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Coeffincient of static friction

  1. Oct 9, 2007 #1
    Kim has her coffee cup on her car's dash when she takes a corner with radius 4m and 20km/hr. What is the minimum coefficient of static friction which would aloow the coffee cup to stay there without slipping?

    this is what i did, but i think it is wrong

    MU= friction force/normal force= ma/mg= a/g= v^2/rg
    so mu= v^2/rg= 30.80/16*9.8=0.19
     
  2. jcsd
  3. Oct 9, 2007 #2

    Doc Al

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    Staff: Mentor

    You equation is correct but your arithmetic is not. Did you square the r by mistake?
     
  4. Oct 9, 2007 #3
    so mu should be = v^2/rg= 30.80/4*9.8=0.78
    that is the minimum coefficient of satic friction which would allow the cup to stay in postition without slipping?
     
  5. Oct 9, 2007 #4

    Doc Al

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    Yep. (If you're dealing with one of those online systems, you may want to double check the accuracy of your calculations lest you be off a bit. Redo v^2.)
     
  6. Oct 9, 2007 #5
    how do you know that it is the minimum mu and tht it will make the cup stay in position?
     
  7. Oct 9, 2007 #6

    Doc Al

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    Answer these questions:
    (1) How much force is required to centripetally accelerate the cup?
    (2) What normal force does the cup exert on the surface of the dash?
    (3) What's the maximum force that static friction can supply for a given mu?

    That should tell you the minimum mu required.
     
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