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Physics (IAL)

Density, Upthrust & Viscous Drag

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Edexcel IAL Physics · Unit 2

Density, Upthrust & Viscous Drag

The big idea: how "packed" something is (density) decides whether it floats or sinks (upthrust), and how "thick" a fluid is decides how fast things can move through it (viscous drag) — and objects falling through fluids settle into a steady speed once these forces balance.

OverviewQuick Summary
  • Density (ρ) = mass per unit volume. It tells you how tightly packed the matter in an object is.
  • Upthrust is the upward push a fluid gives any object submerged (fully or partly) in it — caused by pressure being greater at greater depth.
  • Archimedes' Principle: upthrust = weight of the fluid displaced by the object.
  • An object floats when upthrust can equal its weight before it's fully submerged; it sinks when even fully submerged, upthrust never catches up to weight.
  • Viscous drag is the "friction" a fluid exerts on an object moving through it — governed for small spheres by Stokes' Law: F = 6πηrv.
  • Falling objects reach terminal velocity when weight = upthrust + viscous drag (no more resultant force, so no more acceleration).
  • Core Practical 2 uses falling ball bearings in a viscous liquid to experimentally determine the liquid's viscosity, η.
Topic 1Density

What Density Actually Means

Imagine two boxes of exactly the same size. One is stuffed full of marbles, the other has just a handful rattling around inside. Even though both boxes take up the same space (same volume), the full one is much heavier. That's density in a nutshell — it's a measure of how much mass is squeezed into a given volume.

A balloon and a small lead weight can occupy very different volumes, and yet the lead — despite being physically smaller — is far denser, because its atoms are packed much more tightly and each atom itself is heavier.

ρ = m / V
Density = Mass ÷ Volume
ρ (rho) = density, measured in kg m⁻³ (if mass in kg, volume in m³) or g cm⁻³ (if mass in g, volume in cm³)
m = mass (kg or g)
V = volume (m³ or cm³)

Working Out the Volume

Often the volume isn't handed to you directly — you have to calculate it from the object's shape and dimensions first. The three you'll meet most:

   SPHERE:            CUBE:              CYLINDER:
   V = (4/3)πr³        V = d³             V = πr²l

     .-‾‾-.             ┌────┐              ___
    /      \            │    │d            ( r )
   |   •r   |         d │    │              |  |
    \      /            └────┘              | l|
     ‾-..-‾                d                |__|
        

Unit Conversions — the part everyone slips up on

🍋 Rule of thumb Converting a larger unit → smaller unit: MULTIPLY. Converting a smaller unit → larger unit: DIVIDE.
e.g. 125 m = 125 × 100 = 12 500 cm. e.g. 5 g = 5 ÷ 1000 = 0.005 kg.
But watch out — for area/volume conversions you must square or cube the conversion factor too!
1 mm³ = (1×10⁻³)³ m³ = 1×10⁻⁹ m³. 1 cm³ = (1×10⁻²)³ m³ = 1×10⁻⁶ m³.
Worked Example

A paving slab has a mass of 73 kg and dimensions 40 mm × 500 mm × 850 mm. Calculate its density in kg m⁻³.

Practice Question

A metal cube has sides of 2.0 cm and a mass of 63 g. Find its density in kg m⁻³.

Topic 2Upthrust & Archimedes' Principle

Why Things Float

Push a beach ball underwater and let go — it shoots back up. That upward shove is upthrust, and it exists because pressure in a fluid increases with depth. The bottom of a submerged object feels more pressure pushing up on it than the top feels pushing down, and that pressure difference creates a net upward force.

💜 Archimedes' Principle (learn this word-for-word) An object submerged in a fluid at rest has an upward buoyancy force (upthrust) equal to the weight of the fluid displaced by the object.

Here's the intuition: when you push an object into water, it has to shove the water out of the way — it displaces a volume of water equal to its own submerged volume. Archimedes' Principle says the upthrust you feel is exactly equal to the weight of that displaced water.

Floating vs Sinking

  • An object sinks until the weight of fluid it has displaced equals its own weight.
  • If that balance point happens before the object is fully submerged → it floats.
  • If the object is denser than the fluid, it will still be sinking (accelerating downward) even when fully submerged, because it can never displace enough weight of fluid to match its own weight → it sinks all the way.
              Weight (down, = buoyancy force)
                        |
                        v
                  ┌───────────┐
                  │   SHIP    │
        ~~~~~~~~~~└───────────┘~~~~~~~~~~~
                        ^
                        |
              Buoyancy force (up, = weight of
              displaced water)

     mass of displaced water = mass of ship submerged
        

Calculating Upthrust — the 3-Step Method

  1. Find the volume of the submerged part of the object (= volume of fluid displaced).
  2. Use ρ = m/V (rearranged: m = ρV) to find the mass of that displaced fluid.
  3. Use W = mg to find the weight of the displaced fluid — that weight is the upthrust.
U = W = mg = ρVg
Upthrust = weight of fluid displaced = density of fluid × volume displaced × g
🔵 Why the two-step process trips people up Step 1 (finding volume) isn't the answer — it's just a stepping stone to find out how much fluid was displaced. What you actually want is the weight of that displaced fluid. Chain the equations: m = ρV, then W = mg. Combined: W = ρVg.
Worked Example

Atmospheric pressure at sea level is 100 kPa. The density of sea water is 1020 kg m⁻³. At what depth would the total pressure be 250 kPa?

Worked Example

Icebergs float with a large volume beneath the water. Ice has density 917 kg m⁻³ and volume Vi. Sea water density is 1020 kg m⁻³. What fraction of the iceberg is above the water?

Topic 3Viscous Drag & Stokes' Law

What Is Viscous Drag?

Pour water from a jug and it glugs out easily. Try to pour honey and it crawls out reluctantly. That resistance to flowing — and the resistance a moving object feels as it pushes through the fluid — is viscosity. The frictional force this creates on a moving object is viscous drag.

💜 Definition to memorise Viscous drag is the frictional force between an object and a fluid which opposes the motion between the object and the fluid.

Viscosity itself (η, the Greek letter "eta") is a property of the fluid — how "thick" it is at a given temperature. Low viscosity fluids (water) pour and flow easily. High viscosity fluids (honey, tomato ketchup) resist flowing. Crucially: the rate of flow of a fluid is inversely proportional to its coefficient of viscosity — thicker fluid, slower flow.

Stokes' Law

F = 6πηrv
Viscous drag on a small sphere moving through a fluid
F = viscous drag force (N)
η = coefficient of viscosity of the fluid (N s m⁻² or Pa s)
r = radius of the sphere (m)
v = velocity of the sphere (m s⁻¹)
🌸 Stokes' Law only works under these conditions — an examiner favourite
  • The flow must be laminar (not turbulent)
  • The object must be small
  • The object must be spherical
  • The motion must be at slow speed

Laminar Flow vs Turbulent Flow

As a fluid flows around an object (or the object moves through it), the fluid forms into layers. In laminar flow, every layer moves in the same direction and none of them mix — this happens for slow-moving objects in slow-flowing fluid, and it's the only situation where Stokes' Law applies. In turbulent flow, the layers move in different, chaotic directions and mix together — think of rapids in a river versus a calm stream.

   LAMINAR FLOW                    TURBULENT FLOW
   (layers stay separate,          (layers mix and swirl,
    all same direction)             chaotic directions)

   ──────►  ●  ──────►             ~~~↷  ●  ↶~~~
   ──────►     ──────►             ↷~~     ~~↶
   ──────►  ──────►                  ~~↷↶~~
        

Effect of Temperature on Viscosity

  • Liquids get less viscous as temperature increases (heat makes molecules move more freely past each other).
  • Gases get more viscous as temperature increases (opposite behaviour to liquids — a classic trick question).

Terminal Velocity of a Falling Sphere

Drop a ball bearing into a tall tube of oil. At first it accelerates downward under gravity. But as it speeds up, viscous drag (and upthrust) grow larger and larger, fighting back against its fall. Eventually these forces balance out weight exactly — at that point there's no resultant force, so no more acceleration. The sphere carries on at a constant speed: terminal velocity.

W = U + Fd
At terminal velocity: Weight (down) = Upthrust + Viscous Drag (both up)
              ↑ F_d (viscous drag)
              ↑ U (upthrust)
             ( o )   ← sphere falling through fluid
              |
              ↓ W (weight)

     At terminal velocity: W = F_d + U  (forces balanced)
        

Deriving the Terminal Velocity Equation

This derivation is genuinely worth being able to reproduce — it shows up as "derive an expression for..." in exams. Walk through it slowly:

  1. At terminal velocity: Ws = Wf + 6πηrvterm (sphere's weight = weight of displaced fluid + viscous drag)
  2. Mass of sphere: ms = ρsV = (4/3)πr³ρs, so Ws = (4/3)πr³ρsg
  3. Mass of displaced fluid (same volume as sphere): mf = ρfV = (4/3)πr³ρf, so Wf = (4/3)πr³ρfg
  4. Substitute both into the balance equation: (4/3)πr³ρsg = (4/3)πr³ρfg + 6πηrvterm
  5. Rearrange for vterm, cancelling one factor of r from top and bottom:
vterm = 2r²g(ρs − ρf) / 9η
Terminal velocity of a sphere falling through a fluid
🌿 Two proportionalities worth memorising Terminal velocity is directly proportional to r² (double the radius → four times the terminal velocity) and inversely proportional to η (double the viscosity → half the terminal velocity).
🍋 Don't mix up your subscripts! This is the single biggest source of lost marks in this topic. Always keep clear whether ρ, m, r, W refer to the sphere (subscript s) or the fluid (subscript f). The sphere's radius rs IS the same as the radius of the fluid it displaces (since the fluid takes the shape vacated by the sphere) — but the densities ρs and ρf are almost always different values. Practise writing subscripts consistently before your exam.
Worked Example

A ball bearing of radius 5.0 mm falls at a constant speed of 0.030 m s⁻¹ through an oil with viscosity 0.3 Pa s and density 900 kg m⁻³. Determine the viscous drag acting on the ball bearing.

Practice Question

A steel sphere (density 7800 kg m⁻³, radius 2.0 mm) falls at terminal velocity through glycerol (density 1260 kg m⁻³, viscosity 0.95 Pa s). Calculate its terminal velocity.

Required PracticalCore Practical 2: Investigating Viscosity

Aim & Setup

Aim: drop small spherical ball bearings through a viscous liquid, let them reach terminal velocity, then use the terminal velocity equation (rearranged for η) to calculate the fluid's viscosity.

Independent variable: diameter of ball bearing
Dependent variable: terminal velocity, vterm
Control variables: the fluid being tested, temperature

Equipment

  • Long measuring cylinder
  • Viscous liquid (e.g. thin oil of known density, or washing-up liquid)
  • Stand and clamp
  • Metre rule & rubber bands (as distance markers)
  • Steel ball bearings of different diameters
  • Digital scales & Vernier calipers (to find sphere density)
  • Digital stopwatch
  • Magnet (to retrieve the balls without draining the tube!)

Method — Step by Step

  1. Weigh the balls, measure their radius with Vernier calipers, and calculate their density.
  2. Place three rubber bands around the outside of the tube. The highest band must be far enough below the liquid's surface that the ball has definitely reached terminal velocity by the time it passes it (if the ball is still accelerating when it crosses the first marker, move that marker further down). The remaining two bands should be 10–15 cm apart for accurate timing.
  3. Release the ball and start the timer as it passes the first (highest) band. Use a lap timer to record the time taken to fall distance d₁ (to the middle band) and d₂ (to the lowest band).
  4. Measure and record d₁ (highest to middle band) and d₂ (highest to lowest band) using the metre rule.
  5. Repeat at least 3 times for that diameter, then repeat the whole process for each different ball diameter.
  6. Retrieve the ball bearings from the bottom using the magnet held against the outside of the cylinder.

Analysis — Deriving η from the Data

At terminal velocity the sphere is in equilibrium: Ws = Fd + U. Substituting the weight, drag, and upthrust expressions in (exactly as in the derivation above) and rearranging for η instead of vterm gives:

η = 2r²g(ρs − ρf) / 9v
v here is the measured terminal velocity = distance ÷ time from your data

Evaluating the Experiment

Systematic errors: Ruler must be clamped vertically and positioned close to the tube to avoid parallax error when reading distances. The ball must have already reached terminal velocity before it passes the first marker — otherwise you're timing accelerated motion, not constant-velocity motion.
Random errors: The cylinder needs a large diameter relative to the ball bearing, otherwise the flow becomes turbulent (breaking a Stokes' Law condition). The ball must also fall down the centre of the tube — falling too close to the wall creates pressure differences that alter its velocity.

Safety

  • Measuring cylinders are unstable — clamp them at both top and bottom.
  • Clean up spillages immediately (viscous liquids are extremely slippery underfoot).
  • Avoid getting the fluid in your eyes.
Practice Question

In this practical, why must the two lower rubber bands be spaced well apart (10–15 cm) rather than close together?

Quick ReferenceWhat to Memorise
Density
ρ = m/V — mass per unit volume. Units: kg m⁻³ or g cm⁻³.
Archimedes' Principle
An object submerged in a fluid at rest has an upward buoyancy force (upthrust) equal to the weight of the fluid displaced by the object.
Upthrust formula
U = ρfluidVg, where V is the volume of fluid displaced.
Floating condition
Object floats when upthrust equals its weight before full submersion (mass of displaced fluid = mass of object submerged).
Viscous drag definition
The frictional force between an object and a fluid which opposes the motion between the object and the fluid.
Stokes' Law
F = 6πηrv — only valid for small, spherical objects moving slowly through a fluid in laminar flow.
Terminal velocity condition
W = U + Fd — weight balances upthrust plus viscous drag; no resultant force, constant velocity.
Terminal velocity equation
vterm = 2r²g(ρs − ρf) / 9η — directly proportional to r², inversely proportional to η.
Laminar vs turbulent flow
Laminar: layers move same direction, don't mix (slow-moving). Turbulent: layers move in different directions and mix.
Viscosity & temperature
Liquids: less viscous as temperature rises. Gases: MORE viscous as temperature rises (opposite trend!).
Unit conversion rule
Larger→smaller unit: multiply. Smaller→larger unit: divide. For volume conversions, cube the linear conversion factor.
Self-TestConcepts Checklist
Before the ExamExam Tips & Common Mistakes
Subscript confusion is the #1 mark-loserMixing up ρs (sphere) and ρf (fluid) — or ms and mf — is by far the most common error in upthrust and terminal velocity questions. Always label your working clearly with subscripts, even in rough working.
Unit conversion errorsForgetting to cube the conversion factor when converting volumes (e.g. treating 1 mm³ = 0.001 m³ instead of 1×10⁻⁹ m³) is a very common slip. Always double-check whether you're converting a length, an area, or a volume.
Check the conditions before using Stokes' LawIf a question gives you a large or irregularly-shaped object, or describes fast/turbulent motion, Stokes' Law (F = 6πηrv) doesn't apply — examiners deliberately test whether you recognise this.
"Constant speed" = terminal velocityIf a question says an object is moving at a constant velocity through a fluid, that's your cue that the forces are balanced (W = U + Fd) — you can apply Stokes' Law directly without worrying about acceleration.
Be ready to derive, not just quoteExaminers often ask you to derive vterm = 2r²g(ρs−ρf)/9η from scratch, starting from W = U + Fd. Practise this derivation until you can do it without looking — it's not one to attempt for the first time in the exam hall.
Liquids vs gases and temperatureA classic trick question flips this: liquids get LESS viscous when heated, but gases get MORE viscous when heated. Don't assume they behave the same way.
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