# Archimedes' Law

## Forces acting on a body submerged in a fluid

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}*