11.4 Variation of pressure with depth in a fluid  (Page 2/4)

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 $P_{\text{atm}}$ , measured to be

This relationship means that, on average, at sea level, a column of air above $1.00{\text{m}}^2$ of the Earth’s surface has a weight of $1.01\times {\text{10}}^5\text{N}$ , equivalent to $\text{1 atm}$ . (See [link] .)

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

• Pressure is the weight of the fluid $\text{mg}$ divided by the area $A$ supporting it (the area of the bottom of the container):
  $P=\frac{\text{mg}}{A}.$
• Pressure due to the weight of a liquid is given by
  $P=\mathrm{h\rho g},$

  where $P$ is the pressure, $h$ is the height of the liquid, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.