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Torque about the shoulder?

  1. Jul 21, 2010 #1
    An athlete at the gym holds a 3.0 kg ball in his hand. His arm is .70 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his arm if he holds his arm straight out his side, parallel to the floor. I did T=mgr since theta is zero so both the force of gravity on his arm and ball is down. So 7* 9.8*.70 where the mass is the mass of his arm + mass of the ball. I get 48 N-m but this is wrong the answer is 34 N-m. What am I doing wrong?
     
  2. jcsd
  3. Jul 21, 2010 #2

    ehild

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    This would by right if all mass of the arm were concentrated in the hand.

    ehild
     
  4. Jul 21, 2010 #3
    So is .70 m not the center of gravity. I'm so confused ...
     
  5. Jul 21, 2010 #4
    theta is not zero, it is 90 degrees. And 0.7 m is not the center of gravity. If you have a uniform stick, which length is 0.7 m and has a 4 kg mass, where is then its center of mass?
     
  6. Jul 21, 2010 #5
    .35 thanks :) so Now I add the torques but why do we consider the center of gravity for the ball to be .70 m?
     
  7. Jul 21, 2010 #6
    because you need to calculate torque about the shoulder (well topic name says about shoulder, but in problem statement you say about the arm...). In order to do so, you must take distances from shoulder (its like shoulder is your coordinates starting point), and ball is 0.7 m away from the shoulder.
     
  8. Jul 21, 2010 #7
    Thank you. I'm sorry, I should have said shoulder. Thank you again for your help.
     
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