Answer
The magnitude of the force of the biceps muscle on the lower arm is $~~650~N$
Work Step by Step
To find the upward force $F_b$ exerted on the lower arm by the biceps muscle, we can consider the net torque about a rotation axis located at the elbow contact point.
We can find $F_b$:
$\tau_{net} = 0$
$(4.0~cm)~F_b-(15~cm)(1.8~kg)(9.8~m/s^2)-(33~cm)(7.2~kg)(9.8~m/s^2) = 0$
$(4.0~cm)~F_b = (15~cm)(1.8~kg)(9.8~m/s^2)+(33~cm)(7.2~kg)(9.8~m/s^2)$
$F_b = \frac{(15~cm)(1.8~kg)(9.8~m/s^2)+(33~cm)(7.2~kg)(9.8~m/s^2)}{4.0~cm}$
$F_b = 650~N$
The magnitude of the force of the biceps muscle on the lower arm is $~~650~N$
