Buoyancy, density, and floating
Why does floating depend mainly on a density ratio rather than weight alone?
Change liquid density, object density, and volume; compare weight with buoyant force and observe displacement when the object floats, is neutrally buoyant, or sinks.
Learning goals
- Explain buoyancy with Fb = ρliquid g Vdisplaced.
- Compare object and liquid density to predict floating, neutral buoyancy, or sinking.
- Calculate submerged fraction and displaced volume at floating equilibrium.
Before you begin
- Understand density = mass ÷ volume and recognise kg, m³, L, and N.
- Compare ratios and calculate with decimals.
Buoyancy experiment
- 1Float in water
- 2Change liquid
- 3Neutral
- 4Sink
Observe, then explain
Work through four steps. Each button moves the interactive model to the matching observation state.
-
When the object is less dense than water, partial immersion displaces enough liquid for buoyancy to equal weight.
Your task: Set liquid to 1000 kg/m³ and object to 600 kg/m³; verify 60% immersion.
-
For the same object, a denser liquid provides more buoyancy per displaced volume, so a smaller fraction must be submerged.
Your task: Change liquid density from 800 to 1200 kg/m³ and compare displaced volume.
-
When a fully submerged object has the same density as the liquid, buoyancy equals weight. With no net vertical force in the ideal model, it can remain suspended.
Your task: Set both densities to 1000 kg/m³ and compare the two forces.
-
When fully submerged, maximum buoyancy is less than weight, so the net force is downward. Increasing volume scales both forces together; an unchanged density ratio gives the same outcome.
Your task: Set object density to 1300 kg/m³, then change volume and explain why the status stays the same.
Build an explanation
Buoyancy comes from pressure difference
Pressure in a static liquid increases with depth. The lower surface generally receives a larger upward contribution than the upper surface; the net effect equals the weight of displaced liquid.
Floating balance gives a density ratio
For a uniform floating object, ρliquid g Vdisplaced = ρobject g Vobject, so Vdisplaced/Vobject = ρobject/ρliquid. Mass or total volume changes force magnitudes but not this ratio by itself.
Neutral is not resting on the bottom
Neutral buoyancy means a fully submerged object balances buoyancy and weight without contact support. An object resting on the bottom also experiences a normal force, so its force balance differs.
Check your understanding
Printable worksheet
Complete the worksheet first. Answers and explanations are on a separate page.
Physics
Buoyancy, density, and floating
Why does floating depend mainly on a density ratio rather than weight alone?
- A uniform 750 kg/m³ object floats in a 1000 kg/m³ liquid. What fraction is submerged?
A. 25% B. 75% C. 100%
- A 2.0 L object is fully submerged in a 1000 kg/m³ liquid. With g = 10 N/kg, what is buoyancy?
A. 2 N B. 20 N C. 2000 N
- Which statement is true for a uniform fully submerged neutrally buoyant object?
A. Object and liquid densities match, and buoyancy equals weight B. Object density must be zero C. Buoyancy must exceed weight
- With both densities fixed, what happens when a uniform object's volume doubles?
A. Weight and fully submerged buoyancy both double; the outcome is unchanged B. Only weight doubles C. It must change from sinking to floating
Answer page
Buoyancy, density, and floating
- 75%
While floating, Vdisplaced/Vobject = ρobject/ρliquid = 750/1000 = 0.75.
- 20 N
2.0 L = 0.002 m³, so Fb = 1000 × 10 × 0.002 = 20 N.
- Object and liquid densities match, and buoyancy equals weight
When fully submerged, the force ratio follows the density ratio. Equal densities give equal forces and zero net force in the ideal model.
- Weight and fully submerged buoyancy both double; the outcome is unchanged
Both forces scale with total volume. The density ratio ρobject/ρliquid determines the outcome and has not changed.