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

(a)

The co-efficient of static friction.

Answer to Problem 12.122RP

${\mu }_s=0.9627$

Explanation of Solution

Given information:

At a braking test,

The velocity is reduced from $70mi/h$ to zero

The stopping distance is equal to $170ft$

Co-efficient of kinematic friction is $80\%$ of static friction

For a uniformly accelerated motion,

$v^2=v_0^2+2a\left(x-x_0\right)$

In the above equation,

$v$ - End velocity

$v_0$ - Start velocity

$a$ - Acceleration

$x$ - End position

$x_0$ - Start position

The static friction force is defined as,

$F_{friction}={\mu }_{s}R$

In the above equation,

${\mu }_s$ - Co-efficient of static friction

$R$ - Reaction force

Calculation:

Convert,

$70mi/h=102.667ft/s$

For a uniformly accelerated motion,

$\begin{array}{l}{v}^2=v_0^2+2a\left(x-x_0\right)\\ 0={\left(102.667ft/s\right)}^2+2a\left(170ft\right)\\ a=-31ft/s^2\end{array}$

For the force balance in upwards direction,

$\uparrow \sum F=R-W=0$

Therefore,

$R=W=mg$

Apply Newton’s second law of motion,

$\begin{array}{l}\to \sum F=ma\\ -{\mu }_{s}R=ma\end{array}$

Then,

${\mu }_s=-\frac{ma}{mg}=-\frac{a}{g}$

Therefore,

$\begin{array}{l}{\mu }_s=-\frac{a}{g}=-\frac{\left(-31ft/s^2\right)}{32.2ft/s^2}\\ {\mu }_s=0.9627\end{array}$

Conclusion:

The co-efficient of static friction is equal to ${\mu }_s=0.9627$

To determine

(b)

The stopping distance if a car skids.

Answer to Problem 12.122RP

Stopping distance if car skids is $212.51ft.$

Explanation of Solution

Given information:

At a braking test,

The velocity is reduced from $70mi/h$ to zero.

The stopping distance is equal to $170ft$ without sliding.

Co-efficient of kinematic friction is $80\%$ of static friction.

For a uniformly accelerated motion,

$v^2=v_0^2+2a\left(x-x_0\right)$

In the above equation,

$v$ - End velocity

$v_0$ - Start velocity

$a$ - Acceleration

$x$ - End position

$x_0$ - Start position

The kinematic friction force is defined as,

$F_{friction}={\mu }_{k}R$

In the above equation,

${\mu }_s$ - Co-efficient of kinematic friction

$R$ - Reaction force

Calculation:

Convert,

$70mi/h=102.667ft/s$

According to the given information,

${\mu }_k=0.8{\mu }_s=0.8\left(0.9627\right)=0.77$

Apply Newton’s second law of motion,

$\begin{array}{l}\leftarrow \sum F=ma\\ {\mu }_{k}R=-ma\end{array}$

Therefore,

$a=-\frac{{\mu }_{k}R}{m}=-{\mu }_{k}g$

Substitute,

$a=-{\mu }_{k}g=-\left(0.77\right)\left(32.2ft/s^2\right)=-24.8ft/s^2$

Now,

$v^2=v_0^2+2a\left(x-x_0\right)$

Rearrange to find the stopping distance,

$x-x_0=\frac{v^2-v_0^2}{2a}$

Substitute,

$x-x_0=\frac{0-{\left(102.667ft/s\right)}^2}{2\left(-24.8ft/s^2\right)}=212.51ft$

Conclusion:

For skidding, the stopping distance is equal to $212.51ft.$