28of28 Part A An athlete at the gym holds a 3.0 kg steel ball in his hand His ar

Question

28of28 Part A An athlete at the gym holds a 3.0 kg steel ball in his hand His arm is 70 em long and has a mass ef 3.8 kg with the center of mass at 40% ofthe arm length from the shoulder What is the magnitude of the torque about his shoulder due to the weight of the ball and his am if he holds his am straight out to sside parab to the Boor Express your answer with the appropriate units Laine l Units | Submit Part m at is the magrshude ofthe torque ate. his shoulder chao toto weight of the bal andhi' arm " he holds his arm seagh, b, 45" bb" hoonta Expeess your answer to two significant figures and include the appropriate units

Explanation / Answer

let m1 = 3 kg
m2 = 3.8 kg
L = 70 cm = 0.7 m

A)
Torque about shoulder = m1*g*L*sin(90) + m2*g*0.4*L*sin(90)

= m1*g*L + 0.4*m2*g*L

= 3*9.8*0.7 + 0.4*3.8*9.8*0.7

= 31.0 N.m

A)
Torque about shoulder = m1*g*L*sin(40) + m2*g*0.4*L*sin(45)

= m1*g*L*sin(45) + 0.4*m2*g*L*sin(45)

= 3*9.8*0.7*sin(45) + 0.4*3.8*9.8*0.7*sin(45)

= 21.9 N.m

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