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Acceleration of a system with 2 Blocks and a Wedge

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Block B of mass 10 kg is initially at rest as shown on the upper surface of a 20-kg wedge A which is supported by a horizontal surface. A 2-kg block C is connected to block B by a cord which passes over a pulley of negligible mass. Using computational software and denoting by m the coefficient of friction at all surfaces, use this program to determine the accelerations for values of m $ 0. Use 0.01 increments for m until the wedge does not move and then use 0.1 increments until no motion occurs.
    Capture.png

    2. Relevant equations
    ƩF = ma
    Tc/Tb = e(μ x β) (In this case β = 120*pi/180

    3. The attempt at a solution
    Taking each block as a system, finding Tc then Tb in terms of μ
    Assuming the acceleration of Block C is the same as the acceleration of block B

    Stuck at studying Wedge A. What are the forces acting on it and the equations?
    Is the acceleration of the wedge the same as the acceleration of B?
     
  2. jcsd
  3. May 13, 2013 #2

    haruspex

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    What forces do you think are acting on it?
    If B is sliding on A, it will have a different acceleration; otherwise it will be the same.
     
  4. May 14, 2013 #3
    Can we chooses different axes to study Block B and wedge A ?
     
  5. May 14, 2013 #4

    haruspex

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    You can, but it might be confusing, particularly for anyone trying to check your work.
     
  6. May 14, 2013 #5
    Well no, it is not confusing, it's the opposite. When we choose different axes according to the position of each block, it is much easier to find the Normal force acting on the system and the acceleration since a(y) becomes zero. But since Block B is in contact with the wedge, I wonder if that's allowed. And I really doubt it.
     
  7. May 14, 2013 #6

    haruspex

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    Sounds like you and I mean different things by 'using different axes'. pls post your working so that I can get some idea of what you're doing.
     
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