Question

In Fig. 6-39, a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver’s mass is ${}^{\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}}$.What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

Short answer

The magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley is${}^{1.37\times {10}^3\mathrm{N}}$

Step-by-step solution

Given

$\mathrm{Driver}\text{'}\mathrm{s}\mathrm{mass}=70.0\mathrm{kg}$

Determining the concept

Use the concept of uniform circular motion. Uniform circular motion is a motion of a particle moving at a constant speed on a circle.

Formulae are as follow:

${\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}-{\mathrm{v}}^2/\mathrm{R})$

where, v is the velocity, g is an acceleration due to gravity, m is mass, ${}^{{\mathrm{F}}_{\mathrm{N}}}$is force and R is the radius.

Determining the magnitude of the normal force on the driver from the seat

At the top of the hill, normal force is directed upwards, gravitational force and the centripetal acceleration directed downward. Then by using the Newton’s 2nd law of motion,

${\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}-{\mathrm{v}}^2/\mathrm{R})$

Since, ${\mathrm{F}}_{\mathrm{N}}=0$, as stated in the problem, then${\mathrm{v}}^2=\mathrm{gR}$. Later, at the bottom of the valley, reverse both the normal force direction and the acceleration direction.

Thus,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}=\mathrm{m}(\mathrm{g}+{\mathrm{v}}^2/\mathrm{R}) \\ =2\mathrm{mg} \\ =1372\mathrm{N} \\ \approx 1.37\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley is$1.37\times {10}^3\mathrm{N}$