Consider a gymnast performing the "iron cross" on the rings. what magnitude of moment must be produced at each shoulder joint to maintain this position? assume the weight of an arm acts at a point one-half the length of the arm. let the mass of the gymnast be 80 kg, arm weight be 5 kg, length of arm be 1 m, and gravity be 10 m/s2 . (

375 N*m

Explanation:

First part, statics, forces equilibrium:

The weight of the man (a force going down) will be divided because of the symetry in the two bands equally (2 forces going up) .

Weight=m*g

W=80Kg*10 m/s2=800N

Force in each band:

F=800N/2=400N

Second part. Moment

If we make an imaginary cut in the shoulder of the man, we'll have:

*the force of the band going up aplief in his hand (1m from the shoulder),

*and the weight of the arm going down aplied in the middle point of the arm (0.5m from the shoulder)

Those force generate moments (Torsions) which have to be in equilibrum to be static, to calculate the moment of a force we use:

M=F*d (Moment=Force*distance)

Moment of Band Force

M1=400N*1m=400N*m

Moment of arm weight

Weight=5kg*10m/s2=50N

M2=50N*0.5m=25N*m

As the moments M1 and M2 act in opposite ways we have to know the difference to know the moment that the shoulder have to produce to equilbrate the situation.

M1-M2=400N*m-25N*m=375N*m

So the moment that's the moment that the shoulder have to produce.

answer: 0

explanation: im   14

this is a good question! i'll do some research and come back when i have the answer!