Preview

Buoyancy

Explore Archimedes' Principle with different fluids

A steel nail drops to the bottom of a pond in seconds, yet a hundred-thousand-ton steel ship crosses an ocean. Same metal, same water — so 'heavy things sink' cannot be the rule, and it isn't. Archimedes spotted the real bookkeeping in his bath: anything placed in a fluid pushes fluid aside, and the fluid pushes back with a force equal to the weight of whatever was displaced. That push never asks what the object is made of — only how much room it takes up and how dense the surrounding fluid is. Whether the push wins its contest against gravity is settled by a single ratio: object density over fluid density. Win, and you float with exactly that fraction of yourself underwater — 600 kg/m³ wood in 1000 kg/m³ water rides 60% submerged, no more, no less. Lose, and you settle on the bottom, pushed upward the whole way down. Submarines steer this ratio with ballast tanks; ships cheat it with hulls full of air. In this lab you will drop wood, steel, and a neutrally buoyant sub into the tank, watch the green buoyancy arrow duel the red weight arrow, and catch the exact moment a force balance goes still.

What you'll be able to do

  • State Archimedes' principle and compute the buoyant force F_b = ρ_fluid·V_displaced·g for fully and partially submerged objects, reading the force balance F_net = F_b − W off the sim's dual force arrows and data panel
  • Predict floating, sinking, and neutral-buoyancy outcomes from the density ratio ρ_obj/ρ_fluid alone, and compute a floating object's submerged fraction as exactly that ratio
  • Explain why the buoyant force on a fully submerged object is independent of the object's own weight, material, and depth — it is set only by the fluid density and the displaced volume

Formulas

Fb=ρfluidVdisplacedgF_b = \rho_{fluid} \cdot V_{displaced} \cdot g
Buoyant force (Archimedes)
Float if ρobject<ρfluid\text{Float if } \rho_{object} < \rho_{fluid}
Floating condition
Apparent weight=WFb\text{Apparent weight} = W - F_b
Apparent weight in fluid

Make a prediction

Hit the Sinking Steel preset: a 0.5 L steel sphere (ρ = 7800 kg/m³) sinks through water (ρ = 1000 kg/m³) and comes to rest on the tank floor, fully submerged. Once it is sitting on the bottom, what does the Buoyant Force readout show?

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Your prediction

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Answer: About 4.905 N — fully submerged, the sphere displaces its entire 0.5 L of water no matter what it is made of: F_b = 1000 × 0.0005 × 9.81 = 4.905 N, and sinking never switches that push off

The buoyant force is set entirely by the fluid and the displaced volume: F_b = ρ_fluid × V_displaced × g. Resting on the floor, the sphere is fully submerged, so V_displaced = 0.5 L = 0.0005 m³ and F_b = 1000 × 0.0005 × 9.81 = 4.905 N — exactly what the panel's Buoyant Force row shows, pinned at that value through the fall, the landing, and the rest state. Its weight is W = ρ_obj × V × g = 7800 × 0.0005 × 9.81 = 38.259 N (the Weight row), so Net Force reads −33.354 N in red: the sphere stays down because 4.905 N loses to 38.259 N, not because the 4.905 N went away. Option C inverts the logic of floating: a floating object's weight ends up equal to the buoyant force because the object sinks only until the displaced water's weight matches its own — Frac. Submerged = ρ_obj/ρ_fluid, 0.600 for the Floating Wood preset — the equilibrium adjusts the object; the formula never contains the object's weight. Option B denies a force the panel never stops displaying: a sunken object still displaces its full volume, and the fluid still pays for every cubic centimetre of it. Loverude, Kautz & Heron (Am. J. Phys. 71(11), 2003) document both misreadings in introductory students — the buoyant force tracks the displaced fluid, never the object.

Quiz (0/3)

A 2L object with density 800 kg/m³ in water — does it float? What fraction is submerged?

What is the buoyant force on a fully submerged 0.5m³ object in water?

A steel ship has mass 10,000 kg. What minimum hull volume is needed to float in seawater (1025 kg/m³)?

You can now

  • State Archimedes' principle and compute the buoyant force F_b = ρ_fluid·V_displaced·g for fully and partially submerged objects, reading the force balance F_net = F_b − W off the sim's dual force arrows and data panel
  • Predict floating, sinking, and neutral-buoyancy outcomes from the density ratio ρ_obj/ρ_fluid alone, and compute a floating object's submerged fraction as exactly that ratio
  • Explain why the buoyant force on a fully submerged object is independent of the object's own weight, material, and depth — it is set only by the fluid density and the displaced volume

Step-by-step

  1. Adjust fluid and object densities.
  2. Objects denser than the fluid sink; less dense objects float.
  3. The force diagram shows weight vs. buoyant force.
  4. A floating object settles at the submerged fraction that balances the books — read it live on the panel as you tune the densities.

Key formulas

  • Fb=ρfluidVdisplacedgF_b = \rho_{fluid} \cdot V_{displaced} \cdot gBuoyant force (Archimedes)
  • Float if ρobject<ρfluid\text{Float if } \rho_{object} < \rho_{fluid}Floating condition
  • Apparent weight=WFb\text{Apparent weight} = W - F_bApparent weight in fluid

Frequently asked questions

A 2L object with density 800 kg/m³ in water — does it float? What fraction is submerged?
800 < 1000, so it floats. Fraction = ρ_obj/ρ_fluid = 0.8 = 80%.
What is the buoyant force on a fully submerged 0.5m³ object in water?
F_b = 1000 × 0.5 × 9.8 = 4900 N.
A steel ship has mass 10,000 kg. What minimum hull volume is needed to float in seawater (1025 kg/m³)?
V_displaced = m/ρ_fluid = 10000/1025.