# Determine the moment produced

Determine the moment produced by force $F_C$ about point O. Express the result as a Cartesian vector. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

This question is the same as the previous question. Steps will be skipped as they are the same. Please refer to the previous question if a section is confusing.

Let us express force $F_C$ in Cartesian vector form. To do so, we need to write a position vector from A to C.

$r_{AC}=\left\{(2-0)i+(-3-0)j+(0-6)k\right\}$

$r_{AC}=\left\{2i-3j-6j\right\}$

The magnitude of this position vector is:

magnitude of $r_{AB}\,=\,\sqrt{(2)^2+(-3)^2+(-6)^2}=7$

The unit vector is:

$u_{AB}\,=\,\left(\dfrac{2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)$

Force $F_C$ can now be expressed in Cartesian vector form.

$F_C=420\left(\dfrac{2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)$

$F_C=\left\{120i-180j-360k\right\}$ N

We now need to express a position vector from O to A.

$r_{OA}=\left\{0i+0j+6k\right\}$

To find the moment at O created by force $F_C$, we need to take the cross product of the position vector $r_{OA}$ and force $F_C$.

$M_O=r_{OA}\times F_C$

$M_O=\begin{bmatrix}\bold i&\bold j&\bold k\\0&0&6\\120&-180&-360\end{bmatrix}$

$M_O=\left\{1080i+720j\right\}N\cdot m$

$M_O=\left\{1080i+720j\right\}N\cdot m$