= V γ
= V ρ g (1)
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 )
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
V b ρ b g = V w ρ w g
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
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
The study of fluids - liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.
Calculate the thrust force acting on a submerged surface.
Depth and hydrostatic pressure.
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