What is the Torque on an Outstretched Arm with a 21.5 N Weight?

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SUMMARY

The discussion focuses on calculating the torque produced by a 21.5 N weight held in an outstretched arm. The arm's dimensions include a shoulder-to-elbow length of 33.5 cm and an elbow-to-wrist length of 23.0 cm, with the weight's center located 6.5 cm from the wrist. The torque equation used is T = rF, where the torques about the elbow are analyzed using the equation T1D1 + T2(D1+D2) - T3(D1 + D2 + D3) = 0. The user expresses confusion regarding the placement of the center of torque and the calculations involving the elbow and wrist.

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Homework Statement


A 21.5 N weight is held in the outstretched hand of a person with a shoulder-to-elbow length of 33.5 cm and an elbow-to-wrist length of 23.0 cm. The center of the weight is 6.5 cm from the person's wrist. Determine the magnitude of the torque about the elbow produced by the weight.


Homework Equations


T=rF


The Attempt at a Solution


T1 - Elbow
T2 - Wrist
T3 - Hand w/weight
T1D1 + T2(D1+D2) - T3(D1 + D2 + D3) = 0
T1 = -T2(.565m) + T3(.63)/.335

I know I am doing something wrong with the torques but I can't figure out what? I thought I just put in 21.5 N.
 
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I forgot to mention that I put my center of torque at the shoulder
 
Can I not have the center of torque there?
 

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