How Do You Calculate the Coefficient of Static Friction on an Inclined Plane?

[Hypnos_16]

A 220-kg crate rests on a surface that is inclined above the horizontal at an angle of 18.3°. A horizontal force (magnitude = 501 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

The only real problem I'm having with this one is that i seem to have two variables
so far i have

Fnet = ma
mgsin18.3 + 501 x cos18.3 - Umgcos18.3 = ma
(220 x 9.8 x sin18.3) + (501 x cos18.3) - (U x 220 x 9.8 x cos18.3) = 220a
677 + 475.7 - U2046 = 220a
1142.7 - U2046 = 220a

though i don't even know if that's right, the cosine and sine sometimes mixes me up.

 

[jmb88korean]

Since the applied force is horizantal, you can use the angle they give you to break it up into components. The force of the push going down the ramp plus the force of gravity down the ramp must equal the force of friction. Using that and the normal force, which you can find, you can find the coefficient of static friction.

 

[Hypnos_16]

sooo
(220 x 9.8 x sin18.3) + (501 x cos18.3)
677 + 475.7 = 1152.7
Ff = UFn
1152.7 = (9.81 x 220)u
1152.7 = 2156u
u = 0.534
like that??