forces to determine the magnitudes of the various torques and forces acting on the block. The application of force creates a change in the length of the wire.

Formulae:

The value of the net force at equilibrium,$F_{net}=0$ (i)

The value of the torque at equilibrium,${\tau }_{net}=0$ (ii)

The torque acting on a point,$\tau =r\times F$ (iii)

The block is in equilibrium, so the torques and the forces acting on it are balanced considering equations (i) and (ii).

Hence, using equation (iii) for the same position value, the magnitude of the torque due to wire A is equal to the magnitude of the torque by wire B.

If we consider the torque about the center of mass of the block, the torque by each wire is calculated by the equation (iii).

Now, as given in the problem, the point of the center of mass is closer to the wire B. Hence its moment arm is smaller than the moment arm of wire A. Considering that both have the same torques, we can say that the force by wire B is more than the force by wire A.

As we have seen in part b), the force by wire B is more, and the change in lengthof wireB is more than that in A.

Hence, for the same torques, wire B was shorter originally.