# What does the pressure in the liquid depend on?

## Pressure in liquids

### Pressure and Weight - the basics

The following formula applies to a force that acts on a surface at right angles:

$ \ mathsf {pressure \ = \ \ tfrac {force} {area}} $

If the force is in Newtons (N) and the area in square meters (m^{2}) is given, then the pressure is given in Pascal (Pa).

A liquid is held in its container by its weight. This creates pressure on the container and pressure on any object that is in the liquid.

The following properties apply to any standing liquid in an open container. The experiments show two of them.

#### Pressure works in all directions

The liquid presses on every surface that comes into contact with it, no matter where the surface is pointing. For example, the deep-sea ship below has to withstand the crushing effect of the seawater, which exerts pressure from all sides, not just downwards.

#### Pressure increases with depth

The deeper an object is in a liquid, the greater the weight of the liquid above it and the higher the pressure. Dams at the bottom are thicker to withstand the higher pressure there.

#### Pressure depends on the density of the liquid

The denser the liquid, the higher the pressure at a given depth.

#### Pressure does not depend on the shape of the container

Regardless of the shape or width of a container, the pressure is always the same at a certain depth.

### Calculation of the pressure in a liquid

The container on the right has a base A. It is to a depth *H* filled with a liquid of density $ \ rho $ (Greek letter 'rho'). In order to calculate the prevailing pressure on the base (due to the liquid), you must first know the weight of the liquid:

Volume of liquid =

Base area $ \ cdot $ depth =

$ A \ \ cdot \ h $

Mass of the liquid = density $ \ cdot $ volume =

$ \ rho \ \ cdot \ A \ \ cdot \ h $

Weight of the liquid = mass $ \ cdot $ g =

$ \ rho \ \ cdot \ g \ \ cdot \ A \ \ cdot \ h $

So: force on the base area = pgAh

This force acts on an area A.

Pressure = $ \ mathsf {\ tfrac {force} {area}} \ = \ \ tfrac {\ rho \ cdot g \ cdot A \ cdot h} {A} \ = \ \ rho \ cdot \ g \ cdot h $

In a depth *H* in a liquid of density $ \ rho $, the following applies:

$ \ mathsf {pressure} \ = \ \ rho \ cdot \ g \ cdot h $

#### example

If the density of the water is 1000 $ \ mathrm {\ tfrac {kg} {m ^ 3}} $, what is the water pressure in a 2 m deep swimming pool?

$ \ mathsf {pressure} \ = \ \ rho \ cdot \ g \ cdot h \ = \ \ mathrm {1000 \ \ tfrac {kg} {m ^ 3} \ \ cdot 10 \ \ tfrac {N} {kg} \ \ cdot \ 2 \ m \ = \ 20000 \ Pa} $

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