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I recently solved a problem that for some reason I have no intuition about.

Imagine a box of width $w$, length $l$, and height $h$ filled with a fluid of density $\rho$ on the earth (so under the influence of gravity). Assume that $l > w$, so we can refer to a "long side" of the box (see attached figure). Compare the force on the long side of the box with the weight of the fluid. Rectangular box of width $w$, length $l$, and height $h$ to be filled with fluid.

So, the weight of the fluid in the box is given by $W = \rho g V = \rho g lwh$. Meanwhile, we can find that the force on the long side of the box is given by $F=\frac{1}{2}\rho g lh^2$. (This is just the average pressure on the long side of the box times the area of that side.) Then, looking at the ratio of the force on the long side of the box to the weight of the fluid, we find $F/W = h/(2w)$. This result is what I find puzzling. If I make a very narrow (small $w$) but tall (large $h$) box, for a small amount of fluid, the force on the long side can still be very large (much larger than the weight of the fluid). If I make the long side of the box so it can slide (causing $w$ to change), for small enough $w$ and large enough $h$, I won't be able to hold the side of the box in place, even if the volume of fluid (e.g. water) that the box can hold is very small. We can further decrease $w$ and/or increase $h$ to the point that something like a tractor or bulldozer still won't be able to keep the wall of the box from sliding out.

One thing that makes me feel a little better about this is knowing that for very small $w$, if the wall of the box slides out just slightly, the water level in the box will drop drastically and therefore so will the force that the wall applies on me or the bulldozer or whatever. But still, this idea seems strange. Does anyone have a good way to think about this to make it more intuitive?

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  • $\begingroup$ There is a very popular experiment named "Pascal's barrel". A barrel can explode with very little water. You can look on the net about it. $\endgroup$ – Vincent Fraticelli Jan 21 at 20:02
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The other answer is more detailed and related more specifically to details of your question, but maybe this will create some intuition: It does not take much water to create a great amount of force. For instance, you can float a ship in one cup of water if you make a container that is the same shape as the ship (below the water line) but offset by such an amount that the space between ship and the container has volume one cup. The total weight of the water would be slight, but the upward component of the force over the area of the ship would be equal to the weight of the ship.

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The two values are related, but have no fixed relationship between them.

The total force applied by the pressure of the contained water on the long wall depends only on the long wall's dimensions (assuming the water is filled to the top of the box). It is completely independent of the width $w$ of the box; $w$ could be a millimeter, or a kilometer; it just doesn't matter.

The mass of the contained water, of course, depends on all three dimensions of the box. So, you can easily think up scenarios where the total force on the long wall is far greater than the contained mass, or far less.

So, varying $w$ changes the contained mass, but has no effect on the long wall's force. On the other hand, if you swap the dimensions for $l$ and $h$, you'll end up with a completely different force, but the same contained mass.

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