A man in a gym is holding an 8.0-kg object at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque about his shoulder joint due to the object if his arm is held at 30° below the horizontal?

Question

Brooks Forbes

19 July, 06:20

+2

Answers (1)

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Elijah Le · Answer 1

Answer;

37.4 Nm

Explanation;

W = Mg

= 8 * 9.81 m/s²

= 78.5 N

Torque = Perpendicular Force * distance

= 78.5 * Cos 30 * 0.55 m

= 37.4 Nm