Density, Upthrust & Viscous Drag
Why does a steel ship float when a steel bolt sinks? Why does a marble fall slowly through honey, and faster when the honey is warm? This topic is about how objects behave in fluids (liquids and gases). Three ideas do all the work: density, upthrust and viscous drag. Put them together and you can predict a ball's terminal velocity, which is exactly how Core Practical 2 measures the viscosity of a liquid.
Density
Density tells you how tightly mass is packed into a material: how many kilograms there are in each cubic metre.
Density
A litre of water fills \(1000\ \text{cm}^3 = 1.0\times10^{-3}\ \text{m}^3\) and has a mass of 1.0 kg, so its density is \(1.0 / 1.0\times10^{-3} = 1000\ \text{kg m}^{-3}\). Density belongs to the material, not to the object: saw a steel bar in half and the mass and the volume both halve, so each half still has \(\rho = 7800\ \text{kg m}^{-3}\).
Density is mass per unit volume, \(\rho = m/V\). Its SI unit is kg m⁻³.
| Material | Density / kg m⁻³ | Material | Density / kg m⁻³ |
|---|---|---|---|
| air (20 °C) | 1.2 | sea water | 1030 |
| cork | ≈ 240 | glycerol | 1260 |
| pine wood | ≈ 500 | aluminium | 2700 |
| ice | 917 | steel | ≈ 7800 |
| vegetable oil | ≈ 920 | mercury | 13 600 |
| fresh water | 1000 | gold | 19 300 |
Volume units are the classic trap. \(1\ \text{cm} = 10^{-2}\ \text{m}\), so \(1\ \text{cm}^3 = (10^{-2})^3\ \text{m}^3 = 10^{-6}\ \text{m}^3\), and \(1\ \text{mm}^3 = 10^{-9}\ \text{m}^3\). A handy check: \(1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}\), so water is 1.00 g cm⁻³.
Measuring density
- Regular solid: mass on a balance; measure the sides with a ruler, vernier calipers or micrometer and calculate the volume.
- Irregular solid: mass on a balance; find the volume by displacement. Lower it into a measuring cylinder of water and read the rise in level, or use a displacement (Eureka) can and collect the water that overflows.
- Liquid: weigh an empty measuring cylinder, pour in a measured volume and weigh it again.
A crown has a mass of 1.20 kg. When it is lowered into a full displacement can, 75 cm³ of water overflows. The density of gold is 19 300 kg m⁻³. Is the crown pure gold?
Show solution
The crown displaces its own volume: \(V = 75\ \text{cm}^3 = 75\times10^{-6}\ \text{m}^3\).
\[ \rho = \frac{m}{V} = \frac{1.20}{75\times10^{-6}} = 1.6\times10^{4}\ \text{kg m}^{-3} \]That is well below 19 300 kg m⁻³, so the crown is not pure gold; it has been mixed with a less dense metal. A pure gold crown of the same mass would displace only \(1.20 / 19\,300 = 6.2\times10^{-5}\ \text{m}^3 = 62\ \text{cm}^3\). (This is the problem Archimedes is said to have solved in his bath.)
Upthrust
Push an empty plastic bottle under water and you can feel it shove back up. Every object in a fluid feels an upward force from that fluid. This force is called upthrust, and it obeys a beautifully simple rule.
Archimedes' principle
The second form is just the first one written out: the displaced fluid has mass \(\rho_\text{f} V_\text{disp}\) (from \(\rho = m/V\)), and its weight is that mass times \(g\).
Why the rule works
Here is a lovely thought experiment. Take the object out and imagine the hole it leaves is filled with more of the fluid, the same shape as the object. That blob of fluid just sits there: it doesn't sink or rise, because it's simply part of the fluid. So the surrounding fluid must be pushing up on the blob with a force equal to the blob's weight.
Now put the object back. The surrounding fluid can't tell what is inside that surface: it pushes on the object in exactly the same way it pushed on the blob. So the upthrust on the object equals the weight of the fluid it has displaced, whatever the object is made of.
What upthrust does and doesn't depend on
- It depends on the volume submerged and the density of the fluid. Salt water gives more upthrust than fresh water.
- It does not depend on the object's mass or material. A steel ball and a plastic ball of the same size, both fully under water, feel the same upthrust.
- Once an object is fully submerged, pushing it deeper doesn't change the upthrust (liquids are almost incompressible, so the displaced volume stays the same).
Upthrust is the upward force that a fluid exerts on an object that is wholly or partly immersed in it. It is equal to the weight of the fluid displaced.
Weighing things in water
Hang an object from a newton meter and lower it into water: the reading falls. The newton meter now only has to supply the part of the weight that the upthrust doesn't, so the reading is \(W - U\), sometimes called the apparent weight. A displacement can lets you check Archimedes' principle directly: the water that overflows is the water displaced, so you can weigh it.
A stone hangs from a newton meter. The reading is 5.0 N in air and 3.2 N when the stone is completely submerged in water (density 1000 kg m⁻³). Calculate (a) the upthrust, (b) the volume of the stone and (c) its density.
Show solution
(a) The newton meter reads \(W - U\), so
\[ U = 5.0 - 3.2 = 1.8\ \text{N} \](b) The upthrust is the weight of water displaced, and the stone displaces its own volume:
\[ U = \rho_\text{w} V g \;\Rightarrow\; V = \frac{U}{\rho_\text{w} g} = \frac{1.8}{1000 \times 9.81} = 1.83\times10^{-4}\ \text{m}^3 \](c) Mass from the weight in air, then density:
\[ m = \frac{W}{g} = \frac{5.0}{9.81} = 0.510\ \text{kg}, \qquad \rho = \frac{m}{V} = \frac{0.510}{1.83\times10^{-4}} = 2.8\times10^{3}\ \text{kg m}^{-3} \]Shortcut worth noticing: \(W/U = \rho_\text{stone} V g / \rho_\text{w} V g = \rho_\text{stone}/\rho_\text{w}\), so \(\rho_\text{stone} = 1000 \times 5.0/1.8 = 2800\ \text{kg m}^{-3}\) in one line.
Air is a fluid too, so everything in air feels a small upthrust. A party balloon 30 cm across displaces about 0.014 m³ of air, giving an upthrust of \(1.2 \times 0.014 \times 9.81 \approx 0.17\ \text{N}\). Fill it with helium (much less dense than air) and that is more than its weight, so it rises. For dense objects like a steel ball, upthrust in air is tiny and exam questions let you ignore it, but in a liquid you must include it.
Floating and sinking
Take an object of volume \(V\) and push it completely under. Its weight is \(W = \rho_\text{obj} V g\) and the upthrust is \(U = \rho_\text{f} V g\). The \(V\) and \(g\) are the same in both, so it all comes down to comparing densities:
- \(\rho_\text{obj} > \rho_\text{f}\): \(W > U\), so it sinks. On the bottom, the floor supplies the missing force: \(W = U + R\), where \(R\) is the normal contact force.
- \(\rho_\text{obj} < \rho_\text{f}\): \(U > W\), so it rises until part of it sticks out. As it rises, the submerged volume shrinks and so does \(U\), until \(U = W\) and it floats.
- \(\rho_\text{obj} = \rho_\text{f}\): it stays wherever you put it (neutral buoyancy, like a fish or a submarine).
A floating object is in equilibrium, so the resultant force on it is zero:
\[ U = W \]The upthrust is the weight of the fluid displaced by the submerged volume \(V_\text{sub}\). The weight uses the whole volume \(V\):
\[ \rho_\text{f} V_\text{sub}\, g = \rho_\text{obj} V g \]Cancel \(g\) and divide both sides by \(\rho_\text{f} V\):
\[ \boxed{\frac{V_\text{sub}}{V} = \frac{\rho_\text{obj}}{\rho_\text{f}}} \]\(g\) cancels, so a boat floats at the same level on the Moon.
Ice (917 kg m⁻³) in sea water (1030 kg m⁻³) floats with \(917/1030 = 0.89\) of its volume under the surface, which is why "the tip of the iceberg" is only about a tenth of it. A steel ship floats because what matters is its average density: the hull encloses a huge volume of air, so the ship's total mass divided by the volume of the hull is less than 1030 kg m⁻³. Load it with cargo and it sits lower in the water, displacing more water to balance the extra weight.
A rectangular block of wood measures 0.50 m × 0.20 m × 0.10 m. It floats in water (density 1000 kg m⁻³) with its largest face horizontal and 6.5 cm of its 10 cm thickness under the surface. (a) Calculate the density of the wood. (b) Calculate the smallest mass that could be placed on top of the block to push it completely under.
Show solution
(a) The cross-section is the same all the way down, so the fraction of the volume submerged is the fraction of the thickness submerged:
\[ \frac{\rho_\text{wood}}{\rho_\text{w}} = \frac{6.5}{10} \;\Rightarrow\; \rho_\text{wood} = 0.65 \times 1000 = 650\ \text{kg m}^{-3} \](b) Volume \(V = 0.50 \times 0.20 \times 0.10 = 0.010\ \text{m}^3\). Fully submerged, the upthrust and the block's weight are
\[ U = 1000 \times 0.010 \times 9.81 = 98.1\ \text{N}, \qquad W = 650 \times 0.010 \times 9.81 = 63.8\ \text{N} \]The extra weight needed is \(98.1 - 63.8 = 34.3\ \text{N}\), so the mass is
\[ m = \frac{34.3}{9.81} = 3.5\ \text{kg} \]Viscosity and Stokes' law
Viscosity is how "thick" a fluid is: its resistance to flowing. Honey has a high viscosity and water a low one. When a fluid flows, neighbouring layers slide past each other, and viscosity is like an internal friction between those layers. Its symbol is \(\eta\) (eta).
Now drop a ball into the fluid. The layer of fluid touching the ball is dragged along with it, the next layer out is dragged a little less, and so on: the ball has to keep shearing the fluid around it. That takes a force, so the fluid pulls back on the ball. This is viscous drag. For a small sphere moving slowly, George Stokes worked out exactly how big it is.
Stokes' law: viscous drag on a sphere
Why each factor is there
- \(\eta\): a stickier fluid resists the shearing more.
- \(v\): moving faster means the layers shear past each other faster. At low speeds the drag is simply proportional to the speed.
- \(r\): a bigger sphere drags a bigger region of fluid along with it.
- \(6\pi\): this number comes from the geometry of a sphere. Working it out needs maths well beyond A level, so just use it.
Rearranging gives the unit of viscosity: \(\eta = F/(6\pi r v)\) has units \(\text{N} / (\text{m} \times \text{m s}^{-1}) = \text{N s m}^{-2}\). Since \(1\ \text{N m}^{-2} = 1\ \text{Pa}\), that is the pascal second, Pa s, which in base units is kg m⁻¹ s⁻¹.
When can you use Stokes' law?
Only for small, spherical objects moving at low speeds with laminar flow (no turbulence) around them. A raindrop in air or a ball bearing in thick oil is fine; a football, a car or a skydiver is not. And because viscosity is temperature dependent, the value of \(\eta\) you use must be the one for the temperature of the fluid.
Viscosity depends on temperature
For liquids, viscosity falls steeply as the temperature rises. Golden syrup pours easily after the tin has stood in hot water, and cold engine oil is so thick that engines wear fastest just after starting. Glycerol is about five times less viscous at 40 °C than at 20 °C. So any viscosity value only means something if you also give the temperature, and in an experiment you must measure the temperature and keep it steady.
| Fluid (at 20 °C) | Viscosity \(\eta\) / Pa s |
|---|---|
| air | 1.8 × 10⁻⁵ |
| water | 1.0 × 10⁻³ |
| olive oil | ≈ 0.08 |
| castor oil | ≈ 1.0 |
| glycerol | ≈ 1.4 |
| honey | roughly 2 to 10 (depends on the honey) |
Laminar and turbulent flow
To picture a flow we draw streamlines: lines that show the path a small piece of the fluid follows. Where the streamlines squeeze closer together, the fluid is moving faster (the same amount of fluid is getting through a narrower gap).
- In laminar flow the fluid moves in smooth layers that slide past each other without mixing. The streamlines are smooth and never cross, and the velocity at any point stays the same over time.
- In turbulent flow the fluid swirls into eddies and mixes. The flow pattern keeps changing, and the speed and direction at any point change erratically.
Laminar flow: the fluid flows in layers with no mixing between them; streamlines do not cross and the velocity at each point is constant.
Turbulent flow: the fluid mixes in irregular eddies; the speed and direction of flow at each point change with time.
Flow round an object tends to be laminar when the object is small, moving slowly and the fluid is viscous. Speed it up and a wake of eddies forms behind it. Stokes' law assumes the smooth, layered flow, so it only works in the laminar case. Once there is turbulence the object also has to keep setting the fluid swirling, which costs extra energy, so the drag becomes bigger than \(6\pi\eta r v\).
A favourite question: "Explain why Stokes' law cannot be used to find the drag on a skydiver." Give the conditions it breaks: a skydiver is not small, not spherical and moves fast, so the air flow around them is turbulent.
Terminal velocity
Let's put the three ideas together. A small sphere of radius \(r\) and density \(\rho_\text{s}\) is released from rest in a liquid of density \(\rho_\text{f}\) and viscosity \(\eta\). Three forces act on it:
- Weight \(W = mg = \tfrac{4}{3}\pi r^3 \rho_\text{s} g\), downwards.
- Upthrust \(U = \tfrac{4}{3}\pi r^3 \rho_\text{f} g\), upwards (the weight of liquid displaced).
- Viscous drag \(F = 6\pi\eta r v\), upwards (always opposite to the velocity).
- At the instant of release \(v = 0\), so there is no drag. The resultant force is \(W - U\) and the ball accelerates downwards (with less than \(g\), because of the upthrust).
- As the ball speeds up, the drag grows in proportion to \(v\). The resultant force \(W - U - F\) shrinks, so the acceleration gets smaller and smaller.
- Eventually \(F\) has grown so that \(W = U + F\). The resultant force is zero, so (by Newton's first law) the ball carries on at a constant velocity: its terminal velocity.
On a velocity–time graph this gives a curve that starts steep (gradient = initial acceleration) and flattens off towards a horizontal line at the terminal velocity.
Terminal velocity is the constant, maximum velocity of an object falling through a fluid, reached when the resultant force on it is zero (for a sphere in a liquid, weight = upthrust + viscous drag).
At terminal velocity the resultant force is zero, so the upward forces balance the weight:
\[ W = U + F \]Write the weight using the density of the sphere. Its volume is \(\tfrac{4}{3}\pi r^3\), so its mass is \(\tfrac{4}{3}\pi r^3 \rho_\text{s}\):
\[ W = \tfrac{4}{3}\pi r^3 \rho_\text{s}\, g \]The upthrust is the weight of the liquid displaced, which has the same volume but the liquid's density:
\[ U = \tfrac{4}{3}\pi r^3 \rho_\text{f}\, g \]The drag at terminal velocity \(v_\text{t}\) comes from Stokes' law:
\[ F = 6\pi \eta r v_\text{t} \]Substitute all three into \(W = U + F\) and collect the two "volume" terms:
\[ 6\pi \eta r v_\text{t} = \tfrac{4}{3}\pi r^3 g\,(\rho_\text{s} - \rho_\text{f}) \]\(\rho_\text{s} - \rho_\text{f}\) appears because the upthrust "cancels out" part of the weight.
Divide both sides by \(6\pi\eta r\). The \(\pi\) cancels and one power of \(r\) cancels, and \(\tfrac{4}{3} \div 6 = \tfrac{4}{18} = \tfrac{2}{9}\):
\[ \boxed{v_\text{t} = \frac{2 r^2 g\,(\rho_\text{s} - \rho_\text{f})}{9\eta}} \]Rearranged for the viscosity, which is what Core Practical 2 uses:
\[ \eta = \frac{2 r^2 g\,(\rho_\text{s} - \rho_\text{f})}{9 v_\text{t}} \]The result tells you a lot at a glance:
- \(v_\text{t} \propto r^2\): double the radius and the terminal velocity goes up four times. The weight (minus upthrust) grows with \(r^3\) but the drag only with \(r\), so big balls win.
- \(v_\text{t} \propto 1/\eta\): warm the liquid, its viscosity falls, and balls fall faster.
- If \(\rho_\text{s} < \rho_\text{f}\) the formula gives a negative answer: the sphere rises at a steady speed instead, like a bubble in a fizzy drink.
A steel ball bearing of radius 1.0 mm falls at a terminal velocity of 1.2 cm s⁻¹ through an oil of density 960 kg m⁻³. The density of steel is 7800 kg m⁻³. (a) Calculate the weight of the ball and the upthrust on it. (b) Calculate the viscosity of the oil. (c) State one assumption you have made.
Show solution
(a) Volume \(V = \tfrac{4}{3}\pi (1.0\times10^{-3})^3 = 4.19\times10^{-9}\ \text{m}^3\).
\[ W = \rho_\text{s} V g = 7800 \times 4.19\times10^{-9} \times 9.81 = 3.21\times10^{-4}\ \text{N} \] \[ U = \rho_\text{f} V g = 960 \times 4.19\times10^{-9} \times 9.81 = 3.95\times10^{-5}\ \text{N} \](b) At terminal velocity \(F = W - U = 3.21\times10^{-4} - 0.39\times10^{-4} = 2.81\times10^{-4}\ \text{N}\). From Stokes' law:
\[ \eta = \frac{F}{6\pi r v} = \frac{2.81\times10^{-4}}{6\pi \times 1.0\times10^{-3} \times 0.012} = 1.2\ \text{Pa s} \](The formula \(\eta = 2r^2 g(\rho_\text{s}-\rho_\text{f})/9v_\text{t}\) gives the same answer in one step.)
(c) Any one of: the flow round the ball is laminar (so Stokes' law applies); the ball is a perfect sphere; the ball had reached terminal velocity; the temperature of the oil was constant; the walls of the container don't affect the motion.
Don't write "at terminal velocity, weight = drag". For an object in a liquid the upthrust is a significant part of the balance: weight = upthrust + drag. Only drop the upthrust when the fluid is a gas and the object is much denser than it.
Core Practical 2: viscosity by the falling-ball method
Use a falling-ball method to determine the viscosity of a liquid
A tall measuring cylinder is filled with a viscous, transparent liquid such as glycerol or washing-up liquid. Rubber bands round the cylinder mark out timing points. Small steel ball bearings are dropped in and timed between the bands.
- Measure the diameter of each ball with a micrometer at several orientations and average; halve it to get \(r\).
- Find the densities: \(\rho_\text{s}\) from the mass and volume of a number of balls, and \(\rho_\text{f}\) by weighing a measured volume of the liquid (or use given values).
- Place the top band well below the surface, so the ball is at terminal velocity before timing starts. Measure the distance \(d\) between bands with a metre rule, with your eye level with each band.
- Drop the ball gently in the centre of the cylinder (tweezers, or a short tube held just above the surface). Time it between the bands with a stopwatch, or film it beside a ruler and step through the video frame by frame. Then \(v = d/t\).
- Check terminal velocity has been reached: use three equally spaced bands and confirm the two time intervals are the same.
- Repeat for each ball size, measure the temperature of the liquid with a thermometer, and keep it constant.
Graph: \(v_\text{t} = \dfrac{2g(\rho_\text{s}-\rho_\text{f})}{9\eta}\, r^2\) has the form \(y = mx\). Plot \(v\) against \(r^2\): a straight line through the origin with
\[ \text{gradient} = \frac{2g(\rho_\text{s}-\rho_\text{f})}{9\eta} \quad\Rightarrow\quad \eta = \frac{2g(\rho_\text{s}-\rho_\text{f})}{9 \times \text{gradient}} \]Sources of uncertainty and how to reduce them
- Reaction time when starting and stopping the stopwatch: use bands far apart so the times are long, repeat and average, or use video analysis or light gates.
- Parallax when judging when the ball passes a band: keep your eye level with the band.
- Radius: \(r\) is squared, so its percentage uncertainty is doubled. Use a micrometer, not a ruler.
- Temperature: \(\eta\) changes a lot with temperature, so measure it and avoid warming the liquid (don't hold the cylinder).
- Walls: a ball close to the wall of a narrow tube is slowed down. Use a wide cylinder and drop balls down the middle.
- Air bubbles sticking to the ball add upthrust: wet the balls in the liquid before dropping them.
Safety: wipe up spills straight away (glycerol makes the floor slippery), clamp the tall cylinder so it can't topple, and retrieve the balls with a magnet.
In a falling-ball experiment a student plots terminal velocity \(v\) against \(r^2\) for steel balls (density 7800 kg m⁻³) falling through a liquid of density 1260 kg m⁻³. The best-fit line passes through the origin and has a gradient of \(7.6\times10^{3}\ \text{m}^{-1}\ \text{s}^{-1}\). (a) Calculate the viscosity of the liquid. (b) The student's value is higher than the value in a data book. Suggest one reason why.
Show solution
(a) The gradient is \(2g(\rho_\text{s}-\rho_\text{f})/9\eta\), so
\[ \eta = \frac{2g(\rho_\text{s}-\rho_\text{f})}{9 \times \text{gradient}} = \frac{2 \times 9.81 \times (7800 - 1260)}{9 \times 7.6\times10^{3}} = 1.9\ \text{Pa s} \]Check the units: \(\text{m s}^{-2} \times \text{kg m}^{-3} / (\text{m}^{-1}\,\text{s}^{-1}) = \text{kg m}^{-1}\,\text{s}^{-1} = \text{Pa s}\).
(b) The liquid was colder than the data-book temperature (viscosity rises as temperature falls); or the walls of a narrow cylinder slowed the balls, making \(v\) too small and \(\eta\) too big.
Quick checks
A block floats in water (density 1000 kg m⁻³) with three-quarters of its volume below the surface. What is the density of the block?
For a floating object \(V_\text{sub}/V = \rho_\text{obj}/\rho_\text{f}\), so \(\rho_\text{obj} = 0.75 \times 1000 = 750\ \text{kg m}^{-3}\). 1333 comes from dividing the wrong way round; a floating object can't be denser than the liquid.
A steel ball is fully submerged in water and then lowered deeper. What happens to the upthrust on it?
Upthrust is the weight of fluid displaced, and a fully submerged ball always displaces its own volume. The pressure is bigger on both its top and its bottom, but the difference between them stays the same. A sinking object still feels upthrust; it just isn't enough to balance its weight.
Which of these is a correct unit for viscosity?
\(\eta = F/6\pi r v\) has units N/(m × m s⁻¹) = N s m⁻² = Pa s. Writing N as kg m s⁻² gives kg m s⁻² × s × m⁻² = kg m⁻¹ s⁻¹.
Two steel balls fall through the same glycerol. Ball Q has twice the radius of ball P. Stokes' law applies to both. What is the ratio (terminal velocity of Q) / (terminal velocity of P)?
\(v_\text{t} \propto r^2\). Doubling \(r\) makes the weight minus upthrust 8 times bigger (it goes as \(r^3\)), but the drag at a given speed only doubles (it goes as \(r\)), so the speed must be 4 times bigger for the forces to balance. "Everything falls at the same rate" is only true with no drag and no upthrust.
The glycerol in a falling-ball experiment warms up from 18 °C to 25 °C during the lesson. What happens to the terminal velocity of a ball bearing?
The viscosity of a liquid falls as its temperature rises, and \(v_\text{t} \propto 1/\eta\). The density of glycerol does fall very slightly when it warms, but that effect is tiny compared with the large drop in viscosity.
A small sphere is falling through oil at its terminal velocity. Which statement is correct?
At terminal velocity the resultant force is exactly zero: \(W = U + F\), so \(F = W - U\). If the upthrust equalled the weight, the sphere would not be falling at all.