# Does the magnitude of pressure on the sides (not the top and bottom) of a volume depend on depth?

I understand that at a point in a fluid, such as water, the fluid will be in equilibrium as the sum of the pressure from all directions will equal to zero. For a volume of water, the reasoning should be the same otherwise unequal pressure would mean the volume would move. However, are the magnitudes of different points along the side of the volume the same for all depth or are they different?

A picture would help illustrate my problem

The box can be a solid object or just an imaginary boundary enclosing water (it shouldn't change the question I believe). Since the volume is at rest in the picture, I believe that Pressure - Side.1 and Side.2 will have the same magnitude and cancel out, and the same goes for Side.3 and Side.4. However, is there a difference in magnitude for Side.1 in comparison with Side.3 or Side.4? If so, is there an equation? It's been a long time since I read about it and I tried skimming an introductory physics book but didn't find it.

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Hello user, please don't provide images as links. Of course, SE allows you to insert images in your posts – Waffle's Crazy Peanut Oct 27 '12 at 7:03
Hint to the question(v1): See e.g. Wikipedia. – Qmechanic Oct 27 '12 at 12:29
Simply from an experimental point of view there must be, as submarines' sides are just as tough as their tops and bottoms (that is, more tough than they would have to be to just to separate the top and bottom themselves strongly). – Meow Oct 27 '12 at 15:34

First of all, in liquids at rest the pressure will be the same in every direction (otherwise you could create another of those cubic boundaries, just very small, and due to pressure differences it would move). So the pressure acting on the sides of the cube will be the same that acts due to gravity.

In addition to that, we have Pascal's law, which states

$\Delta P = \rho g h$

So the difference in pressure between two points equals the product of density, gravitational acceleration and the height difference we are analyzing.

Since there is obviously a height difference between your pressures $P_1$ and $P_3$, there is also a difference in the pressure.

I think you can visualize this very intuitively, if you make the top edge of your box line up with the liquid's surface. Now infinitesimally close to the surface, there will be no pressure from the sides (except for the atmospheric pressure), because it is only submerged in infinitesimally shallow water. Whereas at the bottom the box, the points are submerged in water of height $h$ (the height of the box). At this point there is obviously some additional pressure - but because it acts in all directions equally, there is also higher pressure on the sides of the box.

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I see now. Originally I had understood it as the pressures at a point to be in equilibrium and not necessarily of equal magnitude. Thanks! – user1604449 Oct 27 '12 at 18:50