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Torque vs Force confusion

  1. May 17, 2014 #1
    Please look at pictures.

    What is it that Fv-Mg=-Ma? This does not make sense to me. Fv applies torque while Mg is a translational force. How can a force that is applied perpendicularly at a point of other than the center of mass cause translational motion?
     

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  2. jcsd
  3. May 17, 2014 #2

    SammyS

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    attachment.php?attachmentid=69862&d=1400374604.png

    A force is a force. Not translational. Not torque producing.

    The bar's Center of Mass (C.M.) accelerates vertically with an acceleration of [itex]\displaystyle \ -\frac{3g}{4}\ [/itex] .

    The sum of the forces is equal to ##\ M_\text{cm}\,\vec{a}_\text{cm}\ ## .

    The force exerted by the hinge is the only available force besides the weight of the bar.
     
  4. May 17, 2014 #3
    But the Fv force is a torque producing force!
     
  5. May 17, 2014 #4
    I mean "it is not a force because it is a 90° force acting on a rigid object (causes torque)"
     
  6. May 17, 2014 #5

    SammyS

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    Why say that ?

    Torque is always torque about some point.

    FV wasn't needed to find angular acceleration, α , was it ?
     
  7. May 17, 2014 #6
    Hmm. I'm going to ask a question: It may sound dumb: please bare:

    If I'm in outer space, no gravity, no forces around. There is this long rod rigid body flotting around. If I decide to apply a perpendicular force at its end, would that cause rotation, or translational motion?

    ====================
    ^
    |
     
  8. May 17, 2014 #7

    SammyS

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    It would produce both.
     
  9. May 17, 2014 #8
    Ok. Then, why is it that we do not take into account the torque that the hinge gives on the bar?
    (This sound really easy. Please bear with me.)
     
  10. May 17, 2014 #9
    In parallel, with the rod in space, would the translational acceleration of the rod be the same as if I had applied the same force on the center of mass?
     
  11. May 17, 2014 #10

    SammyS

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    In part (a) I'm assuming (because its solution wasn't posted) that the torque about the hinge is what was used to find the angular acceleration of the bar. The reason to choose that point: Any force the hinge applies produces no torque about the hinge, So at that stage in solving the overall problem, FV is not needed. The only force producing a torque about the hinge is gravity, which in this case is treated as if it is applied to the center of mass.


    Once you have found FV, go back and find the torque that the two forces produce about the center of mass. From that determine the angular acceleration of the bar. (Be sure to use the relevant Moment of Inertia).
     
  12. May 17, 2014 #11
    Perfect. Sorry for not posting part a solution!

    Could you help me out on the space rod problem?
     
  13. May 17, 2014 #12

    SammyS

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    Ummm -- in what way? What's the question?
     
  14. May 17, 2014 #13
    "with the rod in space, would the translational acceleration of the rod be the same as if I had applied the same force on the center of mass?"
     
  15. May 17, 2014 #14
    Sorry for my bad english.

    Would the translational acceleration be different if I applied the same force in these two ways

    ==================
    ^
    |



    ==================
    ............... ^
    .... ........... |
     
  16. May 17, 2014 #15

    SammyS

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    F = ma

    It would be the same translational acceleration (of the center of mass).


    To apply the force at the end of the bar, is relatively difficult to do in a practical situation.
     
  17. May 17, 2014 #16
    Would it not consist in only poking the rod with another stick for a fraction of a second?
     
  18. May 17, 2014 #17
    Also, if you kept applying a perpendicular force at that same point until the rod makes a 360 turn, could we be sure that the rod came back to its exact initial position in space even though it rotated and moved by translation?
     
  19. May 17, 2014 #18

    SammyS

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    That's one way. Basically that's applying an impulse.

    Certainly it would rotate through 360° at some point.

    I'm unsure about the rod returning to it's original position. I rather doubt that it would. It would continue rotating faster and faster as long as the force was applied.
     
  20. May 17, 2014 #19

    haruspex

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    There's nothing special about a "torque-producing" force. All forces produce torque. How much torque a force produces depends on the reference point you use to measure it. If the force is F and the reference point is distance d from the line of action of F then the torque is F*d. Regardless of that, the force also features in the equation for linear motion.
    If you apply a force at the middle of a rigid bar, perpendicularly to it, that produces no torque about the centre of the bar but does produce a torque about one end. The bar does not rotate about that end because the bending moment in the bar counters the torque. A springy bar would bend, giving you a (local) rotation about each end.
    If the force is maintained at right angles to the bar it gets a bit hard to analyse. The direction of the force will depend on how the bar has moved up that point. I'm pretty sure the bar would not be back at its original position in space after 360 degrees. Why should it be? If I apply a force to accelerate a stationary particle to the right for some seconds, then switch it to the other direction for the same period of time, it will bring the particle back to rest but not at its original position.
     
  21. May 18, 2014 #20
    So, for rigid objects, I can add whatever forces act on them at whatever point and if the vectorial addition give net 0 force, the object has no acceleration for sure?
     
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