Archimedes' Principle

Equivalent forms

F_b = \rho_{\text{fluid}} V_{\text{displaced}} g

An ancient insight that still floats every modern ship.

Symbol	Name	SI unit	Dimension	Range
$F_b$	Buoyant Forceoutput Upward force on submerged object	N	M·L·T⁻²	0 – 100000
$ρ$	Fluid Density Density of the surrounding fluid	kg/m³	M·L⁻³	1 – 13600
$V$	Displaced Volume Volume of fluid displaced by the object	m³	L³	0.0001 – 10
$g$	Gravity Gravitational acceleration	m/s²	L·T⁻²	9.8 – 9.8

Unit systems

Symbol	SI	CGS	Imperial
$F_b$	N	dyn	lbf
$ρ$	kg/m³	g/cm³	lb/ft³
$V$	m³	cm³	ft³
$g$	m/s²	cm/s²	ft/s²

Where it holds

Static fluid, object fully or partially submerged. Assumes uniform fluid density and gravitational field.

Dimensional analysis

$[\rho Vg] = (M\cdot L^{-3})(L^{3})(L\cdot T^{-2}) = M\cdot L\cdot T^{-2} = [F] \checkmark$

Discovery

Archimedes · -250

Discovered while bathing — the famous 'Eureka!' moment, used to test whether King Hiero II's crown was pure gold by comparing displaced water volumes.

Try this

Why does a steel ship float but a steel coin sink?

A wooden block of volume 0.001 m³ floats half-submerged in water. Find the buoyant force on it.

Research status: stable

Real-world applications

Ship and submarine design
Hot air balloons (displacing colder, denser air)
Hydrometers measuring liquid density
Fish swim bladders for depth control
Iceberg flotation $(\sim 10$ % above water, 90% below)

Common misconceptions

Buoyant force depends on the FLUID's density, not the object's
It depends on the displaced VOLUME, not the object's mass or shape directly
Objects float when their AVERAGE density (including hollow parts) is less than the fluid's

Experimental verification

Originally tested by Archimedes himself with the gold crown. Modern verification: precision densitometry, hydrometers, and ship displacement measurements all confirm

F_b = \rho Vg

to high precision.

Derivation

Consider a submerged object of volume V.

The pressure on its bottom (deeper) face is greater than on its top face

by \rho g\cdot h

, where h is the object's height.

The net upward force from this pressure difference, integrated over the object, equals

\rho gV

— the weight of displaced fluid.

This holds for any shape by the divergence theorem.

Limiting cases

V \to 0

⟶

F_b \to 0

No displaced fluid means no buoyant force.

\rho

_object <

\rho

_fluid⟶ Object floatsWhen the object is less dense than the fluid, buoyancy exceeds weight.

\rho

_object

= \rho

_fluid⟶ Neutral buoyancyObject hovers — used by submarines and fish swim bladders.

What if…

What if you took a steel ship to a freshwater river?

It would sit slightly lower in the water. Freshwater is less dense than seawater $(1000\,\mathrm{vs} 1025\,\mathrm{kg/m}^{3})$ , so a larger volume must be displaced to support the same weight.

What if you weighed an object in air vs in water?

It feels lighter in water by exactly $\rho$ _water $\times V \times g$ — the apparent weight loss equals the buoyant force. This was Archimedes' actual technique with the crown.

What if there were no gravity?

$F_b = \rho Vg = 0$ . No buoyancy in zero-g — astronauts in the ISS experience this, and fluids of different densities won't separate without sedimentation.

1

Half-submerged wooden block

Given ·

ρ:: 1000
V total:: 0.001
V displaced:: 0.0005
g:: 9.8

Find · F_b

Steps

Only the submerged half displaces water: V_displaced $= 0.0005\,\mathrm{m}^{3}$
$F_b = \rho$ _water $\times V$ _displaced $\times g$
$F_b = 1000 \times 0.0005 \times 9.8 = 4.9\,\mathrm{N}$ (equals the block's weight)

Result ·

F_b = \rho \times V

_displaced

\times g = 1000 \times 0.0005 \times 9.8 = 4.9\,\mathrm{N}

2

Iceberg fraction submerged

Given ·

ρ ice:: 917
ρ sea:: 1025

Find · Submerged fraction

Steps

For floating: weight = buoyant force
$\rho _ice \cdot V$ _total $\cdot g = \rho _sea \cdot V_sub \cdot g$
V_sub / V_total $= \rho _ice / \rho _sea \approx 0.895$

Result · Fraction

= \rho _ice / \rho _sea = 917 / 1025 \approx 0.895

(about 89.5% below water)

Hydrostatic Pressure→Bernoulli's Equation→