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Hydrostatic Equilibrium Thought Experiments, for clarification

  1. Mar 24, 2009 #1
    Hey, I'm just trying to make sure I understand the concept of hydrostatic equilibrium, so I devised two thought experiments that (depending on how they resolve) should really help clear up some confusion I have.

    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!
  2. jcsd
  3. Mar 27, 2009 #2

    Doc Al

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    Staff: Mentor

    Your mistake is thinking that the pressure exerted by the water against the glass bottom must equal the weight of the tank+water divided by the area. Not so. It is certainly true that the net force exerted by the glass bottom on the tank+water just equals the weight of the tank+water. However, that force has two components. First there is the force that the glass must exert to balance the water pressure--that just equals the water pressure (which depends on the height of the water in the chimney) times the area of the glass bottom. But there is also the force that the tank walls exert on the glass. The tank walls exert an upward force on the glass, just enough to make up the difference that you point out between water pressure X area and the weight of the tank+water. Realize that as the water is added to the chimney, the water pressure in the tank increases, which pushes up on the top surface of the tank--if it wasn't firmly attached to the glass bottom, it would just lift up and spill the water.

    Same basic issue. The net force that the hand exerts just balances the total weight of tube+water. That's the same in both cases. In the second case, with the spread out base, the water pressure is pretty much the same so the force exerted by the water on the hand is greater. But the walls of that tube must be firmly attached to the hand in order to keep the water inside; like in your first thought experiment, those walls must pull up on the hand.
    I hope this clarifies things a bit.

    Excellent questions, by the way. Good thinking!
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