An athlete at the gym holds a 3.5 kg steel ball in his hand. His arm is 66 cm lo

Question

An athlete at the gym holds a 3.5 kg steel ball in his hand. His arm is 66 cm long and has a mass of 3.6 kg . Assume the center of mass of the arm is at the geometrical center of the arm.

“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 below horizontal?

Explanation / Answer

a)
F1 = 3.5*9.81 = 34.335 N
r1 = 0.66 m

F2 = 3.6*9.81 = 35.316 N
r2 = 0.33 m

T = F1*r1 + F2*r2
= 34.335*0.66 + 35.316*0.33
= 34.32 Nm
Answer: 34.32 Nm

b)
a)
F1 = 3.5*9.81 = 34.335 N
r1 = 0.66* cos 45

F2 = 3.6*9.81 = 35.316 N
r2 = 0.33* cos 45

T = F1*r1 + F2*r2
= (34.335*0.66 + 35.316*0.33) * cos 45
= 34.32* cos 45
= 24.26
Answer: 24.26 Nm

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