How do I find the acceleration of a crate based on weight?

[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?

 

[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/g3. The Attempt at a Solution
a=9.81(.3584-.4201)
a=-.604 m/s2

 

[rock.freak667]

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

 

[der.physika]

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

 

[tiny-tim]

Welcome to PF!

Hi spider3367! Welcome to PF!

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)

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.

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

uhh? what's F_Lorentz ?

 

[spider3367]

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?

 

[berkeman]

(two threads merged)