# Is the moment of a force the same about any point?

I know that when a body is is equilibrium

• 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

• 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?

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 $\vec{R}$

No, you wouldn't. The moment (which is called torque, by the way - at least, the kind of moment you're talking about) of a force around a reference point is

$$\vec{\tau} = \vec{r}\times\vec{F}$$

where $\vec{r}$ is the vector from the reference point to where the force is applied. If the force is zero, then you can tell that the moment (the torque) will be zero, but in any other case it depends on where you put the reference point.

However, any relevant conservation laws or other such equations - $\tau = I\alpha$ or $\vec{\tau} = \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}$, for example - will be valid no matter which reference point you choose.

• So, for example, if there is a resultant moment of 5 Nm clockwise about a point on a body and I calculate the resultant moment using a different reference point, I will get a different answer, say, 3Nm anticlockwise? The rule of same moment about any point only applies to bodies in equilibrium? Apr 21 '14 at 3:31
• Yes, that's what I said. Apr 21 '14 at 4:05