Newtonian mechanics of a metal slab

[uchicago2012]

Homework Statement

A 90 kg metal slab is pulled across the ground by a tractor. The coefficient of friction between the slab and the ground is µ = 0.65. If the force of the tractor on the slab is 600 N and is directed at 35° from the horizontal, what is the acceleration of the slab?

Homework Equations

F_net = ma
F_k = u * F_n
where F_k = friction force and F_n = normal force and u = coefficient of friction

The Attempt at a Solution

So I thought there were four forces acting on the metal slab: F_n, the normal force, F_g, gravity, F, the 600 N force from the tractor, and F_k, the friction force.

F_net,y = F_N + F sin theta - mg - F_k sin theta_Fk = ma
F_net,x = F cos theta - F_k cos theta_Fk = ma

theta = 35 degrees, measured from positive direction of the x axis
theta_Fk = 215 degrees, measured from positive direction of the x axis

Only now that I've arranged everything I can't figure out how to isolate either F_k or F_n so I can solve the thing. I tried:

rearrange equations so:
F_ky = F_k sin theta_Fk = -ma + F_n + F sin theta - F_g
F_kx = F_k cos theta_Fk = -ma + F cos theta

Then, by trig, tan theta_Fk = F_ky/F_kx, so
tan theta_Fk = (-ma + F_n + F sin theta - F_g) / (-ma + F cos theta)

but acceleration doesn't cancel out completely to give Fn so I'm stumped. Am I heading in the completely wrong direction?

 

[PhanthomJay]

There is no y component to the friction force (assuming level ground)...and in the x direction, ther is no angle involved with the friction force (theta = 0) ...and what's the acceleration in the y direction?