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Torque with an arm holding a weight

by wallace13
Tags: holding, torque, weight
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wallace13
#1
Dec23-08, 10:53 AM
P: 31


A man holds a 169 N ball in his hand, with the forearm horizontal (see the drawing). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 21.4 N and has a center of gravity as indicated.



(a) Find the magnitude of M.

(b) Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint.



(Wmuscle*.051)-(21.4*(.089+.051))-(169*.33)= 0

Wmuscle=1152

I got part A correct, but I cannot get part B. I know that The end of the arm (where the ball is) should now be considered the pivot (zero) point.

Here is my attempt:
-(21.4* (.330-.089-.051))+1152- (W elbow*.330)

W elbow= 3503.23N
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PhanthomJay
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Dec23-08, 11:35 AM
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Quote Quote by wallace13 View Post


A man holds a 169 N ball in his hand, with the forearm horizontal (see the drawing). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 21.4 N and has a center of gravity as indicated.



(a) Find the magnitude of M.

(b) Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint.



(Wmuscle*.051)-(21.4*(.089+.051))-(169*.33)= 0

Wmuscle=1152

I got part A correct, but I cannot get part B. I know that The end of the arm (where the ball is) should now be considered the pivot (zero) point.

Here is my attempt:
-(21.4* (.330-.089-.051))+1152 (you forget to multiply 1152 by the distance to the ball) - (W elbow*.330)

W elbow= 3503.23N
Alternatively, or as a check, you should just sum all vertical forces to be sure they add to zero.


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