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Homework Help: Free-Body diagram of block on plane; both move

  1. Sep 15, 2010 #1

    this seems to be an easy enough problem, but for some reason I'm having problems with it.

    1. The problem statement, all variables and given/known data
    A block slides down an inclined plane, and the inclined plane slides horizontally. There is friction at the incline's upper surface, but not at its lower surface. For each body, draw a free body diagram.

    2. Relevant equations
    - none, we only need to draw the free body diagram

    3. The attempt at a solution

    I have already identified the usual suspects:

    Forces on plane:
    - the force of gravity, acting downwards
    - the normal force, acting upwards
    - Additionally, I know that the plane is sliding along a level floor. Because there is no friction (stated in the problem), there is no force resisting the motion, and, assuming that it slides at a constant velocity, that means that there is no horizontal force acting on it at all (because the horizontal component of the net force must be zero)
    - the weight of the block also acts on the plane, so we have another force acting downwards here.

    Here comes the tricky part: I'm not really sure what makes the block move in the same direction as the plane. Is the plane pushing it? What force is that? I understand the other forces on the block (gravity, normal force, friction), I just don't understand what makes it "go along" with the plane.

    Am I making any sense?

  2. jcsd
  3. Sep 15, 2010 #2
    Friction is what makes the block track with the incline plane at its upper part . Normally, the force vector applied would be purely gravity-enforced. Friction would be applied straight upward. In the case of your moving incline plane, the vector is no longer applied straight up but is vector-modified by the horizontal movement of the incline plane. The block fails to resist horizontally applied friction.
  4. Sep 15, 2010 #3


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    If the net force is zero the velocity is constant and if the velocity is constant, the net force is zero, but are you sure that the plane moves with constant velocity?

    The weight of the block acts on the block. But there is a normal force between the block and the incline.

    Why should the block move in the same direction as the incline?

    The plane pushes the block and the block pushes the plane: It is the interaction between them, and the direction of this force is normal to the surface between the block and inclined plane. You should draw a picture.

  5. Sep 15, 2010 #4
    Hi iffydroplight,

    thanks for your reply. I'm not sure, though, if I understand everything.

    This I don't understand. Wouldn't friction point up the slope instead of straight upward (assuming the case where the plane itself doesn't move)? After all, doesn't friction always point in the opposite direction of movement?

    I think I understand what you're saying. It's hard to describe this without using images, but I'll try: let's assume that the plane looks exactly like this one (random picture I found online):


    In this case, Fp would be friction if the plane didn't move.

    Let us further assume that the plane is moving towards the left (after all, the problem doesn't specify). Are you saying now that this movement modifies the Fp vector?
  6. Sep 15, 2010 #5
    No, I'm not. The problem doesn't specify it, so I assumed the simplest case.

    Are you talking about the normal force of the plane pushing on the block? I do understand this force. But what force then, originating from the block, pushes back on the plane? Isn't that the weight of the block?

    I'm sorry, I think I phrased this poorly. What I meant is: not only does the block slide down, but additionally, it also gets pushed horizontally by the motion of the plane.

    That's what I've been trying to do. I guess my main problem is that I don't know how to deal with the fact that the plane is moving. I think I could draw all the forces if it didn't...
  7. Sep 15, 2010 #6
    I made two pictures to illustrate what I mean...

    Attached Files:

  8. Sep 15, 2010 #7


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    Do not assume the simplest case.

  9. Sep 15, 2010 #8
    OK. But even when I don't assume constant velocity, the only thing this is going to change is that there will be an additional horizontal force acting on the plane, correct?
  10. Sep 15, 2010 #9


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    Well, it is not specified if the whole system has a constant horizontal velocity of not. That you can assume zero as the forces remain the same if a system moves with constant velocity with respect to an other inertia system. The horizontal position of the centre of mass of the block-incline system should be in rest as there is no horizontal external force acting on the system. If the block moves down along the incline, the incline has to move horizontally in the opposite direction.

  11. Sep 15, 2010 #10
    Thanks for your patience, ehild.

    This makes sense to me intuitively. What I still don't understand, however, is what the free-body diagram of the block would look like. To be more precise: is the diagram I posted correct?
    Put differently: if I had to compare the free-body diagrams of the block once with the plane being stationary and once with the moving plane, what would be different?

    I feel that I'm getting closer, but I'm still missing something...
  12. Sep 15, 2010 #11


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    Yes it is a question, with a moving constrain. But your forces have directions fixed to the incline, so draw them parallel and normal to it. So your free body diagram on the block is all right. As for the incline, the block and incline interact, so the same normal force and friction acts on the incline, with reversed directions.

  13. Sep 15, 2010 #12
    Ah, that makes sense, thanks!

    Just to help me understand this better: what would change if there was a net force on the plane?
  14. Sep 15, 2010 #13


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    Than you would have include that force in the free body diagram. And if it had a horizontal component the centre of mass would accelerate. And it could happen that the block moved upward on the incline.

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