Brake or turn? Figure 6 - 44 depicts an overhead view of a...

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 $d=107 \mathrm{m},$ and take the car's mass as $m=1400 \mathrm{kg},$ its initial speed as $v_{0}=35 \mathrm{m} / \mathrm{s}$ , and the coefficient of static friction as $\mu_{s}=0.50$ 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 $f_{s, \max } ?(c)$ If the coefficient of kinetic friction between the (sliding) tires and the road is $\mu_{k}=0.40$ , at what speed will the car hit the wall? To avoid the crash, a driver could elect 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 $d$ and at the given speed $v_{0},$ so that the car moves in a quarter circle and then parallel to the wall? (e) Is the required force less than $f_{s, \max }$ so that a circular path is possible?

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 $d=107 \mathrm{m},$ and take the car's mass as $m=1400 \mathrm{kg},$ its initial speed as $v_{0}=35 \mathrm{m} / \mathrm{s}$ , and the coefficient of static friction as $\mu_{s}=0.50$ 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 $f_{s, \max } ?(c)$ If the coefficient of kinetic friction between the (sliding) tires and the road is $\mu_{k}=0.40$ , at what speed will the car hit the wall? To avoid the crash, a driver could elect 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 $d$ and at the given speed $v_{0},$ so that the car moves in a quarter circle and then parallel to the wall? (e) Is the required force less than $f_{s, \max }$ so that a circular path is possible?

Key Concepts

Centripetal Force and Circular Motion

When a car turns, it undergoes circular motion and requires a centripetal force to keep it moving along a curved path. This inward force is provided by the friction between the tires and the road. The required centripetal force is proportional to the square of the velocity divided by the radius of the turn. Analyzing whether the available frictional force can supply the necessary centripetal force is critical for determining if a turn can be executed without skidding.

Deceleration and Braking Distance

Deceleration is the process of reducing speed, which in vehicular dynamics is often achieved through frictional forces during braking. The braking distance is the distance a car travels while decelerating to a stop, and it depends on the initial speed, deceleration rate (which is influenced by friction), and reaction times. Calculating the required frictional force to achieve adequate deceleration is essential to safely stop a vehicle within a given distance.

Kinetic Friction

Kinetic friction comes into play once sliding begins between two surfaces. For vehicles, when tires skid against the road, the kinetic friction force, which is typically lower than static friction, governs the deceleration rate. This concept is important when a car loses traction during braking, affecting the stopping distance and the speed at impact.

Static Friction

Static friction is the force that prevents relative motion between two contacting surfaces when no sliding occurs. In the context of vehicle dynamics, it is the friction between the tires and the road that allows a car to accelerate, brake, or turn without skidding. The maximum static frictional force is defined by the coefficient of static friction multiplied by the normal force, and it sets the limit for the force that can be applied before the tires lose grip.

Transcript

00:01 So let's solve for the deceleration.

00:02 We know v final squared equals v initial squared plus 2 times a, d.

00:08 We can then say that velocity final squared we notice is b0.

00:14 The acceleration would then be equal to negative v initial squared divided by 2 times d.

00:19 This is equaling negative times 35 meters squared, quantity, 35 meters per second quantity squared, divided by 2 times 107 meters.

00:36 This is giving us a deceleration of negative 5 .72 meters per second squared.

00:45 Now, at this point, we can say that the force of friction required to stop the car would be equal to the mass times the absolute value of the accelerators.

01:03 So this would be equal to 1 ,400 kilograms, the mass of the car, multiplied by 5 .72 meters per second squared.

01:13 And we find that the force of friction is going to be equal to approximately 8 .0 times 10 to the 3rd newtons.

01:27 So this would be our answer for part a.

01:29 Unfortunately, we have to round to two significant figures because the mass of the car is only accurate to two significant figures.

01:38 For part b, if we wanted to find the maximum possible static friction, we can say force of friction static max.

01:46 This would be equal to the coefficient of static friction times essentially the weight of the car.

01:50 This would be equal to 0 .50, multiplied by 1 ,400 kilograms, multiplied by 9 .8 .9.

02:04 Meters per second squared.

02:06 This is giving us approximately 6 .9 times 10 to the third noons...

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