Car braking and ratio of force of friction

[smhippe]

Homework Statement

The front brake pads on your car wear out faster than the rear brake pads; in fact,
some cars are specifically designed with large front brake pads because of this. This
results from the fact that when a car is braking the magnitude of the friction force acting
on the front wheels of the car is larger than the magnitude of the friction force acting on
the rear wheels of the car. Consider a car that has a mass of 1500 kg that is decelerating
(by using the brakes) at a rate of 0.25 g. The distance between the front and rear axles is
2.8 m and the center of mass of the car is directly in-between the two axles at a height of
0.6 m above the ground. What is the ratio of the force of friction acting on the front tires
to the force of friction acting on the rear tires? Assume that the coefficients of friction
between the tires and the road are the same for all tires and, for simplicity, assume that
the car is skidding as it decelerates.

The Attempt at a Solution

First I set up a FBD. That didn't seem right...at all. So now I'm thinking about using torque. I don't really know how to set this up.

 

[willem2]

The torque from the frictional force, must be balanced by a torque that comes from the difference in the normal force acting on the front and rear tires.

 

[smhippe]

That means I have to find the normal force on each tire first then right? I can do this by using the FBD and substituting?

 

[willem2]

That means I have to find the normal force on each tire first then right? I can do this by using the FBD and substituting?

You can certainly use the FBD diagram to find out what the torques acting on the car are. (force * distance to center of mass). I don't know what you mean by substituting.
The torques from friction, normal force on the front wheels, normal force on the back wheels has to be equal to 0.

Of course the sum of the normal forces on the wheels has to equal the weight of the car as well.

 

[rdl]

I would approach this problem by first understanding the concept of friction and its role in braking. Friction is a force that opposes motion and is responsible for slowing down or stopping an object. In the case of a car braking, the friction force acts between the tires and the road surface, causing the car to decelerate.

Based on the given information, we can calculate the force of friction acting on the front and rear tires using the formula F = μN, where μ is the coefficient of friction and N is the normal force (equal to the weight of the car in this case). Since the coefficient of friction is the same for all tires, we can ignore it in our calculations.

For the front tires, the normal force is the weight of the car distributed over the front axle, which is given by 1500 kg * 9.8 m/s^2 * 0.6 m / (2.8 m / 2) = 2550 N. Therefore, the force of friction acting on the front tires is 2550 N.

For the rear tires, the normal force is the weight of the car distributed over the rear axle, which is given by 1500 kg * 9.8 m/s^2 * 0.6 m / (2.8 m / 2) = 2550 N. Therefore, the force of friction acting on the rear tires is also 2550 N.

Now, to calculate the ratio of the force of friction acting on the front tires to the force of friction acting on the rear tires, we simply divide the force of friction on the front tires by the force of friction on the rear tires. This gives us a ratio of 2550 N / 2550 N = 1.

This means that the force of friction acting on the front tires is equal to the force of friction acting on the rear tires. This result makes sense because the car is decelerating at a constant rate, meaning the friction forces on both sets of tires must be equal to produce this deceleration.

In conclusion, the ratio of the force of friction acting on the front tires to the force of friction acting on the rear tires is 1. This is due to the equal distribution of weight over both sets of tires and the assumption that the car is skidding as it decelerates.