Forces and Newton-Identifying Acceleration

[312213]

1. The problem statement, all variables and given/known data
A man pulls a rope, with a force of 200N, attached to a box, with a mass of 35kg, 30° from the horizontal. The \muk is 0.3. Find the acceleration of the box.


2. Relevant equations
\SigmaF = ma
Ffr = \muFN

3. The attempt at a solution
FBD
http://img255.imageshack.us/img255/2181/forceandnewtonaccelerat.jpg (http://imageshack.us)

Net Force Equations
\SigmaFy = ma
FN - FG = ma
FN - mg = ma
FN = mg + ma
Acceleration is 0 because the box is not moving up or down so
FN = mg


\SigmaFx = ma
Fappcos\theta - Ffr = ma
Fappcos\theta - \muFN = ma
Fappcos\theta - \mumg = ma
(Fappcos\theta - \mumg) / m = a
((200N)(cos30°) - (0.3)(35kg)(9.8m/s²)) / 35kg = a
2.0087m/s² \approx a

I am told this is wrong and that the correct answer is around 2.87m/s².

What did I do wrong so that I did not arrive at the correct answer?

[slmg_2006]

You are missing a force in the y-dir.

[312213]

It's in there with:

Net Force Equations
\Sigma Fy = ma
FN - FG = ma
FN - mg = ma
FN = mg + ma
Acceleration is 0 because the box is not moving up or down so
FN = mg

[Delphi51]

All your work looks correct to me.
Don't forget to check the question - very easy to copy something incorrectly.

[slmg_2006]

Since the Fapp is at a 30 degrees, isn't there a component of it working in the y-dir?

[312213]

Yes I now see that I forgot to include fapp in the y direction. The correct net equation is:

FN = mg -Fappsin\theta

(Fappcos\theta - \mu(mg -Fappsin\theta) / m = a
((200N)(cos30°) - (0.3)((35kg)(9.8m/s²)-((200N)(sin30°))) / 35kg = a
2.866m/s² = a

This is correct. Thank you.

[Delphi51]

Thanks for catching that, slmg! I forgot about the pull affecting the normal force!