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Finding coefficient of kinetic friction

  1. Apr 19, 2014 #1
    1. The problem statement, all variables and given/known data

    A 100 N force directed 37° below the horizontal is applied to a 30 kg object on a horizontal surface. If the magnitude of the acceleration of the object is 1.3 m/s2, what is the coefficient of kinetic friction (μk) between the object and the surface?

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    I drew the body diagram and tried calculating the net force in the x direction.
    This is what I got:

    F = Fpx (the 100 N force) - Ffr
    Fpx is 100cos37 and Ffr is μmg, so
    ma = 79.86 - 294μ

    I plugged in 1.3 for a and got 0.24 for the answer.
    The correct answer is 0.12.
    Not sure where I went wrong, any help would be appreciated. Thanks!
     
  2. jcsd
  3. Apr 19, 2014 #2

    Simon Bridge

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    ... this is not correct.
    Did you remember to sum the forces in the y direction?
     
  4. Apr 19, 2014 #3
    It said friction force equals u*normal force.. isn't normal force mg?
     
  5. Apr 19, 2014 #4

    Simon Bridge

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    No.

    You have a free body diagram - add up the forces in the y direction.
     
  6. Apr 19, 2014 #5
    ok.
    In the diagram I drew I just have the 100N force, normal force, friction force and mg. Is that correct?
    So the forces in the y direction would be just "100sin37 + Fn - mg" ?
    But can I still set it equal to m*a with a being 1.3?
     
  7. Apr 19, 2014 #6

    Simon Bridge

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    What direction does the magnitude 100sin37 force act in?
    What do these forces add up to? (hint: does the object move in the y direction?)
    Solve for Fn.
     
  8. Apr 20, 2014 #7
    So it's 100sin37 - Fn - mg = 0 ?
    When I solve for Fn I get 233.81. I solved for the forces in the x direction (ma = 100cos37 - Ffr) and I got Ffr to be 40.86.
    So do I just divide that by Fn to get the coefficient?
    It's still wrong, I'm confused
     
  9. Apr 20, 2014 #8

    haruspex

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    No, that's still wrong. The normal force has to balance the sum of the other vertical forces.
     
  10. Apr 20, 2014 #9
    OHHHH I got one of the signs wrong... Just making sure, so it's Fn = 100sin37 + mg? Because then I get 0.115 which should be right
     
  11. Apr 20, 2014 #10

    Simon Bridge

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    Well done.

    Use your free body diagram to get the signs right.
    You should draw the diagonal forces as diagonal arrows - then divide into components on the diagram.
    Thus the 100N force would be diagonally downwards and you cannot get the component directions wrong.

    Note: The forces all add head-to-tail to make a resultant force pointing horizontally.
    If you fiddle with the order of the addition, you should get an easy triangle.
     
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