A 60 kg gymnast holds an iron cross position on the rings. In this position, the

Question

A 60 kg gymnast holds an iron cross position on the rings.  In this position, the gymnast's arms are abducted 90 degrees and his trunk and legs are vertical.  The horizontal distance from each ring to the gymnast's closest shoulder is .60m. The gymnast is in static equilibrium.

a.) What vertical reaction force does each ring exert on each hand?

b.)What torque is exerted by the right ring about the right shoulder joint?

c.) How much torque must the right shoulder adductor muscles produce to maintain the iron cross position?

d.) If the moment arm of the right shoulder adductor muscles about the shoulder joint is 5 cm, how much force must these muscles produce to maintain the iron cross position?

Explanation / Answer

a) The vertical reaction force on each hand is 30 kg.= 30 * 9.8 = 294 N

b) The torque exerted by the right ring about the right shoulder joint is 18kg*m in counterclockwise direction

torque T = Fr = mg r = . 0.60m x 30kg *9.8 = 177 Nm

c) The same amount as in the second question of 18kg*m, but this time in the clockwise direction for it to be in static equilibrium. 177 Nm

d) T = Fr

F 177/0.05 = 3540 N

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