While reading about rotational equilibrium I came across a statement which I was unable to understand:

The torque can be taken about any line perpendicular to the plane of the forces. In general, the torque is different about different lines but it can be shown that if the resultant force is zero, the total torque about any line perpendicular to the plane of the forces is equal.

Can someone help me understand this statement?


closed as unclear what you're asking by ACuriousMind, user36790, Asher, knzhou, Gert Jun 18 '16 at 2:01

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  • 3
    $\begingroup$ Could you tell us what exactly you don't understand about the statement? $\endgroup$ – ACuriousMind Jun 16 '16 at 8:36
  • $\begingroup$ how we can arrive to the statement... $\endgroup$ – Osheen Sachdev Jun 16 '16 at 8:37
  • $\begingroup$ plane of the forces means the plane that is parallel to the forces creating the torque. $\endgroup$ – philip_0008 Jun 16 '16 at 22:33
  • $\begingroup$ line perpendicular to the plane of forces means the rotating axis. $\endgroup$ – philip_0008 Jun 16 '16 at 22:36
  • $\begingroup$ Possible duplicate : Is the moment of a force the same about any point? physics.stackexchange.com/q/109310 $\endgroup$ – sammy gerbil Jun 17 '16 at 15:35

The total torque about some axis is defined as: $\vec{\tau}_\text{tot}=\Sigma \left(\vec{r}_i \times{}\vec{f}_i\right) $

If you change to a parallel axis located $\vec{r}$ away from the first one on the plane of the forces, the new torque will be: $$\vec{\tau}_\text{tot}'=\Sigma \left[(\vec{r}_i+\vec{r}) \times{}\vec{f}_i\right] =\vec{\tau}_\text{tot}+\vec{r}\times{}\Sigma \vec{f}_i=\vec{\tau}_\text{tot}$$

As $\Sigma \vec{f}_i=\vec 0$


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