- #1

uchicago2012

- 75

- 0

## 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?