# If the couple moment acting on the pipe 2

If the couple moment acting on the pipe has a magnitude of 400 N•m, determine the magnitude F of the vertical force applied to each wrench.

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

#### Solution:

The couple moment can be found by taking the cross product between a position vector from A to B and force F.

$M_c=r_{AB}\times F$

Let us first express a position vector from A to B. The coordinates of A and B are:

$A:(0.65i+0.4j+0k)\,\text{m}$

$B:(0.3i+0.2j+0k)\,\text{m}$

The position vector is:

$r_{AB}=\left\{(0.3-0.65)i+(0.2-0.4)j+(0-0)k\right\}$

$r_{AB}=\left\{-0.35i-0.2j+0k\right\}$

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 now need to express from F in Cartesian vector form. Force F is only applied in the z-axis, thus it will only have a k component.

$F=\left\{0i+0j+Fk\right\}$

Let us now take the cross product:

$M_c=\begin{bmatrix}\bold i&\bold j&\bold k\\-0.35&-0.2&0\\0&0&F\end{bmatrix}$

$M_c=\left\{-0.2F\bold i+0.35F\bold j\right\}$

The magnitude of this couple moment is:

magnitude of $M_{c}\,=\,\sqrt{(-0.2F)^2+(0.35F)^2}$

(Remember that the question tells us the magnitude of $M_c$ is 400 N•m)

$400=\sqrt{(-0.2F)^2+(0.35F)^2}$

(square both sides)

$160000=(-0.2F)^2+(0.35F)^2$

$160000=0.04F^2+0.1225F^2$

$160000=0.1625F^2$

(divide both sides by 0.1625)

$F^2=984615.38$

(take the square root of both sides)

$F=992.28$ N

$F=992.28$ N