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Coefficient of static friction.

  • Thread starter alvarez91
  • Start date
  • #1
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Homework Statement


A 1980kg car starts from rest at point A on a 5° incline and coasts through a distance of 158m to point B. The brakes are then applied, causing the car to come to a stop at point C, 21m from point B. Knowing that slipping is impending during the braking period and neglecting air resistance, determine the coefficient of static friction between the tires and the road.


Homework Equations


Vf^2 = V0^2 + 2ad
F = ma

The Attempt at a Solution


I get the velocity at point B by: V = sqrt(0 + 2*(9.81*sin(5))*158) = 16.4371 m/s

I then find the acceleration from point B to point C by: 0 = 16.4371^2 + 2*a*21
∴ a = -6.4328 m/s^2

Then I try and find the coefficient of static friction by:
F = ma = -μmgcos(5)
cancelling the m's: -6.4328 = -μ*9.81*cos(5)
This gives me: μ = .6582 which is not correct. What am I missing? Thanks for any help.
 

Answers and Replies

  • #2
ehild
Homework Helper
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1,854

Homework Statement


A 1980kg car starts from rest at point A on a 5° incline and coasts through a distance of 158m to point B. The brakes are then applied, causing the car to come to a stop at point C, 21m from point B. Knowing that slipping is impending during the braking period and neglecting air resistance, determine the coefficient of static friction between the tires and the road.


Homework Equations


Vf^2 = V0^2 + 2ad
F = ma

The Attempt at a Solution


I get the velocity at point B by: V = sqrt(0 + 2*(9.81*sin(5))*158) = 16.4371 m/s

I then find the acceleration from point B to point C by: 0 = 16.4371^2 + 2*a*21
∴ a = -6.4328 m/s^2

Then I try and find the coefficient of static friction by:
F = ma = -μmgcos(5)
cancelling the m's: -6.4328 = -μ*9.81*cos(5)
This gives me: μ = .6582 which is not correct. What am I missing? Thanks for any help.
When the car stops there is no acceleration. The acceleration you got for the braking period is irrelevant.

ehild
 
  • #3
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Should I only be concerned with the forces of the car when it is stopped?
If so, would it be F = mgsin(5)?

Then I get: mgsin(5) = μmgcos(5)
which would give me: μ = tan(5) = .0875, but that seems too small.
 
  • #4
ehild
Homework Helper
15,478
1,854
You are right, that is too small, but static friction is not a defined quantity. So mgsin(5) ≤ μmgcos(5), μ≥tan(5°). Perhaps they want the coefficient of friction during the braking period, and they say, it is static friction as the tyres do not slip, but they are at the brink of slipping so the static friction is at its maximum.
But you did not calculate that correctly: you need to include both the component of gravity and the force of friction into the resultant force, ma.

ehild
 
Last edited:
  • #5
ehild
Homework Helper
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1,854
∴ a = -6.4328 m/s^2

Then I try and find the coefficient of static friction by:
F = ma = -μmgcos(5)
cancelling the m's: -6.4328 = -μ*9.81*cos(5)
This gives me: μ = .6582 which is not correct. What am I missing? Thanks for any help.
I was confused with the expression "slipping is impending". Read my edited previous post, and include the force of gravity in your equation:

F=ma=mgsin(5)-μmgcos(5)

ehild
 
  • #6
5
0
So if you set the acceleration to zero what happens? ...
You need to slow the block down, so Newton's Law say you have to slow it down.

To actually solve this problem, one method you can use is to set up three equations, one for position, one for velocity, and one for acceleration. You've already found the acceleration, and you can express the other two as functions of time assuming constant acceleration. Solve for your unknowns and you should find find the coefficient of friction.
 

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