The deltoid muscle connects the humerus to the shoulder blade and facilitates outstretching of the arm as shown in the figure. Assume the equivalent weight (W) of the arm to be 30 N and acts vertically down at a horizontal distance of 30 cm. Assume that the deltoid muscle is connected to the humerus at a distance of 15 cm and makes an average angle of $20^\circ$ with the horizontal. The magnitude of tension in the deltoid muscle is _________ N.
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When muscles act at small angles, they must generate very large forces to balance even moderate external loads. This is why lifting the arm sideways requires strong shoulder muscles.
To calculate the tension in the deltoid muscle, we consider the arm to be in static rotational equilibrium about the shoulder joint. This requires that the clockwise and counterclockwise moments acting about the joint balance each other.
Step 1: Compute the clockwise moment caused by the arm weight.
The arm weight is given as
\[
W = 30\ \text{N},
\]
and acts at a horizontal distance of
\[
30\ \text{cm} = 0.30\ \text{m}
\]
from the shoulder.
Thus the clockwise torque is
\[
\tau_W = W \times 0.30 = 30 \times 0.30 = 9\ \text{N·m}.
\]
Step 2: Compute the counterclockwise moment generated by the deltoid muscle.
Let $T$ be the muscle tension.
The muscle attaches at
\[
15\ \text{cm} = 0.15\ \text{m}
\]
from the joint but at an angle of $20^\circ$ to the horizontal.
Only the vertical component of muscle force $T\sin 20^\circ$ contributes to the torque.
Thus the counterclockwise torque is
\[
\tau_T = T \sin(20^\circ)\times 0.15.
\]
Step 3: Apply rotational equilibrium condition.
\[
\tau_T = \tau_W
\]
\[
T\sin(20^\circ)(0.15) = 9
\]
Step 4: Solve for T.
\[
T = \frac{9}{0.15 \sin(20^\circ)}
\]
Evaluate numerically:
\[
\sin(20^\circ) \approx 0.342
\]
\[
0.15 \times 0.342 = 0.0513
\]
\[
T = \frac{9}{0.0513} \approx 175.4\ \text{N}.
\]
Thus, the required deltoid muscle tension is
\[
\boxed{175.4\ \text{N}}.
\]
