An athlete at the gym holds a 4.0 kg steel ball in his hand. His arm is 70 cm lo

Question

An athlete at the gym holds a 4.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

A)What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

B)What is the magnitude of the torque about his shoulder if he holds his arm straight, but 35 below horizontal?

Explanation / Answer

a) torque = r * F sin (theta)
For his arm, we assume the mass is evenly distributed, so we calculate the arms torque component based on the center of mass (which is the center of his arm 0.35 m from his shoulder)

torque = ((0.35 * (4.0*9.8)) + (4 * (0.70*9.8)))(sin (90))
torque = 13.72+27.44
torque = 41.16 Nm

b) torque = r * F sin(theta)

The only thing changing here is theta, which is 55 degrees when his arm is 35 below horizontal

torque = 41.16 * (sin (55) )
torque = 33.72 Nm

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