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I building a tank for water 2ft high, 2ft wide (front to back), 8 feet long (left to right) Common formulas dictate the total force on the long sides is approx 1000lbs (w x h x .43). Where .43 is psi per foot of depth and average psi is at depth 1 foot in this case.

Same formula states that since only depth matters in pressure then if I halve the width of the tank (front to back) then the total force on the long side remains the same 1000 lbs. And if I halve irt again same. And if I theoretically continue to halve until the tank only holds 1 ounce of water the that 1 ounce still exerts 1000 lbs on the long side.

I ask this because the math says that structurally I can double the volume of the tank just by increasing the same materials in the width without any additional support required. I want that to be true cost wise but can't wrap my head around it because it seems to fall apart in the extreme hypothetical case where at some point there just isn't enough water to exert 1000lbs. If the math does break down at some point. Where is that point?

user2624395
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What you observed is known as the Hydrostatic Paradox.

You are confused that the force exerted onto the tank wall corresponds to the weight that is larger than the weight of water contained in the tank. You are probably familiar with levers: you can put a small weight onto the long arm of the lever and thereby exert a large force at the short arm. The gain in force is compensated however by proportionally longer path of the longer arm.

In your example, the water acts as such lever. By the way, this is how Hydraulic machinery works. The difference between the narrow and the wide tank becomes apparent if the lateral wall actually moves by, say, 1 inch. If the tank was 1 inch wide, the force reduce by half. If the tank was much wider, the force will remain the same and probably push the wall further, so the amount of potential energy stored in a wide tank is larger.

gigacyan
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  • It does seem intuitively paradoxical, hence my struggle. Including the example of the dam in the Hydrostatic Paradox link you provided which is exactly my like my example on a larger scale. But the flexible membrane demo pictured in the same link makes it intuitively understandable. Thanks for finding a way that gets through my mental block! Not to dismiss the other answers. Just that gigacyan presented in a way that worked for me. – user2624395 Oct 14 '14 at 00:02
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The point where this pressure law brakes down is when the surface tension of the water starts to play a leading role. You can try this yourself with a narrow tube (I would say make it considerably smaller in inner diameter than a sixteenth of an inch). If you submerge the tube in water and you pull it out, the molecular forces between the water molecules and the tube will keep some water in the tube, even if it is open at the bottom. The thinner the tube, the taller the column of liquid that can be held by the tube.

Spongy materials work the same way, by keeping small volumes of water in their pores. With very fine pores one can make a fairly large volumes of spongy materials that can hold plenty of water without the weight of the water being enough to drain it. If you were to fill a tank with such a sponge, the pressure on the sides of the tank would be partially compensated by the capillary forces of the sponge. The pressure on the bottom, of course, would remain the same as if the same amount of water were in there as without the sponge (plus the weight of the sponge itself). The downside of such a storage method is, that while the walls of the vessels could be made thinner, it would require extra pressure to squeeze the liquid out of the sponge.

CuriousOne
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  • thanks. What if we take the extreme example a little less extreme and say the tank is just wide enough to hold 40 gallons. Keeping the long side 8 x 2 ft I get 2 inch width. That should be wide enough eliminate the tensil factor. 40 gallons weighs 340 lbs yet the math says there is still 1000lbs on the side. What is creating the 1000lbs, is it molecular collision? If so then doesn't temperature factor in? Or does colder water occupy less space therefore more molecules to make 40 gallons canceling out the loss of speed of the molecules to equalize the math? – user2624395 Oct 11 '14 at 17:56
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You are confusing pressure and force. Yes, the pressure is only a function of the water depth. However, the resulting force is pressure times area.

If you have 2 feet of water in the tank, then the pressure at the bottom is .86 PSI. Since the pressure varies linearly and is 0 at the top, the average pressure on any vertical strip of wall is .43 PSI.

Now for the force. If any vertical strip of wall has a pressure of .43 PSI on it, then you have to multiply that average pressure by the area of the strip to get the force. For example, if the strip is 1 inch wide, then the area is 24 in2, and the force on that strip of wall is 10.3 pounds. This will be the force on every 1 inch wide vertical strip of wall. You have to add those up to get the total outward pressure on any wall. For example, a 1 foot wide wall would be 12 of these strips wide, and experience a outward force of 124 pounds.

So making the tank wider doesn't increase the pressure, but it does increase the force on the walls that were made wider. From the above, you should be able to see that any wall has to be able to hold back 124 pounds per foot of wall length. If the wall is 8 feet long, then it has to be able to withstand 991 pounds pushing outwards.

Olin Lathrop
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  • I understand the difference between pressure and force even if just from the difference in units. It just seemed magical that so much less water makes no difference. Until I saw the intuitive flexible membrane demo in gigacyan's Hydrostatic Paradox link. – user2624395 Oct 14 '14 at 00:11
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The answer to your question, as explained to me, is that as long as the dimension in question is large enough to facilitate the fluid's normal behavior as a Newtonian fluid, the same firce will be expended as computed by the standard formula. (My 1st answer was wrong).

john
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My answer that hydro static lateral pressure $Ph=wh$ is applicable until the a total force of same base area equal the same for of vertical pressure $Pv=wh$ (densiryxheight) on base area then apply $Pv$ (the weight stress for lateral force to confirm that suppose $H=10~\text{m}$ filled in thin wall of $100~\text{mm}$ thick and $5~\text{m}$ length if applied hydrostatic equation total force $Fh$ on all side area will be $$Ph=10~\text{m} \times 1000~\frac{\text{kg}}{\text{m}^3} = 10000~\text{kg}- Fh = 1000~\frac{\text{kg}}{\text{m}^3} \times 10~\text{m} \times 5.0~\text{m}/2 = 25000~\text{kg}$$ which is not logic cause the over all weight wall water is $$4~Fw = 1000~\frac{\text{kg}}{\text{m}^3} \times 10~\text{m} \times 0.1~\text{m} \times 5 = 5000~\text{kg}.$$

user12262
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Yasin
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correction for the above the total force by equation for wall 10mH,0.1m Thick , 5 m width filled with water is Fh= 1000 kg/m3x10 mx10 mx5 m/2=250000 kg =250 tone weight of all water at same depth Fw= 1000 kg/m3x10mx0.1mx5m=5000 kg ie pressure by equation is 50 times the weight which is not logic

Yasin
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