Question

The coefficient of static friction between the m = 3.30-kg crate and the 35.0

Answer 1

Let:

m be the mass of the crate,

g be the acceleration due to gravity,

a be the inclination of the slope above the horizontal,

F be the friction force,

P be the applied force,

R be the normal reaction of the slope on the crate,

u be the coefficient of friction.

Parallel to the slope:

mg sin(a) = F ...(1)

Perpendicular to the slope:

mg cos(a) + P = R ...(2)

F / R <= u ...(3)

Dividing (1) by (2) and using (3) to eliminate F and R:

mg sin(a) / (mg cos(a) + P) <= u

mg sin(a) <= u[ mg cos(a) + P ]

mg[ sin(a) - u cos(a) ] <= uP

P >= mg[ sin(a) / u - cos(a) ]

P >= 3.3 * 9.81[ sin(35.0) / 0.325 - cos(35.0) ]

P >= 30.61 N.