You know from Pressure that pressure is force spread over area, $P = F/A$, measured in pascals ($1\,\text{Pa} = 1\,\text{N/m}^2$), and from Density that density is mass per volume, $\rho = m/V$. Put the two together and you can find the pressure anywhere in a still liquid. It is why your ears hurt at the bottom of a pool and why a dam is thickest at its base.
Why it works
Picture a column of water inside the lake, with cross-section $A$, reaching from the surface down to depth $h$. It is at rest, so by the first law the forces on it balance. The water below pushes it up with $PA$. The air above pushes it down with $P_0 A$, and its own weight pulls it down with $mg = \rho (Ah) g$. So $PA = P_0 A + \rho A h g$. Divide by $A$: $$P = P_0 + \rho g h.$$ The area cancelled. Only the depth matters, not how wide the water is.
Key idea
In a still fluid, $P = P_0 + \rho g h$, where $P_0$ is the pressure at the surface (usually the air, about $1.01 \times 10^{5}\,\text{Pa}$) and $h$ is the depth. The gauge pressure $P - P_0 = \rho g h$ leaves out the air. Pressure depends on depth, not on the shape of the container.
| Reading | Includes the air? | Example |
|---|---|---|
| Gauge pressure, $\rho g h$ | No | A tire gauge reads 0 on a flat tire |
| Absolute pressure, $P_0 + \rho g h$ | Yes | Weather reports give the air's absolute pressure |
Worked examples
Worked example: Pressure in a pool
Find the gauge pressure and the absolute pressure $3.0\,\text{m}$ below the surface of a swimming pool. Use $\rho = 1000\,\text{kg/m}^3$, $g = 9.8\,\text{m/s}^2$ and $P_0 = 1.01 \times 10^{5}\,\text{Pa}$.
- Gauge pressure: $\rho g h = (1000\,\text{kg/m}^3)(9.8\,\text{m/s}^2)(3.0\,\text{m}) = 29{,}400\,\text{Pa}$.
- Units: $\text{kg/m}^3 \cdot \text{m/s}^2 \cdot \text{m} = (\text{kg} \cdot \text{m/s}^2)/\text{m}^2 = \text{N/m}^2 = \text{Pa}$.
- Absolute pressure: $1.01 \times 10^{5}\,\text{Pa} + 0.294 \times 10^{5}\,\text{Pa} = 1.30 \times 10^{5}\,\text{Pa}$, about 1.3 times the air pressure.
Answer: $2.94 \times 10^{4}\,\text{Pa}$ gauge; $1.30 \times 10^{5}\,\text{Pa}$ absolute
Worked example: One more atmosphere
How deep must you go in fresh water for the water alone to add one atmosphere, $1.01 \times 10^{5}\,\text{Pa}$?
- Set the gauge pressure equal to one atmosphere: $\rho g h = 1.01 \times 10^{5}\,\text{Pa}$.
- $h = \dfrac{1.01 \times 10^{5}\,\text{Pa}}{(1000\,\text{kg/m}^3)(9.8\,\text{m/s}^2)} = 10.3\,\text{m}$.
- So a diver at about $10\,\text{m}$ feels twice the surface pressure, and each further $10\,\text{m}$ adds about one more atmosphere.
Answer: $10.3\,\text{m}$
Worked example: Same depth, different floors
A wide tank with a floor of $1.5\,\text{m}^2$ and a thin pipe with a floor of $0.010\,\text{m}^2$ are both filled with water to $2.0\,\text{m}$. Compare the gauge pressure and the force of the water on each floor.
- Pressure: both floors are $2.0\,\text{m}$ deep, so both feel $\rho g h = (1000)(9.8)(2.0) = 19{,}600\,\text{Pa}$.
- Force on the tank's floor: $F = PA = (19{,}600\,\text{Pa})(1.5\,\text{m}^2) = 29{,}400\,\text{N}$.
- Force on the pipe's floor: $(19{,}600\,\text{Pa})(0.010\,\text{m}^2) = 196\,\text{N}$. Same pressure, very different forces, because the areas differ.
Answer: Equal pressures, $1.96 \times 10^{4}\,\text{Pa}$; forces $29{,}400\,\text{N}$ and $196\,\text{N}$
Common mistake
Thinking a wider container, or one holding more water, has more pressure at the bottom. Pressure is force per area: the extra water in a wide tank rests on extra floor. In a container that flares out at the top, the slanted walls hold up the extra water; in one that narrows at the top, the walls push down on the water. Check: the formula $P = P_0 + \rho g h$ has no width in it. A second slip is mixing up gauge and absolute pressure; ask whether the air is included.