Chapter 4, Problem 53P

To determine

The minimum length of the uphill escape ramp needed for a runaway truck to come to rest.

Answer to Problem 53P

Solution:

The minimum length of the ramp needed is 4.0×10²m.

Explanation of Solution

A runaway truck traveling with an initial speed $v_0$ climbs up a distance s along the length of the incline which is at an angle $\theta$ and comes to rest.

The free body diagram for the truck is shown in the diagram below.

The weight mg of the truck acts vertically downwards. The normal force $F_N$ acts perpendicular to the incline. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the incline.

Since the truck moves upwards, it experiences a retarding force due to the component of weight $mg\sin \theta$.

    $F=-mg\sin \theta$ ……(1)

By Newton’s second law,

    $F=ma$ ……(2)

From equations (1)and (2),

    $\begin{array}{c}ma=-mg\sin \theta \\ a=-g\sin \theta \end{array}$……(3)

Use the equation of motion,

    $v^2=v_0^2+2as$

Use the expression for a from equation (3)in the above equation.

    $v^2=v_0^2-2\left(g\sin \theta \right)s$

Given:

The initial speed of the truck $v_0=140\text{ km/h}$

The angle of the incline $\theta =11°$

Final speed of the truck $v=0\text{ m/s}$

Formula used:

    $v^2=v_0^2-2\left(g\sin \theta \right)s$

Calculation:

Express the initial speed in m/s

    $\begin{array}{c}{v}_0=140\text{ km/h}\times \frac{1000\text{ m}}{3600\text{ s}}\\ =38.9\text{ m/s}\end{array}$

Substitute the given values of v,$v_0$, $\theta$ in the formula and simplify for s.

    $\begin{array}{c}{v}^2=v_0^2-2\left(g\sin \theta \right)s\\ \left(0\text{ m/s}\right)={\left(38.9\text{ m/s}\right)}^2-2\left(9.8{\text{ m/s}}^2\right)\left(\sin 11°\right)s\\ s=\frac{{\left(38.9\text{ m/s}\right)}^2}{2\left(9.8{\text{ m/s}}^2\right)\left(\sin 11°\right)}\\ =404.6\text{ m}\end{array}$