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Wednesday, November 10, 2010

Pressure due to a liquid

We have seen in this post about the concept of pressure that a solid exerts on a surface.

Today we are going to look at the pressure that a liquid exerts on the bottom of a container in which it is present.

Fig 1 below shows a container containing an unknown orange liquid. The depth of liquid in the container is h. The density of the liquid is denoted by the letter ρ.


Fig 1

Fig 2 below represent the same container but this time in 3 dimensions. The container is in the form of a cuboid. The liquid will thus be in the form of a cuboid with cross-sectional area and height h.


Fig 2

Now how do you obtain the pressure that the liquid is exerting on the bottom of the container?

The equation for pressure is

Pressure = Force Applied /Surface area

              = Force applied by liquid/surface area over which pressure is applied

              = Weight of liquid /Cross-sectional area A

              = mg /A

               = (Volume of liquid *density*g)/A

                = (Ah *ρ)*g/A

                 = h*ρ*g

                 = hρg

Where g is the acceleration due to gravity.


A bottle of water of density 1000 kg/m3 has water of depth 0.10 m in it. If the acceleration due to gravity,g, is 9.81 m/s2 , calculate the pressure due to the water on the bottom of the container.

Pressure due to water =hρg

= 0.10*9.81*1000

=981 Pascal (Pa)

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