A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque about his shoulder joint due to the weight if his arm is held at 30° below
The horizontal?

Question

Quizplus · Accepted Answer

The torque ($\tau$) can be calculated using the formula $\tau = rF\sin(\theta)$, where $r = 0.55$ m is the lever arm distance, $F = mg = 8.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N}$ is the force due to gravity, and $\theta = 30^\circ$ is the angle between the arm and the vertical. Plugging in the values, $\tau = 0.55 \times 78.4 \times \sin(30^\circ) = 0.55 \times 78.4 \times 0.5 \approx 21.56 \, \text{N} \cdot \text{m}$, which rounds to 22 N·m, making the correct answer E, not A as initially indicated. My correction is based on the accurate calculation and understanding of the given physical scenario.

A Man in a Gym Is Holding an 8