Fluid MechanicsTopic #23 of 35

Buoyancy

Archimedes' principle, buoyant force, and floating/sinking objects.

Overview

Buoyancy is the upward force exerted by a fluid on an immersed or floating object. This force allows ships to float and hot air balloons to rise.

Archimedes' Principle

An object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.

Fb=ρfluid×Vdisplaced×gF_b = \rho_{\text{fluid}} \times V_{\text{displaced}} \times g

Where:

  • FbF_b = buoyant force
  • ρfluid\rho_{\text{fluid}} = density of the fluid
  • VdisplacedV_{\text{displaced}} = volume of fluid displaced
  • gg = gravitational acceleration

Floating and Sinking

Condition for Floating

Object floats when:

FbWobjectF_b \geq W_{\text{object}}

Or equivalently:

ρobjectρfluid\rho_{\text{object}} \leq \rho_{\text{fluid}}

Condition for Sinking

Object sinks when:

ρobject>ρfluid\rho_{\text{object}} > \rho_{\text{fluid}}

Neutral Buoyancy

Object hovers when:

ρobject=ρfluid\rho_{\text{object}} = \rho_{\text{fluid}}

Floating Objects

For a floating object in equilibrium:

Fb=WF_b = W ρfluid×Vsubmerged×g=ρobject×Vtotal×g\rho_{\text{fluid}} \times V_{\text{submerged}} \times g = \rho_{\text{object}} \times V_{\text{total}} \times g

Fraction Submerged

VsubmergedVtotal=ρobjectρfluid\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}

Fraction Above Surface

VaboveVtotal=1ρobjectρfluid\frac{V_{\text{above}}}{V_{\text{total}}} = 1 - \frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}

Apparent Weight

The apparent weight of an object in a fluid:

Wapparent=WFb=ρobject×V×gρfluid×V×gW_{\text{apparent}} = W - F_b = \rho_{\text{object}} \times V \times g - \rho_{\text{fluid}} \times V \times g Wapparent=V×g×(ρobjectρfluid)W_{\text{apparent}} = V \times g \times (\rho_{\text{object}} - \rho_{\text{fluid}})

In Terms of True Weight

Wapparent=W(1ρfluidρobject)W_{\text{apparent}} = W\left(1 - \frac{\rho_{\text{fluid}}}{\rho_{\text{object}}}\right)

Density Comparison

MaterialDensity (kg/m³)
Air1.29
Ice917
Water1000
Seawater1025
Iron7870
Mercury13,600
Wood (oak)750
Cork240
Aluminum2700

Stability of Floating Objects

Center of Buoyancy (B)

Centroid of the displaced fluid volume

Center of Gravity (G)

Where the weight acts

Metacenter (M)

Point where the vertical through the new center of buoyancy intersects the original vertical

Stability Conditions

  • M above G: Stable (righting moment)
  • M below G: Unstable (capsizing moment)
  • M at G: Neutral

Examples

Example 1: Buoyant Force

A 0.5 m³ block is fully submerged in water.

Fb=ρVg=1000×0.5×9.8=4900 NF_b = \rho V g = 1000 \times 0.5 \times 9.8 = 4900 \text{ N}

Example 2: Floating Ice

An ice cube (ρ=917\rho = 917 kg/m³) floats in water. What fraction is submerged?

Fraction submerged=ρiceρwater=9171000=0.917=91.7%\text{Fraction submerged} = \frac{\rho_{\text{ice}}}{\rho_{\text{water}}} = \frac{917}{1000} = 0.917 = 91.7\%

About 8.3% is above water (tip of the iceberg!)

Example 3: Floating Wood

A log (ρ=600\rho = 600 kg/m³, volume = 0.1 m³) floats in water. Find submerged volume.

Vsub=V×ρwoodρwater=0.1×6001000=0.06 m3V_{\text{sub}} = V \times \frac{\rho_{\text{wood}}}{\rho_{\text{water}}} = 0.1 \times \frac{600}{1000} = 0.06 \text{ m}^3

Example 4: Ship Loading

A ship has waterline area of 2000 m² and displaces 20,000 m³ of seawater.

Mass of ship:

m=ρV=1025×20000=20.5×106 kgm = \rho V = 1025 \times 20000 = 20.5 \times 10^6 \text{ kg}

If 500,000 kg cargo is added:

ΔV=Δmρ=5000001025=488 m3\Delta V = \frac{\Delta m}{\rho} = \frac{500000}{1025} = 488 \text{ m}^3 Δh=ΔVA=4882000=0.244 m (sinks 24.4 cm deeper)\Delta h = \frac{\Delta V}{A} = \frac{488}{2000} = 0.244 \text{ m (sinks 24.4 cm deeper)}

Example 5: Apparent Weight

A 5 kg iron block (ρ=7870\rho = 7870 kg/m³) is submerged in water.

V=mρ=57870=6.35×104 m3V = \frac{m}{\rho} = \frac{5}{7870} = 6.35 \times 10^{-4} \text{ m}^3 Fb=ρwater×V×g=1000×6.35×104×9.8=6.22 NF_b = \rho_{\text{water}} \times V \times g = 1000 \times 6.35 \times 10^{-4} \times 9.8 = 6.22 \text{ N} Wapparent=mgFb=496.22=42.8 NW_{\text{apparent}} = mg - F_b = 49 - 6.22 = 42.8 \text{ N}

Example 6: Hot Air Balloon

A balloon (volume = 500 m³) contains hot air at 100°C (ρ=0.95\rho = 0.95 kg/m³). Outside air is at 20°C (ρ=1.20\rho = 1.20 kg/m³). Find lift.

Fb=ρoutside×V×g=1.20×500×9.8=5880 NF_b = \rho_{\text{outside}} \times V \times g = 1.20 \times 500 \times 9.8 = 5880 \text{ N} Whot air=ρhot×V×g=0.95×500×9.8=4655 NW_{\text{hot air}} = \rho_{\text{hot}} \times V \times g = 0.95 \times 500 \times 9.8 = 4655 \text{ N} Lift=FbWhot air=1225 N\text{Lift} = F_b - W_{\text{hot air}} = 1225 \text{ N}

This can lift a payload of 125 kg!

Applications

  • Ships and submarines: Controlled buoyancy
  • Hot air balloons: Less dense gas rises
  • Fish swim bladders: Adjust density to hover
  • Hydrometers: Measure fluid density
  • Life jackets: Increase buoyancy to float