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What is the coefficient of Friction?

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    It says: A 20kg object is placed on a horizontal surface. A force of 3N is required to keep the object moving at a constant speed. What is the coefficient of friction between the two surfaces?
    I figured that we know that Ff=uFn and Fn=Fg=mg therefore Fn=20*9.8=196N That's as far as i can go...i am not sure what to do with the 3N force. The answer in the books says that u is .15


    2. Relevant equations
    Fn=Fg=mg
    Ff=u*Fn


    3. The attempt at a solution
    Fn=20*9.8
    =196N
    I'm lost from here...not sure what to do to solve for coefficient of friction given horizontal 3N force etc. Help please?
     
  2. jcsd
  3. Oct 26, 2009 #2
    Did you draw a FBD?
     
  4. Oct 26, 2009 #3
    Yes I drew a free body diagram, but it does not seem to be helping me.
     
  5. Oct 26, 2009 #4
    Are you sure it's 20kg and not 2.0kg?

    Because:
    Fpush - fk = 0
    3 = mu_k*mg
    3 = mu_k*(2.0)(9.8)
    mu_k = .15, which is your answer.

    Now, using 20:
    Fpush - fk = 0
    3 = mu_k*mg
    3 = mu_k*(20)(9.8)
    mu_k = .015

    Also, 20kg(44.1 pounds) would need a bit more than 3 newtons of force to push it constantly, or at least I would think.
     
    Last edited: Oct 26, 2009
  6. Oct 26, 2009 #5
    Yes, it is 20kg in the text, but this text has numerous typing errors so it looks like your mass is correct at 2.0kg based on the answer of .15 for the u value being the correct answer listed in the text. Thanks for your help. Just a few more questions if I may?
    Is the Fpush - fk = ma on the left hand side of the equation the Fnet value? What does mu_k represent? Thank you.
     
  7. Oct 26, 2009 #6
    Yes, Fpush - fk is Fnet, and since the box is moving at a constant speed, ma = 0. So, Fpush - fk = 0, you know the rest.

    mu_k is the co-efficient of friction, μk. μ is mu.
     
  8. Oct 26, 2009 #7
    and fk=uFpush right?
     
  9. Oct 26, 2009 #8
    No, because fk = μkn.
    Where n is the normal force.
    Using the two vertical forces from your FBD:
    (Fnet)y = n - Fg = 0(since your object is not moving up or down.)
    (Fnet)y = n = Fg => n = mg
    So, fk = (μk)(mg)
     
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