Figuring out the pressure exerted by a fluid is a simple calculation. Using the density, the acceleration of gravity, and the depth of the fluid, you can find the fluid pressure.
Fluid Pressure
\(\large\mathsf{ P = \rho gh }\)
Use this table of densities of common substances to answer the questions below. If the substance is a gas, the density provided is at 0°C at 1 atmospheric pressure.
| Substance | Density (kg/m3) |
|---|---|
| Water | 997 |
| Gasoline | 737 |
| Hydrogen | 0.0899 |
| Air | 1.29 |
| Iron | 7.86 x 103 |
| Mercury | 1.36 x 104 |
| Gold | 1.93 x 104 |
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| Calculate the water pressure in the pipes at the bottom of a high-rise building that is fed by a water reservoir 28.2 m above on the roof. |
|
\(\mathsf{ \rho = 997 \text{ kg/m}^3 }\) \(\mathsf{ g = 9.81 \text{ m/s}^2 }\) \(\mathsf{ h = 28.2 \text{ m} }\) \(\mathsf{ P = ? \text{ N/m}^2 }\) |
\(\mathsf{ P = \rho gh }\) | \(\mathsf{ P = 997 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2 \cdot 28.2 \text{ m} }\) \(\mathsf{ P = 2.76 \times 10^5 \text{ N/m}^2 }\) |
| What is the liquid pressure on the bottom of a 2.10 m high tank that is filled with gasoline? |
|
\(\mathsf{ \rho = 737 \text{ kg/m}^3 }\) \(\mathsf{ g = 9.81 \text{ m/s}^2 }\) \(\mathsf{ h = 2.10 \text{ m} }\) \(\mathsf{ P = ? \text{ N/m}^2 }\) |
\(\mathsf{ P = \rho gh }\) | \(\mathsf{ P = 737 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2 \cdot 2.10 \text{ m} }\) \(\mathsf{ P = 1.52 \times 10^4 \text{ N/m}^2 }\) |
| An object that is located 1.52 meters beneath the surface of an unknown substance feels a pressure of 2.03 x 105 N/m2. What is the substance that the object is submerged in? |
|
\(\mathsf{ \rho = ? \text{ kg/m}^3 }\) \(\mathsf{ g = 9.81 \text{ m/s}^2 }\) \(\mathsf{ h = 1.52 \text{ m} }\) \(\mathsf{ P = 2.03 \times 10^5 \text{ N/m}^2 }\) |
\(\mathsf{ P = \rho gh }\) | \(\mathsf{ 2.03 \times 10^5 \text{ N/m}^2 = \rho \cdot 9.81 \text{ m/s}^2 \cdot 1.52 \text{ m}}\) \(\mathsf{ \rho = \frac{2.03 \times 10^5 \text{ N/m}^2}{9.81 \text{ m/s}^2 \cdot 1.52 \text{ m}} }\) \(\mathsf{ \rho = 1.36 \times 10^4 \text{ kg/m}^3 }\) The unknown substance is Mercury. |
