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Accelerating Wedge

  1. Apr 6, 2014 #1

    Radarithm

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    Gold Member

    1. The problem statement, all variables and given/known data

    A 45 degree wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find the block's acceleration. Gravity is directed down.

    2. Relevant equations

    Constraint Equation:
    [tex]\tan\theta=\frac{\ddot{y}}{\ddot{x}-\ddot{X}}[/tex]

    Equations (p is for parallel and p2 is for perpendicular, eg. x and y directions):

    [tex]F_{pb}=N\sin{\theta}=m\ddot{x}[/tex]
    [tex]F_{p^2b}=N\cos{\theta}-mg=m\ddot{y}[/tex]
    [tex]F_{pw}=F-N'\sin{\theta}=MA[/tex]

    3. The attempt at a solution

    It's embarrassing how this problem seems difficult for me; I'm probably making some fundamental error somewhere.

    Newton's 3rd Law: [itex]N=N'[/itex]
    [tex]N=\frac{m\ddot{x}}{\sin{\theta}}=\frac{m(g+\ddot{y})}{\cos{\theta}}[/tex]

    Because the angle is 45 degrees, the tangent of theta is just 1. Solving for the x acceleration, we get:
    [tex]\ddot{x}=(g+\ddot{y})[/tex]

    Now, according to the constraint equation, the acceleration in the y direction is equal to the tangent of theta times the x acceleration of the block minus the x acceleration of the wedge itself. Following the constraint equation, I get:
    [tex]\ddot{y}=(g+\ddot{y})-A[/tex]

    (tangent of 45 is 1). There is no way to solve for the y acceleration here. If, back when I solved for the normal force, I included the tangent of theta, I would've been able to solve for the acceleration but I would have to divide by zero, which means there is no y acceleration. The hint, however, states otherwise: If A = 3g then the y acceleration is g. Where am I going wrong? I'm going to try and solve for the y acceleration instead of the x acceleration first and see where that takes me.
     
  2. jcsd
  3. Apr 6, 2014 #2

    Radarithm

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    Gold Member

    I've reached a rather strange conclusion; I solved for the y acceleration in from the normal force equation (2 of them):
    [tex]\frac{m\ddot{x}}{\cos{\theta}}=\frac{m(g+\ddot{y})}{\sin{\theta}}[/tex]
    [tex]\ddot{y}=\ddot{x}-g[/tex]

    From the constraint equation:
    [tex](\ddot{x}-\ddot{X})\tan{\theta}=\ddot{y}=\ddot{x}-g[/tex]

    Solving for x, we get:
    [tex]\ddot{x}=\frac{-\ddot{X}\tan{\theta}}{(1-\tan{\theta})}-g[/tex]

    If A is 3g, then the y acceleration must be g; this equation shows that, but it is for the x direction. When I plug it into the y equation, things don't make any sense.
     
  4. Apr 6, 2014 #3

    Radarithm

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    I feel like I'm making a mistake that deals with inertial reference frames; bump?
     
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