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?
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Yes, if the torque due to forces in translation equillibriumis 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 $\vec{\text{F}}_1,\ \vec{\text{F}}_2,\ \dots\ \vec{\text{F}}_{\text{i,}}$ etc. Let a force $\vec{\text{F}}_1$ act on a $i^{th}$ particle and torque due to $\vec{\text{F}}_{\text{i}}$ be zeroat a point Q. Since the body is in translation equillibrium, we have: $\sum \vec{\text{F}}_{\text{i}}=0\ \dots(1)$ Again, torque about P is zero. Therefore, we have: $\sum\Big(\vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)=0\ \dots(2)$ Now, torque about point Q will be: $\sum\vec{\text{r}}_{\text{Qi}}\times \vec{\text{F}}_{\text{i}}$
$=\sum\Big( \overrightarrow{\text{r}}_{\text{QP}}+ \overrightarrow{\text{r}}_{\text{pi}}\Big)\times \overrightarrow{\text{F}}_{\text{i}}$ [From fig.] $=\sum\Big( \vec{\text{r}}_{\text{Qp}}\times \vec{\text{F}}_{\text{i}}+ \vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)$
$= \overrightarrow{\text{r}_{\text{Qp}}}\times \overrightarrow{\text{F}}_{\text{i}}+ \overrightarrow{\text{r}_{\text{pi}}}\times \overrightarrow{\text{F}}_{\text{i}}$
$=\sum \overrightarrow{\text{r}_{\text{QP}}}\times\sum \overrightarrow{\text{F}}_{\text{i}}+0$ [From (2)] $ \overrightarrow{\text{r}_{\text{Qp}}}\times0$ [From (1)] $=0$ Thus, $\overrightarrow{\text{F}}$ is zero about any other point Q.
Let us consider a planner lamina of some mass, acted upon by forces $\vec{\text{F}}_1,\ \vec{\text{F}}_2,\ \dots\ \vec{\text{F}}_{\text{i,}}$ etc. Let a force $\vec{\text{F}}_1$ act on a $i^{th}$ particle and torque due to $\vec{\text{F}}_{\text{i}}$ be zeroat a point Q. Since the body is in translation equillibrium, we have: $\sum \vec{\text{F}}_{\text{i}}=0\ \dots(1)$ Again, torque about P is zero. Therefore, we have: $\sum\Big(\vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)=0\ \dots(2)$ Now, torque about point Q will be: $\sum\vec{\text{r}}_{\text{Qi}}\times \vec{\text{F}}_{\text{i}}$
$=\sum\Big( \overrightarrow{\text{r}}_{\text{QP}}+ \overrightarrow{\text{r}}_{\text{pi}}\Big)\times \overrightarrow{\text{F}}_{\text{i}}$ [From fig.] $=\sum\Big( \vec{\text{r}}_{\text{Qp}}\times \vec{\text{F}}_{\text{i}}+ \vec{\text{r}}_{\text{pi}}\times \vec{\text{F}}_{\text{i}}\Big)$
$= \overrightarrow{\text{r}_{\text{Qp}}}\times \overrightarrow{\text{F}}_{\text{i}}+ \overrightarrow{\text{r}_{\text{pi}}}\times \overrightarrow{\text{F}}_{\text{i}}$
$=\sum \overrightarrow{\text{r}_{\text{QP}}}\times\sum \overrightarrow{\text{F}}_{\text{i}}+0$ [From (2)] $ \overrightarrow{\text{r}_{\text{Qp}}}\times0$ [From (1)] $=0$ Thus, $\overrightarrow{\text{F}}$ is zero about any other point Q.
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