Exam Question Help: Solve Force/Acceleration Problem w/ g, Sin30, Cos30

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Homework Statement




Info : g=10ms^-2 Sin30=0.5 Cos30=0.866

http://img227.imageshack.us/img227/8332/xfgdda0.jpg



A box of mass m = 50 kg is initially at rest at point A on a floor with coefficient of kinetic friction μk = 0.1 (see figure below). A person starts pulling the box for a distance Δx = 40 m (from point A to point B) with a force FP = 100 N which acts at a 30o angle as shown in the figure and the box moves horizontally.

a) Draw the free-body diagram of all forces acting on the box

b) Find the normal force FN on the box

c) Find the friction force Ffr on the box

d) Calculate the acceleration a of the box






Homework Equations



– the equation for net force (Fnet = m * a),

– the equation for gravitational force (Fgrav = m * g), and

– the equation for frictional force (Ffrict = µ * Fnorm).

- a=Fnet / m


The Attempt at a Solution



Crap attempts not really worth mentioning.
I can do the Free Body Diagram with No problem.
But can't find the rest, the Equations just don't add up :(


I really need this solved so i can continue studying for my exams :(
 
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The box is in equilibrium in the vertical direction so what does that tell you about the sum of the forces in the vertical direction (The force due to the rope, the force due to gravity, and the normal force)

There is a relationship between the normal force found above and the kinetic friction. The force Ffrict = µ * Fnorm as you stated, so it shouldn't be hard to find.

Once you know the friction force you now know the total force in the horizontal direction and can apply Newton's law F=ma or a=F/m
 
Ffrict=0.1*0? Because the Fnet = 0 right? :S
 
Ok I'm near to solving the exercise.

The only thing i need now is to find the Ffriction which I've been brainstorming for hours now and i can't find.

Can someone give me the answer for ffr? :S