- #1

alejandro7

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"Suppose that a male gymnast wishes to execute an iron cross during a gymnastics session. The total mass of the gymnast is 60 kg. Each ring supports half of the gymnast's weight. Assume that the weight of one of his arms is 5% of his total body weight. The distance from his shoulder joint to where his hands hold the rings is 50 cm. The distance from his hands to the center of mass of his arm is 34 cm. The horizontal distance from his shoulder to the center of mass of his body is 19 cm (from the shoulder to the middle of the chest, not the actual center of mass of the body). If the gymnast is at rest, how much force and torque are at one of his shoulder joints?"

My attempt:

∑F

_{y}: -W+R-W

_{arm}+W/2 = 0

where W = (60 kg)(9.8 m/s

^{2}) = 588 N , R is the reaction force on the shoulder joint, W

_{arm}= (0.05)(60 kg) = 3 kg

Solving, R = 297 N

However, I don't know what the problem means by the torque "at" his shoulder joint. Is it the torque generated by R (i.e. (R)(0.19 m)) or the sum of about the shoulder (0)?

Thanks.