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How do I determine the wretched coefficient of kinetic friction? It's impossible

  1. Oct 1, 2012 #1
    A 3.40 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal, as in the figure below. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between block and ceiling.

    p4-60.gif


    Ugghhh so... this is what I did...

    F = 85.0 N
    m = 3.40 kg
    a = 6.00 m/s2
    fk = μkn

    1) Forces in the x-direction:
    F cos θ
    fk (kinetic friction)

    Forces in the y-direction:
    F sin θ
    n (normal force)

    2) Sum of forces

    ƩFx = F cos θ + fk = 0
    ƩFy = F sin θ + n = 0
    ƩFy = F sin θ + n = 0
    ƩFy = n = -F sin θ

    3) ƩFx = F cos θ + μkn = 0
    3) ƩFx = F cos θ + μk(-F sin θ) = 0
    3) ƩFx = F cos θ + μk(-F sin θ) = 0
    3) ƩFx = 85 cos 55 - μk(85 sin 55) = 0
    3) ƩFx = 85 cos 55 = μk(85 sin 55)
    3) ƩFx = (85 cos 55)/(85 sin 55) = μk
    3) ƩFx = 0.700 = μk

    So I got this wrong, the answer is actually 0.781 (???)

    I don't know what to do!
     
  2. jcsd
  3. Oct 1, 2012 #2
    And THEN, i tried doing this:

    ƩFx = F cos θ + μkn = 6

    85 cos 55 + μk(-85 sin 55) = 6
    85 cos 55 - μk(85 sin 55) = 6
    85 cos 55 = 6 + μk(85 sin 55)
    48.75 - 6 = 69.62μk
    42.75/69.62 = μk
    0.614 = μk

    -____-

    I didn't get that answer!!!
    Can you please show me how to get to that answer please?
     
    Last edited: Oct 1, 2012
  4. Oct 1, 2012 #3

    gneill

    User Avatar

    Staff: Mentor

    I don't see where you've taken gravity into account. Won't it affect the normal force?
     
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