Chapter 13.3, Problem 30PPS

To determine

To show: The force $F=W\tan \left(A+\theta \right)$ .

Explanation of Solution

Given:

The diagram is shown below:

  

The force F necessary to hold the safe in position on the ramp:

  $F=\frac{W\left(\sin A+\mu \cos A\right)}{\cos A-\mu \sin A}$

Referring to the given expression, W is the weight of the safe.

  $\mu =\tan \theta$

Solving the given expression:

  $\begin{array}{c}F=\frac{W\left(\sin A+\mu \cos A\right)}{\cos A-\mu \sin A}\\ =\frac{W\left(\sin A+\tan \theta \cos A\right)}{\cos A-\tan \theta \sin A}\\ =\frac{W\left(\frac{\sin A}{\cos A}+\tan \theta \cos A\right)}{\frac{\cos A}{\cos A}-\tan \theta \frac{\sin A}{\cos A}}\\ =\frac{W\left(\tan A+\tan \theta \right)}{1-\tan \theta \tan A}\end{array}$

Therefore,

  $F=W\left(\tan A+\tan \theta \right)$