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
FB = W
= V γ
= V ρ g (1)
where
FB = buoyant force acting on submerged or floating body (N, lbf)
W = weight (gravity force) of displaced liquid (N, lbf)
V = volume of body below surface of liquid (m3, ft3)
γ = ρ g = specific weight of fluid (weight per unit volume) (N/m3, lbf/ft3)
ρ = density of fluid (kg/m3, slugs/ft3)
g = acceleration of gravity (9.81 m/s2, 32.174 ft/s2)
Example - Density of a Body that floats in Water
A floating body is 95% submerged in water with density 1000 kg/m3.
For a floating body the buoyant force is equal to the weight of the water displaced by the body.
FB = W
or
Vb ρb g = Vw ρw g
where
Vb = volume body (m3)
ρb = density of body (kg/m3)
Vw = volume of water (m3)
ρw = density of water (kg/m3)
The equation can be transformed to
ρb = Vw ρw / Vb
Since 95% of the body is submerged
0.95 Vb = Vw
and the density of the body can be calculated as
ρb = 0.95 Vb (1000 kg/m3) / Vb
= 950 kg/m3
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 /ft3 . The volume of the brick can be calculated
Vbrick = (3 5/8 in) (2 1/4 in) (8 in)
= 65.25 in3
The buoyant force acting on the brick is equal to the weight of the water displaced by the brick and can be calculated as
FB = (( 65.25 in3) / (1728 in/ft3)) ( 1.940 slugs/ft3) ( 32.174 ft/s2)
= 2.36 lbf
The weight or the gravity force acting on the brick - common red brick has specific gravity 1.75 - can be calculated to
WB = (2.36 lbf) 1.75
= 4.12 lbf
The resulting force acting on the brick can be calculated as
W(WB - FB) = (4.12 lbf) - (2.36 lbf)
= 1.76 lbf
