Good afternoon,
I am currently having issues solving a problem:
Let $A$ and $B$ have position vectors $a = 2i – j + 3k$ and $b = 5i – 2j + k$
respectively. The force $F = 3i – 5j – k$ passes through the point $A$.
Show that the vector moment of force, $\vec F$ about the point $B$ is equal to
$3(3i + j + 4k)$.
I was thinking I could simply solve a system of linear equations to find the moment of the force, and once I have obtained it, do the cross product of $\vec r (B)$ for the newly found $\vec M$. Apparently however there are better ways to solve the problem but I can't really find any.
Best Answer
The moment of a force is $\vec M=\vec r\times\vec F$ where $\vec r$ is the position vector from the point about which you are taking the moment to the point of application of the force.
So in this case, the moment about point $B$ of force applied at point $A$ is $\vec M=(\vec a-\vec b)\times\vec F$, because $\vec a-\vec b$ is the position vector from $A$ to $B$.
$$(\vec a-\vec b)\times\vec F=\begin{vmatrix} \vec i & \vec j & \vec k \\ (2-5) & (-1+2) & (3-1) \\ 3 & -5 & -1 \\ \end{vmatrix}=9\vec i +3\vec j +12\vec k$$