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zewei1988 Mar4-10 06:35 AM

Brake System (Torque and Friction)
1. The problem statement, all variables and given/known data
The tires of a 1500 kg car are 0.600 m in diameter and the coefficients of friction with the road surface are µs = 0.800 and µk = 0.600. Suppose the car has a disk brake system. Each wheel is slowed by the frictional force between a single brake pad and the disk-shaped rotor. On this particular car, the brake pad comes into contact with the rotor at an average distance of 18.5 cm from the axis. The coefficients of friction between the brake pad and the disk are s = 0.588 and k = 0.490. Calculate the normal force that must be applied to the rotor such that the car slows as quickly as possible.

2. Relevant equations
Torque = Fd
Friction = n* coefficiant

3. The attempt at a solution
I'm not very sure which values to use. All I did was find the torque required and find the amount of force to apply

0.8 * 1500/4 * 9.8 = 0.185 * F * 0.588

8.11, 1.35 and 7.30 kN are some of the answers that I submitted and are all wrong.

tiny-tim Mar4-10 01:52 PM

Hi zewei1988! :smile:

Hint: you haven't used the radius of the wheel. :wink:

zewei1988 Mar4-10 10:14 PM

Re: Brake System (Torque and Friction)
But which friction coefficient should I use? Static or Kinetic?

tiny-tim Mar5-10 03:25 AM

Which do you think? And why? :smile:

zewei1988 Mar5-10 03:29 AM

Re: Brake System (Torque and Friction)
kinetic? cos it's moving?

tiny-tim Mar5-10 03:35 AM

That's the general idea! :smile:

But there are two friction forces here is there relative motion in both cases? :wink:

zewei1988 Mar5-10 10:00 AM

Re: Brake System (Torque and Friction)
I dun really know.

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