The coefficient of static friction between the m = 2.80 kg crate and the 35.0° incline of the figure below is

The coefficient of static friction between the m = 2.80 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

The answer is 38.1 N

The Picture:
http://www.webassign.net/sercp/p4-41alt.gif

I'm having a hard time getting this answer.

Yes, I've drawn a free body diagram and converted mg into x and y components. for x i have mgsin(35) and for y i have mgcos35.

I know F(sub S)= µ(s)*n
I've set up the equations ∑Fx= µ(s)*n-mgsin(35)=0 and ∑Fy= n-mgcos(35)=0
I have mg=2.80*9.8=27.44

∑Fx= .260*n-15.74=0 and ∑Fy= n-22.48=0

What am I doing wrong?

Weight force pulling downwards = mg sin35 = 15.74 N

this must equal the frictional force for balancing .

Fr = µs x normal force

....= µ x (F + mg cos 37)

....= 0.26 ( F + 22.48)

....= 0.26 F + 5.84

equate Fr and the weight force :

=> 0.26 F + 5.84 = 15.74

=> F = (15.74 - 5.84) / 0.26 = 38.1 N <<<<<<<<<<<<<,