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Quick Torque Question

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    How much torque must the pin exert to keep the rod from rotating ?


    2. Relevant equations

    3. The attempt at a solution

    Tnet=0= -(.500kg)(9.80)(.80m) - (2.0kg)(9.80)(.40m) + torque of pin
    Torque of pin = 11.76 Nm which is correct, but I am confused about something

    To get this answer I found the torque about the pin and said that the torque exerted by the pin must equal that. But I thought that for static equilibrium, the torque around each point had to be zero, and it is not 0 around the pin in this case. Why is this?

    For other problems, my book suggests to pick a point as the pivot point. Then the torque of the forces at that point becomes zero and the sum of the torques of the other forces around this pivot must equal zero. This is what bothers me about this question. If I pick the pin as the pivot, shouldn't sum of torques exerted by gravity on the rod & mass = 0?
  2. jcsd
  3. Apr 14, 2008 #2


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    Science Advisor

    This is indeed interesting... it seems that if you pick any of the points that forces are acting on, you will get an incorrect answer (or something that doesn't make sense). Can anyone shed light onto the situation, because I don't know.
  4. Apr 14, 2008 #3


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    Homework Helper

    If I'm understanding the picture correctly, the torque from the pin is from the frictional force between the sides of the pin and the walls of the hole that it is in. (There are other forces from the pin as well, but I think the point is that if the pin were frictionless it would be impossible for the beam to not rotate.)

    So we can choose a pivot point at the center of the pin, but the frictional forces are not located there; they are a pin's radius away from the center, and so their torque about the center of the pin will not be zero.

    This part of the problem would be rather analogous to sticking a pencil through an index card and spinning it. We can put the pivot point at the center of the pencil to calculate the torque, but the torque from the pencil on the card is not zero because the frictional forces between pencil and card (that cause the torque) are located at the surface of the pencil.
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