I know that when a body is is equilibrium
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There is zero resultant force in any direction, i.e. the sum of all the components of all the forces in any direction is zero
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The sum of the moments about any point is zero (this point does not need to be on the body)
What if the body is not in equilibrium?
Let me clarify: What if there is a resultant moment of $5\text{ Nm}$ clockwise about a point on a body. If I calculated the moment about another random point in or outside the body, would I still get $5\text{ Nm}$ clockwise as my value?
Best Answer
As said before, the answer is no(t always), but there is a simple law which can help you predict whether it will be the case or not, and how the torque is distributed across your solid*.
Using simple algebra and $\times$ distributivity, one can easily prove that
$$ \vec{\tau}_{\vec{p}}=\vec{\tau}_{\vec{o}}+\vec{op}\times\vec{R} $$
where $\vec{R}$ is the resultant vector of all forces in presence. If this resultant is equal to zero, then the torque will be the same at any point on your solid ($\vec{\tau}_{\vec{p}}=\vec{\tau}_{\vec{o}}\:\forall \vec{o},\vec{p}$). This situation with $\vec{\tau}\neq\vec{0}$ is not exotic and you'll often encounter it in real life applications.
On the contrary, if $\lvert\vec{R}\rvert$ is greater than zero, then the torque will be different depending on where you calculate it.
(*)In particular, torque is actually always unchanged by a translation parallel to $\vec{R}$