Question

An athlete at the gym holds a 3.5 kg steel bal in Ns hand His arm is 80 cm long and has a mass of 4.5 kg. Assume the center of mass of the arm is at the geometrical center of the Part A arm What is the magnitude of the sorque about his shoulder due to the welght of the ball and his arm if he holds his arm straight out to his side, paralel to the floor? Express your answer to two signifcant fgures and Include the appropriate units. Value Units T Submit Reauest Answer Part B "What is the magnitude of the torque about his shoulder due to the welght of the bal and his arm if he holds his arm straight, but 45 below horizontal? Express your anawer to two signihcant tgures and include the appropriste units Value Units Submit Beauest Arawet

Answer 1

a)

Torque about the shoulder

= torque due to the weight of the arm + torque due to the weight of the ball

= 4.5*9.81*(0.60/2) + 3.5*9.81*(0.60) Nm

= 13.24 + 20.6 Nm

= ~ 34 Nm....Answer.

b)

Torque when the arm is at 45° below the horizontal

= 33.841 cos45°

= 24 Nm..... Answer.

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