# Homework Help: Newtonian mechanics of a metal slab

1. Sep 19, 2010

### uchicago2012

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?

2. Sep 19, 2010

### 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?