Playground
A block submerged in water with weight (down) and buoyant force (up) arrows. Adjust density and volume to see if it sinks or floats.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Buoyant Forceoutput Upward force on submerged object | N | M·L·T⁻² | 0 – 100000 | |
| Fluid Density Density of the surrounding fluid | kg/m³ | M·L⁻³ | 1 – 13600 | |
| Displaced Volume Volume of fluid displaced by the object | m³ | L³ | 0.0001 – 10 | |
| Gravity Gravitational acceleration | m/s² | L·T⁻² | 9.8 – 9.8 |
Deep dive
Derivation
Consider a submerged object of volume V. The pressure on its bottom (deeper) face is greater than on its top face by ρg·h, where h is the object's height. The net upward force from this pressure difference, integrated over the object, equals ρgV — the weight of displaced fluid. This holds for any shape by the divergence theorem.
Experimental verification
Originally tested by Archimedes himself with the gold crown. Modern verification: precision densitometry, hydrometers, and ship displacement measurements all confirm F_b = ρVg to high precision.
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
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 (~10% above water, 90% below)
Worked examples
Half-submerged wooden block
Given:
- ρ:
- 1000
- V_total:
- 0.001
- V_displaced:
- 0.0005
- g:
- 9.8
Find: F_b
Solution
F_b = ρ × V_displaced × g = 1000 × 0.0005 × 9.8 = 4.9 N
Iceberg fraction submerged
Given:
- ρ_ice:
- 917
- ρ_sea:
- 1025
Find: Submerged fraction
Solution
Fraction = ρ_ice / ρ_sea = 917 / 1025 ≈ 0.895 (about 89.5% below water)
Scenarios
What if…
- scenario:
- What if you took a steel ship to a freshwater river?
- answer:
- It would sit slightly lower in the water. Freshwater is less dense than seawater (1000 vs 1025 kg/m³), so a larger volume must be displaced to support the same weight.
- scenario:
- What if you weighed an object in air vs in water?
- answer:
- It feels lighter in water by exactly ρ_water × V × g — the apparent weight loss equals the buoyant force. This was Archimedes' actual technique with the crown.
- scenario:
- What if there were no gravity?
- answer:
- F_b = ρVg = 0. No buoyancy in zero-g — astronauts in the ISS experience this, and fluids of different densities won't separate without sedimentation.
Limiting cases
- condition:
- V → 0
- result:
- F_b → 0
- explanation:
- No displaced fluid means no buoyant force.
- condition:
- ρ_object < ρ_fluid
- result:
- Object floats
- explanation:
- When the object is less dense than the fluid, buoyancy exceeds weight.
- condition:
- ρ_object = ρ_fluid
- result:
- Neutral buoyancy
- explanation:
- Object hovers — used by submarines and fish swim bladders.
Context
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.
Hook
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.
Dimensions: [ρVg] = (M·L⁻³)(L³)(L·T⁻²) = M·L·T⁻² = [F] ✓
Validity: Static fluid, object fully or partially submerged. Assumes uniform fluid density and gravitational field.