Question

A body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?

Answer 1

Yes , if the torque due to forces in translation equillibrium is zero about a point, it will be zero about other point in the plane.

Let us consider a planner lamina of some mass, acted upon by forces \[\overrightarrow{F_1}, \overrightarrow{F_2}, . . . \overrightarrow{F_i},\] etc.

Let a force \[\overrightarrow{F_i}\] act on a \[i^{th}\] particle and torque due to \[\overrightarrow{F_i}\] be zero at a point Q.

Since the body is in translation equillibrium, we have

\[\Sigma \overrightarrow{F_i} = 0 .............(1)\]

Again, torque about P is zero . Therefore, we have

\[\Sigma\left( \overrightarrow{r_{pi}} \times \overrightarrow{F_i} \right) = 0 ..............(2)\]

Now, torque about point Q will be

\[\Sigma \overrightarrow{r}_{Qi} \times \overrightarrow{F_i} \]
\[ = \Sigma\left( \overrightarrow{r}_{QP} + \overrightarrow{r}_{Pi} \right) \times \overrightarrow{F_i} ............\left[\text{From fig.} \right]\]
\[ = \Sigma\left( \overrightarrow{r}_{QP} \times \vec{F_i} + \overrightarrow{r_{Pi}} \times \overrightarrow{F_i} \right)\]
\[ = \Sigma \overrightarrow{r_{QP}} \times \overrightarrow{F_i} + \Sigma \overrightarrow{r_{Pi}} \times \overrightarrow{F_i} \]
\[ = \overrightarrow{r_{QP}} \times \Sigma \overrightarrow{F_i} + 0 ..........\left[\text{From (2)} \right]\]
\[ = \overrightarrow{r_{QP}} \times 0 .........\left[\text{From (1)} \right]\]
\[ = 0\]

Thus, \[\overrightarrow{F}\] is zero about any other point Q.