Question

Brake or turn? Figure 6- 44 depicts an overhead view of a car’s path as the car travels toward a wall. Assume that the driver begins to brake the car when the distance to the wall is $\mathbf{d}\mathbf{}\mathbf{=}\mathbf{107}\mathbf{}\mathbf{m}$, and take the car’s mass as$\mathbf{m}\mathbf{}\mathbf{=}\mathbf{1400}\mathbf{}\mathbf{kg}$, its initial speed as${\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{35}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, and the coefficient of static friction as${\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{50}$. Assume that the car’s weight is distributed evenly on the four wheels, even during braking.

(a) What magnitude of static friction is needed (between tires and road) to stop the car just as it reaches the wall?

(b) What is the maximum possible static friction${\mathbf{f}}_{\mathbf{s}\mathbf{,}\mathbf{max}}$?

(c) If the coefficient of kinetic friction between the (sliding) tires and the road is${\mathbf{\mu }}_{\mathbf{k}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{40}$, at what speed will the car hit the wall? To avoid the crash, a driver could Select to turn the car so that it just barely misses the wall, as shown in the figure.

(d) What magnitude of frictional force would be required to keep the car in a circular path of radius dand at the given speed${\mathbf{v}}_{\mathbf{0}}$, so that the car moves in a quarter circle and then parallel to the wall?

(e) Is the required force less than${\mathbf{f}}_{\mathbf{s}\mathbf{,}\mathbf{max}}$so that a circular path is possible?

Short answer

1. The magnitude of static friction is$8.0\times {10}^3\text{\hspace{0.17em}N}$.
2. The maximum possible static friction is$6.9\times {10}^3\text{\hspace{0.17em}N}$.
3. The speed at which the car will hit the wall$20\text{\hspace{0.17em}m/s}$.
4. The magnitude of frictional force is$1.6\times {10}^4\text{\hspace{0.17em}N}$.
5. Since ${\mathrm{F}}_{\mathrm{r}}>{\mathrm{f}}_{\mathrm{s},\max }$, no circular path is possible.

Step-by-step solution

Given data

• Distance to wall,$\mathrm{d}=107\text{\hspace{0.17em}m}$.
• Mass of car,$\mathrm{m}=1400\text{\hspace{0.17em}kg}$.
• Initial speed,${\mathrm{v}}_0=35\text{\hspace{0.17em}m/s}$.
• Coefficient of static friction, ${\mathrm{\mu }}_{\mathrm{s}}=0.50$.

To understand the concept

The problem is based on the kinematic equations of motion in which the motion is described at constant acceleration. Also, it deals with the centripetal force. It is a force that makes a body follow a curved path.

Formula:

The velocity in kinematic equations is given by,

${\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{ad}$

The centripetal force is given by,

${\mathrm{F}}_{\mathrm{r}}=\frac{{\mathrm{mv}}_0^2}{\mathrm{r}}$

(a) Calculate the maximum possible static friction fs,max 

Using the kinematic equation, the deceleration of the car is calculated as,

${\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{ad}$

Substitute the values, and we get,

$\begin{array}{c}0={\left(35\mathrm{m}/\mathrm{s}\right)}^2+2\mathrm{a}\left(107\text{\hspace{0.17em}m}\right)\\ \mathrm{a}=-5.72{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

Thus, the force of friction required to stop by car is given by,

$\mathrm{f}=\mathrm{m}\left|\mathrm{a}\right|$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{f}=\left(1400\text{\hspace{0.17em}kg}\right)(5.72{\text{\hspace{0.17em}m/s}}^{\text{2}})\\ \mathrm{f}\approx 8.0\times {10}^3\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of static friction is $8.0\times {10}^3\text{\hspace{0.17em}N}$.

(b) the maximum possible static friction fs,max

The maximum possible static friction is given by,

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}$

Substitute the values, and we get,

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}=\left(0.50\right)(1400\text{kg})(9.80{\text{m/s}}^{\text{2}})$

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}\approx 6.9\times {10}^3\text{N}$.

Thus, the maximum possible static friction is$6.9\times {10}^3\text{\hspace{0.17em}N}$.

(c) Find out at what speed will the car hit the wall if the coefficient of kinetic friction between the (sliding) tires and the road is μk = 0.40

If ${\mathrm{\mu }}_{\mathrm{k}}=0.40$, then${\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mg}$, and the deceleration is$\mathrm{a}=-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}$.

Therefore, the speed of the car when it hits the wall is,

$\mathrm{v}=\sqrt{{\mathrm{v}}_0^2+2\mathrm{ad}}$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{{\left(35\text{\hspace{0.17em}m/s}\right)}^2-2\left(0.40\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\left(107\text{\hspace{0.17em}m}\right)}\\ \mathrm{v}\approx 20\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed at which the car will hit the wall $20\text{\hspace{0.17em}m/s}$.

(d) Calculate the magnitude of frictional force required to keep the car in a circular path of radius d and at the given speed  v0, so that the car moves in a quarter circle and then parallels the wall

The force required to keep the motion circular is,

${\mathrm{F}}_{\mathrm{r}}=\frac{{\mathrm{mv}}_0^2}{\mathrm{r}}$

Substitute the values, and we get,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{r}}=\frac{\left(1400\text{\hspace{0.17em}kg}\right){\left(35.0\text{\hspace{0.17em}m/s}\right)}^2}{107\text{\hspace{0.17em}m}}\\ {\mathrm{F}}_{\mathrm{r}}=1.6\times {10}^4\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of frictional force is$1.6\times {10}^4\text{\hspace{0.17em}N}$ .

(e) Find out if the required force is less than fs,max  so that a circular path is possible 

From the calculations, we can say that since ${\mathrm{F}}_{\mathrm{r}}>{\mathrm{f}}_{\mathrm{s},\max }$that means no circular path is possible.