91Ó°ÊÓ

• The distance between the hand and the shoulder joint is, L = 64 cm = 0.64 m .
• The distance of the deltoid muscle from the shoulder joint is, a = 15 cm = 0.15 m .
• The angle of the deltoid muscle with the humerus bone is, $\mathrm{θ}={12}^{∘}$.
• The tension in the cable of the machine is, $T_m=36N$.

When any muscle in a human body is stretched, it experiences a certain amount of tensile force. So, the tensile force acting on a body muscle increases muscle length till the force is applied.

In sign convention for a system of forces, the magnitude of the tensile forces is considered positive while the compressive forces are considered negative.

The free-body diagram of the arm is given by,

Here$T_d$is the tension in each deltoid muscle.

Calculating torque about the shoulder joint, we get,

$$\begin{gathered} (T_m\cos {35}^{∘})×L=\left(T_d\sin \mathrm{θ}\right)×a \\ {T}_d=\frac{(T_m\cos {35}^{∘})×L}{a\sin \mathrm{θ}} \end{gathered}$$

Putting the values, and we get,

$$\begin{gathered} T_d=\frac{36N×0.819×0.64m}{0.15m×\sin {12}^{∘}} \\ {T}_d=\frac{36N×0.819×0.64m}{0.15m×0.208} \\ {T}_d=604.8N \end{gathered}$$

Hence, the tension in each deltoid muscle is 604.8 N.