Question

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?

Answer 1

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.

Answer 2

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