Chapter 5, Problem 28P

(a)

To determine

The frictional force on the car.

Answer to Problem 28P

The frictional force on the car is $\underset{\_}{2300\text{N}}$.

Explanation of Solution

Given that the radius of the curved path is $410\text{m}$, the speed of the car is $32\text{m/s}$, the banking angle of the road is $5.0°$, and the mass of the car is $1400\text{kg}$.

Let the the $x$-axis point horizontally toward the the center of curvature and $y$-axis point upward.

The Free-body diagram of the car in the given situation is shown in Figure 1.

Write the condition for equilibrium of forces in the $x$ and $y$ directions.

${\displaystyle \sum F_x}=0$ (I)

${\displaystyle \sum F_y}=0$ (II)

Apply the condition in equation (II) to the free-body diagram of the car.

$N\cos \theta -mg-f\sin \theta =0$ (III)

Here, $N$ is the magnitude of normal force, $\theta$ is the angle of banking, $m$ is the mass of the car, $g$ is the acceleration due to gravity, and $f$ is the magnitude of the frictional force.

Solve equation (III) for $N$.

$N=\frac{f\sin \theta +mg}{\cos \theta }$ (IV)

Apply the condition in equation (I) to the free-body diagram of the car.

$N\sin \theta +f\cos \theta -\frac{m{v}^2}{r}=0$ (V)

Here, $m{v}^2/r$ represents the centripetal force.

Use equation (IV) in (V) and solve for $f$.

$\begin{array}{c}\left(\frac{f\sin \theta +mg}{\cos \theta }\right)\sin \theta +f\cos \theta -\frac{m{v}^2}{r}=0\\ \left(f\sin \theta +mg\right)\left(\frac{\sin \theta }{\cos \theta }\right)+f\cos \theta =\frac{m{v}^2}{r}\\ f=\frac{\left(m{v}^2/r\right)-mg\left(\frac{\sin \theta }{\cos \theta }\right)}{\left(\sin \theta \left(\frac{\sin \theta }{\cos \theta }\right)+\cos \theta \right)}\\ =m\left(\frac{v^2\cos \theta }{r}-g\sin \theta \right)\end{array}$ (VI)

Conclusion:

Substitute $32\text{m/s}$ for $v$, $5.0°$ for $\theta$, $410\text{m}$ for $r$, $9.80{\text{m/s}}^2$ for $g$, and $1400\text{kg}$ for $m$ in equation (VI) to find $f$.

$\begin{array}{c}f=\left(1400\text{kg}\right)\left(\frac{{\left(32\text{m/s}\right)}^2\cos 5.0°}{410\text{m}}-\left(9.80{\text{m/s}}^2\right)\sin 5.0°\right)\\ =2287\text{N}\\ \approx 2300\text{N}\end{array}$

Therefore, the frictional force on the car is $\underset{\_}{2300\text{N}}$.

(b)

To determine

The speed with which one must drive the car so that the force of friction is zero.

Answer to Problem 28P

The speed with which one must drive the car so that the force of friction is zero is $\underset{\_}{19\text{m/s}}$.

Explanation of Solution

Write the expression for the force of friction for the given track.

$f=m\left(\frac{v^2\cos \theta }{r}-g\sin \theta \right)$

The force of friction to be zero, the term $f$ in the above equation has to be replaced by $0$.

$0=m\left(\frac{v^2\cos \theta }{r}-g\sin \theta \right)$ (VII)

Solve equation (VII) for $v$.

$\begin{array}{l}{v}^2=\frac{rg\sin \theta }{\cos \theta }\\ v=\sqrt{rg\tan \theta }\end{array}$ (VIII)

Conclusion:

Substitute $410\text{m}$ for $r$, $9.80{\text{m/s}}^2$ for $g$, and $5.0°$ for $\theta$ in equation (VIII) to find $v$.

$\begin{array}{c}v=\sqrt{\left(410\text{m}\right)\left(9.80{\text{m/s}}^2\right)\left(\tan 5.0°\right)}\\ =18.74\text{m/s}\\ \approx 19\text{m/s}\end{array}$

Therefore, the speed with which one must drive the car so that the force of friction is zero is $\underset{\_}{19\text{m/s}}$.