Archimedes' Law
Forces acting on bodies submerged in fluids.
Archimedes' principle states that:
"If a solid body floats or is submerged in a liquid  the liquid exerts an upward thrust force  a buoyant force  on the body equal to the gravitational force on the liquid displaced by the body."
The buoyant force can be expressed as
F _{ B } = W
= V γ
= V ρ g (1)
where
F _{ B } = buoyant force acting on submerged or floating body (N, lb_{f} )
W = weight (gravity force) of displaced liquid (N, lb_{f} )
V = volume of body below surface of liquid (m^{3}, ft^{3} )
γ = ρ g = specific weight of fluid (weight per unit volume) (N/m^{3}, lb_{f} /ft^{3} )
ρ = density of fluid (kg/m^{3}, slugs/ft^{3} )
g = acceleration of gravity (9.81 m/s^{2}, 32.174 ft/s^{2})
Example  Density of a Body that floats in Water
A floating body is 95% submerged in water with density 1000 kg/m ^{3}.
For a floating body the buoyant force is equal to the weight of the water displaced by the body.
F _{ B } = W
or
V _{ b } ρ _{ b } g = V _{ w } ρ _{ w } g
where
V _{ b } = volume body (m^{3} )
ρ _{ b } = density of body ( kg/m^{3} )
V _{ w } = volume of water (m^{3} )
ρ _{ w } = density of water ( kg/m^{3} )
The equation can be transformed to
ρ _{ b } = V _{ w } ρ _{ w } / V _{ b }
Since 95% of the body is submerged
0.95 V _{ b } = V _{ w }
and the density of the body can be calculated as
ρ _{ b } = 0.95 V _{ b } (1000 kg/m^{3} ) / V _{ b }
= 950 kg/m^{3}
Example  Buoyant force acting on a Brick submerged in Water
A standard brick with actual size 3 5/8 x 2 1/4 x 8 (inches) is submerged in water with density 1.940 slugs /ft^{3} . The volume of the brick can be calculated
V _{ brick } = (3 5/8 in) (2 1/4 in) (8 in)
= 65.25 in^{3}
The buoyant force acting on the brick is equal to the weight of the water displaced by the brick and can be calculated as
F _{ B } = (( 65.25 in^{3} ) / (1728 in/ft^{3} )) ( 1.940 slugs/ft^{3} ) ( 32.174 ft/s^{2})
= 2.36 lb_{f}
The weight or the gravity force acting on the brick  common red brick has specific gravity 1.75  can be calculated to
W _{ B } = (2.36 lb_{f} ) 1.75
= 4.12 lb_{f}
The resulting force acting on the brick can be calculated as
W _{ (WB  FB) } = (4.12 lb_{f} )  (2.36 lb_{f} )
= 1.76 lb_{f}
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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