. ⇒√( Brake or turn? Figure 6) 44 depicts an overhead view...

$. \Rightarrow \sqrt{\text { Brake or turn? Figure } 6}$ 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 disFigure $6-44$ tributed evenly on the four wheels, Problem 58 . 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 $f_{s, \max } ?(\mathrm{c})$ 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?

$. \Rightarrow \sqrt{\text { Brake or turn? Figure } 6}$ 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 disFigure $6-44$
tributed evenly on the four wheels,

Problem 58 .
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 $f_{s, \max } ?(\mathrm{c})$ 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

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. In vehicle dynamics, static friction is crucial as it enables tires to grip the road, allowing the car to decelerate when braking and to change direction when turning without skidding. Its maximum value, determined by the normal force and the coefficient of static friction, sets an upper limit on the forces that can be applied before slipping occurs.

Kinetic Friction

Kinetic friction comes into play once there is relative motion between two surfaces—in this case, when a car's tires are sliding on the road. It is generally lower than static friction, meaning that if the tires lock and skid, the car will decelerate less effectively. This concept is important when analyzing scenarios where the maximum available grip is exceeded, affecting the stopping distance and control during braking.

Work-Energy Principle

The work-energy principle states that the work done by the forces acting on an object is equal to the change in its kinetic energy. In the context of vehicle braking, the work done by friction (force times distance) is used to dissipate the car’s initial kinetic energy, and this relationship allows for the calculation of required friction forces or stopping distances.

Centripetal Force

Centripetal force is the inward force necessary to keep an object moving along a circular path. When a car makes a turn, the friction between the tires and the road must provide this centripetal force to change the direction of the vehicle’s velocity without causing it to skid outward. This concept is key to understanding the limits of safe turning at a given speed.

Transcript

00:01 For this problem on the topic of force and motion, we are shown in the figure an overhead view of a car's path as it travels toward a wall.

00:07 The driver begins to break when the wall is 107 meters from the car, and we are given that the car's mass is 1 ,400 kilograms and its initial speed 35 meters per second.

00:19 The coefficient of static friction is 0 .5, and we want to assume that the car's weight is distributed evenly on the four wheels.

00:27 We want to find the magnitude of static friction required to stop the.

00:30 Car just as it reaches the wall, the maximum possible static friction, the coefficient of kinetic friction between the tires and the road is 0 .4, and we want to find the speed that the car will hit the wallward, and we want to find the magnitude of the frictional force that would be required to keep the car in a circular path of radius d at the given speed v0, so that the car moves in a quarter circle and then parallel to the wall.

00:55 And lastly, we want to know if the required force is less than the static frictional the maximum static frictional force so that a circular path is possible.

01:05 So firstly we'll use kinematics to find the deceleration of the car and we'll use the equation v squared is equal to v0 squared plus 2a d.

01:19 And so the final speed of the car is zero.

01:22 This is equal to its initial speed 35 meters per second squared plus two times the acceleration a into 107 meters this gives the deceleration of the car a to be minus 5 .72 meters per square second.

01:45 And so the force of friction required to stop the car, little f is equal to m times the magnitude of the acceleration by newton's second law.

01:57 This is the mass of the car 1 ,400 kg times the magnitude of the acceleration 5 .7.

02:05 2 meters per square second.

02:09 And so the friction of force has magnitude 8 times 10 to the power 3 newtons...

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