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This question already has an answer here:

When there are forces acting on a rigid body, the two conditions that has to be satisfied for the body to be in static equilibrium are:
1. The sum of all forces must be zero (translational equilibirum): $$\sum_i F_i = 0$$
2.The sum of all torques relative to any point must be zero (rotational equilibrium): $$\sum_i \tau_i=0$$

Why does the torque have to be zero relative to any point? My understanding is that as long as torque is zero with respect to the pivot, the body will not be in rotational motion. What does it mean when you talk about torque with no reference to a pivot?

Also, how can you makes sure that torque is zero relative to any point? Do you consider all the possible points?

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marked as duplicate by sammy gerbil, Kyle Kanos, stafusa, Jon Custer, rob Jun 18 '18 at 22:35

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When condition (1) is met, then condition (2) is equivalent to zero torque through the pivot (which may not be well defined for a rigid body that can move unconstrained). The following theorem holds:

Under the condition that the sum of the acting forces is zero the torque is zero relative to all points if and only if it is zero relative to one point.

The proof done by computing the change of torque under a change of the basis point (so that $\vec r \mapsto \vec r' = \vec d + \vec r$): \begin{align*} \sum_i \vec \tau'_i &= \sum_i \vec r'_i \times \vec F_i = \sum_i (\vec r_i + \vec d) \times \vec F_i \\ &= \sum_i \vec r_i \times \vec F_i + \vec d \times \underbrace{\left(\sum_i \vec F_i\right)}_{\vec F_T} \\ &= \sum_i \vec \tau_i + \vec d \times \left( \sum_i \vec F_i \right). \end{align*} So the even stronger statement holds, that the sum of the torques changes by $\vec d \times \vec F_T$ under a change of reference point. If the sum of the forces acting on the body is zero, this means that the total torque does not change when the reference point is changed.

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