1.7 Density, upthrust and viscous drag
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Unit 1 Mechanics and Materials, AS content, including Core Practical 2. Free-body diagrams and the general idea of terminal velocity are in 1.2; fluid pressure is used here only to explain where upthrust comes from.
What this page covers
8 points
- Density, \(\rho = m/V\), and how to measure it for regular and irregular solids and for liquids
- Upthrust as the weight of fluid displaced, and why it exists
- Floating and sinking, and the fraction of a floating body below the surface
- Hydrometers and hot-air balloons
- Viscosity, laminar and turbulent flow, and how viscosity depends on temperature
- Stokes' law, \(F = 6\pi\eta rv\), and when it can be used
- Terminal velocity of a sphere in a liquid: weight = upthrust + viscous drag
- Core Practical 2: measuring viscosity by the falling-ball method
In one minute
6 points
- A fluid exerts two forces on a body inside it, upthrust (always up, equal to the weight of fluid displaced) and viscous drag (opposing motion), and with the weight these explain floating, balloons and terminal velocity in a liquid.
- Density \(\rho = m/V\), in kg m⁻³; 1 g cm⁻³ = 1000 kg m⁻³.
- Upthrust = weight of fluid displaced = \(\rho_{\text{fluid}}V_{\text{displaced}}\,g\). It exists because fluid pressure increases with depth.
- A body floats when upthrust = weight; the fraction submerged is \(\rho_{\text{body}}/\rho_{\text{fluid}}\).
- Stokes' law \(F = 6\pi\eta rv\) gives the drag on a small sphere moving slowly in laminar flow. The viscosity \(\eta\) (Pa s) of a liquid falls steeply as it warms.
- At terminal velocity in a liquid \(W = U + F\), which gives \(v = 2r^2g(\rho_s - \rho_f)/9\eta\); Core Practical 2 finds \(\eta\) from the gradient of \(v\) against \(r^2\).
Contents
13 sections
- What is density, and how do you measure it?
- What is upthrust, and why does it exist?
- Why do some objects float?
- How do hydrometers and hot-air balloons use upthrust?
- What is viscosity, and what is Stokes' law?
- How does a sphere reach terminal velocity in a liquid?
- How do you measure viscosity with a falling ball (Core Practical 2)?
- Exam tips for 1.7
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is density, and how do you measure it?
Because 1 cm³ = 10⁻⁶ m³ and 1 g = 10⁻³ kg, 1 g cm⁻³ = 10⁻³ ÷ 10⁻⁶ = 1000 kg m⁻³. One litre is 1000 cm³ = 10⁻³ m³.
| Substance | Typical density / kg m⁻³ | Substance | Typical density / kg m⁻³ |
|---|---|---|---|
| air (room temperature) | 1.2 | glycerol | 1260 |
| pine wood | about 550 | aluminium | 2700 |
| ice | 917 | steel | 7800 |
| fresh water | 1000 | lead | 11 300 |
| sea water | about 1030 | gold | 19 300 |
Measuring density
- Regular solid: mass on a top-pan balance; dimensions with a ruler, vernier calipers or a micrometer; then calculate the volume (cuboid \(lwh\), sphere \(\tfrac43\pi r^3\), cylinder \(\pi r^2h\)).
- Irregular solid: find the volume by displacement. Lower it into water in a measuring cylinder and record the rise in level, or use a displacement (eureka) can and collect the overflow.
- Liquid: weigh an empty measuring cylinder, add a measured volume of liquid, weigh again; the difference is the mass of the liquid.
- Read a measuring cylinder at eye level, from the bottom of the meniscus, to avoid parallax.
A pebble of mass 64.8 g is lowered into a measuring cylinder containing water. The water level rises from 50.0 cm³ to 74.0 cm³. Calculate the density of the pebble in kg m⁻³.
Volume of pebble = volume of water displaced: \(V = 74.0 - 50.0 = 24.0\ \text{cm}^3 = 24.0\times10^{-6}\ \text{m}^3\). Mass: \(m = 64.8\times10^{-3}\ \text{kg}\).
\(\rho = m/V = 64.8\times10^{-3} \div 24.0\times10^{-6}\) → ρ = 2.70 × 10³ kg m⁻³.
Quick check: \(64.8 \div 24.0 = 2.70\ \text{g cm}^{-3}\), and × 1000 gives the same answer.
Converting cm³ to m³ by dividing by 100. A cube 1 m across is 100 cm × 100 cm × 100 cm, so 1 m³ = 10⁶ cm³. A density that comes out a thousand times too big or too small usually means this step went wrong.
What is upthrust, and why does it exist?
A fluid (liquid or gas) pushes on every surface in contact with it, at right angles to the surface. The pressure in a fluid increases with depth, because there is more fluid above pushing down. So the bottom of an immersed body is pushed up harder than its top is pushed down. The sideways pushes at each depth cancel. The resultant of all these pushes is the upthrust, acting vertically upwards.
The upthrust does not depend on what the body is made of, only on the volume it displaces and the density of the fluid. That is why it equals the weight of the fluid displaced: the same pushes would exactly hold up a "body" made of the fluid itself, which simply sits in equilibrium.
For a fully submerged body \(V_{\text{displaced}}\) is the whole volume of the body. For a floating body it is only the volume below the surface.
An aluminium block measures 5.0 cm × 4.0 cm × 3.0 cm. It hangs from a newton-meter and is lowered until it is completely under water. Density of aluminium = 2700 kg m⁻³; density of water = 1000 kg m⁻³. Calculate the reading on the newton-meter when the block is fully submerged.
Volume: \(V = 5.0\times4.0\times3.0 = 60\ \text{cm}^3 = 6.0\times10^{-5}\ \text{m}^3\).
Weight: \(W = \rho Vg = 2700\times6.0\times10^{-5}\times9.81 = 1.589\ \text{N}\). Upthrust = weight of water displaced: \(U = 1000\times6.0\times10^{-5}\times9.81 = 0.589\ \text{N}\).
Equilibrium, so the newton-meter supplies the rest: \(T = W - U\) → T = 1.00 N.
The block's weight has not changed; the newton-meter reads less because the water now carries part of the weight.
Why do some objects float?
When a body is released in a fluid, compare its weight \(W\) with the upthrust \(U_{\max}\) it would feel if fully submerged:
- \(U_{\max} > W\): it rises, and floats with part of it above the surface.
- \(U_{\max} = W\): it stays wherever it is put, like a submarine trimmed for neutral buoyancy.
- \(U_{\max} < W\): it sinks.
Both forces depend on the same volume, so this is the same as comparing densities: a body floats if its average density is less than the fluid's. A steel ship floats because its average density, steel plus the air inside the hull, is less than that of water.
A floating body is in equilibrium, \(U = W\). It sinks just far enough to displace its own weight of fluid. Setting \(\rho_{\text{fluid}}V_{\text{sub}}g = \rho_{\text{body}}Vg\) and cancelling \(g\) gives the fraction submerged:
\[\frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}}\]A cylindrical navigation buoy has a mass of 85 kg and a radius of 0.30 m. It floats upright in sea water of density 1030 kg m⁻³. (a) Calculate the depth of the buoy below the surface. (b) Calculate the extra mass that must be attached on top so that its base is 0.50 m below the surface.
(a) Floating, so weight of sea water displaced = weight of buoy, and \(g\) cancels: \(\rho(\pi r^2h) = m\), so \(h = 85 \div (1030\times\pi\times0.30^2) = 85 \div (1030\times0.283) = 0.292\ \text{m}\) → h = 0.29 m.
(b) Mass of sea water displaced at 0.50 m depth: \(1030\times0.283\times0.50 = 145.6\ \text{kg}\). This must equal the total floating mass, so the extra mass \(= 145.6 - 85 = 60.6\ \text{kg}\) → 61 kg.
For any floating body, start by writing "floating, so upthrust = weight". Then write each side as \(\rho Vg\) and cancel \(g\). Make sure the volume on the upthrust side is only the part below the surface.
How do hydrometers and hot-air balloons use upthrust?
The hydrometer
A hydrometer is a sealed glass float: a weighted bulb with a narrow, uniform stem marked with a density scale. Its weight is fixed, so it always floats with the same weight of liquid displaced. In a denser liquid a smaller volume is enough, so it floats higher. The density is read where the liquid surface crosses the stem, which means higher densities are marked lower down the stem.
The scale is not uniform. Each equal step in density needs a smaller change of submerged volume at higher densities, so the marks crowd together towards the bulb. A narrow stem makes it sensitive.
The hydrometer in the figure is placed in a sample of fruit juice. It floats with 6.2 cm of its stem below the surface. Calculate the density of the juice in kg m⁻³.
Volume of juice displaced = bulb + submerged part of the stem: \(V = 16.0 + 0.50\times6.2 = 19.1\ \text{cm}^3\).
Floating, so mass of juice displaced = mass of hydrometer: \(\rho = 20.0 \div 19.1 = 1.047\ \text{g cm}^{-3}\) → ρ = 1.05 × 10³ kg m⁻³.
The hot-air balloon
The envelope displaces its own volume of the cool outside air, so the upthrust is \(\rho_{\text{cool}}Vg\). Heating the air inside makes it expand; some leaves through the open bottom, so the air left inside is less dense and weighs \(\rho_{\text{hot}}Vg\). The balloon rises when the upthrust is greater than the total weight: hot air + envelope + basket + burner + people.
The useful lift per cubic metre is \((\rho_{\text{cool}} - \rho_{\text{hot}})g\), only a few newtons, which is why envelopes are so large. On a hot day, or at altitude, the possible load is smaller.
A balloon envelope holds 2800 m³ of air. The outside air has a density of 1.20 kg m⁻³ and the hot air inside has a density of 0.950 kg m⁻³. Calculate (a) the upthrust on the balloon and (b) the maximum total mass of envelope, basket, burner and passengers for the balloon to lift off. Ignore the volume of everything except the envelope.
(a) Upthrust = weight of cool air displaced: \(U = 1.20\times2800\times9.81 = 32\,962\ \text{N}\) → U = 3.30 × 10⁴ N.
(b) Weight of the hot air: \(0.950\times2800\times9.81 = 26\,095\ \text{N}\). Lift-off when \(U \ge\) weight of hot air + \(Mg\): \(M = (32\,962 - 26\,095)\div9.81\) → M = 700 kg.
Shortcut: \(M = (\rho_{\text{cool}} - \rho_{\text{hot}})V = (1.20 - 0.950)\times2800 = 700\ \text{kg}\).
Forgetting the weight of the hot air. The upthrust comes from the cool air displaced, but the balloon must also carry the hot air inside it. Using \(U\) alone would suggest a load of 3360 kg, almost five times too big.
What is viscosity, and what is Stokes' law?
Viscosity is a measure of a fluid's resistance to flow, how "thick" it is. Treacle and glycerol have high viscosity; water and air have low viscosity. When a body moves through a fluid, the fluid next to its surface moves with it and drags on the layers beyond. The result is a viscous drag force opposing the motion.
The coefficient of viscosity \(\eta\) (eta) has the unit Pa s (equivalently N s m⁻² or kg m⁻¹ s⁻¹). Typical values at 20 °C: air 1.8 × 10⁻⁵ Pa s, water 1.0 × 10⁻³ Pa s, glycerol about 1.4 Pa s.
Laminar and turbulent flow
| Laminar (streamline) flow | Turbulent flow | |
|---|---|---|
| Description | the fluid moves in smooth layers that slide past one another; streamlines never cross | the fluid swirls in eddies; the layers mix and the velocity at a point changes irregularly |
| Velocity at a fixed point | constant in time | changes randomly in size and direction |
| When it happens | low speeds, small bodies, viscous fluids, streamlined shapes | high speeds, large bodies, low-viscosity fluids, blunt shapes |
| Drag | relatively small; proportional to speed | much larger; grows faster than speed |
\(F\) = viscous drag (N), \(\eta\) = viscosity of the fluid (Pa s), \(r\) = radius of the sphere (m), \(v\) = speed of the sphere relative to the fluid (m s⁻¹). Valid only for a small sphere moving at low speed so that the flow is laminar, in a fluid that extends far from the sphere (well away from the container walls). Use the radius, not the diameter.
Viscosity depends on temperature
For liquids, viscosity decreases sharply as the temperature rises: glycerol is about 1.4 Pa s at 20 °C but only about 0.6 Pa s at 30 °C. (Gas viscosity rises slowly with temperature.) So any viscosity value must be quoted with its temperature, and in an experiment the temperature must be measured and kept constant.
How does a sphere reach terminal velocity in a liquid?
Three forces act on a sphere falling through a liquid: weight \(W\) down, upthrust \(U\) up, and viscous drag \(F\) up, opposing the motion. \(W\) and \(U\) are constant, but \(F\) starts at zero and increases with speed. The resultant force \(W - U - F\) therefore shrinks as the sphere speeds up, so the acceleration falls. When \(F\) has grown until \(W = U + F\), the resultant force is zero and the sphere moves at its terminal velocity.
With Stokes' law and \(V = \tfrac43\pi r^3\):
\[\tfrac43\pi r^3\rho_s g = \tfrac43\pi r^3\rho_f g + 6\pi\eta rv \quad\Rightarrow\quad v = \frac{2r^2g(\rho_s - \rho_f)}{9\eta}\]Here \(\rho_s\) is the density of the sphere, \(\rho_f\) the density of the fluid and \(v\) the terminal velocity. This result is not in the formulae list: derive it from \(W = U + F\), and learn the volume of a sphere. Terminal velocity is proportional to \(r^2\): double the radius and (weight − upthrust) goes up 8 times, while the drag at a given speed goes up only 2 times, so \(v\) goes up 4 times.
A steel ball bearing of radius 1.0 mm falls through glycerol. Density of steel = 7800 kg m⁻³; density of glycerol = 1260 kg m⁻³; viscosity of glycerol = 1.4 Pa s. Calculate the terminal velocity of the ball bearing.
Volume: \(V = \tfrac43\pi(1.0\times10^{-3})^3 = 4.19\times10^{-9}\ \text{m}^3\).
Weight: \(W = 7800\times4.19\times10^{-9}\times9.81 = 3.21\times10^{-4}\ \text{N}\). Upthrust: \(U = 1260\times4.19\times10^{-9}\times9.81 = 5.18\times10^{-5}\ \text{N}\).
At terminal velocity \(F = W - U = 2.69\times10^{-4}\ \text{N}\). Stokes' law: \(v = F/(6\pi\eta r) = 2.69\times10^{-4} \div (6\pi\times1.4\times1.0\times10^{-3}) = 0.0102\ \text{m s}^{-1}\) → v = 1.0 × 10⁻² m s⁻¹.
Check with the formula: \(2\times(1.0\times10^{-3})^2\times9.81\times(7800 - 1260)\div(9\times1.4) = 0.0102\ \text{m s}^{-1}\). The speed is low and the ball is small, so the flow really is laminar and Stokes' law is justified.
Draw the three forces before writing any equation, and decide whether upthrust must be included. In air, the upthrust on a dense sphere is tiny and usually ignored; in a liquid it is a significant fraction of the weight (here 16%) and must be included.
How do you measure viscosity with a falling ball (Core Practical 2)?
Idea: measure the terminal velocity of small spheres falling through the liquid, then use \(W = U + F\) with Stokes' law to find \(\eta\).
- Apparatus: a tall, wide tube of the liquid with marks (rubber bands) at equal spacings, small steel balls, a stopwatch, a metre rule, a micrometer, a thermometer, a balance and a measuring cylinder.
- Diameter: micrometer, several readings at different orientations, averaged; halve for the radius.
- Densities: the steel from the mass and volume of a batch of balls; the liquid from the mass of a measured volume.
- Terminal velocity: set the top band far enough below the surface that the ball is already at terminal velocity. Check that the times for successive equal intervals are the same, then \(v\) = distance ÷ time.
- Release each wetted ball at the centre of the surface, away from the walls; repeat each size several times, use five or six sizes, and keep the temperature constant.
- Safety: tube clamped in a tray; wipe up slippery spills at once.
Analysing the results
Rearranged, \(v = \left[\dfrac{2g(\rho_s - \rho_f)}{9\eta}\right] r^2\), so a graph of \(v\) against \(r^2\) should be a straight line through the origin, with gradient \(2g(\rho_s - \rho_f)/9\eta\). Hence \(\eta = 2g(\rho_s - \rho_f)/(9\times\text{gradient})\). The graph uses every ball and shows up any anomaly, such as a large ball slowed by the walls.
| Diameter d / mm | r² / mm² | Mean time for 0.200 m / s | v / mm s⁻¹ |
|---|---|---|---|
| 1.50 | 0.563 | 29.34 | 6.82 |
| 2.00 | 1.000 | 17.18 | 11.6 |
| 2.50 | 1.563 | 10.67 | 18.7 |
| 3.00 | 2.250 | 7.56 | 26.5 |
| 3.50 | 3.063 | 5.42 | 36.9 |
| 4.00 | 4.000 | 4.27 | 46.8 |
Use the graph above to determine the viscosity of the liquid.
Gradient from the large triangle: \(41.4\ \text{mm s}^{-1} \div 3.50\ \text{mm}^2 = 11.83\ \text{mm s}^{-1}\,\text{mm}^{-2}\).
Convert to SI: 1 mm s⁻¹ ÷ 1 mm² = 10⁻³ ÷ 10⁻⁶ = 10³ m⁻¹ s⁻¹, so gradient = \(11.83\times10^{3}\ \text{m}^{-1}\,\text{s}^{-1}\).
\(\eta = \dfrac{2g(\rho_s - \rho_f)}{9\times\text{gradient}} = \dfrac{2\times9.81\times(7800 - 1260)}{9\times11.83\times10^{3}} = 1.205\ \text{Pa s}\) → η = 1.21 Pa s.
Uncertainties
- Diameter: a micrometer reads to 0.01 mm, so the uncertainty is half of that, 0.005 mm. For a 2.00 mm ball: 0.005 ÷ 2.00 × 100 = 0.25%.
- Time: repeat readings vary by more than the stopwatch resolution (reaction time), so use half the range. For times of 7.62 s, 7.50 s and 7.56 s: mean 7.56 s, uncertainty ± 0.06 s, which is 0.8%.
- The timing is usually the largest source of uncertainty. Longer distances between the bands, or filming the fall with a timer in view, reduce it.
If you are asked how you know the ball has reached terminal velocity, give evidence from a measurement, not an assumption: the times for two (or more) equal distances are the same, so the speed is constant.
Exam tips for 1.7
Convert every quantity to SI before substituting: g to kg, cm³ to m³ (× 10⁻⁶), mm to m, mPa s to Pa s. A power-of-ten slip in a unit conversion is the easiest mistake to make in this topic.
Stokes' law and the volume of a sphere both use the radius. Measurements are usually given as diameters, so halve them first; \(r^3\) makes a diameter mistake eight times too big.
For a sphere in a liquid write \(W = U + F\) in words first: "weight = upthrust + viscous drag".
When asked when Stokes' law applies, give all the conditions: a small sphere, at low speed, in laminar flow (and far from the container walls). "It works for spheres" is not enough.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Heavy objects sink and light objects float. | A large ship floats and a small pebble sinks. | What matters is average density compared with the fluid's, or weight compared with the maximum upthrust. |
| Upthrust depends on the material of the body. | The fluid pushes in the same way on any body of the same shape at the same depth. | Upthrust depends only on the volume displaced and the density of the fluid. |
| A body loses weight when it is put in water. | Its mass and the Earth's pull on it are unchanged. | The upthrust supports part of the weight, so a newton-meter reads less (apparent weight). |
| Upthrust grows the deeper a submerged body goes. | The pressure on every face rises by the same amount, so the difference does not change. | For a fully submerged body in a liquid of constant density, upthrust is the same at any depth. |
| Stokes' law gives the air resistance on any falling object. | Large or fast bodies cause turbulent flow, with much larger drag. | Stokes' law holds only for small spheres at low speed in laminar flow. |
| A warmer liquid is "thicker" because the molecules move faster. | For liquids, viscosity falls as temperature rises. | Warm oil and warm syrup flow more easily; terminal velocities in them are higher. |
How this topic connects
1.2 Forces and Newton's laws
Free-body diagrams, resultant force and the chain of reasoning for terminal velocity all carry over; here the drag has a formula and upthrust is added to the diagram.
1.8 Hooke's law and 3.3 The AS core practicals
Density and viscosity describe materials in bulk; the next pages describe how solids stretch. The falling-ball analysis (rearrange to \(y = mx\), plot \(v\) against \(r^2\), read a large gradient, find percentage uncertainties) is a pattern that recurs throughout practical work.
Thermal physics (Unit 5)
Hot-air balloons rely on air becoming less dense as it is heated at constant pressure, which the gas laws describe quantitatively.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\rho = m/V\) | density; \(\rho\) in kg m⁻³ | yes |
| upthrust = weight of fluid displaced | upthrust on any immersed body | yes |
| \(U = \rho_{\text{fluid}}V_{\text{displaced}}\,g\) | upthrust as a calculation; \(U\) in N | no — derive it from the line above with \(W = mg\) |
| \(F = 6\pi\eta rv\) | viscous drag on a small sphere in laminar flow (Stokes' law); \(\eta\) in Pa s | yes |
| \(V = \tfrac43\pi r^3\) | volume of a sphere | no — learn it |
| \(W = U + F\) | sphere at terminal velocity in a fluid | no — learn it |
| \(v = 2r^2g(\rho_s - \rho_f)/9\eta\) | terminal velocity of a sphere (Stokes' law) | no — derive it |
| \(V_{\text{sub}}/V = \rho_{\text{body}}/\rho_{\text{fluid}}\) | fraction of a floating body below the surface | no — derive it |
Check your understanding
1. 250 cm³ of olive oil has a mass of 229 g. Calculate its density in kg m⁻³.
\(\rho = 229 \div 250 = 0.916\ \text{g cm}^{-3} = 916\ \text{kg m}^{-3}\).
2. Explain why a steel ship floats although steel is much denser than water.
The hull encloses a large volume of air, so the average density of the ship is less than that of water. It sinks until the water it displaces weighs as much as the ship; then upthrust = weight and it floats.
3. A glass bead of radius 0.50 mm moves at 4.0 mm s⁻¹ through an oil of viscosity 0.80 Pa s. Calculate the viscous drag on it.
\(F = 6\pi\times0.80\times0.50\times10^{-3}\times4.0\times10^{-3} = 3.02\times10^{-5}\ \text{N} \approx 3.0\times10^{-5}\ \text{N}\).
4. In Core Practical 2 the liquid warms up during the lesson. Explain the effect on the measured times.
The viscosity of the liquid decreases as it warms, so the drag at a given speed is smaller and the terminal velocity is larger. The times between the bands become shorter, and the result for \(\eta\) applies to a temperature that was not constant.
5. A hydrometer floats with more of its stem below the surface in liquid X than in water. Is X more or less dense than water? Explain.
Less dense. The hydrometer must displace its own weight of liquid; it displaces a larger volume of X, so each cubic metre of X weighs less than a cubic metre of water.
Specification checklist
- (23) I can use density = mass ÷ volume, convert between g cm⁻³ and kg m⁻³, and describe how to find the density of solids and liquids.
- (24) I can explain where upthrust comes from and use upthrust = weight of fluid displaced, for submerged and for floating bodies.
- (24) I can apply upthrust to floating and sinking, hydrometers and hot-air balloons.
- (25a) I can use Stokes' law, \(F = 6\pi\eta rv\), to calculate the viscous drag on a sphere.
- (25b) I can state that Stokes' law applies only to small spheres at low speeds in laminar flow, describe laminar and turbulent flow, and explain how the viscosity of a liquid changes with temperature.
- (25b) I can explain, using weight = upthrust + viscous drag, how a sphere reaches terminal velocity in a liquid, and calculate that velocity.
- (26) I can describe and analyse the falling-ball method for the viscosity of a liquid (Core Practical 2), including how to confirm terminal velocity and how to use a graph of \(v\) against \(r^2\).