Upthrust and flotation

What you'll learn

This revision guide covers upthrust and flotation as required for AQA GCSE Physics (forces and motion topic). You'll understand how fluids exert upward forces on submerged objects, why objects float or sink, and how to calculate upthrust using pressure differences. These concepts link directly to density, pressure, and Newton's laws of motion.

Key terms and definitions

Upthrust — the upward buoyant force exerted by a fluid (liquid or gas) on an object that is fully or partially submerged in it.

Fluid — a substance that can flow, including both liquids and gases; fluids exert pressure and upthrust on objects within them.

Displacement — the volume of fluid pushed out of the way by a submerged object; this displaced fluid has mass and weight.

Buoyancy — the tendency of an object to float or rise when submerged in a fluid, caused by upthrust acting upward.

Density — mass per unit volume of a substance, measured in kg/m³ or g/cm³; calculated using ρ = m/V.

Archimedes' principle — the upthrust acting on an object submerged in a fluid equals the weight of the fluid displaced by that object.

Neutral buoyancy — the state when an object's weight exactly equals the upthrust, causing it to remain suspended at a constant depth without rising or sinking.

Core concepts

Understanding upthrust

When an object is placed in a fluid, the fluid exerts pressure on all surfaces of the object. Pressure in a fluid increases with depth according to the equation:

pressure = height × density × gravitational field strength

or p = h × ρ × g

Because pressure increases with depth, the upward pressure on the bottom surface of a submerged object is greater than the downward pressure on the top surface. This pressure difference creates a net upward force — the upthrust.

Key points about upthrust:

• Upthrust always acts vertically upward
• Upthrust acts on objects in both liquids and gases
• The magnitude of upthrust depends on the volume of fluid displaced
• Upthrust does not depend on the mass or density of the object itself
• Upthrust does depend on the density of the fluid

Archimedes' principle and calculating upthrust

Archimedes' principle states that the upthrust on an object equals the weight of the fluid it displaces. This is a fundamental principle for calculating upthrust.

To calculate upthrust:

1. Find the volume of fluid displaced (equals the volume of the submerged part of the object)
2. Calculate the mass of this displaced fluid using: mass = density × volume
3. Calculate the weight of this displaced fluid using: weight = mass × gravitational field strength

Therefore: upthrust = weight of displaced fluid = volume × fluid density × g

In equation form: F = V × ρ_fluid × g

Where:

• F = upthrust (N)
• V = volume of fluid displaced (m³)
• ρ_fluid = density of the fluid (kg/m³)
• g = gravitational field strength (N/kg), typically 10 N/kg on Earth

For a fully submerged object, the volume displaced equals the object's total volume. For a partially submerged (floating) object, only the submerged portion displaces fluid.

Floating, sinking and density

Whether an object floats or sinks depends on the balance between two forces:

1. Weight — acting downward on the object (W = m × g)
2. Upthrust — acting upward from the fluid (F = V × ρ_fluid × g)

Objects sink when:

• Weight > Upthrust
• The object's density > fluid density
• The object is denser than the surrounding fluid

Objects float when:

• Weight = Upthrust (when floating at equilibrium)
• The object's average density < fluid density
• The object is less dense than the surrounding fluid

Objects rise through the fluid when:

• Upthrust > Weight
• Such as a bubble rising through water

When an object floats in equilibrium on the surface:

• The upthrust from the displaced fluid exactly balances the object's weight
• Only part of the object is submerged
• The fraction submerged depends on the ratio of object density to fluid density

For a floating object: ρ_object × V_object × g = ρ_fluid × V_submerged × g

This simplifies to: ρ_object × V_object = ρ_fluid × V_submerged

Practical applications of upthrust

Ships and boats: Ships are made of steel (density approximately 7800 kg/m³), which is much denser than water (1000 kg/m³). However, ships are designed with a hollow hull containing large volumes of air. This gives the ship a low average density overall. The ship displaces a large volume of water, creating enough upthrust to support the weight of the steel and cargo.

Submarines: Submarines control their buoyancy using ballast tanks. To dive, they fill ballast tanks with seawater, increasing their average density. To surface, they blow compressed air into the tanks, forcing water out and decreasing average density. At neutral buoyancy, the submarine maintains a constant depth.

Hot air balloons: Hot air is less dense than cold air. A hot air balloon displaces a large volume of cold air, creating upthrust. When the upthrust exceeds the total weight (balloon fabric, basket, passengers, and hot air), the balloon rises. Pilots control altitude by heating or cooling the air inside the envelope.

Hydrometers: These devices measure liquid density by floating at different depths in liquids of different densities. A hydrometer floats higher in denser liquids and lower in less dense liquids. They're used to test battery acid concentration or alcohol content in brewing.

Upthrust in gases

Air and other gases also exert upthrust on objects, though the effect is usually small because gas densities are low compared to liquids.

Density of air at sea level: approximately 1.2 kg/m³ Density of water: 1000 kg/m³

Since air is roughly 800 times less dense than water, upthrust in air is approximately 800 times smaller than in water for the same volume displaced.

However, upthrust in air becomes significant for:

• Large volume objects (hot air balloons, airships)
• Very low density objects (helium balloons)
• Precision mass measurements using balances

A helium balloon rises because:

• Helium density (0.18 kg/m³) < air density (1.2 kg/m³)
• Upthrust from displaced air > weight of helium plus balloon material
• Net upward force causes the balloon to rise

Pressure difference and upthrust derivation

Consider a cubic object of side length 'h' fully submerged in a fluid:

The pressure on the top surface: p_top = h_top × ρ × g The pressure on the bottom surface: p_bottom = h_bottom × ρ × g

The difference in depth between top and bottom = h (the height of the cube)

Force on bottom (upward) = p_bottom × A = h_bottom × ρ × g × A Force on top (downward) = p_top × A = h_top × ρ × g × A

Upthrust = Force on bottom - Force on top = (h_bottom - h_top) × ρ × g × A = h × ρ × g × A = V × ρ × g (since V = A × h)

This equals the weight of fluid displaced, confirming Archimedes' principle.

Worked examples

Example 1: Calculating upthrust on a submerged object

Question: A metal block with volume 0.004 m³ is fully submerged in water. The density of water is 1000 kg/m³ and g = 10 N/kg. Calculate the upthrust on the block. [3 marks]

Solution:

Step 1: Identify the relevant equation Upthrust = volume × density of fluid × g [1 mark]

Step 2: Substitute values Upthrust = 0.004 m³ × 1000 kg/m³ × 10 N/kg [1 mark]

Step 3: Calculate Upthrust = 40 N [1 mark]

Mark scheme notes: Students must use the density of water, not the metal. The upthrust depends only on the volume displaced and the fluid density.

Example 2: Explaining floating and sinking

Question: A wooden block has a density of 600 kg/m³ and a volume of 0.002 m³. It is placed in water (density 1000 kg/m³).

(a) Calculate the weight of the wooden block. g = 10 N/kg [3 marks] (b) Will the block float or sink? Explain your answer. [2 marks]

Solution:

(a) Step 1: Calculate mass of block density = mass / volume, so mass = density × volume mass = 600 kg/m³ × 0.002 m³ = 1.2 kg [1 mark]

Step 2: Calculate weight Weight = mass × g Weight = 1.2 kg × 10 N/kg [1 mark] Weight = 12 N [1 mark]

(b) The block will float [1 mark] because the density of wood (600 kg/m³) is less than the density of water (1000 kg/m³), so the upthrust when fully submerged (20 N) would be greater than its weight (12 N) [1 mark].

Alternative acceptable answer for (b): The block will float because when it is fully submerged, the upthrust (20 N) exceeds its weight (12 N), causing it to rise until it reaches equilibrium floating on the surface.

Example 3: Volume submerged when floating

Question: An ice cube floats in water with 90% of its volume submerged. The density of water is 1000 kg/m³. Calculate the density of ice. [4 marks]

Solution:

Step 1: Apply principle that weight = upthrust for floating objects ρ_ice × V_ice × g = ρ_water × V_submerged × g [1 mark]

Step 2: Cancel g from both sides and note that V_submerged = 0.90 × V_ice ρ_ice × V_ice = ρ_water × 0.90 × V_ice [1 mark]

Step 3: Simplify by canceling V_ice ρ_ice = 0.90 × ρ_water [1 mark]

Step 4: Calculate ρ_ice = 0.90 × 1000 kg/m³ = 900 kg/m³ [1 mark]

Common mistakes and how to avoid them

• Confusing the density to use in upthrust calculations: Always use the density of the fluid, not the object. Upthrust depends on the weight of displaced fluid, which requires the fluid's density.

• Forgetting that upthrust acts on objects in air too: Many students only think of upthrust in water. Remember that air exerts upthrust, though it's usually negligible. This explains why helium balloons rise and why hot air balloons float.

• Mixing up when objects float versus sink: An object floats if its average density is less than the fluid density, not when it's heavier or lighter. Think about density comparison, not just weight or size.

• Not using the submerged volume for partially submerged objects: For floating objects, only the submerged portion displaces fluid. The upthrust equals the weight of this displaced volume, not the total object volume.

• Incorrect unit conversions: Ensure density is in kg/m³, volume in m³, and g in N/kg for consistency. Converting cm³ to m³ requires dividing by 1,000,000 (or 10⁶), a common source of errors.

• Stating that upthrust depends on object mass or density: Upthrust depends only on the volume of fluid displaced and the fluid's density. Two objects of the same volume but different masses experience the same upthrust when fully submerged in the same fluid.

Exam technique for "Upthrust and flotation"

• Identify 'explain' versus 'calculate' questions: When asked to explain floating or sinking, compare densities or compare upthrust to weight. For calculations, use equations systematically: show the formula, substitute values with units, and calculate.

• Show all working in multi-step problems: AQA awards marks for method even if the final answer is incorrect. Write the equation first, substitute values on a new line, then calculate. This can secure 2 out of 3 marks even with an arithmetic error.

• Use comparison statements for explanations: Strong answers compare quantities explicitly: "The upthrust (20 N) is greater than the weight (12 N), so..." or "Wood density (600 kg/m³) < water density (1000 kg/m³), therefore...". Vague statements like "it's lighter" won't earn marks.

• Check your answer makes physical sense: If you calculate an upthrust smaller than the object's weight but conclude it floats, you've made an error. If density comes out negative, re-check your working. Quick sense-checks prevent lost marks.

Quick revision summary

Upthrust is the upward force exerted by fluids on submerged objects, caused by pressure increasing with depth. By Archimedes' principle, upthrust equals the weight of displaced fluid, calculated as V × ρ_fluid × g. Objects float when their average density is less than the fluid density, making upthrust equal to or greater than their weight. Objects sink when denser than the fluid. Applications include ship design, submarine ballast systems, and hot air balloons. Always use fluid density in upthrust calculations, not object density.