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What is the effect of multiple forces on a 2D rigid rectangle?

  1. Jun 23, 2013 #1
    Say I have a rectangle of width w and height h centred at (0, 0) on a 2D coordinate plane.
    The rectangle is free to move on a frictionless medium. It is not attached at any point.

    Case 1: A single force F = (Fx)i + (Fx)j acts at a point (x, y)
    How does the rectangle move? I looked it up on some websites and I think that the tangential component of the force rotates the rectangle around it's centre of gravity and the parallel component of the force causes translational motion. Is this correct?

    Case 2: Multiple forces F1, F2, ..., Fn act on points (x1, y1), (x2, y2), ..., (x3, y3)
    I have searched everywhere for a solution to this problem but I cannot find it. Is there a way to figure out mathematically how the rectangle rotates and/or translates?

    Is there a general method to figure out the rotational and translational motion in both cases?
     
  2. jcsd
  3. Jun 23, 2013 #2

    Doc Al

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    Partly right. The main error is in thinking that only the "parallel" component contributes to translational acceleration. Not so: The entire net force will determine the translational acceleration of the center of mass.

    For rotation about the center of mass, consider the torque about that point from each force. Use that to determine the rotational acceleration about the center of mass.

    See my comments above.

    Yes. See my comments above.

    Realize that as the body moves and rotates, the torques and thus the rotational acceleration changes.
     
  4. Jun 23, 2013 #3
    But does the rectangle always rotate about its center or mass? Remember, the rectangle is not fixed. It's free to move. It's like a rectangular plate floating freely through 2D space.

    I've attached an example. In this case, would the square still rotate around it's centre of gravity because common sense doesn't say so. Although common sense has proven wrong quite a lot of times in Physics.
     

    Attached Files:

  5. Jun 23, 2013 #4
    And also, yes, in the case of a single force, the whole force does contribute but if we were to calculate the acceleration vector of the centre of mass then we would use a = F/m where F is the parallel component of the force, right?
     
  6. Jun 23, 2013 #5

    Doc Al

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    You can always treat the motion of a body as the sum of the translational motion of its center of mass plus its rotation about the center of mass. "Rotating about its center of mass" does not mean that the center of mass remains fixed.

    The forces shown in your diagram indicate that there will be rotational acceleration about the center of mass. (We are talking about the instantaneous acceleration at the moment the square and the applied forces are as shown. Things change as time goes on.)

    Why do you think it won't rotate about its center of mass?
     
  7. Jun 23, 2013 #6

    Doc Al

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    No.

    To find the acceleration of the center of mass you would use the entire force, not just one component.

    In your diagram above, what do you think the instantaneous acceleration of the center of mass will be?
     
  8. Jun 23, 2013 #7
    Um. I'm unsure. The net force on the object above is zero, right? So there should be no translational acceleration of the centre of mass?
     
  9. Jun 23, 2013 #8

    Doc Al

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    Exactly.
     
  10. Jun 23, 2013 #9
    And the total torque of the object is the sum of the torques of each force on the object?
     
  11. Jun 23, 2013 #10

    Doc Al

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    Right.
     
  12. Jun 23, 2013 #11
    What if the net force is not zero? Then how can we figure out the translational acceleration of the object? Do we find the net force and its point of application and then find the parallel component of that?
     
  13. Jun 23, 2013 #12

    D H

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    No. You find the net force and divide by mass. Period. The point of application is irrelevant as far as translational motion is concerned. The relevant equation is [itex]\vec F=m\vec a[/itex].
     
  14. Jun 23, 2013 #13
    No no. There is both rotational and translational motion. Imagine this square is floating in 2D space with only these forces acting on it. There is both rotational and translational acceleration, right?
     
  15. Jun 23, 2013 #14

    Doc Al

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    Use the full net force and Newton's 2nd law. (Stop with the parallel component already!)
     
  16. Jun 23, 2013 #15

    Doc Al

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    Not given the forces in your diagram. The net force is zero, thus the translational acceleration of the center of mass at that moment is zero.
     
  17. Jun 23, 2013 #16
    I don't understand. In the diagram, how is it possible for the square to not rotate? Doesn't the square have both a translational and rotational acceleration?
     
  18. Jun 23, 2013 #17

    Doc Al

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    Who says it doesn't rotate?

    No. What's the net force on it, according to your diagram. (You've already answered that. Stick to it.)
     
  19. Jun 23, 2013 #18
    Okay. I think I understand the translation. What about the rotation? Does it rotate around the centre of mass? Or another point?
     
  20. Jun 23, 2013 #19

    D H

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    Yes.


    Pick any fixed point on your rigid body. A corner, the middle of a side, the center of mass. Regardless of your choice, you can *always* describe the motion of a rigid body as a combination of the translation of that selected point and a rotation of the body about that point.

    The math is going to be easier with some choices. The center of mass is convenient because its the choice that decouples translational and rotational motion. Another convenient choice is the point about which motion is purely rotational. Things get hairy otherwise. That doesn't mean that its invalid. It just means that describing the motion is more complex.
     
  21. Jun 23, 2013 #20
    If we choose any arbitrary point, the net force acts on that point and the moment is around that point. But will the motion be the same regardless of the point chosen?
     
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