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Homework Help: Car braking and ratio of force of friction

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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.
  2. jcsd
  3. May 6, 2010 #2
    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.
  4. May 6, 2010 #3
    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?
  5. May 7, 2010 #4
    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.
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