Revision notes for Edexcel A Level Physics Density, upthrust and viscous drag. Open the guide for explanations and worked examples. Written against the Edexcel A Level Physics (9PH0) specification, so the content matches what's examinable rather than general Physics background.

Density, upthrust and viscous drag

What you'll learn

  • How density links mass and volume, including common unit conversions.
  • Why an object in a fluid experiences an upward force called upthrust.
  • How viscous drag changes with speed and leads to terminal velocity.
  • How Stokes’ law is used in falling-ball viscosity experiments.

Starting point: forces in fluids

A fluid is a substance that can flow: a liquid or a gas. When an object is in a fluid, the forces on it may include:

  • its weight, acting vertically downwards
  • upthrust, acting vertically upwards
  • drag, acting opposite to the direction of motion

Your usual mechanics still applies. The resultant force is the vector sum of all the forces, and if the resultant force is zero the object has no acceleration.

Definition

Weight

The weight of an object is the gravitational force on it:

W=mgW = mgW=mg

where mmm is mass in kilograms, ggg is gravitational field strength in newtons per kilogram, and WWW is weight in newtons.

Density

Density tells you how much mass is packed into a given volume. A small object can still have a high density if it has a large mass for its size.

Definition

Density

Density is mass per unit volume:

ρ=mV\rho = \frac{m}{V}ρ=Vm​

where ρ\rhoρ is density in kilograms per cubic metre, mmm is mass in kilograms, and VVV is volume in cubic metres.

For A-Level calculations, use SI units unless the question clearly says otherwise. The most common trap is volume conversion:

1 cm3=1×10−6 m31\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^31 cm3=1×10−6 m3

because each centimetre is converted in three dimensions.

Example

Finding the density of an irregular solid

A stone has mass 74.0 g. When lowered into a measuring cylinder, it displaces 28.0 cm³ of water. Find its density.

  1. Convert the measurements into SI units:

    m=74.0 g=0.0740 kgm = 74.0\ \text{g} = 0.0740\ \text{kg}m=74.0 g=0.0740 kg V=28.0 cm3=28.0×10−6 m3=2.80×10−5 m3V = 28.0\ \text{cm}^3 = 28.0 \times 10^{-6}\ \text{m}^3 = 2.80 \times 10^{-5}\ \text{m}^3V=28.0 cm3=28.0×10−6 m3=2.80×10−5 m3

  2. Substitute into the density equation:

    ρ=mV=0.07402.80×10−5\rho = \frac{m}{V} = \frac{0.0740}{2.80 \times 10^{-5}}ρ=Vm​=2.80×10−50.0740​

  3. Calculate and quote a sensible number of significant figures:

    ρ=2.64×103 kg m−3\rho = 2.64 \times 10^3\ \text{kg m}^{-3}ρ=2.64×103 kg m−3

    The stone is denser than water, so you would expect it to sink.

Common Mistake

Converting cubic centimetres

Do not use 1 cm3=10−2 m31\ \text{cm}^3 = 10^{-2}\ \text{m}^31 cm3=10−2 m3. The correct conversion is 1 cm3=10−6 m31\ \text{cm}^3 = 10^{-6}\ \text{m}^31 cm3=10−6 m3.

Upthrust

When an object is partly or fully immersed in a fluid, the fluid pushes on it from all directions. The pressure is greater lower down, so the upward push on the bottom is larger than the downward push on the top. The result is an upward force called upthrust.

Schematic showing upthrust on an immersed block due to greater pressure on the lower face and the displaced fluid volume

Definition

Upthrust

Upthrust is the upward force exerted by a fluid on an object that is partly or fully immersed in it.

The key result is Archimedes’ principle.

Key Idea

Archimedes’ principle

The upthrust on an object is equal to the weight of the fluid it displaces:

U=ρfVdisplacedgU = \rho_f V_{\text{displaced}} gU=ρf​Vdisplaced​g

where ρf\rho_fρf​ is the fluid density and VdisplacedV_{\text{displaced}}Vdisplaced​ is the volume of fluid displaced.

If an object is fully submerged, the displaced volume is the object’s full volume. If it is floating, the displaced volume is only the submerged volume.

Example

Finding apparent weight in water

A solid object has weight 6.0 N and volume 2.0×10−4 m32.0 \times 10^{-4}\ \text{m}^32.0×10−4 m3. It is fully submerged in water of density 1000 kg m−31000\ \text{kg m}^{-3}1000 kg m−3. Find the upthrust and the apparent weight.

  1. Since the object is fully submerged, the displaced volume is the object’s volume:

    Vdisplaced=2.0×10−4 m3V_{\text{displaced}} = 2.0 \times 10^{-4}\ \text{m}^3Vdisplaced​=2.0×10−4 m3

  2. Calculate the upthrust:

    U=ρfVdisplacedgU = \rho_f V_{\text{displaced}} gU=ρf​Vdisplaced​g U=1000×2.0×10−4×9.81=1.96 NU = 1000 \times 2.0 \times 10^{-4} \times 9.81 = 1.96\ \text{N}U=1000×2.0×10−4×9.81=1.96 N

  3. The apparent weight is the actual weight minus the upthrust:

    Wapparent=6.0−1.96=4.04 NW_{\text{apparent}} = 6.0 - 1.96 = 4.04\ \text{N}Wapparent​=6.0−1.96=4.04 N

    So the object would appear to weigh about 4.0 N when submerged.

Floating and sinking

An object floats when the upthrust can balance its weight. For a floating object:

U=WU = WU=W

This means the object displaces just enough fluid for the weight of displaced fluid to equal its own weight.

A useful comparison is between the object’s average density and the fluid density:

  • if ρobject<ρf\rho_{\text{object}} < \rho_fρobject​<ρf​, the object floats
  • if ρobject=ρf\rho_{\text{object}} = \rho_fρobject​=ρf​, the object is neutrally buoyant
  • if ρobject>ρf\rho_{\text{object}} > \rho_fρobject​>ρf​, the object sinks
Tip

Floating fraction

For a floating object of uniform density:

VsubmergedVtotal=ρobjectρf\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{object}}}{\rho_f}Vtotal​Vsubmerged​​=ρf​ρobject​​

This is why a low-density object floats with only part of its volume underwater.

Viscous drag

When an object moves through a fluid, the fluid exerts a resistive force on it. This force is called drag. In liquids, especially for small slow-moving objects, the drag is often described as viscous drag.

Definition

Viscosity

Viscosity is a measure of how strongly a fluid resists flow. A fluid with high viscosity, such as glycerol, flows less easily than a fluid with low viscosity, such as water.

Viscous drag acts opposite to the object’s motion. For a ball falling down through a liquid, viscous drag acts upwards.

As the ball speeds up, the viscous drag increases. This changes the resultant force, so the acceleration is not constant.

Schematic showing the forces on a sphere falling through a viscous liquid and a velocity-time graph approaching terminal velocity

Terminal velocity

At first, a falling sphere may have very little drag because its speed is small. The downward force is greater than the upward forces, so it accelerates downwards.

As its speed increases, the viscous drag increases. Eventually:

mg=U+Fdmg = U + F_dmg=U+Fd​

The resultant force is then zero, so the sphere continues at a constant speed. This constant speed is called terminal velocity, usually written as vtv_tvt​.

Definition

Terminal velocity

Terminal velocity is the constant velocity reached when the resultant force on a moving object is zero because the driving force is balanced by resistive forces.

Key Idea

Force balance at terminal velocity

For a sphere falling through a liquid at terminal velocity:

weight=upthrust+viscous drag\text{weight} = \text{upthrust} + \text{viscous drag}weight=upthrust+viscous drag

so

mg=U+Fdmg = U + F_dmg=U+Fd​

Stokes’ law

For a small sphere moving slowly through a fluid, the viscous drag is given by Stokes’ law:

Fd=6πηrvF_d = 6 \pi \eta r vFd​=6πηrv

where:

  • FdF_dFd​ is viscous drag in newtons
  • η\etaη is viscosity in pascal seconds
  • rrr is the radius of the sphere in metres
  • vvv is the speed of the sphere in metres per second

At terminal velocity, v=vtv = v_tv=vt​, so Stokes’ law can be combined with the force balance equation.

For a sphere:

V=43πr3V = \frac{4}{3}\pi r^3V=34​πr3

The effective downward force is weight minus upthrust:

mg−U=43πr3g(ρs−ρf)mg - U = \frac{4}{3}\pi r^3 g\left(\rho_s - \rho_f\right)mg−U=34​πr3g(ρs​−ρf​)

At terminal velocity:

6πηrvt=43πr3g(ρs−ρf)6\pi \eta r v_t = \frac{4}{3}\pi r^3 g\left(\rho_s - \rho_f\right)6πηrvt​=34​πr3g(ρs​−ρf​)

Rearranging gives:

η=2r2g(ρs−ρf)9vt\eta = \frac{2r^2 g\left(\rho_s - \rho_f\right)}{9v_t}η=9vt​2r2g(ρs​−ρf​)​
Common Mistake

When Stokes’ law applies

Stokes’ law is only valid for small spheres moving slowly through a fluid with smooth, laminar flow. It becomes unreliable if the flow is turbulent, the sphere is too close to the container walls, or the sphere has not yet reached terminal velocity.

Example

Calculating viscosity from terminal velocity

A steel ball of radius 1.50×10−3 m1.50 \times 10^{-3}\ \text{m}1.50×10−3 m falls through oil at terminal velocity 0.180 m s−10.180\ \text{m s}^{-1}0.180 m s−1. The density of steel is 7800 kg m−37800\ \text{kg m}^{-3}7800 kg m−3 and the density of the oil is 900 kg m−3900\ \text{kg m}^{-3}900 kg m−3. Calculate the viscosity of the oil.

  1. Use the terminal-velocity viscosity equation:

    η=2r2g(ρs−ρf)9vt\eta = \frac{2r^2 g\left(\rho_s - \rho_f\right)}{9v_t}η=9vt​2r2g(ρs​−ρf​)​

  2. Find the density difference:

    ρs−ρf=7800−900=6900 kg m−3\rho_s - \rho_f = 7800 - 900 = 6900\ \text{kg m}^{-3}ρs​−ρf​=7800−900=6900 kg m−3

  3. Substitute the values:

    η=2(1.50×10−3)2×9.81×69009×0.180\eta = \frac{2\left(1.50 \times 10^{-3}\right)^2 \times 9.81 \times 6900}{9 \times 0.180}η=9×0.1802(1.50×10−3)2×9.81×6900​

  4. Calculate the viscosity:

    η=0.188 Pa s\eta = 0.188\ \text{Pa s}η=0.188 Pa s

    To two significant figures, the oil’s viscosity is 0.19 Pa s0.19\ \text{Pa s}0.19 Pa s.

Practical link: falling-ball method

In the falling-ball method, you release a small sphere into a tall tube of liquid and measure its terminal velocity. Usually you time how long it takes to travel between two marks after it has already fallen some distance, so it is likely to be moving at terminal velocity.

Good practical technique includes:

  • measuring the ball diameter with a micrometer, then halving it to find rrr
  • using a long tube so the ball reaches terminal velocity before timing starts
  • timing over a large distance to reduce percentage uncertainty
  • repeating timings and calculating a mean
  • keeping temperature constant, because viscosity changes with temperature
  • avoiding measurements too close to the tube walls, where wall effects increase drag
Tip

Graph method

If you measure the time for several different distances after terminal velocity is reached, a graph of distance against time should be a straight line. The gradient gives vtv_tvt​.

Exam technique

In the exam

  1. Draw a force diagram before writing equations: for a falling sphere in liquid, show weight downwards, upthrust upwards, and drag upwards.
  2. Check whether the object is fully submerged or floating before choosing VdisplacedV_{\text{displaced}}Vdisplaced​.
  3. For terminal velocity questions, set the resultant force to zero: usually mg=U+Fdmg = U + F_dmg=U+Fd​.
Self review

Check yourself

  • Why does pressure in a fluid produce an upward resultant force on a submerged object?
  • A wooden block floats with 70% of its volume underwater. What does that tell you about its density compared with the liquid?
  • In a falling-ball viscosity experiment, why should timing start only after the ball has already fallen some distance?

Density, upthrust and viscous drag Revision Guide