Problem $4[2]$

A gymnast holds the "iron cross" position as shown, in which his arms are abducted $90.$ The rings $+$ cables are angled $$ off the vertical, the distance from the rings to his shoulders is $0\mathrm{.}7$ m $,$and the distance between his shoulders is $0\mathrm{.}3$ m $.$

a$.$ The muscle grouping responsible for resisting the continued abduction of the shoulder is the shoulder adductors $-$ for the purposes of this problem we'll give them a collective moment arm of $4$ cm attaching at some points distal to the shoulder joint. Find the

magnitude of the force which this muscle group must generate to maintain equilibrium, including a FBD in your process. Assume the weight of each arm is equivalent to $5\%$ of the mass of the gymnast and the center of gravity of the arm is $50\%$ of the distance between the shoulder and the hand.

b$.$ If the man's left arm $($shoulder to hand$)$ defines the

positive $x-$axis and the trunk of his body $($toe to head$)$ defines

the positive $y-$axis, imagine that someone pushes on his foot

with a force vec$(F)=10()+15()-20(k).$ If the vertical

$($perpendicular$)$ distance from his feet to the axis of his

shoulders is $1\mathrm{.}4$ m $,$ What is the moment generated about the

line formed by his arms?

Hint: you'll need two free body diagrams, one for the external

and body forces alone and another that exposes the internal

forces of note

Body weight $=80kg$

$=8$