# Homework Help: Force below horizon

1. Feb 22, 2010

### jwxie

1. The problem statement, all variables and given/known data

Find the acceleration of the crate when it is pushed by the same force at an angle of 30 degree below the horizon.

Given mass = 2.0kg Applied Force = 10 N, and uk = 0.047

2. Relevant equations

Fx/y = ma = sum of all force vectors
Fk = ukN

3. The attempt at a solution

So I did a free body diagram. Here is my attempt

<Fx = ma
<Fy = 0

Fx = ma = Fcos(-30) * Fk
Fy = 0 = Fsin(-30)+N-mg

Is that it? Anything tricky?

I just want to make sure all my setups are correct.

2. Feb 22, 2010

### ideasrule

That's right, but to prevent confusing yourself, you might want to express everything in terms of magnitudes. So ma=Fcos(30)*Fk: the magnitude of ma is equal to the magnitude of Fcos(-30)*Fk. N=mg+Fsin(30), because the ground pushes up against both gravity and the y component of the force F. I find that this approach is much more intuitive than using Fcos(-30) and Fsin(-30).

EDIT: Oops, that's not right. You forgot to account for the contribution of gravity to the normal force.

3. Feb 22, 2010

### xcvxcvvc

Why are you multiplying the x-component of the applied force by the force of friction?

didn't he do that here?

4. Feb 22, 2010

### jwxie

yeah that was a typo. it was suppose to be minus
Fx = F - Fk
and Fk also has x, y components, which is along the x axis

Last edited: Feb 22, 2010
5. Feb 22, 2010

### xcvxcvvc

Looks good to me. Just make sure you keep your signs straight -- I'd write F + Fk with Fk being a negative vector to denote direction. I do, however, believe you meant the right thing.

6. Feb 22, 2010