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×g
Where:
- Fb = buoyant force
- ρfluid = density of the fluid
- Vdisplaced = volume of fluid displaced
- g = gravitational acceleration
Floating and Sinking
Condition for Floating
Object floats when:
Fb≥Wobject
Or equivalently:
ρobject≤ρfluid
Condition for Sinking
Object sinks when:
ρobject>ρfluid
Neutral Buoyancy
Object hovers when:
ρobject=ρfluid
Floating Objects
For a floating object in equilibrium:
Fb=W
ρfluid×Vsubmerged×g=ρobject×Vtotal×g
Fraction Submerged
VtotalVsubmerged=ρfluidρobject
Fraction Above Surface
VtotalVabove=1−ρfluidρobject
Apparent Weight
The apparent weight of an object in a fluid:
Wapparent=W−Fb=ρobject×V×g−ρfluid×V×g
Wapparent=V×g×(ρobject−ρfluid)
In Terms of True Weight
Wapparent=W(1−ρobjectρfluid)
Density Comparison
| Material | Density (kg/m³) |
|---|
| Air | 1.29 |
| Ice | 917 |
| Water | 1000 |
| Seawater | 1025 |
| Iron | 7870 |
| Mercury | 13,600 |
| Wood (oak) | 750 |
| Cork | 240 |
| Aluminum | 2700 |
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 N
Example 2: Floating Ice
An ice cube (ρ=917 kg/m³) floats in water. What fraction is submerged?
Fraction submerged=ρwaterρice=1000917=0.917=91.7%
About 8.3% is above water (tip of the iceberg!)
Example 3: Floating Wood
A log (ρ=600 kg/m³, volume = 0.1 m³) floats in water. Find submerged volume.
Vsub=V×ρwaterρwood=0.1×1000600=0.06 m3
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 kg
If 500,000 kg cargo is added:
ΔV=ρΔm=1025500000=488 m3
Δh=AΔV=2000488=0.244 m (sinks 24.4 cm deeper)
Example 5: Apparent Weight
A 5 kg iron block (ρ=7870 kg/m³) is submerged in water.
V=ρm=78705=6.35×10−4 m3
Fb=ρwater×V×g=1000×6.35×10−4×9.8=6.22 N
Wapparent=mg−Fb=49−6.22=42.8 N
Example 6: Hot Air Balloon
A balloon (volume = 500 m³) contains hot air at 100°C (ρ=0.95 kg/m³). Outside air is at 20°C (ρ=1.20 kg/m³). Find lift.
Fb=ρoutside×V×g=1.20×500×9.8=5880 N
Whot air=ρhot×V×g=0.95×500×9.8=4655 N
Lift=Fb−Whot air=1225 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