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Work in two dimensions?

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A worker pushed the Piano weighing 93N up a ramp into a moving van, pushing horizontally, parallel to the ground. The ramp extends 4m over the ground and 3m high. The top, sloped surface of the ramp holding the piano is 5m long. The worker exerts a force of 85 N. How much work does she do?


    2. Relevant equations
    W = Fd


    3. The attempt at a solution
    I suppose I could use arctan or something to find the angle and use cosine, but would the angle for the physical situation be the same as the angle for the force vector?
     
  2. jcsd
  3. Nov 7, 2009 #2
    By using vectors, you can break the problem down into horizontal and vertical components.

    You don't describe whether or not friction is considered in your problem. Is the ramp assumed to be frictionless? If it is, try calculating work using only the vertical component of your force vector, because no work is done to translate an object horizontally, since there would technically be no opposing forces.
     
  4. Nov 7, 2009 #3

    rock.freak667

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    the piano has a component parallel to the surface. FP

    The force of the man also has a component parallel to the surface. FM

    What are the values of these forces?
     
  5. Nov 7, 2009 #4
    That's the problem exactly as it's written. I assumed it's to be considered frictionless, and it says that the worker pushes horizontally and that the force she uses is 85N, so I kind of assumed that that force must be parallel to the horizontal; however, I haven't a clue how to work it from there.
     
  6. Nov 7, 2009 #5

    tiny-tim

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    Hi kashiark! :smile:
    "assumed"?

    what else can it possibly mean??
    I don't really understand what you're asking here,

    but anyway you don't need to find the actual angle.

    Just use cos (and the fact that this is a 3 4 5 triangle). :smile:
     
  7. Nov 7, 2009 #6
    So I should use the ratio 4:5? How can the force up the slope be more than the force the worker put on it? It's very possible that I'm completely misunderstanding what you're telling me to do...
     
  8. Nov 7, 2009 #7

    tiny-tim

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    I don't understand …

    work done = force "dot" displacement = force times displacement times cosine.

    And cosine always ≤ 1.
     
  9. Nov 7, 2009 #8
    Why would I multiply the force times the cosine? Then I'd have adjacent²/hypotenuse.
     
  10. Nov 8, 2009 #9

    tiny-tim

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    I don't understand :redface:

    where does your other "adjacent" come from? :confused:
     
  11. Nov 8, 2009 #10

    ideasrule

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    The formula for work is W=Fdcos(theta): in other words, the component of F parallel to d, times d.
     
  12. Nov 8, 2009 #11
    Are we supposed to be using d of the hypotenuse or d parallel to the horizontal? Either way, Fdcos(theta) makes no sense. If we were finding work using the distance parallel to the horizontal, it would just be 85*4; if we were finding work using the distance of the slope, Fdcos(theta) would yield 5*85²/hypotenuse.
     
  13. Nov 8, 2009 #12

    tiny-tim

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    The vector d is the displacement …

    displacement is where the thing goes from and to …

    in this case, it goes up the slope, so d is along the slope, and it has magnitude d = 5.

    F is horizontal, so cos = 4/5.
     
  14. Nov 8, 2009 #13
    so, Fd*cos would be equivalent to 4*85?
     
  15. Nov 8, 2009 #14

    tiny-tim

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    Yup! :biggrin:

    Happy, or still worried? :smile:
     
  16. Nov 8, 2009 #15
    So we were just looking for the work done in the dimension parallel to the horizontal?
     
  17. Nov 9, 2009 #16

    tiny-tim

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    (just got up :zzz: …)

    Yes. :smile:

    For example, if the force is gravity, then we're looking for the work done parallel to the vertical direction …

    in other words, we're only interested in the difference in height! :wink:
     
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