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Determining coefficient of kinetic friction help!

  1. Sep 30, 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. Sep 30, 2012 #2
    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
    -----------------------
    It is accelerating, net Fx not zero.
    The force applied is pressing the block to the ceiling. The weight is pulling it away.
    I got this value - 0.78093138
     
    Last edited: Sep 30, 2012
  4. Oct 1, 2012 #3

    So is it supposed to be 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 what you did?
     
    Last edited: Oct 1, 2012
  5. Oct 1, 2012 #4
    Force of kinetic friction:
    fk = μkn

    Horizontal force of acceleration in the x-direction/Value of the bottom leg of the triangle:
    Facceleration-x = Fa(cos θ)
    Facceleration-x = 85(cos 55)
    Facceleration-x = 48.75 N

    Normal force n (value of the upright, vertical leg of the triangle):
    n = Facceleration-y(sin θ) + mg
    n = 85(sin 55) + (3.40 kg)(6.00 m/s2)
    n = 69.62 + 20.4
    n = 90.02 N

    fk = μkn
    48.75 = μk(90.02)
    48.75/90.02 = μk
    0.541 = μk

    aksjflkjadsfjds -____-
    Help?
     
  6. Oct 1, 2012 #5
    You were completely wrong.
    Net force doesn't equal 6.
    Thanks for nothing.
     
  7. Oct 1, 2012 #6
    I not sure whether you read the question.
    I just copy and paste the original question you posted and just underlined it.
     
  8. Oct 1, 2012 #7
    Here your original question and i underlined it.
     
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