# Physics Ramp Problem

1. Oct 7, 2007

### ehump20

1. The problem statement, all variables and given/known data
A student pulls on a box up an incline of 10 degrees. The student pulls with a force of 180 N directed at 30 degrees above the line of the incline. The box has a mass of 28 kg, and the coefficient of friction between the box and floor is 0.299. The acceleration of gravity is 9.8 m/s^2.

What is the acceleration of the box?

2. Relevant equations
Fnet = ma

3. The attempt at a solution
From Fnet = ma I get:
Fp - Ffr = ma, where Fp is the force of the person pulling and Ffr is the force of friction resisting.

Ffr = MkFn
Ffr = (.299)(Fn)

I then solved for Fn:
Fn = mg/cos10
Fn = (28)(9.8) / cos10
Fn = 278.63

So Ffr = (.299)(278.63)
Ffr = 83.31 N

So we have Fp - Ffr = ma,
so Fp - 83.31 = (28)a

We need Fp.
Fp = cos30 * 180
Fp = 155.88

So Fp - 83.31 = (28)a,
so 155.88 - 83.31 = (28)a

Solving for a i get a = 2.59 m/s^2

I submitted this to my online homework system and it said the answer was wrong. Where did I go wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 7, 2007
2. Jul 2, 2009

### deezy911

...you forgot mgsin(theta) in your original "Fnet" equation. Include mgsin(theta) and you should get the acceleration. --------> ma = T - F - mgsin(theta)