Show that the moment of a couple does not depend on the point about which you take the moments.

(N/A) Consider a couple acting on a rigid body as shown in the figure. The forces $F$ and $-F$ act respectively at points $B$ and $A$. These points have position vectors $r_1$ and $r_2$ with respect to the origin $O$.
Let us calculate the moments of the forces about the origin $O$.
The moment of the couple is the sum of the moments of the two forces forming the couple:
$\text{Moment} = r_1 \times (-F) + r_2 \times F$
$= r_2 \times F - r_1 \times F$
$= (r_2 - r_1) \times F$
From the triangle law of vector addition,we have $r_1 + AB = r_2$,which implies $AB = r_2 - r_1$.
Substituting this into the expression for the moment,we get:
$\text{Moment} = AB \times F$
Since $AB$ is the vector representing the separation between the two forces,the moment of the couple depends only on the forces and their separation,not on the origin $O$ chosen to calculate the moments. Thus,the moment of a couple is independent of the point about which the moments are taken.