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19 July, 17:56

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?

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  1. 19 July, 19:36
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    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
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