1. The problem statement, all variables and given/known data 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? 2. Relevant equations Fnet = ma Fk = u * Fn where Fk = friction force and Fn = normal force and u = coefficient of friction 3. The attempt at a solution So I thought there were four forces acting on the metal slab: Fn, the normal force, Fg, gravity, F, the 600 N force from the tractor, and Fk, the friction force. Fnet,y = FN + F sin theta - mg - Fk sin thetaFk = ma Fnet,x = F cos theta - Fk cos thetaFk = ma theta = 35 degrees, measured from positive direction of the x axis thetaFk = 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 Fk or Fn so I can solve the thing. I tried: rearrange equations so: Fky = Fk sin thetaFk = -ma + Fn + F sin theta - Fg Fkx = Fk cos thetaFk = -ma + F cos theta Then, by trig, tan thetaFk = Fky/Fkx, so tan thetaFk = (-ma + Fn + F sin theta - Fg) / (-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?