Forces and Newton-Identifying Acceleration

[312213]

Homework Statement

A man pulls a rope, with a force of 200N, attached to a box, with a mass of 35kg, 30° from the horizontal. The $$\mu$$_k is 0.3. Find the acceleration of the box.

Homework Equations

$$\Sigma$$F = ma
F_fr = $$\mu$$F_N

The Attempt at a Solution

FBD
http://img255.imageshack.us/img255/2181/forceandNewtonaccelerat.jpg

Net Force Equations
$$\Sigma$$F_y = ma
F_N - F_G = ma
F_N - mg = ma
F_N = mg + ma
Acceleration is 0 because the box is not moving up or down so
F_N = mg


$$\Sigma$$F_x = ma
F_appcos$$\theta$$ - F_fr = ma
F_appcos$$\theta$$ - $$\mu$$F_N = ma
F_appcos$$\theta$$ - $$\mu$$mg = ma
(F_appcos$$\theta$$ - $$\mu$$mg) / 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
F_N - F_G = ma
F_N - mg = ma
F_N = mg + ma
Acceleration is 0 because the box is not moving up or down so
F_N = 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 F_app 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 f_app in the y direction. The correct net equation is:

F_N = mg -F_appsin$$\theta$$

(F_appcos$$\theta$$ - $$\mu$$(mg -F_appsin$$\theta$$) / 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!