# Fluid Statics

## Variation of Pressure with Depth in a Fluid


### Learning Objectives

By the end of this section, you will be able to:
1. Define pressure in terms of weight.
2. Explain the variation of pressure with depth in a fluid.
3. Calculate density given pressure and altitude.
If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense.
Consider the container in . Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid  divided by the area  supporting it (the area of the bottom of the container):
We can find the mass of the fluid from its volume and density:
The volume of the fluid  is related to the dimensions of the container. It is
where  is the cross-sectional area and  is the depth. Combining the last two equations gives
If we enter this into the expression for pressure, we obtain
The area cancels, and rearranging the variables yields
This value is the pressure due to the weight of a fluid. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation  represents the pressure due to the weight of any fluid of average density  at any depth  below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered.  illustrates this situation.
Atmospheric pressure is another example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”).  However, the average pressure at sea level is given by the standard atmospheric pressure , measured to be
This relationship means that, on average, at sea level, a column of air above  of the Earth’s surface has a weight of , equivalent to . (See .)
What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the total pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way.

### Section Summary

1. Pressure is the weight of the fluid  divided by the area  supporting it (the area of the bottom of the container):
    
2. Pressure due to the weight of a liquid is given by
    where 

### Conceptual Questions


### Problems & Exercises
