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9 July, 04:56

A post is wrapped two full turns around with a belt. The tension in the belt is 7500 N by exerting a force of 150 N on its free end. Determine the coefficient of static friction between the belt and the post.

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  1. 9 July, 06:37
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    Coefficient of friction is

    Ū = 0.31

    Explanation:

    T2 = T1 * e^ (ūơ)

    Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,

    ū = coefficient of friction

    Ơ = 2pai * N

    Where N = number of turns = 2

    Ơ = 4pai

    7500 = 150e^ (ū*4pai)

    50 = e^ (ū * 4pai)

    lin 50 = 4pai * ū

    Ū = 3.91/4pai

    Ū = 0.31
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