Calculating Torque: A Gym Athlete's Challenge

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Simple Torque Q :)

1. Homework Statement [/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. Homework 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 don't get it :*(
 
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leslie113 said:
His arm is 65.0 cm long and has a mass of 5.59 kg.
Where does the weight of his arm act?
 
Doc Al said:
Where does the weight of his arm act?

This is where i think the problem is too. I don't 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 doesn't 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?
 
leslie113 said:
so i can clump it together with the weight of the ball?
Absolutely not! In calculating the torque due to any force, the distance from the axis is critical. The weight of anybody 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.
 
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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