What is the magnitude of the torque about his shoulder

[leisiminger]

An athlete at the gym holds a 4.0 steel ball in his hand. His arm is 70 long and has a mass of 4.0
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?
What is the magnitude of the torque about his shoulder if he holds his arm straight, but 50 below horizontal?

I'm lost

 

[LowlyPion]

An athlete at the gym holds a 4.0 steel ball in his hand. His arm is 70 long and has a mass of 4.0
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?
What is the magnitude of the torque about his shoulder if he holds his arm straight, but 50 below horizontal?

I'm lost

What is the definition of Torque?

That would be a good start.

 

[thomate1]

The magnitude of torque is a measure of the force that causes an object to rotate around a fixed point. In this scenario, the fixed point is the athlete's shoulder. To calculate the magnitude of torque, we need to know the force applied, the distance from the force to the fixed point, and the angle between the force and the distance.

In the first scenario, where the athlete holds the steel ball straight out to his side, the force applied is the weight of the steel ball, which is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2). The distance from the force to the shoulder is 70 cm, or 0.7 m. Since the arm is held parallel to the floor, the angle between the force and the distance is 90 degrees. Using the formula for torque (T = F x d x sinθ), we can calculate the magnitude of torque to be:

T = (4.0 kg x 9.8 m/s^2) x 0.7 m x sin(90°) = 27.44 Nm

In the second scenario, where the athlete holds his arm straight, but 50 degrees below horizontal, the force applied is still the weight of the steel ball, but now the distance from the force to the shoulder is different. Using trigonometry, we can calculate that the distance is now 0.7 m x cos(50°) = 0.45 m. The angle between the force and the distance is still 90 degrees. Plugging these values into the torque formula, we get:

T = (4.0 kg x 9.8 m/s^2) x 0.45 m x sin(90°) = 17.64 Nm

Therefore, the magnitude of torque about the shoulder is lower in the second scenario because the distance from the force to the shoulder is shorter. This shows that the angle and distance from the force to the fixed point can greatly affect the magnitude of torque exerted on an object.