(adsbygoogle = window.adsbygoogle || []).push({}); 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

F_{net}= ma

F_{k}= u * F_{n}

where F_{k }= friction force and F_{n}= 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: 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?

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# Homework Help: Newtonian mechanics of a metal slab

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