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Force and momentum in gymnastics position

  1. Aug 8, 2015 #1
    Hi. I'm trying to solve this problem but I'm not sure my solution is right:

    "Suppose that a male gymnast wishes to execute an iron cross during a gymnastics session. The total mass of the gymnast is 60 kg. Each ring supports half of the gymnast's weight. Assume that the weight of one of his arms is 5% of his total body weight. The distance from his shoulder joint to where his hands hold the rings is 50 cm. The distance from his hands to the center of mass of his arm is 34 cm. The horizontal distance from his shoulder to the center of mass of his body is 19 cm (from the shoulder to the middle of the chest, not the actual center of mass of the body). If the gymnast is at rest, how much force and torque are at one of his shoulder joints?"

    My attempt:

    ∑Fy : -W+R-Warm+W/2 = 0
    where W = (60 kg)(9.8 m/s2) = 588 N , R is the reaction force on the shoulder joint, Warm = (0.05)(60 kg) = 3 kg

    Solving, R = 297 N

    However, I don't know what the problem means by the torque "at" his shoulder joint. Is it the torque generated by R (i.e. (R)(0.19 m)) or the sum of about the shoulder (0)?

    Thanks.
     
  2. jcsd
  3. Aug 8, 2015 #2

    CWatters

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    I would assume they mean the torque that the man has to generate about the shoulder. However I think I would begin my answer with.. The person is at rest so the net torque about any point sums to zero. The torque that the man has to generate about is shoulder to stop his arms rotating is...
     
  4. Aug 8, 2015 #3
    The torque that the man has to generate about is shoulder to stop his arms rotating is the reaction force (R) times the distance. But which distance?
     
  5. Aug 8, 2015 #4

    haruspex

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    If you take moments about one shoulder joint, you will have a set of forces between that joint and the corresponding hand, each exerting a torque at his shoulder. This will be equal and opposite to the sum of torques from the forces between that shoulder and the other hand. They are asking for the magnitude that these two torques have (each).
     
  6. Aug 9, 2015 #5
    Is my result for R correct or should I include the other arm in the calculation (so the forces would be 2 x W/2, 2 x Warm, 2 x R and W)?
     
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