Chapter 6.2, Problem 47E

Solution

i.

To calculate: The magnitude of the force that is required for rolling the car.

  $415.82\text{ lb}$

Given:

Weight of car = $2000\text{ pounds}$

The angle formed by the car with horizontal $={12}^{\circ }$

Explanation:

Consider the following figure:

  

Gravity is acting vertically downward so the force because of gravity can be represented as $\text{F}=〈0,-2000〉$ .

From the figure, it is understood that the ${\text{F}}_{\text{1}}$ is acting along the street. So, the amount of force required for rolling the car is,

  $\begin{array}{l}\sin {12}^{\circ }=\frac{\left|{\text{F}}_{\text{1}}\right|}{\left|\text{F}\right|}\\ \left|{\text{F}}_{\text{1}}\right|=2000\sin {12}^{\circ }\end{array}$

Thus,

  $\left|{\text{F}}_{\text{1}}\right|\approx 415.82$ lb

Conclusion:

The required force is $415.82\text{ lb}$ .

ii.

To find: The force perpendicular to the street.

  $1956.30\text{ lb}$

Given:

Weight of car = $2000\text{ pounds}$

The angle formed by the car with horizontal $={12}^{\circ }$

Explanation:

From the figure, it is understood that the ${\text{F}}_{\text{2}}$ is acting perpendicular to the street. So, the force is,

  $\begin{array}{l}\cos {12}^{\circ }=\frac{\left|{\text{F}}_2\right|}{\left|\text{F}\right|}\\ \left|{\text{F}}_{\text{2}}\right|=2000\cos {12}^{\circ }\end{array}$

Thus,

  $\left|{\text{F}}_{\text{2}}\right|=1956.30$ lb

Conclusion:

The force perpendicular to the street is $1956.30\text{ lb}$ .