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Aceeleration of a vertex

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Four similar rods of uniform density are connected with frictionless hinges, and this frame is placed to a horizontal smooth tabletop, such that its shape is a square. Vertex P is acted upon by a horizontal force in the direction of the diagonal, and due to this force it begins to move at an acceleration of aP. What is the acceleration of the opposite vertex Q at which it begins to move?


    2. Relevant equations



    3. The attempt at a solution
    I have no idea how to begin with this question. I can't visualize how the body would move when only one one of the vetex is given an acceleration.

    Any help is appreciated. Thanks!
     

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  2. jcsd
  3. Feb 19, 2013 #2

    SammyS

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    Each rod has mass (inertial). Each rod has a moment of inertia (rotational inertia).

    Draw a free body diagram for each rod.
     
  4. Feb 21, 2013 #3
    Sorry for being so late.

    I don't really have a clue what forces are acting on each rod.
    The vertex P has acceleration in the left direction. So the force at P acts in the left direction. It rotates about the top most point but that point isn't fixed. After some time it will also move. There are many things going on simultaneously which is confusing me.
     
  5. Feb 21, 2013 #4

    haruspex

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    By symmetry, you can parameterise the movement by a single angle.
    There are three unknown forces. The forces between the rods at Q are clearly in the y direction, likewise at P. The two at the other joints may be in any direction, but are mirror images of each other. In terms of these forces and angles, you can write the equations of motion of the rods. Throw in the fact that two rods must move in the same trajectory where they meet and see what you get.
     
  6. Feb 23, 2013 #5
    How? :confused:
    The force at P is in the left direction (or x direction).
     
  7. Feb 23, 2013 #6

    haruspex

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    The applied force is in the x direction, but the two rods there will move in the same way in that direction, so the force between the two rods is in the y direction.
     
  8. Feb 24, 2013 #7
    I still don't get it, can you please show it through a diagram? :uhh:
     
  9. Feb 24, 2013 #8

    haruspex

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    Label the other two vertices R, S.
    At P, rods PR, PS are each subjected to a force F/2 towards Q. They also exert a force G normally to this on each other. Suppose that's a tension, so it acts towards the line PQ on each. Then there'll be a force parallel to PQ acting between PR and QR at R. Probably a compression - call its magnitude H. Similarly at S. Finally, a force K parallel to RS at Q between QR and QS.
    Each of the four points accelerates. R and S will accelerate equally and oppositely. In terms of the accelerations aP and aR, you can write the linear and angular accelerations of rod PR. Likewise the other rods.
    You can now write out the two linear and one angular force/acceleration equations for each rod. By symmetry, it's only worth doing this for, say, PR and QR. Pls try to to write those out.
     
  10. Feb 24, 2013 #9
    Have I mentioned the forces correctly?
    2gx0j9k.png

    How did you figure out these two forces?
    Wouldn't there be any force in the y-direction at R? Why there is no horizontal force acting at Q?

    About what point do I need to take moments? R and S are not fixed points so I don't think I can calculate moments about those points.
     

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  11. Feb 24, 2013 #10

    haruspex

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    In terms of your diagram, F acts to the left on the whole system, F/2 on each of PR, PS.
    G acts vertically, down on PR, up on PS. This is a force that PR, PS exert on each other. By symmetry, it can only be straight up-and-down.
    H acts to the left on QR, QS and to the right on PR, PS.
    K acts vertically, down on QR, up on QS. This is a force that QR, QS exert on each other. By symmetry, it can only be straight up-and-down.
    (Note: G and K might turn out negative.)
    You're only concerned with instantaneous acceleration, so the points are effectively fixed.
     
  12. Feb 24, 2013 #11
    Please confirm that I have mentioned the forces correctly.
    After your confirmation, I will begin making equations.

    Thank you for your help so far. :)
     

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  13. Feb 25, 2013 #12

    haruspex

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    That is indeed what I wrote, but I realise now I did make one mistake. At junctions R and S there will be a vertical force as well. It will be upwards on RQ, SP and downwards on RP, SQ. Sorry about that.
     
    Last edited: Feb 25, 2013
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