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Physics: Buoyancy (Archimedes) Calculator

Calculate buoyant force based on Archimedes' Principle. Input fluid density and submerged volume to find upward force.

F_b = ρVg

Fluid Density Presets

Fluid Density (ρ)

kg/m³

Submerged Volume (V)

m³

Gravity (g)

m/s²

Buoyant Force (F_b)

9,810

Newtons (N)

Displaces 1,000 kg of fluid

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Buoyancy (Archimedes' Principle)

Archimedes' Principle states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces: F_b = ρ V g.

Variables

ρ (rho): Density of the fluid (kg/m³)
V: Displaced (submerged) volume of the object (m³)
g: Acceleration due to gravity (≈ 9.81 m/s²)
F_b: Buoyant force (Newtons)

Float or Sink?

An object floats if the buoyant force is greater than or equal to its weight ($F_b \ge mg$). This is equivalent to saying the object floats if its average density is less than the fluid's density.

What is Archimedes' Principle — The Physics of Displaced Fluid?

Every object immersed in a fluid experiences an upward buoyant force exactly equal to the weight of the fluid it displaces. This is Archimedes' Principle, and it governs everything from whether a steel ship floats to how a submarine changes depth, how a hydrometer measures liquid density, and why a hot air balloon rises. The key insight is that buoyancy depends on the volume and density of the displaced fluid, not the material or weight of the object itself.

Mathematical Foundation

Archimedes Buoyant Force

F_b = \rho_{fluid} \times V_{sub} \times g

F_b

= Buoyant force (N) — upward force exerted by displaced fluid on submerged object

\rho_{fluid}

= Density of the fluid (kg/m3): fresh water 998.2, seawater 1,025, glycerin 1,261, mercury 13,534 at 15C

V_{sub}

= Volume of object submerged in the fluid (m3); equals total volume V when fully immersed

g

= Gravitational acceleration: 9.807 m/s2 at sea level (9.780 at equator, 9.832 at poles)

Apparent Weight (Submerged Weight)

W_{app} = W_{true} - F_b = (\rho_{obj} - \rho_{fluid}) \times V \times g

W_{app}

= What a scale reads when object is weighed while submerged (N); negative means object pushes upward off the scale

\rho_{obj}

= Density of the submerged object (kg/m3); determines whether object sinks, hovers, or rises relative to fluid

Fraction Submerged at Floating Equilibrium

\frac{V_{sub}}{V_{total}} = \frac{\rho_{obj}}{\rho_{fluid}}

V_{sub}/V_{total}

= Fraction of total object volume below the fluid surface when floating freely at equilibrium

\rho_{obj}/\rho_{fluid}

= Density ratio — human body (approx 985 kg/m3) floats at 985/998 = 98.7% submerged in fresh water

Laws & Principles

Archimedes' Principle (287–212 BC): The buoyant force on an object equals the weight of fluid displaced, regardless of the object's composition. This is why a hollow steel sphere (average density < water) floats while a solid steel ball (density 7,874 kg/m³) sinks. The critical threshold is always the average density of the object compared to the fluid density.
The Three Buoyancy States: (1) Positive: ρ_object < ρ_fluid → object floats (F_b > W_true); (2) Neutral: ρ_object = ρ_fluid → object hovers at any depth (F_b = W_true); (3) Negative: ρ_object > ρ_fluid → object sinks but with reduced apparent weight. Submarines achieve neutral buoyancy by adjusting ballast tank volume to match displaced water weight exactly to total sub weight.
Fluid Density Temperature Dependence: Water reaches maximum density at 4°C (999.97 kg/m³). At 100°C it is 958.4 kg/m³ — a 4.2% density drop that proportionally reduces buoyant force. In marine engineering, this matters for vessels operating between cold Arctic waters (denser) and warm tropical waters (less dense): a cargo ship loaded to its Plimsoll line in cold North Atlantic water may exceed its draft limit when the same load enters warmer Mediterranean or tropical waters. The International Load Line Convention (1966) mandates different Plimsoll marks for different ocean zones precisely because of this effect.

Step-by-Step Example Walkthrough

" A rescue diver weighs 85 kg and has a volume of 0.082 m³ (82 liters). They are submerged in seawater (ρ = 1,025 kg/m³). Find: (1) buoyant force, (2) apparent weight, (3) required weight belt mass to achieve neutral buoyancy. "

1. Buoyant force: F_b = 1,025 × 0.082 × 9.807 = 824.0 N (upward). This equals the weight of 84.03 kg of seawater displaced.
2. True weight of diver: W = 85 × 9.807 = 833.6 N. Diver density = 85 / 0.082 = 1,036.6 kg/m³ (slightly denser than seawater).
3. Apparent weight submerged: W_app = 833.6 - 824.0 = 9.6 N. The diver would sink slowly but exert almost no force on a scale if weighed underwater.
4. Neutral buoyancy without gear: Diver needs F_b = W_true, so displaced volume must equal 85/1,025 = 0.08293 m³. Diver has 0.082 m³ — deficit of 0.92 liters. Inflate BCD by 0.92 L to achieve neutral buoyancy.
5. Weight belt (negative buoyancy): Each kg of lead (ρ_lead = 11,340 kg/m³) adds W = 9.807 N downward but only displaces 1/11,340 m³ × 1,025 × 9.807 = 0.882 N upward. Net force per kg of lead = 9.807 - 0.882 = 8.932 N downward.

Final Result: The diver has an apparent submerged weight of only 9.6 N (about 0.98 kg) — nearly neutrally buoyant with no equipment adjustment. Inflating the BCD by 0.92 liters achieves perfect neutral buoyancy. A 2 kg lead weight belt provides 17.9 N of net negative buoyancy, giving the diver a comfortable controlled descent at 9.6 + 17.9 = 27.5 N apparent negative buoyancy.

Quick Answer: What is buoyant force and how is F_b = ρVg calculated?

Archimedes' Principle states that any object fully or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces: F_b = ρ_fluid × V_submerged × g, where ρ_fluid is the fluid density (kg/m³), V_submerged is the volume of fluid displaced (m³), and g is gravitational acceleration (9.80665 m/s² standard; 9.807 m/s² for most engineering calculations). The force is always vertical and upward — it is a pressure-gradient effect, not a surface phenomenon. Whether an object floats, sinks, or is neutrally buoyant depends entirely on the comparison between F_b and the object's weight (W = m × g = ρ_object × V_total × g): if F_b > W, the object rises; if F_b < W, it sinks; if F_b = W, it is neutrally buoyant. For a solid steel cube (density 7,850 kg/m³) 0.10 m per side submerged in seawater (1,025 kg/m³): F_b = 1,025 × 0.001 × 9.807 = 10.05 N upward; weight = 7,850 × 0.001 × 9.807 = 76.98 N downward — net sinking force = 66.93 N. The steel sinks despite the substantial buoyant force.

Common Fluid Densities for Buoyancy Calculations

Fluid	ρ (kg/m³)	ρ (lb/ft³)	Temperature / Notes
Fresh water	998.2	62.32	20°C; density peaks at 999.97 at 3.98°C
Seawater (ocean average)	1,025	63.99	Salinity ~35 ppt; varies 1,020–1,030 by location/depth
Seawater (Dead Sea)	1,240	77.44	~340 ppt salinity; humans float effortlessly
Glycerin (glycerol)	1,261	78.74	25°C; used in viscosity standards and medical applications
Mercury (liquid)	13,534	845	20°C; extremely dense — steel (7,850 kg/m³) floats in mercury
Gasoline	720–740	45–46	Floats on water; important for spill containment
Ethanol (100%)	789	49.3	20°C; beer/wine ≈ 990–1,010 kg/m³ (near water density)
Air (sea level)	1.204	0.0752	20°C, 1 atm; buoyancy in air is critical for balloon and airship lift calculations
All densities at standard atmospheric pressure unless noted. Ocean density varies with temperature (cold water denser), salinity, and depth (pressure). Use 1,025 kg/m³ for typical ocean calculations; 1,000 kg/m³ for freshwater engineering. Air buoyancy is usually negligible for liquids but significant for precision mass weighing (>0.1% correction for dense objects).

Pro Tips & Critical Buoyancy Calculation Mistakes

Do This

✓Use V_submerged (displaced volume), not V_total, when the object is only partially submerged. F_b = ρ_fluid × V_submerged × g — the buoyant force depends only on the volume of fluid displaced, not the total volume of the object. For a floating log: if 70% of its volume is below the waterline, use V_submerged = 0.70 × V_total. At the equilibrium float position, F_b = W → ρ_fluid × V_submerged × g = ρ_object × V_total × g → V_submerged/V_total = ρ_object/ρ_fluid. A human body (~985 kg/m³ average) floats with ~96% submerged in freshwater (998 kg/m³) and ~4% above surface — which explains why only the very top of the head floats above the waterline. In salt water (1,025 kg/m³), the submerged fraction reduces to 96.1% → slightly more body above the surface.
✓Calculate apparent weight (weight in fluid) to understand the load on lifting equipment or scales. W_apparent = W_true − F_b = (ρ_object − ρ_fluid) × V × g. This is critical for underwater construction: an 18,000 kg concrete block (ρ = 2,300 kg/m³) submerged in seawater (1,025 kg/m³): V = 18,000/2,300 = 7.826 m³; F_b = 1,025 × 7.826 × 9.807 = 78,670 N = 8,019 kg-force; W_true = 18,000 kg-force; W_apparent = 18,000 − 8,019 = 9,981 kg-force — the crane only needs to lift ~9,981 kg instead of 18,000 kg. Failure to account for buoyancy in underwater crane calculations can overload lifting equipment.

Avoid This

✗Don't confuse buoyancy stability with buoyancy magnitude — a heavy ship floats but can capsize if the center of buoyancy and center of gravity are misaligned. Buoyancy force acting upward through the center of buoyancy (B) must pass above the center of gravity (G) when the vessel tilts for the vessel to be stable. The metacentric height (GM) determines righting moment: GM = BM − BG (where BM = I/V, I = second moment of waterplane area, V = displaced volume). A positive GM (>0) means the vessel is stable and will self-right when tilted; GM = 0 is neutral; GM < 0 means the vessel is unstable and will capsize under a small disturbance. Most cargo ships require GM ≥ 0.15–0.5 m; cruise ships require GM ≥ 0.3 m. The buoyancy force from Archimedes' principle tells you if a ship floats — metacentric analysis tells you if the ship will stay upright.
✗Don't neglect the buoyancy of gas pockets in underground tanks, buried pipelines, or basement structures during flooding — buoyancy forces can be catastrophically large. An empty buried underground storage tank (UST) with volume 40,000 L (40 m³) submerged in a saturated soil water table: F_b = 1,000 × 40 × 9.807 = 392,280 N = 40,000 kg upward force. If the tank weighs only 4,000 kg (steel shell), the net upward force is 36,000 kg — sufficient to lift the tank out of the ground, rupturing fuel lines, electrical conduit, and structural connections. The 2010 Nashville floods damaged dozens of USTs this way. Anti-buoyancy anchor straps or soil overburden calculations are mandatory for all buried tanks below the water table.

Frequently Asked Questions

How do submarines control buoyancy using ballast tanks?

Submarines use ballast tanks to actively change V_submerged (and therefore their average density) without changing their external dimensions. When surfaced: ballast tanks are filled with air → average density < seawater → positive buoyancy, floats. To dive: compressed air is vented, seawater floods ballast tanks → average density ≈ seawater → neutral buoyancy at depth. To surface: high-pressure air (from on-board storage at 2,000–3,000 psi) is blown into ballast tanks → expels seawater → average density < seawater → positive buoyancy, rises. Fine depth control is achieved with trim tanks (small fore/aft tanks for precise pitch adjustment) and diving planes (hydroplane surfaces generating hydrodynamic lift/downforce). At depth, water compressibility is negligible but the submarine's hull compresses slightly under pressure → average density increases slightly → trim tanks compensate. Modern nuclear submarines maintain neutral buoyancy to within approximately ±50 kg at operating depth.

How does Archimedes' Principle apply to hot air balloons?

A hot air balloon floats because the heated air inside the envelope is less dense than the surrounding cool air — exactly Archimedes' Principle applied to gases. F_b = ρ_{cool air} × V_envelope × g. The balloon lifts off when F_b > total weight (envelope + basket + payload + hot air inside). Example: balloon envelope V = 2,800 m³; cool air density at 20°C sea level = 1.204 kg/m³; F_b = 1.204 × 2,800 × 9.807 = 33,054 N (3,370 kg-force upward). Hot air at 100°C inside envelope: ρ_hot = 1.204 × (273+20)/(273+100) = 0.946 kg/m³; weight of hot air = 0.946 × 2,800 × 9.807 = 25,964 N. Net lift available = 33,054 − 25,964 = 7,090 N ≈ 723 kg for envelope, basket, and payload. Heating the air to 120°C (ρ = 0.901 kg/m³) increases net lift by ~130 kg. Altitude reduces both ρ_{cool air} and thus F_b — there is a ceiling altitude where lift equals total weight even at maximum burner temperature.

How does a hydrometer use Archimedes' Principle to measure liquid density?

A hydrometer is a weighted glass float that sinks to a depth where F_b = W_hydrometer. Since W is fixed (the hydrometer is a calibrated instrument): ρ_fluid × V_submerged × g = constant → V_submerged ∝ 1/ρ_fluid. In a denser fluid, less volume is submerged → the hydrometer floats higher. In a less dense fluid, more volume is submerged → it sinks lower. The graduated stem reads fluid density (or specific gravity) directly at the waterline. Applications: auto mechanics use a battery hydrometer (ρ = 1.265 g/mL at full charge → 1.12 g/mL at fully discharged, for lead-acid batteries); winemakers measure must specific gravity (1.080–1.120 pre-fermentation to calculate potential alcohol); brewers use specific gravity to calculate alcohol by volume: ABV ≈ (OG − FG) × 131.25, where OG = original gravity and FG = final gravity in specific gravity units (both measured with a hydrometer). Each ΔSG of 0.001 corresponds to approximately 1.1 kg/m³ in density.

Why does a steel ship float even though steel is ~8× denser than water?

A steel ship floats because its average density (steel hull + hollow interior filled with air, cargo, crew, machinery) is less than water. The key is not the density of the material but the density of the entire displacement volume. For a simplified example: a steel box-ship 100 m × 20 m × 10 m (L×W×D) with 2 cm steel walls: External volume = 20,000 m³. Steel volume = outer shell area × 0.02 m thickness ≈ 480 m³ (rough estimate). Steel mass = 480 m³ × 7,850 kg/m³ = 3,768,000 kg. Average density = 3,768,000 ÷ 20,000 m³ = 188 kg/m³ — far below water's 1,000 kg/m³. The ship will float with only 188/1,000 = 18.8% of its volume below the waterline when unloaded. Loading cargo increases average density toward 1,000 kg/m³ — the ship's draft (submerged depth) increases. The maximum loaded draft (“Plimsoll line”) is marked on every ocean-going vessel and corresponds to the maximum safe displacement volume before freeboard (height above waterline) becomes insufficient to prevent waves from swamping the deck.