91Ó°ÊÓ

1. Distance between the elbow point of contact and the lower arm CM, $D=0.15\text{m},d=0.04\text{m}$.
2. Distance between the elbow point of contact and the palm, L =0.33m .
3. Mass of the ball,M =7.2 kg .
4. Mass of the lower arm,m =1.8 kg .

The ball is held in hands and balanced. It is neither moving linearly nor rotating about any pivot point. So, it is in static equilibrium condition. By selecting a pivot point and applying the static equilibrium conditions, you can write the torque equation in terms of force and distance. Solving this equation, you would get the unknown tension and force.

Using the figure given in the problem andcondition of static equilibrium, we can write torque and force equations as,

$$\begin{gathered} ∑Torque=0 \\ \left(d×T\right)+\left(0×F\right)-\left(D×mg\right)-\left(L×Mg\right)=0...........................\left(1\right) \\ ∑F_x=0 \\ F+\left(M+m\right)g-T=0.........................\left(2\right) \end{gathered}$$

Substitute the values in equation 1, and we get,

$$\begin{gathered} \left(0.04×T\right)+0-\left(0.15×1.8×9.8\right)-\left(0.33×7.2×9.8\right)=0 \\ T×0.04=25.9308 \\ T=648\text{N} \end{gathered}$$

Hence, the magnitude of the force of the bicep muscle on the lower arm is$6.5×{10}^2\text{N}$ .

Substitute the values in equation 2, and we get,

$$\begin{gathered} F+\left(7.2+1.8\right)×9.8-648=0 \\ F+88.2-648=0 \\ F≈560\text{N} \end{gathered}$$

Hence, the force between the bony structures at the elbow contact point is $5.6×{10}^2\text{N}$.