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Simple Torque Q

  1. Jan 24, 2008 #1
    Simple Torque Q :)

    1. The problem statement, all variables and given/known data[/b]

    An athlete at the gym holds a 3.64 kg steel ball in his hand. His arm is 65.0 cm long and has a mass of 5.59 kg. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?

    2. Relevant equations[/b]

    t= rfsin theta

    3. The attempt at a solution[/b]

    t= (0.65)[9.81 (3.64+5.59)] sin 90
    = 58.8N*m

    WHAT AM I DOING WRONG? i dont get it :*(
  2. jcsd
  3. Jan 24, 2008 #2

    Doc Al

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    Staff: Mentor

    Where does the weight of his arm act?
  4. Jan 24, 2008 #3
    This is where i think the problem is too. I dont know how to account for the weight of the arm in the solution. But i know we have to, since if i just do without, the answer is wrong as well.

    But doesnt the weight like the ball just acts to pull the torque downwards towards the pull of gravity? so i can clump it together with the weight of the ball?
  5. Jan 24, 2008 #4

    Doc Al

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    Staff: Mentor

    Absolutely not! In calculating the torque due to any force, the distance from the axis is critical. The weight of any body acts at the body's center of mass. If you aren't given any details about where the center of mass of the arm is, just assume it's at the center of the arm.

    The total torque will be the sum of the torque due to the weight of the ball (which acts at the end of the arm, since it's in the hand) and the torque due to the weight of the arm.
  6. Jan 24, 2008 #5
    WOW! you are amazing. i totally get what I'm doing wrong now. thanks for the ultra fast replies and being so clear and concise with your hints! :)

    I realllly appreciate your help!
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