An athlete at the gym holds a 5.29 kg steel ball in his hand. His arm is 67.0 cm long and has a mass of 4.57 kg. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?
You have to assume a cg for the arm. Assuming midway...
4.57*67/2+5.29*67 in N-cm of torque
What is the magnitude of the torque about his shoulder if he holds his arm straight, but 43.8° below horizontal?
I tried multiplying the torque force by sin43.8 but it was wrong :(
To determine the magnitude of the torque about the athlete's shoulder, we need to calculate the force exerted by the steel ball and the leverage provided by the athlete's arm.
1. First, let's calculate the force exerted by the steel ball. We can use the formula F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity. In this case, the mass of the steel ball is 5.29 kg. The acceleration due to gravity is approximately 9.8 m/s^2. So the force exerted by the steel ball is:
F = 5.29 kg * 9.8 m/s^2 = 51.842 N (rounded to three significant figures).
2. Next, we need to calculate the leverage provided by the athlete's arm. The leverage is given by the formula L = r * F, where L is the leverage, r is the length of the arm, and F is the force. In this case, the length of the arm is given as 67.0 cm, which is equal to 0.67 m. So the leverage provided by the arm is:
L = 0.67 m * 51.842 N = 34.738 N·m (rounded to three significant figures).
Therefore, the magnitude of the torque about the athlete's shoulder, when he holds his arm straight out to his side, parallel to the floor, is approximately 34.738 N·m.