Chapter 6.4, Problem 74E

To determine

Find the force required to keep the vehicle from rolling down the hill. Find the force perpendicular to the hill.

Answer to Problem 74E

  $\boxed{\Vert F\Vert =937.7pounds}$ and $\boxed{\Vert N\Vert =5317.96pounds}$ .

Explanation of Solution

Given information:

A sport utility vehicle with a gross weight of $5400$ pounds is parked on a slope of $10°$ . Assume that the only force to overcome is the force of gravity. Find the force required to keep the vehicle from rolling down the hill. Find the force perpendicular to the hill.

Calculation:

As shown on the diagram, the vehicle is parked on a slope of $10°$ , now, $W_x$ is the component of the weight of the vehicle along the slope and $F$ is the breaking force vector opposite to this. The weight of the vehicle is $5400$ pounds, that is,

  $\Vert W\Vert =5400$

The component form of the weight along and perpendicular to the slope is,

  $\begin{array}{l}W=\langle \Vert W\Vert \sin 10°,\Vert W\Vert \cos 10°\rangle \\ =\langle 5400\sin 10°,5400\cos 10°\rangle \\ =\langle 937.7,5317.96\rangle \end{array}$

  $N$ is the force normal to the ground as shown in figure.

Since the vehicle does not move, $W_x$ is balanced by the breaking force $F$ .

Therefore,

  $W_X=\Vert F\Vert$

Since, $W_x$ is the component of the weight along the slope,

Therefore,

  $W_x=937.7$

Hence, the force required to keep the vehicle from rolling down the hill is,

  $\boxed{\Vert F\Vert =937.7pounds}$

Similarly, the force perpendicular to the slope of the hill, $N$ , is equal in magnitude and opposite to the direction of the component of the weight normal to the hill.

Hence,

  $\boxed{\Vert N\Vert =5317.96pounds}$