Upthrust and Flotation: Worked Examples

Upthrust and flotation: start with the displaced volume

An object in liquid can experience upthrust even when it sinks. Floating is a particular balance of forces; upthrust is a force that also acts on many objects which are fully immersed. Separating those ideas helps you decide when to use the whole object volume, when to use only the submerged part and when an additional supporting force is needed.

These worked examples of upthrust and flotation use g = 10 N kg⁻¹, uniform-density liquids and rigid, impermeable objects unless stated otherwise. Ignore air buoyancy, surface tension and fluid motion. The object must displace liquid in the way described; an object sealed against the container floor may not satisfy the usual freely immersed model.

Weight, upthrust and apparent weight

Actual weight W = mg is the gravitational force on the object. Upthrust U = ρ_liquidgV_displaced equals the weight of the displaced liquid. Use the liquid’s density in this equation, not the object’s density. The volume is measured in m³ and the result is in newtons.

Apparent weight commonly means the force registered by a support or spring balance. For an object held at rest by an upward tension T in a liquid, T + U = W, so T = W − U. The gravitational force has not vanished; the support simply supplies less upward force because the liquid supplies some.

If U is greater than W, an upward-only support cannot hold the object stationary. A negative value from W − U signals that the assumed force direction is wrong. A downward restraint would be required, or the object would accelerate upward when released. Do not report a negative spring-balance reading without examining the setup.

Three force diagrams described in words
State	Vertical forces on the object	Conclusion
Floating freely at rest	Upward U; downward W	U = W; only part may be submerged
Fully immersed, held by a string	Upward U and T; downward W	U + T = W for equilibrium
Fully immersed, released with W greater than U	Upward U; downward W	Initial resultant W − U is downward

A. FLOATING AT REST
        ↑ U = 4.8 N
air     |
~~~~[ BLOCK ]~~~~ water surface
water   |
        ↓ W = 4.8 N

B. HELD, FULLY IMMERSED
~~~~~~~~~~~~~~~ water surface
water   ↑ T = 6 N     ↑ U = 3 N
         \           /
            [SOLID]
               |
               ↓ W = 9 N

C. RELEASED, INITIALLY AT REST
~~~~~~~~~~~~~~~ water surface
water          ↑ U = 3 N
               |
            [SOLID]
               |
               ↓ W = 9 N

Three original schematic force diagrams, using the data in examples 1 and 2. Arrows identify force directions, not scaled magnitudes or exact application points. A crosses the surface; B and C are completely below it. C has an initial downward resultant of 6 N.

The last row describes release from rest before significant drag develops. A sinking object moving at constant terminal speed has an additional upward drag force. Its constant speed does not mean that upthrust alone equals its weight.

Worked example 1: a fully immersed solid

A solid has mass 0.90 kg and external volume 300 cm³. It hangs motionless from a spring balance, completely immersed in water of density 1,000 kg m⁻³, without touching the container. Calculate its weight, upthrust and balance reading.

Convert volume first: 1 cm³ = 10⁻⁶ m³, so 300 cm³ = 0.000300 m³. Weight = 0.90 × 10 = 9.0 N. Complete immersion means displaced water volume equals the external object volume. Upthrust = 1,000 × 10 × 0.000300 = 3.0 N.

The balance provides T = 9.0 − 3.0 = 6.0 N upward. Check equilibrium: 6.0 + 3.0 = 9.0 N. If the string is removed at that instant, the initial resultant is 6.0 N downward, so the object starts to sink despite experiencing upthrust.

Its mean density is 0.90/0.000300 = 3,000 kg m⁻³, greater than the water’s density. This independently agrees with the force comparison. The displaced water has mass 0.300 kg and weight 3.0 N; confusing these numbers would give the right digits but the wrong physical quantity.

Worked example 2: a freely floating block

A uniform block has volume 800 cm³ and density 600 kg m⁻³. It floats at rest in water of density 1,000 kg m⁻³. Find its submerged volume and submerged fraction.

Its mass is 600 × 0.000800 = 0.480 kg and its weight is 4.80 N. Floating equilibrium requires U = W. Thus displaced volume = 4.80/(1,000 × 10) = 0.000480 m³ = 480 cm³. The submerged fraction is 480/800 = 0.60, or 60%.

Combining the equations directly gives V_submerged/V_object = ρ_object/ρ_liquid = 600/1,000. That shortcut requires free floating equilibrium with no additional vertical support. The remaining 320 cm³ is above the water. Using all 800 cm³ in the upthrust formula would describe complete immersion, not this floating state.

For a uniform rectangular block floating upright with constant horizontal cross-section, 60% of its height is also submerged. That height conclusion is not automatic for an irregular object: the volume fraction and height fraction can differ.

Worked example 3: loading a floating container

A rigid watertight container has mass 0.70 kg. In its specified upright orientation, the largest displaced volume before water reaches the rim is 2.00 dm³. It floats in fresh water of density 1,000 kg m⁻³. Find the theoretical additional mass at the rim limit.

Convert 2.00 dm³ to 0.00200 m³. Maximum displaced water mass = 1,000 × 0.00200 = 2.00 kg. At the limiting equilibrium, the combined mass of container and load is 2.00 kg. Additional mass = 2.00 − 0.70 = 1.30 kg.

This is a theoretical limit, not a safe carrying capacity. Stability, waves, uneven loading, material strength and required freeboard are excluded. Exactly at the rim, there is no allowance for disturbance. The 2.00 dm³ refers to excluded water volume, including the sealed space; it is not merely the volume of the container’s solid walls.

A metal container can therefore float even though its wall material is denser than water. The relevant average density includes the mass and water-excluding volume of the complete floating object. If water enters, both the mass accounting and displaced-volume model need reconsideration.

Worked example 4: the same object in two liquids

A sealed object weighs 12.0 N in air under our negligible-air-buoyancy assumption. Its apparent weight is 7.0 N when fully immersed in water. Find its volume and its apparent weight when fully immersed in oil of density 800 kg m⁻³.

Water upthrust = 12.0 − 7.0 = 5.0 N. Hence V = 5.0/(1,000 × 10) = 0.000500 m³. In oil, U = 800 × 10 × 0.000500 = 4.0 N. Apparent weight = 12.0 − 4.0 = 8.0 N.

The oil gives a larger balance reading because its lower density produces less upthrust for the same displaced volume. The object has not gained mass. Once the same rigid object is completely immersed in a uniform liquid, lowering it farther does not increase its displaced volume; within these assumptions, its upthrust remains unchanged.

Three solved practice tasks

Practice 1: convert volume and name the density

A fully immersed object displaces 250 cm³ of a liquid with density 1,200 kg m⁻³. Its weight is 5.0 N. Find the upthrust and upward supporting tension for equilibrium.

Solution: V = 0.000250 m³. U = 1,200 × 10 × 0.000250 = 3.0 N. T = 5.0 − 3.0 = 2.0 N. The given liquid density belongs in the upthrust equation; no object-density calculation is needed.

Practice 2: test whether floating is possible

A rigid uniform object has density 900 kg m⁻³. Could it float freely at the surface of oil of density 750 kg m⁻³ with no air space or other support?

Solution: The proposed submerged fraction is 900/750 = 1.20. A fraction above one is impossible for this object. Even complete immersion cannot displace enough oil to balance its weight, so it sinks in the stated model.

Practice 3: neutral buoyancy

A fully submerged object has mass 0.36 kg and volume 400 cm³ in a liquid of density 900 kg m⁻³. Compare its weight and upthrust.

Solution: W = 3.6 N. U = 900 × 10 × 0.000400 = 3.6 N. It is neutrally buoyant. If initially at rest and undisturbed, it has no resultant vertical force and can remain submerged; neutral buoyancy does not require part of it to emerge.

Frequently asked questions

Does a sinking stone experience upthrust?

Yes, when freely immersed. It sinks because its weight exceeds the upthrust at release, not because the upthrust is absent.

Which volume belongs in Archimedes’ principle?

Use displaced-fluid volume. For a partly immersed impermeable solid this is its submerged volume; for complete immersion it is its water-excluding external volume.

Is upthrust measured in kilograms?

No. Upthrust is a force in newtons. The displaced-fluid mass is in kilograms and must be multiplied by g.

Is the object’s mass smaller underwater?

No. In this model its mass and actual weight remain the same. A supporting scale reads less because upthrust supplies part of the support.

Does higher liquid density always mean more upthrust?

For equal displaced volume, yes. A freely floating object instead adjusts its submerged volume so that upthrust still balances its unchanged weight.

Can I use the density fraction for a tied-down object?

Not directly. Include the tether force in equilibrium. The familiar density-ratio shortcut assumes that weight and upthrust are the only relevant vertical forces.

Continue practising

The existing Physics topical revision collection requires login for its resources. For the physical principle behind these original examples, see OpenStax on Archimedes’ principle.