Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 8–45. All forces are shown. Calculate about (a) point C, the CM, and (b) point P at one end.

The length of the beam is \(l = 2\;{\rm{m}}\).

The magnitude of the first force is\({F_1} = 56\;{\rm{N}}\).

The magnitude of the second force is\({F_2} = 52\;{\rm{N}}\).

The magnitude of the third force is \({F_3} = 65\;{\rm{N}}\).

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.

Suppose the forces oriented in the anti-clockwise direction are positive and those in the clockwise direction are negative. Each force exerted at C has a lever arm of \(r = \frac{l}{2} = 1\;{\rm{m}}\).

Therefore, the torque about the center of mass C is given by the following:

\(\begin{align}{\tau _{\rm{C}}} = \sum {rF\sin \theta } \\ &= - \left( {\frac{l}{2}} \right){F_1}\sin {32^{\rm{o}}} + \left( {\frac{l}{2}} \right){F_2}\sin {32^{\rm{o}}}\\ &= - \left( {\frac{{2\;{\rm{m}}}}{2}} \right)\left( {56\;{\rm{N}}} \right)\sin {32^{\rm{o}}} + \left( {\frac{{2\;{\rm{m}}}}{2}} \right)\left( {52\;{\rm{N}}} \right)\sin {32^{\rm{o}}}\\ &= 14\;{\rm{m}}\;{\rm{N}}\end{align}\)

This torque is directed anti-clockwise.

The force exerted at C has a lever arm of \({r_1} = 1\;{\rm{m}}\), and the force exerted at the end has a lever arm of \({r_2} = 2\;{\rm{m}}\). Thus, the torque about point P is given by the following:

\(\begin{align}{\tau _{\rm{P}}} &= \sum {rF\sin \theta } \\ &= - r{F_1}\sin {32^{\rm{o}}} + r{F_2}\sin {32^{\rm{o}}}\\ &= - \left( {2\;{\rm{m}}} \right)\left( {56\;{\rm{N}}} \right)\sin {32^{\rm{o}}} + \left( {1\;{\rm{m}}} \right)\left( {65\;{\rm{N}}} \right)\sin {45^{\rm{o}}}\\ &= - 13\;{\rm{m}}\;{\rm{N}}\end{align}\)

The negative sign indicates a clockwise torque.