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Calculating acceleration on an inclined plane (with friction)

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A 38kg block is on a slope that is 50 degrees above the horizon. The box is attached to a 9kg weight that rests on a flat surface (connected to the top of the slope). The coefficient of friction for both surfaces is 0.12. Calculate the acceleration of the system.


    2. Relevant equations
    Fnet = Fapp - Ff
    Fgx = (sin[angle])(gravity)(mass)
    F = ma

    3. The attempt at a solution

    Fnet = Fapp - Ff
    Fnet = (38 * 9.81 * (sin50)) - (9*9.81*.12) - ((Cos50)*9.81*38*0.12)
    Fnet = 246.1

    F = ma

    a = f/m
    a = 246.21/(38+9)
    a = 5.24m/s^2

    My professor got 2.9m/s^2 and I can't see where (if) I'm going wrong. Help would be absolutely fantastic. Thanks!
     
  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    I can't find a fault in your reasoning or calculation either.
     
  4. Nov 6, 2012 #3
    He was using the formula "Fnet = Fapp - Ff (of the 38kg object) - T2" where T2 is tension in the rope that is connecting the two weights, if that helps. I can't really explain much about about it though, since it was the first time I had ever seen it used in my life and he didn't really explain it either.

    I have a test about this kind of thing tomorrow and it's freaking me out right now. >_> Thanks so much!
     
  5. Nov 6, 2012 #4

    haruspex

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    That's a perfectly reasonable approach, but it should produce the same answer:
    M = 38, m = 9, kinetic friction coefficient = μ, slope = α, accn = a.
    Ma = Mg sin(α) - T - Mgμ cos(α)
    ma = T - mgμ
    Summing:
    Ma + ma = Mg sin(α) - mgμ - Mgμ cos(α)
    a = g(M sin(α) - mμ - Mμ cos(α))/(M+m)
    = 9.81*(38*sin(50*pi()/180) - 9*.12 - 38*.12*cos(50*pi()/180))/(38+9) = 5.24
     
  6. Nov 6, 2012 #5
    Well I'll show that to him before the test, then. Thank you so much; I really appreciate it!
     
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