Chapter 12, Problem 12.122RP

In the braking test of a sports car, its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance.

To determine

Find the coefficient of static friction.

Answer to Problem 12.122RP

The coefficient of static friction is $\underset{\_}{0.963}$.

Explanation of Solution

Given information:

The initial velocity $\left(v_0\right)$ of sports car is $70\text{mi/h}$.

The final velocity (v) of the sports car is 0.

The distance $\left(s-s_0\right)$ travelled by the sports car is $170\text{ft}$

Calculation:

Write the general equation of weight of the car (W).

$W=mg$

Here, m is the mass of the car and g is the acceleration due to gravity.

Sketch the free body diagram and kinetic diagram of the sports car as shown in Figure (1).

Refer Figure (1).

Consider the vertical equilibrium.

$\begin{array}{l}\Sigma F_y=0\\ N-W=0\\ N=W\end{array}$

Here, N is the normal force on the car.

Substitute mg for W.

$N=mg$

Substitute $32.2{\text{ft/s}}^2$ for g.

$\begin{array}{c}N=m\left(32.2\right)\\ =32.2m\end{array}$

Find the deceleration of the car using the equation:

$\frac{v^2}{2}-\frac{v_0^2}{2}=a_t\left(s-s_0\right)$

Substitute 0 for v, 70 mi/h for $v_0$, and 170 ft for $s-s_0$.

$\begin{array}{l}\frac{0}{2}-\frac{{\left(70\frac{\text{mi}}{\text{h}}\times \frac{5,280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3,600\text{s}}\right)}^2}{2}=a_t\left(170\right)\\ a_t=-31.001\text{ft/s}\end{array}$

Apply coefficient of static friction for braking without skidding.

Refer Figure 1.

Find the coefficient of static friction.

$\begin{array}{l}\Sigma F_t=m{a}_t\\ -{\mu }_{s}N=m{a}_t\end{array}$

Substitute 32.2m for N and $-31.001\text{ft/s}$ for $a_t$.

$\begin{array}{l}-{\mu }_s\left(32.2m\right)=m\left(-31.001\right)\\ {\mu }_s=\frac{31.001}{32.2}\\ {\mu }_s=0.963\end{array}$

Thus, the coefficient of static friction is $\underset{\_}{0.963}$.

To determine

Find the stopping distance for the same initial velocity if the car skids.

Answer to Problem 12.122RP

The stopping distance for the same initial velocity if the car skids is $\underset{\_}{212\text{ft}}$.

Explanation of Solution

Given information:

The coefficient of kinetic friction is 80 percent of the coefficient of static friction.

Calculation:

Find the coefficient of kinetic friction using the equation:

${\mu }_k=0.8{\mu }_s$

Substitute 0.963 for ${\mu }_s$.

$\begin{array}{c}{\mu }_k=0.8\left(0.963\right)\\ =0.7704\end{array}$

Apply coefficient of kinetic friction for braking with skidding.

Refer Figure (1).

Find the deceleration of the sports car $\left(a_t\right)$.

$\begin{array}{l}\Sigma F_t=m{a}_t\\ -{\mu }_{k}N=m{a}_t\end{array}$

Substitute 0.7704 for ${\mu }_k$ and32.2m for N.

$\begin{array}{l}-\left(0.7704\right)\left(32.2m\right)=m\left(a_t\right)\\ a_t=0.7704\times 32.2\\ a_t=-24.807{\text{ft/s}}^2\end{array}$

The deceleration is constant.

Find the stopping distance for the same initial velocity if the car skids using the equation:

$\frac{v^2}{2}-\frac{v_0^2}{2}=a_t\left(s-s_0\right)$

Substitute 0 for v, 70 mi/h for $v_0$, and $-24.807{\text{ft/s}}^2$ for $a_t$.

$\begin{array}{l}\frac{0}{2}-\frac{{\left(70\frac{\text{mi}}{\text{h}}\times \frac{5,280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3,600\text{s}}\right)}^2}{2}=\left(-24.807\right)\left(s-s_0\right)\\ -5270.22=-24.807\left(s-s_0\right)\\ s-s_0=212.45\text{ft}\\ s-s_0\approx 212\text{ft}\end{array}$

Thus, the stopping distance for the same initial velocity if the car skids is $\underset{\_}{212\text{ft}}$.