Question

a man in a gym is holding an 8 kg kettlebell 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 gravitational force on the kettlebell if his arm is held at 30 degrees below the horizontal?

Answer 1

The torque about the shoulder joint due to the gravitational force on the kettlebell is approximately 23.46 N·m.

Step-by-step explanation:

To find the torque about the shoulder joint due to the gravitational force on the kettlebell, we need to calculate the moment arm and the gravitational force on the kettlebell. The moment arm is the perpendicular distance from the shoulder joint to the line of action of the gravitational force, which is given as 0.55 m.

The gravitational force on the kettlebell can be calculated using the formula F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity. Here, the mass of the kettlebell is 8 kg and the acceleration due to gravity is approximately 9.8 m/s².

Therefore, the torque about the shoulder joint can be calculated as Torque = F * d * sin(θ), where F is the gravitational force, d is the moment arm, and θ is the angle below the horizontal. Substituting the values, we get Torque = (8 kg * 9.8 m/s²) * 0.55 m * sin(30°).

Calculating this, we find that the torque about the shoulder joint due to the gravitational force on the kettlebell is approximately 23.46 N·m.