# Homework Help: How do I find the acceleration of a crate based on weight?

1. Feb 24, 2010

### spider3367

I'm a little confused on how to complete this problem. Any help would be appreciated

A 32 kg crate is pushed on a floor with a kinetic friction coefficient of .45. If the crate is pushed with a force of 220 N at an angle of 21 degrees below the horizontal, then what is the acceleration of the crate?

2. Feb 24, 2010

### spider3367

Acceleration of a Crate

1. A 32 kg crate is pushed on a floor with a kinetic friction coefficient of .45. If the crate is pushed with a force of 220 N at an angle of 21 degrees below the horizontal, then what is the acceleration of the crate?

Relevant equations
a=g(sinθ-μkcosθ)

μk=-a/g

3. The attempt at a solution
a=9.81(.3584-.4201)
a=-.604 m/s2

Last edited: Feb 24, 2010
3. Feb 24, 2010

### rock.freak667

Draw the free body diagram for the object. Split the force into its x and y components. Then simply use Newton's 2nd law

4. Feb 24, 2010

### der.physika

Re: Acceleration of a Crate

$$\vec F_{Lorentz}=0$$

5. Feb 24, 2010

### tiny-tim

Welcome to PF!

Hi spider3367! Welcome to PF!
erm … that's completely the wrong formula (it looks like the formula for steady motion on a slope, of angle θ).

This is a flat floor, and the normal force is greater than usual because the applied force (of 220 N) is increasing it.
uhh? what's FLorentz ?

6. Feb 24, 2010

### spider3367

Re: Acceleration of a Crate

I forgot to add that I am given the answer: .865m/s2. I just need to show how to arrive at it. I assumed the 220cos (21)-.45(32)(29)/32 was the correct way to do it as mentioned above, but clearly that does not come out to .865. What am I doing wrong???

7. Feb 24, 2010