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Mass on a moving wedge

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    2eofudg.jpg

    This is a frictionless system with the wedge on a frictionless horizontal surface. When the system is released, the horizontal wedge (mass M) with diagonal angle theta moves to the left with constant acceleration a. What is it?

    I hope I'm right when I say this is not a very easy question, because I spent ages on it but I'm still not sure what the answer is. Can someone please help me check the answer?

    2. Relevant equations

    The mass (=m) does not leave the wedge. Taking the perpendicular component of the wedge's acceleration,

    [tex]ma sin \theta = mg cos \theta - N[/tex] where N is the normal reaction force between the wedge and the mass, directed perpendicular to the plane.

    Applying Newton's second law, the horizontal acceleration on the wedge is solely created by the normal force. Therefore [tex]Ma = N sin \theta[/tex]. Then applying algebra,

    [tex]a = \frac{mg sin\theta cos\theta}{M + m {sin}^2 \theta}[/tex].
     
  2. jcsd
  3. Mar 4, 2009 #2
    I think you have your answer. No values have been given for any of the variables so the best you can get is an expression for a.

    Edit: I just worked it out myself and I got the same expression for a as you
     
    Last edited: Mar 4, 2009
  4. Mar 5, 2009 #3
    All right... thanks :)
     
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