In Fig. 12-48, the driver of a car on a horizontal road...

In Fig. $12-48,$ the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between tires and road is $0.40 .$ The separation between the front and rear axles is $L=4.2 \mathrm{m},$ and the center of mass of the car is located at distance $d=1.8 \mathrm{m}$ behind the front axle and distance $h=0.75 \mathrm{m}$ above the road. The car weighs 11 $\mathrm{kN}$ . Find the magnitude of $(\mathrm{a})$ the braking acceleration of the car, (b) the normal force on each rear wheel, (c) the normal force on each front wheel, (d) the braking force on each rear wheel, and (e) the braking force on each front wheel. (Hint: Although the car is not in translational equilibrium, it is in rotational equilibrium.)

In Fig. $12-48,$ the driver of a car on a horizontal road makes
an emergency stop by applying the brakes so that all four wheels
lock and skid along the road. The coefficient of kinetic friction between tires and road is $0.40 .$ The separation between the front and
rear axles is $L=4.2 \mathrm{m},$ and the center of mass of the car is located at distance $d=1.8 \mathrm{m}$ behind the front axle and distance $h=0.75 \mathrm{m}$
above the road. The car weighs 11 $\mathrm{kN}$ . Find the magnitude of $(\mathrm{a})$
the braking acceleration of the car, (b) the normal force on each rear wheel, (c) the normal force on each front wheel, (d) the braking force on each rear wheel, and (e) the braking force on each
front wheel. (Hint: Although the car is not in translational equilibrium, it is in rotational equilibrium.)

Key Concepts

Kinetic Friction

Kinetic friction is the resistive force that acts between surfaces in relative motion. In the context of emergency braking, this friction force between the tires and the road is what decelerates the vehicle, and its magnitude is determined by the product of the coefficient of kinetic friction and the normal force (i.e., the component of the car’s weight normal to the road). This concept is fundamental in understanding how the braking system utilizes friction to slow down the car.

Dynamic Weight Transfer

Dynamic weight transfer refers to the redistribution of a vehicle's weight during acceleration or deceleration. When a car brakes, inertia causes a forward shift of the center of mass, increasing the normal force on the front wheels and decreasing it on the rear wheels. This phenomenon is crucial for accurately analyzing the braking forces on different axles, as it affects both the braking performance and stability of the vehicle.

Rotational Equilibrium

Rotational equilibrium in a vehicle under braking implies that the net torque about any axis is zero, even though the car is decelerating. This assumption allows one to set up moment equations to solve for unknown forces (such as the normal forces on each axle) by considering the positions of the forces relative to the center of mass. It ensures that the sum of the moments caused by the deceleration and gravitational forces balance out.

Center of Mass

The center of mass of a vehicle is the point where its mass can be considered to be concentrated. Its location relative to the axles plays a critical role in both translational and rotational dynamics during braking. A higher or more rearward center of mass will result in different load transfers during deceleration, thus affecting the distribution of normal forces and, consequently, the braking force available at each wheel.

Force Distribution on Axles

Force distribution on a vehicle’s axles is the division of the overall frictional braking force among the front and rear wheels. This distribution is influenced by the dynamic weight transfer and the vehicle’s geometry, such as the distances of the center of mass from the front axle and the ground. Understanding this concept is key to determining how much braking force each wheel can exert and is essential for ensuring the stability and effectiveness of the braking system.

Transcript

00:01 Here, knowing that for part a, knowing that the force is going to be equal to the mass times acceleration, and here this would simply be equal to the negative coefficient of friction times mg, we can find the absolute value of the acceleration or the magnitude of the acceleration.

00:16 This would be equal to the coefficient of kinetic friction times the acceleration due to gravity.

00:22 So this would be 0 .40 multiplied by 9 .8 meters per second squared, and this is giving us 3 .92 meters per second squared.

00:32 This would be your answer for part a.

00:36 For part b, we can use equation 129, and we can evaluate the torques about the car's center of mass.

00:47 We are going to bear in mind that the friction forces are acting horizontally at the bottom of the wheels, so we can say that the total kinetic frictional force would be equal to the coefficient of kinetic friction times m.

01:00 And this is going to be approximately 3 .92m.

01:05 We can say that then the vertical distance of the total frictional force would be a vertical distance of 0 .75 meters below the center of mass.

01:23 So we can say that the equilibrium of the torts, the sum of the torques, we know this is going to be zero because the car is in static equilibrium.

01:30 And this would be equal to 3 .92m multiplied by 0 .75 meters plus the force normal sub r so the force normal on the right we can then say that here this would be times 2 .4 meters minus force normal sub f this would be times 1 .8 and again this is equaling 0 we know that here we know that here equation 128 holds also holds we know that the acceleration is vertical and the acceleration is horizontal and not vertical so we can say equation 128 we can then say this is going to be f sub nr plus f sub n f this would equal the total weight and so we know the mass is obtained from the car's weight the mass would be equal to 11 ,000 newtons which would be the car's weight divided by 9 .8 meters per second squared this is equaling uh rather we can actually just keep it like this and then um substitute in for equation 12 8 so once we substitute in for that mass here we find that f sub nr is going to be equal to approximately 3 ,929 newtons we can say that this is going to be going to be approximately 3 ,930 newtons.

03:16 And we can then say that for part b, because since each involves two wheels, we can then say that we have roughly 2 .0 times 10 to the third neutens on each rear wheel.

03:42 So this would be your answer for part b...

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