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A cylinder with a radius of $10 \;\mathrm{m}$ and a height of $10 \;\mathrm{m}$ is resting on top of a body of water. The cylinder is less dense than water. At the center of the base (equal to water level), a circular plane extends out to a radius of $100 \;\mathrm{m}$. At the end large circle, the surface curves upwards in a concave shape (imagine a sombrero with a vertical cylinder at the center).The base of the cylinder ($r= 10 \;\mathrm{m}$) is open to the water below it.

Water is added on top of the sombrero, forcing the object down into the water. The cylinder at the center has a volume $3140 \;\mathrm{m^3}$. That volume of water has a mass of $3140000 \;\mathrm{kg}$. The surface area of the large circle is $31400 \;\mathrm{m^3}$. Subtract the area of the cylinder ($314 \;\mathrm{m^3}$) and you have a total surface area of $31000 \ m^3$ on top of the sombrero. Add water until the sombrero is pushed down into the body of water (a depth of $1 \;\mathrm{m}$ of water has a mass of $3100000 \;\mathrm{kg}$ ). Water is displaced as the sombrero sinks until the downwards force is counteracted by buoyancy. Imagine there is no air in the cylinder and water is pressed against the top of the cylinder.

The atmosphere of the Earth pushes against the cylinder, sombrero, and the body of water. The top cylinder is now $9 \;\mathrm{m}$ above the body of water. Two meters below the top of the cylinder directly at the center, there is a mouth of a hose surrounded by water (mouth is facing downwards). This point is located $7 \;\mathrm{m}$ above the body of water. The hose extends outside the cylinder and leads over the edge of sombrero (imagine $101 \;\mathrm{m}$ away from center). This end of the hose is currently closed. Relative to the water level, this point is $5 \;\mathrm{m}$ above water level. The difference in height of the two ends is $2 \;\mathrm{m}$.

All of the sudden, the closed end transforms into an opening. The water at the end of the hose is now exposed to air at 1 atmosphere. Could you siphon water from the cylinder slowly where the water is replaced from the water from the base of the cylinder?

For example, $1 \;\mathrm{m^3}$ of water has a mass of $1000 \;\mathrm{kg}$. Water pressure is greater at the base of the cylinder than the top. Water is incompressible.

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The water level in the cylinder remains the same as the level outside the sombrero. If it were higher, the weight of the water in the cylinder would push down and move the water beneath until the levels equalise.

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  • $\begingroup$ I thought a lot about your answer and one thing that confuses me. For example, if you take a 1 liter bottle of water, submerge the mouth in a pool of water, and turn it upside down, then the water does not pour out even if that pool of water is a fraction of the amount in the bottle (more mass in the bottle than the pool). What allows this at a small scale but not a bigger size? Thanks $\endgroup$
    – user259268
    Commented Dec 2, 2021 at 20:45
  • $\begingroup$ @ Doug Interesting question. Ok, assuming that no air can get into the cylinder, you are right that atmospheric pressure could support a column of about 10m inside the cylinder. The end of the siphon 2m down from the top of the cylinder would have a pressure $h\rho g$ = $20,000$Pa, but the other end would be at atmospheric pressure of $101,000$Pa, so the water wouldn't flow out of the cylinder (to be replaced). If you tried a taller cylinder, e.g. 20m, only a 10m water column would be supported inside the cylinder, with a vacuum at the top $\endgroup$
    – John Hunter
    Commented Dec 2, 2021 at 22:35
  • $\begingroup$ Thanks for the update John. Siphons make no sense! I guess it depends on whether you believe atmospheric pressure plays a role (siphons work in vacuums!) or fluid cohesion is the principle force acting on the liquid. Amazing that we have known about siphons for thousands of years and still don't know the primary mechanism. $\endgroup$
    – user259268
    Commented Dec 2, 2021 at 23:58