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Without detailed information about the shape and mass distribution of his arm, we have to assume that it's
homogeneous and has uniform cross section, so the center mass of his arm is at (71.8 / 2) = 35.9 cm from
the shoulder.

The torque due to the steel ball is (M G R) = (3.61 x 9.78 x 0.718) = 25.35 newton-meters. (rounded)

The torque due to the weight of his arm is (4.7 x 9.8 x 0.359) = 16.54 more newton-meters. (rounded)

The sum of the two torques is 41.89 newton-meters.

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Q: An athlete at the gym holds a 3.61 kg steel ball in his hand His arm is 71.8 cm long and has a mass of 4.70 kg What is the magnitude of the torque about his shoulder if he holds his arm straight out?
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A man holds a 4.0kg steel ball in hand His arm is 70cm long and has a mass of 4.0kg What is the magnitude of the torque about his shoulder if he holds his arm straight but 50 degrees below horizontal?

29 N-m


A man holds a 4.0kg steel ball in hand His arm is 70cm long and has a mass of 4.0kg What is the magnitude of the torque about his shoulder if he holds his arm straight but 30 degrees below horizontal?

ball in hand = 4 * 9.807 = 39.228 nlever length = cos 30deg. * 0.7 = 0.6062 metrestorque = 39.228 * 0.6062 = 23.78 newton - metre.torque generated by arm = 23.78 / 2 = 11.89 newton - metre.total torque = 23.78 + 11.89 = 35.67 newton - metre


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Related questions

A man holds a 4.0kg steel ball in hand His arm is 70cm long and has a mass of 4.0kg What is the magnitude of the torque about his shoulder if he holds his arm straight but 50 degrees below horizontal?

29 N-m


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A man holds a 4.0kg steel ball in hand His arm is 70cm long and has a mass of 4.0kg What is the magnitude of the torque about his shoulder if he holds his arm straight but 30 degrees below horizontal?

ball in hand = 4 * 9.807 = 39.228 nlever length = cos 30deg. * 0.7 = 0.6062 metrestorque = 39.228 * 0.6062 = 23.78 newton - metre.torque generated by arm = 23.78 / 2 = 11.89 newton - metre.total torque = 23.78 + 11.89 = 35.67 newton - metre


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No