Answer
a) $M=1212N$
b) The force has a magnitude of $1012N$ and is directed downward.
Work Step by Step
We call the force applied by the upper arm bone to the forearm $E$. Since all the other forces are vertical forces, it will not be possible for $E$ to have a horizontal component, because there are no other horizontal forces to balance. That means $E$ is a vertical force, which, we will assume, points upward.
Therefore, on the vertical side, $E$ and $M$ point upward while the weight of the forearm $W_f$ and the weight of the ball $W_b$ point downward. So, $$\sum F_y=E+M-W_f-W_b=0$$ $$E+M=W_f+W_b=178+22=200N (1)$$
- $W_f$ produces a clockwise torque $-W_f(0.051m+0.089m)=-(22N)(0.14m)=-3.08N.m$
- $W_b$ produces clockwise torque $-W_b(0.33m)=-(178N)(0.33m)=-58.74N.m$
- $M$ produces counterclockwise torque $+(0.051m)M$
$$\sum\tau=0.051M-58.74-3.08=0$$ $$M=1212N$$
Plugging back into (1), we have $E=200N-1212N=-1012N$
The negative sign means $E$ is directed downward.
