# $P$ = $ρgh$ - intuitive understanding of the equation?

I've come across this equation recently which relates pressure with the product of density, gravitational acceleration and height difference in a medium.

I understand that

$P$ = $ρgh$ expands to $\frac{m}{V}g(h_2-h_1)$

Therefore the $V - m^3$ in the denominator gets "reduced" by $h - m$ and becomes $A - m^2$, a surface, breaking down to the definition of pressure:

$P = \frac{F}{A}$

So, it's clear that mathematically it works, of course... But in my head, not so much. Could someone give me some intuitive way how to think about it?

How do density, grav. acceleration and height of a column of a medium "give" pressure?

$\frac{mg}{V}h$ -> $\frac{F}{V}h$ -> $\frac{Fh}{V}$ -> $\frac{W}{V}$

I tried doing this up here and I ended up with energy density, it seems... I'm at a loss how to think about this, as you can see. Hah, I've just checked Wikipedia, it seems that pressure and energy density indeed share the same units and at times could be considered synonyms. Well, that's neat, but still doesn't help me.

Much obliged!

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This tells you how much pressure there is in a fluid of density $\rho$ at a depth of $h$, on the Earth's surface. The reason it works is because the fluid at depth h has to hold up the fluid above it. If you look at a column of fluid with cross section area A stretching from h up to 0, the mass of the fluid is $\rho Ah$, its weight is $\rho A g h$, and this is the force holding it up. Dividing by the area gives the pressure at depth h.

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Beautiful, thank you! –  Curiousman Sep 7 '11 at 2:54
We know that under 10 m of water the pressure in our lungs roughly doubles. So a 10 m water column ($\rho_{water}=1 g/cm^3, \rho_{air} =$ dependent on altitude) is equal in weight to the atmosphere above it.
To estimate the total weight of the atmosphere (if you e.g. want to compare the observed annual rise in $CO_2$ of 2 ppm with the total worldwide emissions), you just multiply the surface of the earth with 10m and use the density of water.