# The pipe assembly is subjected to

The pipe assembly is subjected to the 80-N force. Determine the moment of this 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:

When trying to find the moment in 3-D coordinates,  the first step is to express the force in Cartesian vector form. Let us express the 80 N in Cartesian form.

To express this force in Cartesian form, we first need to find F’. F’ is the force that lies on the x-y plane. We then use that to calculate $F_y$ and $F_x$. Note that this is a step that must always be performed if a force is shown in that manner. From the diagram, we see that:

$F'=80\cos30^0=69.28$ N

Notice how F’ is now the hypotenuse of the new triangle that lies on the x-y plane. Again, using F’ as the hypotenuse, we can easily calculate $F_y$ and $F_x$.

$F_x=69.28\sin40^0=44.5$ N

$F_y=69.28\cos40^0=53.1$ N

$F_z$ can be found by taking the $\sin30^0$ of 80 N.

$F_z=-80\sin30^0=-40$ N
(Why is $F_z$ negative? Notice how $F_z$ is downwards. In other words, it’s along the negative z-axis.)

We can now express the force in Cartesian form:

$F=\left\{44.5i+53.1j-40k\right\}$

The next step is to express a position vector from where we want to calculate the moment to where the force is being applied. In this question, we are trying to find the moment at A, thus our position vector begins at A. The force is applied at C. Therefore, our position vector would be from A to C.

To express the position vector, we need to figure out where points A and C lie in our diagram. We write them both down in Cartesian form. From the diagram, we see that:

$A:(0i+0j+0k)\text{m}$

$C:((300+250)i+400j-200k)\text{mm}=(0.55i+0.4j-0.2k)\text{m}$

Thus, $r_{AC}$ is:

$r_{AC}=\left\{(0.55-0)i+(0.4-0)j+(-0.2-0)k\right\}$

$r_{AC}=\left\{0.55i+0.4j-0.2k\right\}$

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 $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

The final step is to take the cross product of the position vector and the force, which will give us the moment produced at A.

$M_A=r_{AC}\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\0.55&0.4&-0.2\\44.5&53.1&-40\end{bmatrix}$

$M_A=\left\{-5.38i+13.1j+11.4k\right\}\,\text{N}\cdot\text{m}$

$M_A=\left\{-5.38i+13.1j+11.4k\right\}\,\text{N}\cdot\text{m}$