# Acts at the end of the beam

The force $F=\left\{600i+300j-600k\right\}$ N acts at the end of the beam. Determine the moment of the force about point A.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

The moment at A can be found by taking the cross product of $r_{AB}\times \vec F$. Here, $r_{AB}$ represents a position vector from A to B, in other words, the location of the point where we are calculating the moment and the path from that point to the location where the force is being applied.

Let us write the locations of points A and B in Cartesian vector form.

From the image, we see that the locations of points A and B are at the following places:

$A:(0i+0j+0.4k)$ m

$B:(0.2i+1.2j+0.4k)$ m

The position vector, $r_{AB}$ is:

$r_{AB}\,=\,\left\{(0.2-0)i+(1.2-0)j+(0.4-0.4)k\right\}$

$r_{AB}=\left\{0.2i+1.2j+0k\right\}$

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

We can now calculate the moment by taking the cross product. Remember that force $F=\left\{600i+300j-600k\right\}$.

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\0.2&1.2&0\\600&300h&-600\end{bmatrix}$

$M_A=\left\{-720i+120j-660k\right\}N\cdot m$

$M_A=\left\{-720i+120j-660k\right\}N\cdot m$