BIO A Good Work out. You are doing exercises on a Nautilus machine in a gym to strengthen your deltoid (shoulder) muscles. Your arms are raised vertically and can pivot around the shoulder joint, and you grasp the cable of the machine in your hand 64.0 cm from your shoulder joint. The deltoid muscle is attached to the humorous 15.0 cm from the shoulder joint and makes a ${\mathbf{12}}^{\mathbf{\circ }}$angle with that bone (Fig. E11.22). If you have set the tension in the cable of the machine to 36.0 N on each arm, what is the tension in each deltoid muscle if you simply hold your outstretched arms in place? (Hint: Start by making a clear free-body diagram of your arm.)

• 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 humorous bone is: $\theta ={12}^{\circ }$.

• 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,

$$\begin{gathered} (T_m\cos {35}^{\circ })\times L=\left(T_d\sin \theta \right)\times a \\ {T}_d=\frac{(T_m\cos {35}^{\circ })\times L}{a\sin \theta } \end{gathered}$$

Putting the values, and we get,

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

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