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25 October, 14:54

Walter Arfeuille of Belgium lifted a 281.5 kg load off the ground using his teeth. Suppose Arfeuille can hold just three times that mass on a 30.0° slope using the same force. What is the coefficient of static friction between the load and the slope?

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  1. 25 October, 16:21
    0
    Gravitational, g = 9.81 m/s2

    Theta = 30 degrees, using the same force.

    Let u be the coefficient of static friction

    Force applied F = mg

    Static Friction Force = u3mgcos (theta)

    The gravitational Force = 3mgsin (theta)

    So the force equation = 3mgsin (theta) - (u3mgcos (theta) + mg) = 0

    So 3mgsin (theta) = (u3mgcos (theta) + mg) = > 3sin (theta) = (u3cos (theta) + 1)

    => u = (3sin (30) - 1) / 3cos (30) = > u = (3 (1/2) - 1) / (3 x 0.866)

    => u = (1.5 - 1) / 2.59 = 0.5 / 2.59 = 0.5 / 2.598

    Coefficient of Static Friction u = 0.19.
  2. 25 October, 17:27
    0
    Let's call

    m = 281.5Kg

    The force that Walter can make is by definition:

    F = m * g

    where,

    g: acceleration of gravity.

    Suppose Walter is at the top of the slope:

    By making a free-body diagram we have the following forces in the direction of the slope:

    Friction force = (μ3mgcosø)

    Force of gravity = (3mgsinø),

    Force exerted by Walter = mg

    We have then:

    mg + μ3mgcosø-3mgsinø = 0.

    Clearing the friction coefficient:

    μ = (3sinø-1) / (3cosø).

    Substituting the angle:

    μ = 0.19.

    answer

    the coefficient of static friction between the load and the slope is 0.19
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