An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 76 cm lo

Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 76 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder. What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight out to his side, parallel to the floor? What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight, but 45 degree below horizontal?

Explanation / Answer

a)

Center of mass at 30.4 cm from the shoulder

Torque about the shoulder
= torque due to the weight of the arm + torque due to the weight of the ball
= 3.8 *9.81*(0.304) + 3.0 *9.81*(.0.76) Nm
=33.69 Nm.

b)
Torque when the arm is at 45 degrees below the horizontal
= 33.69 cos(45 degrees)
= 23.83 Nm.

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