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Recoil of wedge when block slides down

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass 3.1 kg is initially at rest on a wedge of mass 19 kg, height 0.20 m, and an incline angle of 30° as shown in the figure below. There is no friction between the wedge and the floor. Starting at the top of the incline, the block is released and slides toward the bottom of the wedge. At the same time, the wedge "recoils" and slides some distance L to the right. Find L when the block has reached the bottom of the wedge.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 26, 2009 #2

    rl.bhat

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    In the problem it is stated that there is no friction between the wedge and the floor. I presume that there is no friction between wedge and block.
    The block fall freely vertically through a distance Y. As a recoil the wedge moves through X horizontally. Resolve Y into two components: y, perpendicular to inclined plane and x, parallel to the inclined plane. If you draw the free body diagram, you can see that y = X*sinθ. Hence acceleration ay = aX*sinθ,....(1) where aX is the acceleration of wedge.
    For the block, mg*cosθ - Ν = m*ay.......(2) and mg*sinθ = m*ax...(3)
    For wedge N*sinθ = M*aX......(4)
    From equation (1), (2) and (4) solve for aX.
    From equation(3), find the time t taken by the block to slide down the inclined plane.
    Using the kinematic equation find the distance L traveled by the wedge in time t.
     
  4. Oct 26, 2009 #3
    I'm totally confused. Can you make the variables for acceleration a little clearer. And wouldn't the normal be applied to the y and not the x components when finding the sum of forces?
     
  5. Oct 26, 2009 #4

    rl.bhat

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    Consider the block and wedge separately.
    Forces acting on wedge:
    Weight of the wedge Mg towards the base.
    Normal reaction N1 on the base in the upward direction.
    These two forces won't contribute to the motion of the wedge.
    Normal reaction N due to the block on the wedge acts normal to the inclined plane.
    Its component parallel to the base, ( N*sinθ ), pushes the wedge towards right with acceleration aX. So ( N*sinθ ) = M*aX....(4)
    Forces acting on the block:
    Froce mg acting vertically downwards.
    Normal reaction N due to wedge, normally away from the inclined plane.
    mg*sinθ accelerates the block along the inclined plane ( along x direction). So mg*sinθ = ax.
    The net force (mg*cosθ - N) pushes the block towards the inclined plane ( along y direction). So (mg*cosθ - N) = m*ay ...(2)
    If aX is the acceleration of the wedge in the horizontal direction, then ay = aX*sinθ
     
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