The vector sum of a system of non-collinear f | vedclass

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(B) No,it is not necessarily zero about any arbitrary point.The torque $\vec{	au}$ about a point $P$ is given by $\vec{	au}_P = \sum (\vec{r}_i - \v

Vedclass · Accepted Answer

No

Vedclass · Answer

(B) No,it is not necessarily zero about any arbitrary point.The torque $\vec{	au}$ about a point $P$ is given by $\vec{	au}_P = \sum (\vec{r}_i - \vec{r}_P) 	imes \vec{F}_i$.If we shift the point to $P'$,the new torque is $\vec{	au}_{P'} = \sum (\vec{r}_i - \vec{r}_{P'}) 	imes \vec{F}_i = \sum (\vec{r}_i - \vec{r}_P + \vec{r}_P - \vec{r}_{P'}) 	imes \vec{F}_i$.This simplifies to $\vec{	au}_{P'} = \vec{	au}_P + (\vec{r}_P - \vec{r}_{P'}) 	imes \sum \vec{F}_i$.Since the net force $\sum \vec{F}_i 
eq 0$,the torque $\vec{	au}_{P'}$ will only be zero if the vector $(\vec{r}_P - \vec{r}_{P'})$ is parallel to the net force vector $\sum \vec{F}_i$.

The vector sum of a system of non-collinear forces acting on a rigid body is given to be nonzero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero,does this mean that it is necessarily zero about any arbitrary point?

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