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.

#### Solution:

Show me the final answer↓

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:

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

Position vector r_{AB} is then:

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.

(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:

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

Substitute the values we know:

(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

#### Final Answer: