Free Pearson Edexcel International A Level Physics lesson

Density, upthrust and floating

Compare mass with volume, then account for the displaced liquid.

What you’ll learn

  • Determine density from measurements with consistent volume units.
  • Calculate upthrust using the fluid density and the submerged volume.
  • Explain floating and apparent weight with a force balance.

Density compares equal volumes

Density ρ (rho) is mass per unit volume. It lets us compare materials without confusing a large sample with a dense one. A large foam block can have more mass than a small metal bead while having much lower density. For a uniform material at a stated temperature, doubling the sampled volume doubles its mass, so their ratio stays the same. For an object containing cavities or several materials, total mass divided by its external volume gives its average density.

Use ρ = m/V with mass in kilograms and volume in cubic metres to obtain kg m⁻³. A centimetre is 10⁻² m, so a cubic centimetre is (10⁻²)³ m³ = 10⁻⁶ m³. Consequently 1 g cm⁻³ is 1000 kg m⁻³. Convert the volume as a cube; dividing a volume in cm³ by 100 converts it incorrectly. Rearrange to m = ρV or V = m/ρ before substitution when the missing quantity changes.

Measure mass using a zero-checked balance. For a regular solid, measure suitable dimensions at several positions and calculate its volume. An irregular solid that does not dissolve, absorb the liquid or trap bubbles can be submerged in a measuring cylinder: its volume is the final reading minus the initial reading. Read the meniscus at eye level and use a cylinder whose resolution is suitable for the change. Uncertainty in both volume readings contributes to the uncertainty in their difference. Repeat observations reveal scatter; they do not remove a balance zero error or air trapped on the object.

ρ = m / V 1 cm³ = 10⁻⁶ m³ 1 g cm⁻³ = 1000 kg m⁻³

Symbols and units

  • ρ is density in kg m⁻³; m is mass in kg; V is volume in m³.
  • 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³.

How the equation explains the idea

  1. Density compares mass per unit volume: ρ = m/V. At fixed density, m = ρV is a line through the origin with gradient ρ.
  2. Convert both numerator and denominator: 1 g cm⁻³ = (10⁻³ kg)/(10⁻⁶ m³) = 1000 kg m⁻³.
  3. Changing sample size leaves density unchanged only when composition and physical conditions remain comparable.

Worked example

Identify an irregular solid

  1. Mass = 0.324 kg. Cylinder readings are 45.0 cm³ and 165.0 cm³.
  2. Solid volume = 120.0 cm³ = 1.200 × 10⁻⁴ m³.
  3. Density = 0.324 / (1.200 × 10⁻⁴) = 2700 kg m⁻³. This measurement alone is not a unique chemical identification.

The measured density is 2.70 × 10³ kg m⁻³.

Keep this idea. Choose a volume measurement appropriate to the sample, convert cubed units, and divide mass by volume.

The displaced fluid determines upthrust

A fluid exerts pressure on a surface from all directions. In a stationary liquid, pressure increases with depth. The upward pressure force on the lower part of an immersed object is therefore greater than the downward force on its upper part; the horizontal contributions cancel in the simple symmetric case. Their resultant is upthrust. For an object surrounded by fluid, Archimedes’ principle gives upthrust equal to the weight of the fluid displaced, not the weight of the object itself.

Use U = ρfluid g Vsubmerged. The relevant density belongs to the displaced fluid; the relevant volume is only the part of the object below the surface. For a rigid, completely submerged object in a liquid of approximately constant density, moving deeper does not change its displaced volume or upthrust, even though the pressure everywhere increases. This model excludes contact sealing an object against the bottom and significant changes in fluid density or object volume.

Draw the forces before calculating an apparent weight. A dense object held stationary by a vertical thread has upward tension T and upthrust U and downward weight W. Balance gives T = W − U. A force meter in the thread reads tension, so its reading falls on immersion while the gravitational weight is unchanged. A less dense object held fully submerged needs a downward restraining force instead. Upthrust is present even when an object sinks; sinking means the other forces have not balanced it.

U = ρfluid g Vsubmerged stationary suspended object: T + U = mg

Symbols and units

  • U is upthrust in N; ρfluid is fluid density in kg m⁻³; Vsubmerged is displaced fluid volume in m³.
  • T is support tension in N; m is object mass in kg; g is field magnitude in m s⁻².

How the equation explains the idea

  1. Displaced fluid would have mass ρfluidVsubmerged and weight ρfluidgVsubmerged. Archimedes’ principle gives this weight as the upthrust.
  2. For a stationary suspended object, T + U = mg, hence T = mg − U. Submerging more volume reduces required tension until the relevant volume limit is reached.
  3. Use the displaced volume, which need not equal the whole object volume for partial immersion.
Upthrust is proportional to displaced fluid volume

Conditions. Computed U = ρgV with g = 10 m s⁻², water-model density 1000 kg m⁻³ and oil-model density 800 kg m⁻³. Volume coordinates in cm³ are multiplied by 10⁻⁶ to obtain m³.

Displaced volume / cm³; Upthrust / N.

At 100 cm³ displacement, upthrust is 1.0 N in the denser fluid and 0.8 N in the less dense fluid. Both lines pass through zero; their gradient is ρg after accounting for the plotted volume unit.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A force meter in water

  1. A 0.500 kg object displaces 200 cm³ = 2.00 × 10⁻⁴ m³ of water.
  2. With water density 1000 kg m⁻³ and g = 9.81 N kg⁻¹, U = 1.962 N.
  3. Weight = 4.905 N. For a stationary object suspended by a thread, T = 4.905 − 1.962 = 2.943 N.

The meter reads about 2.94 N; the object’s weight is still 4.905 N.

Keep this idea. Use the fluid density and immersed volume, then include upthrust in the chosen object’s force diagram.

Floating adjusts the displaced volume

A freely floating object at rest has upthrust equal to its weight when surface tension and air buoyancy are negligible. It settles to a depth that displaces enough liquid to provide that upthrust. For a uniform solid with external volume V and density ρobject, balancing ρfluid g Vsubmerged = ρobject g V gives Vsubmerged/V = ρobject/ρfluid. The same relation can use an object’s average density, including sealed cavities, if the fluid does not enter those cavities.

If the average density is lower than the liquid density, the required immersed fraction is less than one and partial immersion is possible. If the densities are equal, a completely submerged object can be neutrally buoyant under the ideal assumptions. If the average density is larger, even full submersion cannot provide sufficient upthrust for free static equilibrium. A large hollow metal hull can float because its total mass divided by the water-excluding volume is small; flooding changes that situation.

Distinguish capacity from equilibrium. Adding cargo increases the weight and requires a greater displaced volume, so a floating craft sits lower. A maximum safe displacement is a constraint set by the vessel design and conditions; it is not automatically the entire geometrical hull volume. These examples use ideal static balances, not real vessel safety ratings. Moving an unchanged floating object into a denser liquid reduces the immersed fraction while its weight and equilibrium upthrust remain the same.

Floating freely: ρfluid g Vsubmerged = mg Vsubmerged / V = ρaverage / ρfluid

Symbols and units

  • V is total object volume and Vsubmerged is displaced volume, both in m³.
  • ρaverage = m/V and ρfluid are object-average and fluid densities in kg m⁻³.

How the equation explains the idea

  1. For free floating with no other vertical support, ρfluidgVsubmerged = mg. Replace m by ρaverageV.
  2. Cancel g and divide by ρfluidV: Vsubmerged/V = ρaverage/ρfluid. An average density half the fluid density gives half the volume submerged.
  3. A predicted fraction above one cannot describe free floating in this simple model; full immersion cannot supply enough upthrust.

Worked example

A sealed model float

  1. The float has total external volume 0.0040 m³ and mass 2.4 kg.
  2. Its average density is 2.4 / 0.0040 = 600 kg m⁻³.
  3. In water of density 1000 kg m⁻³, the immersed fraction is 600/1000 = 0.60. The displaced volume is 0.0024 m³.

Sixty percent of its volume is below the surface at rest.

Keep this idea. A floating object changes its immersion until the displaced fluid’s weight matches its own weight.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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