Chapter 5, Problem 90P

(a)

To determine

To Find:The force of static friction exerted on the car by the runway surface.

The minimum coefficient of static friction necessary for the car to sustain this speed.

Answer to Problem 90P

  $500\text{N}$ ; $0.0255$

Explanation of Solution

Given Information:

Mass of the car, $m=2000\text{kg}$

Air drag on the car, $F_D=500\text{N}$

Speed of the car, $v=100\text{km/h}$

Formula Used:

Newton’s second law:

  $F_{net}=ma$

Frictional force

  $f_s={\mu }_{s}N$

Calculation:

The following is the free body diagram of the car as the car travels on the runway.

  

Assume the mass of the car as $M$ . In the above figure, normal force on the car is $N$ , the frictional force between the car and the road is, $f_s$ and the drag force is $F_D$ .

Convert the units of the speed of the car from $\text{km/h to m/s}$

Speed of the car, $v=100\text{km/h}$

  $\begin{array}{l}\text{=}\left(\text{100}\frac{\text{km}}{\text{h}}\right)\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)\left(\frac{\text{1 h}}{\text{3600}\text{s}}\right)\\ \text{=27}\text{.8 m/s}\end{array}$

Since, the car is moving with constant velocity, the acceleration of the car is zero.

Net force acting on the car along z axis,

  $\begin{array}{l}{F}_{net,x}=M{a}_x\\ \Rightarrow f_s-F_D=0\mathrm{...}\left(1\right)\\ \Rightarrow f_s=F_D\end{array}$

Substitute $500\text{ N}$ for $F_D$ in equation $\left(1\right)$ to solve for the static frictional force exerted by the runway on the car, $f_s=500\text{N}$

Thus, the required static frictional force on the car is $500\text{N}$ .

Net force acting on the car along $y$ axis:

  $\begin{array}{l}{F}_{net,y}=M{a}_y\\ \Rightarrow N-Mg=0\\ \Rightarrow N=Mg\end{array}$

Rewrite equation $\left(1\right)$

  $\begin{array}{l}{f}_s=F_D\\ \Rightarrow {\mu }_{s}N=F_D\left[\text{Since, }N=Mg\right]\\ \Rightarrow {\mu }_{s}Mg=F_D\\ \Rightarrow {\mu }_s=\frac{F_D}{Mg}\end{array}$

Here ${\mu }_s$ is the coefficient of static frictional force

Substitute $500\text{ N for }{F}_D\text{ and 2000 kg for }M$ in this equation

  $\begin{array}{c}{\mu }_s=\frac{\text{500 N}}{\left(\text{2000}\text{kg}\right)\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right)}\\ =0.0255\end{array}$

Conclusion:

Thus, the required coefficient of static frictional force is $0.0255$ .

(b)

To determine

To Find: The minimum coefficient of static friction necessary for the car to hold the last radius of curvature without skidding.

Answer to Problem 90P

The radius of the curve at $\theta ={45}^{\text{o}}\text{is 3}\text{.09 km}$ .

The radius of the curve at $\theta ={88}^{\text{o}}\text{is 108 m}$ .

The required coefficient of friction for the car without skidding is $0.73$ .

Explanation of Solution

Given Information:

Mass of the car, $m=2000\text{kg}$

Air drag on the car, $F_D=500\text{N}$

Speed of the car, $v=100\text{km/h}$

Formula Used:

Newton’s second law:

  $F_{net}=ma$

Frictional force

  $f_s={\mu }_{s}N$

Calculation:

The following is a free body diagram of the car on a circular path.

  

In the above figure, $r$ represents the radius of the curve, $f_s$ represents the static frictional force exerted by the road on the tires of the car and $\theta$ is the angle made by the frictional force with the speed of the car.

Apply Newton’s second law to the motion of the car and calculate the net force acting on the car

The radial force acting on the car along x-axis.

  $\begin{array}{l}{\displaystyle \sum F_x}=M{a}_c\\ f_s\sin \theta =\frac{M{v}^2}{r}\mathrm{....}\left(2\right)\end{array}$

Here, $a_c$ is the centripetal acceleration and $v$ is the speed of the car.

The tangential force acting on the car along x-axis.

  $\begin{array}{l}{\displaystyle \sum F_{\tan }}=M{a}_1\\ f_s\cos \theta -F_D=0\mathrm{....}\left(3\right)\\ f_s\cos \theta =F_D\end{array}$

Divide equation $\left(2\right)$ by equation $\left(3\right)$ .

  $\tan \theta =\frac{M{v}^2}{r{F}_D}$

Rearrange this equation for $r$ .

  $r=\frac{M{v}^2}{\tan \theta F_D}\mathrm{..........}\left(4\right)$

Substitute $500\text{N for }{F}_D,2000\text{ kg for }M,{45}^{\text{o}}\text{ for }\theta \text{ and 27}\text{.8}\text{m/s for }v$ in this equation,

  $\begin{array}{c}r=\frac{M{v}^2}{\tan \theta F_D}\\ =\frac{\left(2000\text{kg}\right){\left(\text{27}\text{.8}\text{m/s}\right)}^2}{\tan {45}^{\text{0}}\left(500\text{N}\right)}\\ =\left(3.09\times {10}^3\text{m}\right)\left(\frac{\text{1}\text{km}}{{10}^3\text{m}}\right)\\ =3.09\text{km}\end{array}$

Thus, the radius of the curve at $\theta ={45}^{\text{o}}\text{ is 3}\text{.09 km}$ .

Similarly, use equation $\left(3\right)$ to find the radius of the curve at $\theta ={88}^{\text{o}}$ .

  $r=\frac{M{v}^2}{\tan \theta F_D}$

Substitute $500N\text{ for }{F}_D,2000\text{ kg for }M,{88}^{\text{o}}\text{ for }\theta \text{ and 27}\text{.8}\text{m/s for }v$ in this equation.

  $\begin{array}{c}r=\frac{M{v}^2}{\tan \theta F_D}\\ =\frac{\left(2000\text{kg}\right){\left(\text{27}\text{.8}\text{m/s}\right)}^2}{\tan {88}^{\text{0}}\left(500\text{N}\right)}\\ =107.9\text{m}\\ =108\text{m}\end{array}$

Thus, the radius of the curve at $\theta ={88}^{\text{o}}\text{ is }108\text{m}$ .

Calculate the minimum coefficient of friction required for the car to hold at radius $108\text{ m}$ without skidding by making use of Newton’s second law of motion.

  $\begin{array}{l}{\displaystyle \sum F}=Ma\\ f_s=\frac{M{v}^2}{r}\\ \Rightarrow {\mu }_{s}Mg=\frac{M{v}^2}{r}\\ \Rightarrow {\mu }_s=\frac{v^2}{rg}\end{array}$

Substitute $27.8\text{ m/s for }v\text{ and 108 m for }r$ in this equation,

  $\begin{array}{c}{\mu }_s=\frac{v^2}{rg}\\ =\frac{{\left(\text{27}\text{.8}\text{m/s}\right)}^{\text{2}}}{\left(\text{108}\text{m}\right)\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right)}\\ =0.73\end{array}$

Conclusion:

Thus, the required coefficient of friction for the car without skidding is $0.73$ .