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Moving Wedge

  1. Aug 31, 2006 #1

    "A smooth wedge of mass M has a triangular cross section with a side inclined at an angle theta to the horizontal base. The wedge can slide without friction along a horizontal support. Placed on the side of the wedge is a mass m that can slide with no friction along the side. Find vectors of the acceleration for the wedge and for m after the bodies are released from rest."

    I'm taking the x coordinate as the direction along the horizontal support and the y coordinate as the direction perpendicular to the horizontal support.

    Ok. I tried using the Lagrangian method and the problem then appeared to become extremely complicated. So I'm using Newton's laws to solve this.

    It just seems like the normal force which is exerted on the mass m by the block M will not be the same as that when the wedge at rest, since the mass m accelerates partially with the acceleration of the mass M but I'm not sure.

    Any ideas here?
  2. jcsd
  3. Aug 31, 2006 #2

    Doc Al

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    Staff: Mentor

    Newton's laws should work just fine. :smile:

    Start by labeling the forces that act on each object. What force do they exert on each other? What's the net force on the entire system?

    Take advantage of the constraints. For example: You know that the wedge will only have a horizontal component of acceleration. And assume that the sliding mass maintains contact with the wedge.
  4. Aug 31, 2006 #3
    Yes but can we assume that the normal force to the wedge on the block is mgcos(theta) when we have a system where the wedge is moving? It just seems like if the wedge is accelerating then the normal force on the block will not be mgcos(theta). Almost as though we have to work with the system in a nonintertial reference frame (the reference frame of the wedge as compared to a point at rest) to make the problem more easier.
  5. Aug 31, 2006 #4

    Doc Al

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    You are correct on both counts. (1) The normal force is not just mg cos(theta) and (2) looking at the problem from a noninertial frame will help.
  6. Aug 31, 2006 #5
    What would be the horizontal force pushing the wedge? If the wedge was at rest, the total force on the wedge from the block would be mgcos(theta), just like the normal force that the wedge exerts on the block (Newton's 3rd law). However the wedge accelerates and even though Newton's 3rd law still holds, the normal force F_N does not take on the same form.

    If the wedge were at rest, the horizontal force would be something like mgcos(theta)*sin(theta).

    However, the wedge is moving with an accleration that we are supposed to find.

    If we could find this acceleration, we could find our answer.

    This problem is getting confusing.
  7. Aug 31, 2006 #6

    Doc Al

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    Your thinking is exactly on target. Don't give up! You are going to have to solve for the normal force, if you want to know it.

    I'll start you off. Let's start by focusing on the horizontal accelerations of each object. Let's call the horizontal acceleration of the sliding mass a1 (to the right, say) and the acceleration of the wedge a2 (to the left).

    Let's call the normal force F_n.

    (1) What's the relationship between a1 and a2?
    (2) What's the horizontal force on the sliding mass? Apply Newton's 2nd law.

    These will give you two equations. To get a third, view things from the noninertial frame of the wedge. Consider forces perpendicular to the surface.

    Give it a shot.
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