Answer
The magnitude of the force on the forearm from the humerus is $~~2100~N$
Work Step by Step
Let "up" be the positive direction.
We can consider the vertical forces on the forearm to find the force on the forearm from the humerus $F_h$:
$\sum F_y = 0$
$F_h+1900~N+(15~kg)(9.8~m/s^2)-(2.0~kg)(9.8~m/s^2) = 0$
$F_h = (2.0~kg)(9.8~m/s^2)-1940~N-(15~kg)(9.8~m/s^2)$
$F_h = -2100~N$
The magnitude of the force on the forearm from the humerus is $~~2100~N$.
