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Torque - In Depth

  1. May 31, 2005 #1
    I have a few years of physics experience, and as I was thinking about a few things the other day, something about the concept of torque seemed a little weird to me.

    Simple Forces are easy to understand - gravity pulls down on an object, the ground pushes back up with an equal and opposite force, etc, etc. Now torque - If you think about it from the perspective of work it makes sense - If you apply a force at the end of a long arm to rotate it, the force is greater closer to the axis of rotation because it moves a shorter distance, but performs the same amount of work.

    However, consider a rigid bar in static equilibrium, with a fixed axis of rotation at one end. There is a weight in the middle, say 100 kg. You push up farther away from the axis so you only have to push with a 50kg force to keep it from falling. At that instant, there is no motion, so you cannot say your hand travels a greater distance than the weight so the force needed is less.

    With no motion, how does the system "know" (for lack of a better word) to take your 50kg force at the end, and apply 100kg of upward force where the weight is?
  2. jcsd
  3. May 31, 2005 #2
    This sounds fallacious but I think I know what you mean. To apply a torque of equal magnitude at a closer distance requires a larger force.

    I'm assuming the bar is horizontal. Firstly, you are mistaking weights, masses and forces. A weight is a force due to gravity, and a mass is just a mass for the purposes of this argument. Having a 100kg weight at a distance 'r' from the AoR will yield a torque of 100kg*9.8m/s^2*r. To counter this torque, you would need to apply a torque of equal magnitude in the opposite direction. Lets say your arm can apply a force of 100N

    [tex] \tau_{net} = 0 = \tau_{cw} + \tau_{ccw} \ and \ so \ \tau_{cw} = -\tau_{ccw}[/tex]

    Say the weight provides the clockwise torque, then

    [tex] \tau_{cw} = mgr = 980r_1 [/tex]

    [tex] \tau_{ccw} \mbox{ will need to be of the same magnitude,} 100r_2 [/tex]

    [tex] 980r_1 = 100r_2 \ and so \ r_2 = 9.8r_1 [/tex]. Only with radiuses described by this relationship will the system be in equilibrium.
    The system doesnt 'know' per sai, it will simply follow through any actions that forces acting on it dictate. If your arm wasnt pushing, the entire system would begin rotating. You deliberately position your arm at a certain radius so that the system is not rotating.

    I hope this is along the liens of what you were looking for.
  4. May 31, 2005 #3


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    Staff: Mentor

    Think about it a different way, using center of gravity. Take a rigid beam and support it on each end. Put a weight at the center - it is distributed evenly between the two supports. Put it 2/3 of the way to one side and one support takes double the force of the other. How does the support "know" how much force with which to push up?

    For some reason, people have no problem with that scenario, but they do with a lever. But from a standpoint of statics (the way you analyze the system), this situation and the lever situation you describe are exactly the same (just flipped over).
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