# Ok all you engineers/math geeks

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

1. ### dveitWell-Known MemberVIP

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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:

FBD:

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. ### AirjockieWatanabe Whore!!!

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3. ### reikosheaHS Troll...And ModModeratorVIP

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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. ### dveitWell-Known MemberVIP

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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. ### corvetteguy78Well-Known MemberVIP

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To quote a classic line

Name that movie

6. ### SlushboxTeggyIt's only stupid if it doesn't workVIP

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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_SolSiMember of the 20 nut clubModeratorVIP

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back to school

8. ### corvetteguy78Well-Known MemberVIP

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And we have a winner!

9. ### dveitWell-Known MemberVIP

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Nevermind, I figured it out..

10. ### reikosheaHS Troll...And ModModeratorVIP

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

11. ### dveitWell-Known MemberVIP

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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. ### TurboMirageYEEAAAHHHVIP

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so, what is the depth?

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