# Archimedes' Law

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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