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Basic torque/static equilibrium problem

  1. Jul 16, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-16_15-16-11.png

    2. Relevant equations
    sum of forces = 0N
    sum of torque = 0Nm

    3. The attempt at a solution

    Net force has to be already zero (T = mg). But, the torque isn't zero, otherwise this question would have another answer choice. But the gravitational torque should cancel with the tension torque right?

    I looked at the answer and it says "the tension does not exert a torque." But, I've done so many of those pulley problems with pulleys w/ masses using forces and torque in which there are two tensions in the rope and the torque is (T1-T2) = I*alpha.

    Why does the tension not apply in this problem as torque but does in the pulley problem?
     
  2. jcsd
  3. Jul 16, 2017 #2

    haruspex

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    Torque is (nearly) always only meaningful in respect of a specified axis. A force exerts a torque about any axis that is not in its line of action. So the tension in the rope exerts a torque about the pivot, a different torque about the mass centre, but no torque about the point where it is tied to the beam.
     
  4. Jul 16, 2017 #3
    The answer says that the tension doesn't provide torque though. I understand that it doesn't provide torque if the axis was at the place where the rope touched the plank, but the key says that it doesn't provide torque even if the axis was the pivot. If it did provide torque, the net torque would be 0Nm since r = r and T = mg. Why is there no torque by the tension w/ the axis as the pivot?

    thanks
     
  5. Jul 16, 2017 #4

    haruspex

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    Well, that is simply not true. Are you sure it explicitly says that, or does it merely omit to say which axis it is using?
    Can you post the whole solution text, maybe as an image?
     
  6. Jul 16, 2017 #5
    I think I might've misunderstood the answer sheet, but I still don't get it:

    upload_2017-7-16_15-38-2.png

    If the axis was the place where the rope touched the plank, then tension torque would cancel with the grav torque. But the axis is the pivot.
     
  7. Jul 16, 2017 #6

    haruspex

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    No, it quite clearly says it is taking the point where the rope attaches to the beam as the axis.
    "Consider the torque about the point where the rope is attached. The tension provides no torque [there]. .... the pivot force must exert a clockwise torque [there]."
     
  8. Jul 16, 2017 #7
    But then... wouldn't the gravitational torque also be 0 since r = 0? Neither the tension or weight forces would apply torque.. I'm so confused....
     
  9. Jul 16, 2017 #8

    haruspex

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    No. Call the pivot point P and the rope attachment point R. The beam need not be uniform, so let its mass centre be at G. Say the force at the pivot is F. Moments about R are:
    • Zero for the tension
    • Mg x GR anticlockwise for the weight of the beam
    • F x PR for pivot force.
     
  10. Jul 16, 2017 #9
    Yeah, I'm officially dumb. I completely forgot about center of gravity.....
     
  11. Jul 16, 2017 #10

    haruspex

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    So is it all clear now?
     
  12. Jul 16, 2017 #11
    Yup, it's all good. Thanks for the help!

    Quick confirmation: an object in static equilibrium will have a net torque of 0 given ANY coordinate system, right?
     
  13. Jul 16, 2017 #12

    haruspex

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    Yes.
    And note I wrote that "Torque is (nearly) always only meaningful in respect of a specified axis." The exception is when you have a couple (aka a screw). This is a system forces with no net force and yet a net moment. The simplest is two equal and opposite forces with different lines of action. In this case, the torque is the same about whatever axis you choose.

    You might find it useful to read my article https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/.
     
  14. Jul 16, 2017 #13
    Thanks. I'll give that a read :)
     
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