Hydrostatic Force acting on Submerged Surface
Calculate the thrust force acting on a submerged surface.
The thrust force acting on a surface submerged in a liquid can be calculated as
F = p_{a} A
=ρ g h_{a} A (1)
where
F = thrust force (N)
p_{a} = ρ g h_{a} = average pressure on the surface (Pa)
A = area of submerged surface (m^{2})
h_{a} = average depth (m)
ρ = density (kg/m^{3}) (water 1000 kg/m^{3})
g = acceleration of gravity (9.81 m/s^{2})
Example  The thrust force acting on the side of a container
The trust force acting on the submerged vertical side of a container can be calculated as
F = p_{a} A
= ((p_{t} + p_{b}) / 2) A
= (ρ g (h_{t }+ h_{b}) / 2) A (2)
where
p_{t} = pressure at the top of the submerged surface (Pa)
p_{b} = pressure at the bottom of the submerged surface (Pa)
h_{t} = depth at the top of the submerged surface (m)
h_{b} = depth at the bottom of the submerged surface (m)
The thrust on a surface with width 1 m and height from 0 m to 2 m  in a waterfilled container can be calculated as
F = (ρ g (h_{t }+ h_{b}) / 2) A
= (1000 kg/m^{3}) (9.81 m/s^{2}) ((0 m) + (2 m)) / 2) ((1 m) (2 m))
= 19620 N
= 19.6 kN
Example  The thrust force acting on the bottom of a container
The trust force acting on the horizontal bottom of a submerged container can be calculated as
F = p_{b }A
= ρ g h_{b} A (2b)
The thrust on a bottom with width 1 m and length 2 m  on depth 1 m  in a water filled container can be calculated as
F = ρ g h_{b }A
= (1000 kg/m^{3}) (9.81 m/s^{2}) (1 m) ((1 m) (2 m))
= 19620 N
= 19.6 kN
Related Topics

Fluid Mechanics
The study of fluids  liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.
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