Calculate the net torque about the axle of the wheel shown in Fig. 8–42. Assume that a friction torque of\(0.60\;{\rm{m}}\;{\rm{N}}\)opposes the motion.

The frictional torque is \({\tau _{\rm{f}}} = 0.60\;{\rm{m}}\;{\rm{N}}\).

The magnitude of the first force is\({F_1} = 18\;{\rm{N}}\).

The magnitude of the second force is\({F_2} = 35\;{\rm{N}}\).

The magnitude of the third force is\({F_3} = 28\;{\rm{N}}\).

The inner radius of the axle is\({r_{\rm{1}}} = 12\;{\rm{cm}} = 0.12\;{\rm{m}}\).

The outer radius of the axle is\({r_{\rm{2}}} = 24\;{\rm{cm}} = 0.24\;{\rm{m}}\).

Torque is defined as the measure of the force that causes an object to rotate about an axis. It is also called the rotational equivalent of linear force.

Mathematically, torque is represented as follows:

\(\begin{align}\vec \tau &= \vec r \times \vec F\\\left| \tau \right| &= rF\sin \theta \end{align}\) … (i)

Here, F is the applied force, and r is the perpendicular distance between the point about which the torque is calculated and the point of application of force.

Each applied force is perpendicular to the lever arm, i.e., \(\theta = {90^{\rm{o}}}\). Suppose the forces oriented in the anti-clockwise direction are negative and those directed in the clockwise direction are positive.

Therefore, the torque due to the three applied forces is given by the following:

\(\begin{align}{\tau _{{\rm{applied}}\;{\rm{forces}}}} &= {r_2}{F_1} + {r_1}{F_2} + {r_2}{F_3}\\ &= \left( {0.24\;{\rm{m}}} \right)\left( {18\;{\rm{N}}} \right) + \left( {0.12\;{\rm{m}}} \right)\left( {35\;{\rm{N}}} \right) - \left( {0.24\;{\rm{m}}} \right)\left( {28\;{\rm{N}}} \right)\\ &= 1.8\;{\rm{m}}\;{\rm{N}}\end{align}\)

This torque is clockwise, so assume that the wheel is rotating clockwise.

The frictional force will act in the anti-clockwise direction. Therefore, the net torque will be given by the following:

\(\begin{align}{\tau _{{\rm{net}}}} &= {\tau _{{\rm{applied}}\;{\rm{forces}}}} - {\tau _{\rm{f}}}\\ &= 1.8\;{\rm{m}}\;{\rm{N}} - 0.60\;{\rm{m}}\;{\rm{N}}\\ &= 1.2\;{\rm{m}}\;{\rm{N}}\end{align}\)

Thus, the net torque exerted on the axle is 1.2 m N (clockwise).