Final answer:
The torque about the man's shoulder joint, when holding an 8.0-kg weight 0.55 m away at a 30.0-degree angle below the horizontal, is 21.46 N·m.
Step-by-step explanation:
The subject of this question is Physics, specifically focusing on the concept of torque. In the scenario, a man is holding an 8.0-kg weight at his side, which we're assuming is 0.55 m from his shoulder joint, and his arm is held at a 30.0 degrees angle below the horizontal. To find the torque about his shoulder joint due to the weight, we use the formula for torque (τ), which is τ = r * F * sin(θ), where r is the lever arm's length (0.55 m), F is the force applied (weight of the mass), and θ is the angle between the force and the lever arm.
First, we need to calculate the force exerted by the weight due to gravity, which is F = m * g, where m is the mass (8.0 kg) and g is the acceleration due to gravity (9.8 m/s²). The force is therefore F = 8.0 kg * 9.8 m/s² = 78.4 N.
To find the torque, we then apply the sine of the angle, which is 30.0 degrees below the horizontal. Since torque is the mechanical work done by a force acting at a distance, it is important to use the sine of the angle to find the component of force that acts perpendicular to the lever arm.
The torque is thus calculated as τ = 0.55 m * 78.4 N * sin(30.0 degrees). The sin(30.0 degrees) is 0.5, so the torque is τ = 0.55 m * 78.4 N * 0.5 = 21.46 N·m.