For the gymnast to hold his body up as shown, each arm must support 1/2 his tota

Question

For the gymnast to hold his body up as shown, each arm must support 1/2 his total weight of 750N. To accomplish this, his shoulder joint and back muscle for each aim work together to provide the balancing torque and force. Take the total length of his arm to be L=52cm. Balance forces and torques to solve for: a. the shoulder force magnitude |F_s| b. the angle theta it makes with the arm c. the muscle force magnitude |F_m |. First set up the static equilibrium equations (without solving or simplifying) in the boxes provided.

Explanation / Answer

Apply net torque abour left end, Tnet = 0

Fm*4*sin(45) - 375*52 = 0

Fm = 375*52/(4*sin(45))

= 6894 N <<<<<<<-------------------Answer answer for part c

Apply, Fnetx = 0

Fsx - Fm*cos(45) = 0

Fsx = Fm*cos(45)

= 6894*cos(45)

= 4875 N

now Apply, Fnety = 0

Fsy + 750 - Fm*sin(45) = 0

Fsy = Fm*sin(45) - 750

= 6894*sin(45) - 750

= 4125 N

so, Fs = sqrt(Fsx^2 + Fsy^2)

= sqrt(4875^2 + 4125^2)

= 6386 N <<<<<<<-------------------Answer answer for part a

theta = tan^-1(Fsy/Fsx)

= tan^-1(4125/4875)

= 40.2 degrees <<<<<<<-------------------Answer for part b

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