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Friction Forces help

  1. Sep 18, 2012 #1
    1. A 25.0-kg block is initially at rest on a horizontal surface.
    A horizontal force of 75.0 N is required to set the block in
    motion, after which a horizontal force of 60.0 N is required
    to keep the block moving with constant speed. Find (a) the
    coefficient of static friction and (b) the coefficient of
    kinetic friction between the block and the surface.



    2. Relevant equations



    3. i need help setting up the problem. i would apply the net force model for the x direction and equilibrium model for the y model. i need to know if this is right.
    Fx = 75.0N - F(static) - 60.0N = ma
    Fy = n - mg = 0

     
  2. jcsd
  3. Sep 18, 2012 #2

    rl.bhat

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    What is the expression for the coefficient of static friction?
     
  4. Sep 18, 2012 #3
    is the expression for static friction just
    -Fs = ma
     
  5. Sep 18, 2012 #4

    rl.bhat

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    Not the above expression.
    The expression for μ in terms of normal force and maximum frictional force.
     
  6. Sep 18, 2012 #5


    oh okay. so i got

    μ(s)n = ma

    than i got n = mg for the second equation.

    do i plug in "n" to the expression μ(s)n = ma ?

    which would give me μ(s)mg = ma
     
  7. Sep 18, 2012 #6

    rl.bhat

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    Correct. Now in the given problem, what is the normal force and ma?
     
  8. Sep 18, 2012 #7
    hmm would mg = -245?
    i don't know what the normal force is. i thought that mg and n always canceled to make it equal to zero.

    opps. i meant would ma = 75
     
  9. Sep 18, 2012 #8

    rl.bhat

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    Normal reaction is equal and opposite to mg. mg acts on the floor and normal reaction acts on the block. Now you can find the coefficient of static friction.
     
  10. Sep 18, 2012 #9
    THANKS! i found it!:smile:
     
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