The drawback to this approach of concentrating on liquids is that liquids are very dense compared with gases and so we do not have to go very far down into a liquid before the pressure builds up enormously due to the weight of all the mass of liquid above us. This variation of pressure with depth is almost insignificant in gases.

If you were to climb to the top of any mountain in the world, then you would not notice any difference in air pressure even though your altitude may have increased by about 1000 metres. In a liquid, however, the difference in pressure is very noticeable in just a few metres of height (or depth) difference.

Anyone who has ever tried to swim down to the bottom of a swimming pool will have noticed the pressure building up on the ears after just a couple of metres. This phenomenon is, of course, caused by gravity which makes the water at the top of the swimming pool press down on the water below, which in turn presses down even harder on the ears.

In order to quantify this increase of pressure with depth we need to look at the force balance on a submerged surface, so let us make that surface an ear drum, as shown in Figure 3.1.7

We are dealing with gravitational forces, which always act vertically, and so we only need to consider the effect of any liquid, in this case water, which is vertically above the ear drum. Water which is to either side of the vertical column drawn in the diagram will not have any effect on the pressure on the ear drum, it will only pressurize the cheek or the neck, etc.

The volume of water which is pressing down on the ear drum is the volume of a cylinder of height h,\ equal to the depth of the ear, and end area A, equal to the area of the ear drum,

Volume = hA

Therefore the mass of water involved is
Volume × density = #hA where # is the density in kg/m3 and the weight of this water is
Mass × gravity = #ghA

We are interested in the pressure p rather than this force, so that we can apply the result to any shaped surface. This pressure will be uniform across the whole of the area of the ear drum and we can therefore rewrite the force due to the water as pressure × area. Hence: pA = #ghA

Cancelling the areas we end up with:
p = #gh

Since the area of the eardrum cancelled out, this result is not specific to the situation we looked at; this equation applies to any point in any liquid. We can therefore apply this formula to calculate the pressure at a given depth in any liquid in an engineering situation.

There are two further important features that need to be stressed:

  1. Two points at the same depth in the same liquid must be at the same pressure even if one of them is not directly underneath the full depth.
  2. # The same pressure due to depth can be achieved with a variety of different shaped columns of a liquid since only the vertical depth matters   

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