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

  1. Feb 24, 2010 #1
    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. jcsd
  3. Feb 24, 2010 #2
    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
  4. Feb 24, 2010 #3

    rock.freak667

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    Draw the free body diagram for the object. Split the force into its x and y components. Then simply use Newton's 2nd law
     
  5. Feb 24, 2010 #4
    Re: Acceleration of a Crate

    [tex]
    \vec F_{Lorentz}=0[/tex]
     
  6. Feb 24, 2010 #5

    tiny-tim

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    Welcome to PF!

    Hi spider3367! Welcome to PF! :smile:
    erm :redface: … 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 ? :confused:
     
  7. Feb 24, 2010 #6
    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???
     
  8. Feb 24, 2010 #7

    berkeman

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    Staff: Mentor

    (two threads merged)
     
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