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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.

Respuesta :

Answer:

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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