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Minimum Angle to Prevent Slip

  1. Dec 2, 2015 #1
    1. The problem statement, all variables and given/known data
    9.12)
    The mass of the box on the left is 30kg and the mass of the box on the right is 40kg. The coefficient of static friction between each box and the inclined plane is μs = 0.2. Determine the minimum angle α for which the boxes will remain stationary.
    20151202_052804.jpg

    2. Relevant equations
    μsN = Fs
    Equillibrium Equations

    3. The attempt at a solution
    As you can hopefully see, Box 1 is the yellow box on the right and box 2 is the blue box on the left. My problem is finding N in the second equation. It's just looking too nasty...
    Box 1:
    Fx = T - Fs - Mgcos(30)
    Fy = N - MgSin(30)
    Mg = 392.4 => N = 196.2 => Fs39.24 => T = 379.04

    Box 2:
    Fx = T - Fs - Mgcos(α)
    Fy = N - MgSin(α)
    Mg = 294.3 => α = Cos-1(N / 294.3)
    This answer isn't very useful to me... You can see at the bottom of my notes I tried making substituting alpha into Fx, but it gives a nasty thing to factor. Historically this means I screwed up somewhere rather than the problem being this complicated...?
    20151202_052450.jpg
     
  2. jcsd
  3. Dec 2, 2015 #2

    CWatters

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    I've only have a quick look at this so I might be wrong but I think I can see a few issues with this equation...

    1) The box isn't accelerating so what does that say about the net force in any direction?
    2) Which way are you assuming friction acts on Box 1?
    3) Check you haven't used Cos when it should be Sin?
     
  4. Dec 2, 2015 #3

    haruspex

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    Judging from the working, whitejac has substituted that the net force is zero. But I second your points 2 and 3.
    Also, whitejac, I strongly advise against plugging in numerics until the final step. Keep everything symbolic until then. There are many advantages.
     
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