Figure 1 depicts an arm holding a ball in static equilibrium. Figure 2 shows the free-body diagram of the lower arm where \(F\) is the force applied by the biceps muscle at an angle of \(\theta = 60\) degrees with respect to the lower arm, and \(R_1\) and \(R_2\) are the reaction forces acting at point O. The weights of the ball and the lower arm are \(W_1 = 50\) Newtons (N) and \(W_2 = 20\) N, respectively. The force exerted by the biceps muscle is \(F =\) N. (Round off to the nearest integer)
Assume the lower arm to be a rigid body.
In the free-body diagram, measured along the lower arm from the pivot O: the biceps force \(F\) acts at \(4\) cm from O; the forearm's weight \(W_2\) acts a further \(12\) cm beyond that (i.e. \(16\) cm from O); and the ball's weight \(W_1\) acts a further \(16\) cm beyond that (i.e. \(32\) cm from O).
Show Hint
Take moments about O; only \(F\sin\theta\) contributes a moment for the biceps force, and \(R_1, R_2\) act at O so they contribute none.
Instead of reasoning about moment arms directly, set up the problem using position and force vectors and compute torques as cross products. Place the origin at O, with the arm lying along the positive $x$-axis.
The point where the biceps force acts is at position $\vec{r}_F = (4, 0)$ cm. The force itself makes an angle $\theta = 60^\circ$ with the arm (the $x$-axis), so its components are:
\[
\vec{F} = (F\cos 60^\circ,\ F\sin 60^\circ)
\]
The torque of a 2D force about the origin is the scalar $\tau = xF_y - yF_x$. Since $\vec{r}_F$ has $y = 0$, this simplifies to $\tau_F = 4 \times F\sin 60^\circ$, acting to rotate the arm counterclockwise (positive).
The forearm weight acts at $\vec{r}_{W_2} = (16, 0)$ cm with force $\vec{W}_2 = (0, -20)$ N (downward, negative $y$). Its torque is $\tau_{W_2} = 16 \times (-20) - 0 = -320$ N.cm (clockwise, negative).
The ball's weight acts at $\vec{r}_{W_1} = (32, 0)$ cm with force $\vec{W}_1 = (0, -50)$ N. Its torque is $\tau_{W_1} = 32 \times (-50) - 0 = -1600$ N.cm (clockwise, negative).
The reaction forces $R_1$ and $R_2$ act at the origin itself, so their position vector is $(0,0)$ and they contribute zero torque about O, no matter their magnitude.
For equilibrium the net torque about O must vanish:
Writing torque as the cross product $\tau = xF_y - yF_x$ automatically picks out only the perpendicular component of each force, without needing to separately reason about "moment arms".
Forces applied exactly at the pivot, like $R_1$ and $R_2$, always drop out of the torque equation since their position vector is zero.
The same balance of torques gives the biceps force needed to hold the arm and ball still.
The force exerted by the biceps muscle is about $555$ N.
