Answer
The magnitude of the force on the forearm from the triceps muscle is $~~1900~N$
Work Step by Step
Note that the tension $T$ in the rope is equal to the weight of the block.
We can use the sum of the torques about the rotation axis to find the torque $\tau_t$ due to the force $F$ from the triceps muscle:
$\sum \tau_i = 0$
$\tau_t - (15~cm)(2.0~kg)(9.8~m/s^2)~sin~120^{\circ}+(35~cm)(15~kg)(9.8~m/s^2)~sin~60^{\circ} = 0$
$\tau_t = (15~cm)(2.0~kg)(9.8~m/s^2)~sin~120^{\circ}-(35~cm)(15~kg)(9.8~m/s^2)~sin~60^{\circ}$
$\tau_t = -4201~N\cdot cm$
We can find the magnitude of the force $F$ from the triceps muscle:
$r~F~sin~120^{\circ} = 4201~N\cdot cm$
$F = \frac{4201~N\cdot cm}{(2.5~cm)~sin~120^{\circ}}$
$F = 1900~N$
The magnitude of the force on the forearm from the triceps muscle is $~~1900~N$
