ciencia

Buoyant Force Calculator — Archimedes' Principle

Calculator Free · Private
Reviewed by: (política editorial ) · Last reviewed:
Was this calculator helpful?

The Buoyant Force Calculator computes the upward force exerted on any object partially or fully submerged in a fluid, using Archimedes' principle (c. 250 BCE): F_b = ρ × V × g, where ρ is the fluid density (kg/m³), V is the submerged volume (m³), and g is gravitational acceleration (9.81 m/s² on Earth). This force determines whether an object sinks, floats, or remains neutrally buoyant — making it essential in naval engineering, hydrodynamics, atmospheric science, and everyday physics problems.

Last reviewed: June 3, 2026 Verified by Source: NIST – Standard Acceleration of Gravity (g = 9.80665 m/s²), NIST Chemistry WebBook – Fluid Properties (Water, Air, etc.), Wikipedia – Archimedes' Principle, NOAA – Ocean Water Density and Salinity 100% private

Buoyant force (Archimedes' principle): **F_b = ρ × V × g**, where ρ is the fluid density (kg/m³), V is the submerged volume (m³), and g = 9.81 m/s². Example: a 0.1 m³ object fully submerged in freshwater (1000 kg/m³) → F_b = 1000 × 0.1 × 9.81 = **981 N** upward. Seawater (1025 kg/m³) gives 1005.5 N for the same volume.

When to use this calculator

  • Determining whether a ship hull design displaces enough water volume to support its total weight without sinking.
  • Calculating the lifting capacity of a submarine's ballast tanks when transitioning between dive depths.
  • Estimating the buoyant force on a weather balloon filled with helium rising through air at a given altitude and air density.
  • Figuring out how much of an iceberg volume remains above the water surface given the density of ice (917 kg/m³) vs. seawater (1025 kg/m³).
  • Designing concrete blocks or anchor systems that must overcome buoyancy in underwater construction projects.
  • Checking whether a foam-filled life jacket provides sufficient upward force to keep an average adult (750 N weight) above water.

Worked Example

  1. Fluid: freshwater (ρ = 1000 kg/m³)
  2. Submerged volume V = 0.1 m³
  3. g = 9.81 m/s² (standard Earth gravity)
  4. F_b = 1000 × 0.1 × 9.81 = 981 N
Result: 981 N upward buoyant force (≈ weight of 100 kg on Earth)

How it works

3 min read

Formula: F_b = ρ × V × g

Archimedes' principle states that any object immersed 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:
  F_b            = Buoyant Force (Newtons, N)
  ρ_fluid        = Density of the fluid (kg/m³)
  V_submerged    = Volume of the object that is submerged (m³)
  g              = Gravitational acceleration (m/s²)
                   Standard: 9.80665 m/s² (NIST defined)
                   Common approximation: 9.81 m/s²

Buoyancy condition:
  If F_b > W_object  → object FLOATS (net upward force)
  If F_b = W_object  → object is NEUTRALLY BUOYANT
  If F_b < W_object  → object SINKS (net downward force)

Float fraction (for floating objects):
  Fraction submerged = ρ_object / ρ_fluid

---

Reference Table — Buoyant Force by Fluid (per 1 m³, g = 9.81 m/s²)

FluidDensity (kg/m³)F_b per 1 m³ (N)F_b per 1 m³ (lbf)
Air (sea level, 15 °C)1.22512.02.70
Air (10 km altitude)0.4144.060.91
Gasoline (15 °C)~7207,0631,588
Ethanol (20 °C)7897,7401,740
Diesel fuel~8508,3391,875
Freshwater (20 °C)998.29,7922,201
Freshwater (4 °C, max dens.)10009,8102,205
Seawater (avg, 3.5% salinity)102510,0552,260
Seawater (Dead Sea ~33%)~124012,1642,734
Whole blood (37 °C)~106010,3992,338
Glycerin (20 °C)126112,3702,781
Mercury (20 °C)13,534132,76929,838

Density values from NIST Chemistry WebBook and standard engineering references.

---

Quick-Reference Table — Common Submerged Volumes in Freshwater (ρ = 1000 kg/m³)

Submerged Volume (m³)Buoyant Force (N)Equivalent Lifted Mass (kg)
0.001 (1 litre)9.811.0
0.01 (10 litres)98.110.0
0.05 (50 litres)490.550.0
0.1 (100 litres)981.0100.0
0.5 (500 litres)4,905500.0
1.0 (1,000 litres)9,8101,000.0
5.049,0505,000.0
10.098,10010,000.0

---

Typical Use Cases with Numbers

Example 1 — Steel Block Partially Submerged in Freshwater


A solid steel block (ρ_steel ≈ 7850 kg/m³) with total volume 0.02 m³ is half-submerged:
  • V_submerged = 0.01 m³

  • F_b = 1000 × 0.01 × 9.81 = 98.1 N

  • Weight of block = 7850 × 0.02 × 9.81 = 1540.1 N

  • Since F_b (98.1 N) ≪ W (1540.1 N), the block sinks rapidly.
  • Example 2 — Helium Weather Balloon in Air


    A helium balloon has a submerged (in air) volume of 1.5 m³. Air density at sea level ≈ 1.225 kg/m³:
  • F_b = 1.225 × 1.5 × 9.81 = 18.03 N

  • Mass of helium inside = 0.1786 kg/m³ × 1.5 m³ = 0.268 kg → Weight ≈ 2.63 N

  • Net lift available = 18.03 − 2.63 = ~15.4 N (before accounting for envelope mass)
  • Example 3 — Iceberg Fraction Above Water


    Ice density ≈ 917 kg/m³; seawater density ≈ 1025 kg/m³.
  • Fraction submerged = 917 / 1025 = 0.895 → ~89.5% of the iceberg is below sea level.

  • Only ~10.5% is visible above the surface — the origin of the "tip of the iceberg" expression.
  • ---

    Common Mistakes

    1. Using object volume instead of submerged volume. If an object is only 60% submerged, V_submerged = 0.6 × V_total. Using the full volume overestimates F_b significantly.

    2. Using the wrong fluid density. Freshwater (1000 kg/m³) and seawater (1025 kg/m³) differ by 2.5% — enough to change whether a marginally buoyant object floats or sinks. Always specify the fluid.

    3. Treating g as exactly 9.8 or 10 m/s². NIST defines standard gravity as 9.80665 m/s². Using 10 m/s² introduces a ~2% error — problematic in precision engineering.

    4. Ignoring fluid temperature effects on density. Water at 4 °C is 1000 kg/m³, but at 100 °C it drops to 958.4 kg/m³ — a 4.2% reduction that measurably affects buoyancy in hot-water systems.

    5. Confusing buoyant force with net force. F_b is an upward force only. Net force = F_b − W_object. An object "feeling lighter" in water still has the same gravitational weight; only the apparent weight changes.

    6. Neglecting fluid compressibility at depth. For deep-sea applications, seawater density increases slightly with depth (up to ~1050 kg/m³ at 10,000 m), affecting submarine buoyancy calculations.

    Frequently asked questions

    What is Archimedes' principle and what is the buoyancy formula?

    Archimedes' principle states that any object immersed in a fluid is pushed upward by a force equal to the weight of the fluid it displaces. The formula is F_b = ρ_fluid × V_submerged × g. For 0.5 m³ submerged in freshwater: F_b = 1000 × 0.5 × 9.81 = 4,905 N — that upward push equals the weight of 500 kg of water, regardless of the object's own material.

    What value of g should I use for buoyancy calculations?

    NIST defines standard gravitational acceleration as 9.80665 m/s² (commonly rounded to 9.81 m/s²). On the Moon, g ≈ 1.62 m/s², which drastically reduces buoyant force even in the same fluid. For Earth-based engineering calculations, use 9.80665 m/s² for precision work, or 9.81 m/s² for general physics — never round to 10 m/s² in safety-critical design.

    What is the density of freshwater vs. seawater, and does it matter for buoyancy?

    Freshwater at 20 °C has a density of approximately 998.2 kg/m³, while average ocean seawater (3.5% salinity) is about 1025 kg/m³ — roughly 2.7% denser. For a 1 m³ submerged volume, that difference equals 263 N of extra buoyant force. In the Dead Sea (~33% salinity, ~1240 kg/m³), buoyancy is so strong that it is nearly impossible for a person to submerge voluntarily.

    How do I calculate what fraction of a floating object is above the waterline?

    For a freely floating object, the submerged fraction equals the ratio of object density to fluid density: fraction_submerged = ρ_object / ρ_fluid. Ice (917 kg/m³) floating in seawater (1025 kg/m³) submerges 917/1025 ≈ 89.5%, leaving 10.5% above the surface. A piece of oak (~700 kg/m³) floating in freshwater submerges 70%, with 30% above water.

    Can buoyancy occur in gases, not just liquids?

    Yes — buoyancy works in any fluid, including gases. A helium-filled balloon in air experiences an upward buoyant force because helium (0.1786 kg/m³) is far less dense than air (1.225 kg/m³ at sea level). The net lift per cubic meter of helium is (1.225 − 0.1786) × 9.81 ≈ 10.27 N/m³. Hot-air balloons work by reducing the air density inside the envelope below ambient levels.

    What does 'neutrally buoyant' mean and how is it achieved?

    An object is neutrally buoyant when its average density exactly equals the surrounding fluid density, so F_b = W_object and the net force is zero. Submarines achieve this by flooding or pumping ballast tanks to adjust their average density. Scuba divers aim for neutral buoyancy (average human body density ≈ 985 kg/m³, very close to freshwater) by adjusting their buoyancy compensator (BCD) and weight belts.

    How does water depth affect buoyancy?

    For most practical purposes with liquids, buoyancy depends only on the volume of fluid displaced and the fluid density at that location — not on the depth per se. However, seawater density increases slightly with depth due to compression (from ~1025 kg/m³ at the surface to ~1050 kg/m³ at 10,000 m). This means submarines experience marginally greater buoyancy at extreme depths, a factor that must be accounted for in deep-sea vehicle design.

    What is apparent weight and how does buoyancy relate to it?

    Apparent weight is the measured weight of a submerged object: W_apparent = W_actual − F_b. A steel block weighing 500 N in air, submerged in freshwater with V = 0.004 m³, experiences F_b = 1000 × 0.004 × 9.81 = 39.24 N, giving an apparent weight of 500 − 39.24 = 460.76 N. This principle underlies hydrostatic weighing, a gold-standard body composition assessment technique.

    Why does a steel ship float if steel is denser than water?

    A ship floats because its average density — total mass divided by total enclosed volume (including air spaces) — is less than water. A typical ocean freighter might have a hull steel mass of 20,000 tonnes, but its total enclosed volume is large enough that the average density falls below 1025 kg/m³. The buoyant force equals the weight of seawater displaced by the entire hull volume below the waterline.

    How does buoyancy differ between an object floating versus fully submerged?

    When an object floats, only part of its volume is submerged, and buoyancy exactly equals its weight (equilibrium). The submerged fraction = ρ_object / ρ_fluid. When fully submerged, the buoyant force is fixed at ρ_fluid × V_total × g regardless of depth. If F_b > weight, the object rises; if F_b < weight, it sinks. The calculator uses the submerged volume you enter — for floating objects, enter only the submerged fraction of the total volume.

    Sources and references