Archimedes' law and floating of bodies

The buoyant (Archimedes) force $F_A = \rho_{\text{l}}\, g\, V_{\text{sub}}$, conditions for floating, the waterline and displacement.

Terms

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Formulas

Archimedes force

F_A = \rho_{\text{ж}}\, g\, V_{\text{погр}}

Quantities
Symbol	Name	Unit
$F_A$	buoyant (Archimedes) force	N
$\rho_{\text{ж}}$	density of the liquid (gas)	kg/m³
$g$	acceleration due to gravity	m/s²
$V_{\text{погр}}$	volume of the submerged part of the body	m³

Solved for another quantity:

V_{\text{погр}} = \dfrac{F_A}{\rho_{\text{ж}}\, g}

\rho_{\text{ж}} = \dfrac{F_A}{g\, V_{\text{погр}}}

Note. Here $g \approx 9.8\ \text{m/s}^2$ , in problems $10\ \text{m/s}^2$ is often used. Example ( $g = 10\ \text{m/s}^2$ ): a body of volume $200\ \text{cm}^3 = 0.0002\ \text{m}^3$ fully immersed in water: $F_A = 1000 \cdot 10 \cdot 0.0002 = 2\ \text{N}$ ; in kerosene ( $800\ \text{kg/m}^3$ ) — $1.6\ \text{N}$ . The force depends on the density of the liquid and the volume of the submerged part, not on the material of the body or the depth.

Conditions for floating through densities

\rho_{\text{тела}} \gtrless \rho_{\text{ж}}

Quantities
Symbol	Name	Unit
$\rho_{\text{тела}}$	density of the body (uniform)	kg/m³
$\rho_{\text{ж}}$	density of the liquid	kg/m³

Solved for another quantity:

\dfrac{V_{\text{погр}}}{V} = \dfrac{\rho_{\text{тела}}}{\rho_{\text{ж}}}\ (\rho_{\text{тела}} < \rho_{\text{ж}})

Note. Example: iron ( $7800\ \text{kg/m}^3$ ) sinks in water but floats in mercury ( $13\,600\ \text{kg/m}^3$ ); pine ( $400\ \text{kg/m}^3$ ) and ice ( $900\ \text{kg/m}^3$ ) float in water. A floating uniform body is submerged to the fraction $\dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_{\text{body}}}{\rho_{\text{l}}}$ : an ice floe in water is $0.9$ submerged (the fraction follows from $F_A = F_{\text{g}}$ ). The rule holds for solid uniform bodies; an iron ship floats because it is hollow and its average density is less than that of water.

Conditions for floating through forces

F_{\text{тяж}} \gtrless F_A

Quantities
Symbol	Name	Unit
$F_{\text{тяж}}$	force of gravity on the body, $F_{\text{g}} = m g$	N
$F_A$	Archimedes force when the body is fully immersed	N

Solved for another quantity:

m g = \rho_{\text{ж}}\, g\, V_{\text{погр}}

Note. Example ( $g = 10\ \text{m/s}^2$ ): an iron block of $1\ \text{dm}^3$ in water: $F_{\text{g}} = 7800 \cdot 10 \cdot 0.001 = 78\ \text{N}$ , $F_A = 1000 \cdot 10 \cdot 0.001 = 10\ \text{N}$ ; $78 > 10$ — sinks. A pine block of the same volume: $F_{\text{g}} = 4\ \text{N} < 10\ \text{N}$ — it rises and floats, and the force on the floating body is $F_A = F_{\text{g}} = 4\ \text{N}$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A body in a liquid

The green arrow is the Archimedes force (up), the red one is the force of gravity (down). If the body floats, these forces are equal.

Graphs

The line F_A(V_sub) is drawn for the chosen liquid, the marker is the submerged part of the body. Click the graph to change the volume.

F_A(V_{\text{sub}})

\rho_{\text{body}},\ \rho_{\text{l}}

400 < 1,000 kg/m³

V_{\text{sub}}/V

0.4 (80 cm³)

F_A

0.8 N

Floats on the surface

Formulas with your numbers

F_{\text{g}} = \rho_{\text{body}} g V =400\cdot 10\cdot 0.0002 = 0.8\ \text{N}

F_{A,\,\text{full}} = \rho_{\text{l}} g V =1000\cdot 10\cdot 0.0002 = 2\ \text{N}

F_{\text{g}} < F_{A,\,\text{full}}

F_A = F_{\text{g}} = 0.8\ \text{N},\quad \dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_{\text{body}}}{\rho_{\text{l}}} = \dfrac{400}{1000} = 0.4

How to read the graphs

The $F_A(V_{\text{sub}})$ graph is a straight line through the origin: the Archimedes force is directly proportional to the volume of the submerged part of the body.
The slope of the line is $\rho_{\text{l}}\, g$ : in a denser liquid (for example, mercury) the line is steeper than in water or kerosene. The slope does not depend on the density of the body.
If the body sinks, it is fully submerged and the point on the graph sits at $V_{\text{sub}} = V$ . If it floats, the point is further left: $V_{\text{sub}} = V \cdot \rho_{\text{body}} / \rho_{\text{l}}$ , and at that point $F_A$ equals the force of gravity.
The fraction of a floating body that is submerged does not depend on its volume, only on the ratio of the densities: pine in water is always $0.4$ submerged, ice $0.9$ .

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Terms

Studied 0 of 5

Formulas

Archimedes force

F_A = \rho_{\text{ж}}\, g\, V_{\text{погр}}

Quantities
Symbol	Name	Unit
$F_A$	buoyant (Archimedes) force	N
$\rho_{\text{ж}}$	density of the liquid (gas)	kg/m³
$g$	acceleration due to gravity	m/s²
$V_{\text{погр}}$	volume of the submerged part of the body	m³

Solved for another quantity:

V_{\text{погр}} = \dfrac{F_A}{\rho_{\text{ж}}\, g}

\rho_{\text{ж}} = \dfrac{F_A}{g\, V_{\text{погр}}}

Conditions for floating through densities

\rho_{\text{тела}} \gtrless \rho_{\text{ж}}

Quantities
Symbol	Name	Unit
$\rho_{\text{тела}}$	density of the body (uniform)	kg/m³
$\rho_{\text{ж}}$	density of the liquid	kg/m³

Solved for another quantity:

\dfrac{V_{\text{погр}}}{V} = \dfrac{\rho_{\text{тела}}}{\rho_{\text{ж}}}\ (\rho_{\text{тела}} < \rho_{\text{ж}})

Conditions for floating through forces

F_{\text{тяж}} \gtrless F_A

Quantities
Symbol	Name	Unit
$F_{\text{тяж}}$	force of gravity on the body, $F_{\text{g}} = m g$	N
$F_A$	Archimedes force when the body is fully immersed	N

Solved for another quantity:

m g = \rho_{\text{ж}}\, g\, V_{\text{погр}}

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A body in a liquid

The green arrow is the Archimedes force (up), the red one is the force of gravity (down). If the body floats, these forces are equal.

Graphs

The line F_A(V_sub) is drawn for the chosen liquid, the marker is the submerged part of the body. Click the graph to change the volume.

F_A(V_{\text{sub}})

\rho_{\text{body}},\ \rho_{\text{l}}

400 < 1,000 kg/m³

V_{\text{sub}}/V

0.4 (80 cm³)

F_A

0.8 N

Floats on the surface

Formulas with your numbers

F_{\text{g}} = \rho_{\text{body}} g V =400\cdot 10\cdot 0.0002 = 0.8\ \text{N}

F_{A,\,\text{full}} = \rho_{\text{l}} g V =1000\cdot 10\cdot 0.0002 = 2\ \text{N}

F_{\text{g}} < F_{A,\,\text{full}}

F_A = F_{\text{g}} = 0.8\ \text{N},\quad \dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_{\text{body}}}{\rho_{\text{l}}} = \dfrac{400}{1000} = 0.4

How to read the graphs

The

F_A(V_{\text{sub}})

graph is a straight line through the origin: the Archimedes force is directly proportional to the volume of the submerged part of the body.

The slope of the line is

\rho_{\text{l}}\, g

: in a denser liquid (for example, mercury) the line is steeper than in water or kerosene. The slope does not depend on the density of the body.

If the body sinks, it is fully submerged and the point on the graph sits at

V_{\text{sub}} = V

. If it floats, the point is further left:

V_{\text{sub}} = V \cdot \rho_{\text{body}} / \rho_{\text{l}}

, and at that point

F_A

equals the force of gravity.

The fraction of a floating body that is submerged does not depend on its volume, only on the ratio of the densities: pine in water is always

0.4

submerged, ice

0.9