# The board is used to hold

The board is used to hold the end of a four-way lug wrench in position. If a torque of 30 N•m about the x axis is required to tighten the nut, determine the required magnitude of the force F that the man’s foot must apply on the end of the wrench in order to turn it. Force F lies in a vertical plane.

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

#### Solution:

Let us first draw a simplified diagram.

Notice the position vector, starting from location A to location B. Let us define this position vector in Cartesian vector form. The locations of points A and B are:

$A:(0.25i+0j+0k)$ m

$B:(0.25i+0.25j+0k)$ m

Position vector $r_{AB}$ is then:

$r_{AB}=\left\{(0.25-0.25)i+(0.25-0)j+(0-0)k\right\}$

$r_{AB}=\left\{0i+0.25j+0k\right\}$ m

We will now express the force F in Cartesian form. Remember that force F will have two components, a y-component and a z-component.

$F=\left\{0i+F\cos60^0j-F\sin60^0k\right\}$

(simplify)

$F=\left\{0i+0.5F\,j-0.866F\,k\right\}$

Note that our z-component is negative because the z-component of the force is going down (or lies in the negative z-axis).

We can now find the required force by finding the moment created along the x-axis. This value is given to us in the question as 30N•m. The moment along the x-axis is:

$M_x=i\cdot r_{AB}\times F$

(Remember that the x-axis has a unit vector of i)

Substitute the values we know:

$30=\begin{bmatrix}1&0&0\\0&0.25&0\\0&0.5F&-0.866F\end{bmatrix}$

(Solve by taking the cross product)

$30=1\left[(0.25)(-0.866F)-(0.5F)(0)\right]+0+0$

$30=-0.2165F$

The negative sign indicates that the moment created is towards the negative x-axis.

$30=0.2165F$

$F=138.57$ N

$F=138.57$ N