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Ok all you engineers/math geeks

Discussion in 'Members' Lounge' started by dveit, Sep 17, 2009.

  1. dveit

    dveit Well-Known Member VIP

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    I need your help on this fluid mechanics problem.

    Problem Statement:
    An open tank has a vertical partition and contains gasoline (ρ=700 kg/m3 ) to the right at a depth of 4m as sketched. The partition contains a rectangular gate that is 4m high and 2m wide and is hinged at the bottom with a stop at the top against outflow of gasoline. Water is slowly added to the left side of the tank that was originally empty. At what depth h will the gate start to open?

    Diagram:

    [​IMG]

    FBD:

    [​IMG]

    So my friend already worked out this analysis for me:

    FRg = γghcgAg
    where g refers to gasoline..
    FRg = (700 kg/m^3)(9.81 m/s^2)(2m)(4m * 2 m)
    = 110*10^3 N
    = 110 kN
    FRw = γwhcwAw
    where w refers to water...
    FRw = (9.80*103 N/m^3)(h/2)(2m * h)
    where h is depth of the water...

    FRw = (9.80*103)h^2

    For Equilibrium,

    ΣMH = 0
    so that,
    FRw*lw = FRg*lg
    with lw = h/3 and lg = 4/3 m

    Thus,
    (9.80*10^3)(h^2)(h/3) = (110*10^3 N)(4/3 m)
    = h = 3.55 m



    This all makes sense to me except one part. Where does "with lw = h/3 and lg = 4/3m" come from? How are these moment arms derived?

    Thanks to anyone who can help.

     
    Last edited: Sep 17, 2009
  2. Airjockie

    Airjockie Watanabe Whore!!!

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  3. reikoshea

    reikoshea HS Troll...And Mod Moderator VIP

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    I dont think that makes a helluva a lot of sense. i would think the answer would be:

    (9.80*103)(h3) = 4(110*103 N)

    I haven't taken an engineering class in 4 years so take that with a grain of salt.
     
  4. dveit

    dveit Well-Known Member VIP

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    That's what I thought too but I didn't think he would make that simple of a mistake. It gets the right answer, I just thought perhaps I was missing something.
     
  5. corvetteguy78

    corvetteguy78 Well-Known Member VIP

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    To quote a classic line

    "The answer is....4?"

    Name that movie
     
  6. SlushboxTeggy

    SlushboxTeggy It's only stupid if it doesn't work VIP

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    That's not Little Big League, is it?

    And to quote another awesome character in response to that problem....

    "Seriously.... da fuck???"
     
  7. E_SolSi

    E_SolSi Member of the 20 nut club Moderator VIP

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    back to school
     
  8. corvetteguy78

    corvetteguy78 Well-Known Member VIP

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    And we have a winner!
     
  9. dveit

    dveit Well-Known Member VIP

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    Nevermind, I figured it out..
     
  10. reikoshea

    reikoshea HS Troll...And Mod Moderator VIP

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    im curious....whered the 3's come from?
     
  11. dveit

    dveit Well-Known Member VIP

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    You need to calculate the center of pressure for the moment arms, you can't just use h and 4.

    Cp = (I / (Yc*A)) + Yc

    where I is interia, Yc is the distance to center of mass, and A is area

    Cp for water = ((1/12) * 2 * h^3)/((h/2) * (h*2)) + h/2 = 2h/3
    Cp for gas = ((1/12) * 2 * 4^3)/((4/2)*(4*2)) + (4/2) = 8/3

    Multiply by a factor of 1/2 and you get h/3 and 4/3.

    Even though you can get the same answer using 4 and h, technically thats not the correct way to do it and I guess its just a coincidence they're the same.
     
  12. TurboMirage

    TurboMirage YEEAAAHHH VIP

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    so, what is the depth? :)
     
  13. reikoshea

    reikoshea HS Troll...And Mod Moderator VIP

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