Tough hydrostatic pressure problem?

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A hemisphere shell with mass m and radius R rests on a horizontal table. At the top of the shell there is a hole. The shell is filled with a liquid with density ρ.1 up to height h<R, The air has density ρ.a.

What is the heigth h, measured from the table, when the shell lifts from the table because of the pressure from the liquid?

[I ran across this recently. Made my brain hurt a little trying to figure it out. I thought some of you might like to try the challenge, too.]

Update:

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I liked this problem. It has a surprizingly simple and beautiful result. I got:

h = (3m/(π(ρ1 – ρ.a)))^1/3

Which is in accord with Dr D and where Civ Air is going. I can’t say for certain that it is correct, but I believe so.

I believe that Jaz_Wil’s error was the failure to properly figure the fluid pressure at the hemi-sphere surface: It varies with

p = (h – y)*(ρ1 – ρ.a) g

where y is the height of the hemisphere surface above the table.

The “ρ1 – ρ.a” factor is simply the density of the fluid taking into the account of the bouancy factor of air.

Civ Air — you don’t need to worry about the differential air density across the height. It will be de minimus.

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✅ Answers

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  • If you neglect the density of the air, I’m getting

    h^3 = 3m / (ρπ)

    ρ = density of liquid.

    Dimensionally this works out. I think jaz_will should not have the R in his final answer.

    Measure x from the table. At a height x (< h), the pressure is

    ρg(h-x).

    Note the pressure is atmospheric (zero) at the top of the liquid.

    Measure θ from the horizontal

    sinθ = x/R

    Elemental area on which liquid pressure acts is

    2πRcosθ * dx/cosθ

    = 2πR dx

    Elemental force (at angle θ) = ρg(h-x)*2πR dx

    Vertical component of this force = ρg(h-x)*2πRsinθ dx

    = ρg*x*(h-x)*2πdx

    Finally the total vertical force comes from integrating this from 0 to h.

    Fv = π/3 * ρg h^3 = mg

    h^3 = 3m / (ρπ)

  • I am not sure if this is right: it has been almost a decade since my physics competition days. But the answer I got is so simple it ought to be right 🙂

    Very cool question, by the way.

    Equation used: pressure at depth H from the surface is ρ g H (this is the only part I am not sure about).

    Force exerted on an area dA is equal to p dA, in the direction of the normal to the area element. So we work in spherical coordinates.

    It is clear by symmetry that the the forces in the directions parallel to the table cancel out completely, so we only need to work with force in the vertical direction. Let θ be the azimuthal angle measured to that it is 0 at the table top and π/2 at the top of the bowl. The infinitesimal force in the vertical direction is given by pressure * area * trig component for the vertical part. In other words

    dF(θ) = p dA sin θ = p . 2π R cos θ sin θ dθ

    In the spherical coordinate, the depth H is given by

    H = R ( sin Θ – sin θ )

    where R sin Θ = h, the total height of the fluid. So

    dF(θ) = 2 π ρ g R^2 (sin Θ – sin θ) sin θ cos θ dθ

    Now we integrate in θ from 0 to Θ And so the force due to liquid pressure is, after a small computation:

    (1/3) π ρ g R^2 (sin Θ)^3 = (1/3) π ρ g h^3 / R

    Now, the effective fluid pressure is ρ = ρ.1 – ρ.a, and the weight of the shell is mg, so setting the forces to balance, we have that

    π (ρ.1 – ρ.a) h^3 = 3 m R

    from which h can be solved by simple algebra.

  • I agree with most of the methodology of the previous answers, however, the pressure the air exerts on the top fluid will have an effect. I have a helicopter to go look at right now, but I will try to include that effect in my derivation if I can get to it tonight (I get off work in about an hour).

    ***

    Ok, had to reject the repair back to inspection for an error on their part so I got some more time

    ***

    Pressure at depth H from the surface is ρ.1 g H + Patmosphere

    Patmoshpere = ρ.a*R*T (R = gas constant of air)

    Ptotal = ρ.1 g H + ρ.a*R*T

    H = r ( sin Θ – sin θ )

    dF(θ) = p dA sin θ

    = p . 2π r cos θ sin θ dθ

    = 2 π (ρ.1 g H + ρ.a*R*T) r sin θ cos θ dθ

    = 2 π (ρ.1 g r ( sin Θ – sin θ ) + ρ.a*R*T) r sin θ cos θ dθ

    = 2 π (ρ.1 g r^2 ( sin Θ – sin θ ) + ρ.a R T r) sin θ cos θ dθ

    F = 2 π (ρ.1 g r^2 (sin(Θ)^3/2 – sin(Θ)^3/3) + ρ.a R T r sin(Θ)^2/2

    F = 1/3 π ρ.1 g r^2 sin(Θ)^3 + 1/2 ρ.a R T r sin(Θ)^2

    This is the vertical force the fluid exerts upwards onto bowl.

    We need to find the force the air exerts down on the bowl over the same effective area the fluid is pushing up. Same process

    I will try to derive the down force tomorrow when I get a chance.

    I’ve been racking my brain trying to think of a way to get atmoshperic pressure from a given density, but the only way I can think of right now is through the ideal gas law, so I’m stuck using the temperature variable.

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