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Dynamics Block

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 N and a vertical force P are then applied to the block (the vertical force is upwards). The coefficients of friction for the block and surface are Ms = 0.40 and Mk = 0.25. Determine th emagnitude of the frictional force acting on the block if the magnitude of P is a) 8.0 N, b) 10 N, and c) 12 N


    2. Relevant equations
    fs = Mk * Fn
    Fnet = ma
    Fg - Fn - P = 0
    3. The attempt at a solution
    I tried using the third equation to get Fn and plug this value into the first equation to get force of friction. I didn't get the right answer... for P = 8.0 N the answer should be 6.0 N. How?
     
  2. jcsd
  3. Feb 22, 2009 #2

    Doc Al

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    Staff: Mentor

    Show exactly what you did. Hint: First you must determine whether the relevant friction is static or kinetic.
     
  4. Feb 22, 2009 #3
    Fg - Fn - P = 0
    24.5 = Fn + 8
    Fn = 16.5

    fs = 16.5 * 0.25 = 4.1 N
     
  5. Feb 22, 2009 #4

    Doc Al

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    Good.

    This assumes that the block moves. Does it?
     
  6. Feb 22, 2009 #5
    well Yea if the force applied is 6.0 N and the friction is 4.1 then yes it should move.
     
  7. Feb 22, 2009 #6
    Oh wait no it doesn't because Fn * Ms = 8.25 which is greater than 6 N so it doesn't move. Now what?
     
  8. Feb 22, 2009 #7

    Doc Al

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    How does static friction work? Note that for static friction, μN is the maximum possible static friction that the surfaces can generate before being forced to move. But static friction will be only what it needs to be to prevent slipping (up to the maximum). How much does static friction need to be in this case to prevent slipping?
     
  9. Feb 22, 2009 #8
    6 to prevent slipping?
     
  10. Feb 22, 2009 #9

    Doc Al

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    Staff: Mentor

    Right!
     
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