Answer
Part A) The force required of the deltoid muscle is about $230N$
Part B) The force on the shoulder joint is about $220N$ and acts at an angle of $8.4^{\circ}$ from the horizontal
Work Step by Step
A) We sum the torques about the shoulder joint to solve for $F_M$;
$(+ \circlearrowleft) \sum \tau=0$
$F_M\sin15^{\circ}*0.12m-mg*.24m=0$
$F_M=\frac{mg*.24m}{\sin15^{\circ}*0.12m}$
$F_M=\frac{3.0kg*9.8m/s^2*.24m}{\sin15^{\circ}*0.12m}$
$F_M\approx230N$
The force required of the deltoid muscle is about $230N$.
B) Sum the forces horizontally to relate $\theta$ to $F_J$;
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$\cos\theta*F_J-\cos15^{\circ}*F_M=0$
$F_J=\frac{\cos15^{\circ}*F_M}{\cos\theta}$
We sum the torques about the point on the arm where $F_M$ acts to solve for $F_J$:
$(+ \circlearrowleft) \sum \tau=0$
$.12m*\sin\theta*F_J-mg*.12m=0$
$\sin\theta*F_J-mg=0$
Substitute $\frac{\cos15^{\circ}*F_M}{\cos\theta}$ for $F_J$
$\sin\theta*(\frac{\cos15^{\circ}*F_M}{\cos\theta})-mg=0$
$\tan\theta*\cos15^{\circ}*F_M-mg=0$
$\theta=\arctan(\frac{mg}{\cos15^{\circ}F_M})$
$\theta=\arctan(\frac{3.3kg*9.8m/s^2}{\cos15^{\circ}230N})$
$\theta\approx8.4^{\circ}$
Now we can use the angle to solve for the force required.
$F_J=\frac{\cos15^{\circ}*230N}{\cos8.4^{\circ}}$
$F_J\approx220N$
The force on the shoulder joint is about 220N and acts at an angle of $8.4^{\circ}$ from the horizontal.
