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Coefficient of Rolling Friction

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Two bicycle tires are set rolling with the same initial speed of 3.00 along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 and goes a distance of 17.2 ; the other is at 105 and goes a distance of 92.3 . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be = 9.80 .

    What is the coefficient of rolling friction for the tire under low pressure?



    2. Relevant equations



    3. The attempt at a solution

    I found the acceleration to be -0.19 I am stuck at that part I dont know what to do from there.
     
  2. jcsd
  3. Oct 14, 2007 #2

    PhanthomJay

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    Yes, good so far, don't forget your units. Now that you have solved for a, why not try Newton 2?
     
  4. Oct 14, 2007 #3
    Ive tried that but Fnet=ma... they dont give mass in the problem. I cant find the coefficient without mass. Im really stuck...I know I have to use newtons second law but I dont know how to start...
     
  5. Oct 14, 2007 #4

    PhanthomJay

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    You may not need to know the mass. What's the formula for the friction force, which is given as Fnet in this problem?
     
  6. Oct 14, 2007 #5
    Friction force= coefficient of friction * Normal force...?
     
  7. Oct 14, 2007 #6

    PhanthomJay

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    Yes, and what is the normal force in this problem?
     
  8. Oct 14, 2007 #7
    Hmmm that is what I am stuck on calculating normal force. Is there a way to calculate normal force with the information provided?
     
    Last edited: Oct 14, 2007
  9. Oct 14, 2007 #8

    PhanthomJay

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    To calculate the normal force, look in the y direction for all forces, and apply Newton 1, since there is no acceleration in the y direction..
     
  10. Oct 14, 2007 #9
    Fn= 0?
     
    Last edited: Oct 14, 2007
  11. Oct 14, 2007 #10

    PhanthomJay

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    Yes, Fnet = 0 in the y direction. The normal force acts up on the tire in the y direction. What other force acts in the y direction?
     
  12. Oct 14, 2007 #11
    Force of gravity. 9.8m/s^2
     
  13. Oct 14, 2007 #12

    PhanthomJay

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    The acceleration of gravity is 9.8m/s^2;the force of gravity is the tires weight, which is what?
     
  14. Oct 14, 2007 #13
    Sorry I dont really know...
     
  15. Oct 14, 2007 #14

    PhanthomJay

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    Oh,you should be familiar with the equation for weight : W = mg. So if the tire weighs 'mg', then the normal force must be ????
     
  16. Oct 14, 2007 #15
    n=mg...
     
  17. Oct 14, 2007 #16

    PhanthomJay

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    Yes! Now you've got everything you need. Go back into the horizontal direction and solve for the friction coefficient. The mass term should cancel out......
     
  18. Oct 14, 2007 #17
    I believe thats where Im stuck... I cant seem to continue from there. I would be missing the friction force to calculate the coefficient of friction.
     
  19. Oct 14, 2007 #18

    PhanthomJay

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    No, you already stated what the friction force, F_f is: it's the coef of friction, u, times the normal force, N, that is F_f = Fnet = u*N, and since N=mg, then the friction force is u*mg. And since F_net = ma, can you now solve for u?
     
  20. Oct 14, 2007 #19
    Thank you for your help. I dont know why I couldnt put that all together. Thanks again.
     
  21. Oct 14, 2007 #20

    PhanthomJay

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    It's OK, you toughed it out.
     
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