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?

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.

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