Brake System (Torque and Friction)

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SUMMARY

The discussion focuses on calculating the normal force required for a 1500 kg car with a disk brake system to achieve maximum deceleration. The car's tires have a diameter of 0.600 m, and the coefficients of friction with the road are µs = 0.800 and µk = 0.600. The brake pads have coefficients of friction µs = 0.588 and µk = 0.490, with contact occurring at an average distance of 18.5 cm from the axis. Participants emphasize the importance of using the correct friction coefficient, specifically kinetic friction, due to the motion of the vehicle.

PREREQUISITES
  • Understanding of torque calculations (Torque = Fd)
  • Knowledge of friction coefficients (static and kinetic)
  • Basic physics of forces and motion
  • Familiarity with brake system mechanics
NEXT STEPS
  • Study the application of torque in braking systems
  • Learn about the differences between static and kinetic friction in practical scenarios
  • Explore the physics of deceleration in vehicles
  • Investigate the role of brake pad materials in friction performance
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Students studying physics, automotive engineers, and anyone interested in understanding vehicle braking dynamics and friction mechanics.

zewei1988
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Homework Statement


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 µs = 0.588 and µk = 0.490. Calculate the normal force that must be applied to the rotor such that the car slows as quickly as possible.


Homework Equations


Torque = Fd
Friction = n* coefficiant


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.
 
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Hi zewei1988! :smile:

Hint: you haven't used the radius of the wheel. :wink:
 
But which friction coefficient should I use? Static or Kinetic?
 
Which do you think? And why? :smile:
 
kinetic? cos it's moving?
 
That's the general idea! :smile:

But there are two friction forces here … is there relative motion in both cases? :wink:
 
I dun really know.
 

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