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Coeff of friction

  1. May 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A force of 10N is applied downward at an angle of 30 deg with respect to the horizontal on a block. There is friction between the block and the floor and the block remains stationary. M=2kg
    (a) What is the magnitude of the normal force acting on the block?

    (b) The coefficient of friction between the floor and block is μ = 0.8. What is the magnitude of the frictional force acting on the block?

    2. Relevant equations
    Fn = mg + 10sin30
    Fn(mu) = Fk


    3. The attempt at a solution
    I got part A -- Fn = mg + 10sin30 = 19.6 + 5.0 = 24.6

    But for part B, I thought it would be 24.6*.8 = 19.7, but according to the answer key, it should be 8.7.
     
    Last edited: May 11, 2009
  2. jcsd
  3. May 11, 2009 #2
    The block is stationary on the floor so the sum of the horizontal forces must equal 0. So draw the free body diagram along with the horizontal forces acting on it.
     
  4. May 11, 2009 #3
    So, Fμ = 10 cos 30 ?
     
  5. May 11, 2009 #4
    Correct.
     
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