1. A tank of water, 1000 meters in the x and y directions but only 1 meter vertically, is full of water (a billion liters of water). There is a chimney sort of thing at a corner, connected to the main tank so water can flow freely between them, but it is only one meter to a side, but tall (how tall doesn't matter). When the water level is at just 1 meter (so no water is in the chimney), the hydrostatic pressure at all points along the wall is easy. Along the bottom, for example, the gauge pressure is [tex]\rho[/tex]*g*z, or 1000 kg/m[tex]^{3}[/tex]*9.8 m/s[tex]^{2}[/tex]*1 m = 9800 Pa, and the total hydrostatic pressure is P[tex]_{atm}[/tex]+[tex]\rho[/tex]*g*z, or 101,325 Pa + 9,800 Pa = 111,125 Pa.

Now, say we added 1000 L of water, to raise the level of water in the chimney to the 2 meter mark. Now, we've only added 1000 L of water, so our total amount of water in the system is 1,000,001,000 L. The gauge pressure along the bottom of the whole container is twice as great: [tex]\rho[/tex]*g*z = 1000 kg/m[tex]^{3}[/tex]*9.8 m/s[tex]^{2}[/tex]*2 m = 19,600 Pa.

So far, I'm pretty sure I'm right about all this. However, now we add a balloon with some amount of air to the inside of the tank, right along the bottom, which will shrink as the hydrostatic pressure increases. As we continue adding water in 1000 L increments (increasing the water depth by 1 m each time), the balloon will shrink linearly as hydrostatic pressure increases linearly. Lets say (at the same time) that I replace the whole bottom wall of the tank with a sheet of glass that connects to the rest of the tank (still waterproof), and that the sheet of glass can only stand so much pressure before it breaks. If the total pressure the water exerts on it exceeds 150 kPa, it will break.

So, hydrostatic pressure along the bottom will reach 150 kPa when the water level of the chimney is at 4.967 meters (simple arithmetic). We'll approximate to 5 meters. However, one could also calculate the pressure exerted on the glass as the weight of the whole container (which is weightless) and its water (total) divided by the area of the glass. With the chimney filled to 5 meters, there are now 1,000,004,000 L of water in the system. The weight is thus 1,000,004,000 kg. This divided over the surface area of the glass bottom (1000 m * 1000 m = 1,000,000 m[tex]^{2}[/tex]) is 1,000.004 Pa, or just over a kilopascal. So add that to the atmospheric pressure (101.325 kPa) and the total pressure is just 102.325 kPa, nowhere close to the breaking point of the glass!

So which is right? I'm pretty confident that the balloon will accurately measure hydrostatic pressure, but I also feel like the glass wall should feel that hydrostatic pressure and have to push back that amount to maintain equilibrium, so it should be a measure of the hydrostatic equilibrium too! Where did I go wrong?

A shorter situation, to further clarify...

2. A man has a tube 1000 m tall on the palm of his hand, but only 1 cm in diameter. The bottom wall of the water tube is, in fact, his hand. The water at the bottom of this tube feels a lot of gauge pressure (1000 kg/m[tex]^{3}[/tex]*9.8 m/s[tex]^{2}[/tex]*1,000 m = 9800 kPa), and even more total hydrostatic pressure (9800 kPa + 101.325 kPa = 9901.325 kPa). All that pressure is pushing against his hand (and I'm sure its enough to drill a hole right through!). However, what if he tapers the bottom of the tube out quickly, to 1 m diameter. Say he's also a giant, so his palm is that big. Common sense says that the weight of the water (which admittedly is less high, but not by very much if he tapers it out quickly) is now distributed over a much larger area, so his hand stands a good chance of now NOT being crushed. However, his hand is still under (about) the same hydrostatic pressure, so why wouldn't it?

Any responses would be much appreciated!