Pressure Question -- How do you get P = ρgh?

[askor]

How do you get P = ρgh?

 

[BvU]

Consider a volume of material and make a force balance.
Or look it up in a textbook at high school level

 

[Mech_Engineer]

The equation you've referenced is a simplified fundamental hydrostatics equation, see here: https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_pressure

The hydrostatic pressure can be determined from a control volume analysis of an infinitesimally small cube of fluid. Since pressure is defined as the force exerted on a test area (p = F/A, with p: pressure, F: force normal to area A, A: area), and the only force acting on any such small cube of fluid is the weight of the fluid column above it, hydrostatic pressure can be calculated according to the following formula...

For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions: Since many liquids can be considered incompressible, a reasonably good estimation can be made from assuming a constant density throughout the liquid. (The same assumption cannot be made within a gaseous environment.) Also, since the height h of the fluid column between z and z0 is often reasonably small compared to the radius of the Earth, one can neglect the variation of g. Under these circumstances, the integral is simplified into the formula...

 

[jack action]

$$P = \frac{F}{A} = \frac{mg}{A} = \frac{(\rho V)g}{A} = \rho g \frac{V}{A} = \rho gh$$

 

[Randy Beikmann]

$$P = \frac{F}{A} = \frac{mg}{A} = \frac{(\rho V)g}{A} = \rho g \frac{V}{A} = \rho gh$$

Or, in words, divide the weight of a column of liquid (with constant cross-sectional area A) by the area at the bottom of the column (again, A). Even though this formulation doesn't prove it, the answer is the same whether or not the area is constant, or even the actual shape of the vessel.