# Forces and Newton-Identifying Acceleration

1. Feb 23, 2009

### 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 $$\mu$$k is 0.3. Find the acceleration of the box.

2. Relevant equations
$$\Sigma$$F = ma
Ffr = $$\mu$$FN

3. The attempt at a solution
FBD
http://img255.imageshack.us/img255/2181/forceandnewtonaccelerat.jpg [Broken]

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

$$\Sigma$$Fx = ma
Fappcos$$\theta$$ - Ffr = ma
Fappcos$$\theta$$ - $$\mu$$FN = ma
Fappcos$$\theta$$ - $$\mu$$mg = ma
(Fappcos$$\theta$$ - $$\mu$$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?

Last edited by a moderator: May 4, 2017
2. Feb 23, 2009

### slmg_2006

You are missing a force in the y-dir.

3. Feb 23, 2009

### 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

4. Feb 23, 2009

### Delphi51

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

5. Feb 23, 2009

### slmg_2006

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

6. Feb 23, 2009

### 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.

Last edited: Feb 23, 2009
7. Feb 23, 2009

### Delphi51

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