Physiworld · Fluid Mechanics revision notes · www.physiworld.com/notes/fluid/
All revision notes Revision notes · Fluid MechanicsFluid Mechanics revision notes
Density, floating and sinking, upthrust, pressure in liquids and the air, hydraulics, and fluids in motion from continuity to Bernoulli. Each section follows one lesson: what you need to know, key terms, the equations with units, a worked example, common mistakes and an exam tip.
1 What Is a Fluid?
What you need to know
Fluids
- A fluid is any substance that can flow: every liquid and every gas. A solid keeps its own shape.
- In a gas the particles are far apart, so a gas is easy to compress. In a liquid the particles already touch, so a liquid hardly compresses at all.
- Liquids and gases flow because their particles can slide past each other. Even something as stiff as pitch is a fluid if it flows, however slowly.
Density
- Density is the mass of each unit of volume: ρ = m ÷ V, in kg/m³ or g/cm³. 1 g/cm³ = 1000 kg/m³; water is 1.0 g/cm³.
- Density belongs to the material, not to the size of the piece: cut a block in half and its density does not change.
- For a regular solid, measure its sides and calculate the volume. For an irregular solid, lower it into a measuring cylinder (or a displacement can) and use the rise in the water level as its volume. Measure the mass on a balance.
Float or Sink
- An object floats in a fluid that is denser than it and sinks in a fluid that is less dense. With equal densities it neither rises nor sinks.
- Changing the fluid can change the result: dissolving salt makes water denser, so an egg that sank in tap water floats in strong brine.
- Ice (about 0.92 g/cm³) is less dense than water, so it floats and lakes freeze from the top down.
Layers
- Liquids that do not mix form layers: the densest at the bottom, the least dense on top, whatever order they are poured in.
- On a mass–volume graph each material gives a straight line through the origin, and its gradient is the density. The steepest line is the densest material.
Key terms
- Fluid
- a substance that can flow: a liquid or a gas
- Density
- mass per unit volume, in kg/m³ or g/cm³
- Displacement
- the rise in liquid level when an object is lowered in, equal to the object's volume when it is fully under
- Compressible
- able to be squashed into a smaller volume, like a gas
Equations
| Equation | What it says, symbols and units |
|---|---|
ρ = mV | Density is the mass of each cubic metre of a material; a solid less dense than a liquid floats in it.
1 g/cm³ = 1000 kg/m³. 1 cm³ = 10−6 m³. |
Diagram to know
A measuring cylinder before and after an irregular stone is lowered in, with the two water levels read at the bottom of the meniscus, at eye level. The difference between the readings is the stone's volume.
Worked example
A piece of granite has a mass of 52 g. A measuring cylinder reads 30.0 cm³ before it goes in and 50.0 cm³ after. Find its density in g/cm³ and in kg/m³. Does it float in water?
- Volume = rise in level = 50.0 − 30.0 = 20.0 cm³.
- ρ = m ÷ V = 52 ÷ 20.0 = 2.6 g/cm³.
- In kg/m³: 2.6 × 1000 = 2600 kg/m³.
- 2.6 g/cm³ is more than water's 1.0 g/cm³, so it sinks.
Answer: 2.6 g/cm³ = 2600 kg/m³; it sinks in water
Common mistakes
- Using the final reading of the measuring cylinder as the volume instead of the rise in level.
- Mixing units: a mass in g with a volume in m³. Keep g with cm³, or kg with m³.
- Saying a heavy object sinks because it is heavy. Floating depends on density, not on mass: a huge ship floats and a small pebble sinks.
Exam tip
When you describe the density practical, name every instrument (balance, ruler, measuring cylinder or displacement can) and say what you do with each reading. "Read the meniscus at eye level" is a common extra mark.
AP Physics and NGSS: AP Physics 1 (Unit 8) uses density the same way, ρ = m/V in kg/m³, and moves straight on to pressure and buoyancy. The particle model of solids, liquids and gases is also NGSS MS-PS1-4.
2 Buoyancy & Archimedes
What you need to know
Upthrust
- An object in a fluid feels an upward force, the upthrust (buoyant force).
- Archimedes' principle: the upthrust equals the weight of the fluid the object pushes aside (displaces).
- Once an object is fully under, pushing it deeper does not increase the upthrust: it cannot push aside any more fluid.
Volume, Not Weight
- The upthrust depends only on the volume under the surface and the density of the fluid: Fb = ρfluidVg. It does not depend on what the object is made of.
- Two cubes of the same size, one of lead and one of aluminium, get the same upthrust in water. A denser fluid, such as brine, gives more upthrust.
- The object pushes down on the fluid as hard as the fluid pushes up on it (a third-law pair), so a beaker on a balance reads more when something hangs in it.
Floating
- A floating object pushes aside its own weight of fluid: upthrust = weight, and the resultant force is zero.
- Load a boat more and it sits deeper, until the extra water it pushes aside weighs as much as the extra load. The volume under water is V = m ÷ ρfluid.
Average Density
- An object made of several parts (glass and an air bubble, a ship's steel hull and the air inside it) floats if its average density = total mass ÷ total volume is less than the fluid's.
- Submarines and fish change their average density by letting water into tanks or gas into a swim bladder. A floating object's fraction under the surface equals its average density divided by the fluid's density.
Key terms
- Upthrust
- the upward force a fluid exerts on an object in it, in N
- Archimedes' principle
- the upthrust equals the weight of fluid displaced
- Displaced fluid
- the fluid pushed aside by the part of the object below the surface
- Average density
- total mass ÷ total volume of an object made of several materials
Equations
| Equation | What it says, symbols and units |
|---|---|
Fb = ρfVg | The upthrust is the weight of the fluid pushed aside.
|
V = mρ | A floating object pushes aside its own mass of fluid; this is the volume under the surface.
|
Diagram to know
A free-body diagram of a buoy held under water by a chain: the upthrust up, and the weight and the chain's tension both down. For a floating boat: only the weight down and an equal upthrust up.
Worked example
A float of mass 3.0 kg and volume 0.020 m³ is held fully under fresh water (1000 kg/m³) by a rope tied to the bottom. Find the upthrust and the tension in the rope.
- Upthrust = ρVg = 1000 × 0.020 × 9.8 = 196 N, upwards.
- Weight = mg = 3.0 × 9.8 = 29.4 N, downwards.
- The float is in equilibrium: upthrust = weight + tension, so tension = 196 − 29.4 = 167 N (to 3 s.f.), downwards.
Answer: Upthrust 196 N; tension 167 N
Common mistakes
- Using the object's density instead of the fluid's density in ρVg.
- Thinking a heavier object of the same size gets more upthrust. Same volume under, same upthrust.
- Saying a floating object pushes aside its own volume of water. It pushes aside its own weight of water.
Exam tip
In any upthrust question, draw the free-body diagram first and decide whether the object is floating, held under, or sinking. Floating means upthrust = weight; held under means upthrust = weight + the holding force.
AP Physics and NGSS: AP Physics 1 (Unit 8) writes the buoyant force as Fb = ρVg and uses it in Newton's second law. In GCSE the upthrust is a Higher-tier, Physics-only idea; Cambridge 0625 only asks you to predict floating and sinking from densities.
3 Pressure & Pascal's Principle
What you need to know
Pressure
- Pressure is the force acting at right angles on each unit of area: p = F ÷ A, in pascals (1 Pa = 1 N/m²).
- The same force on a smaller area gives a bigger pressure (a nail tip, a knife edge, a stiletto heel); a bigger area gives a smaller pressure (snowshoes, tractor tyres, a bed of many nails).
- A mass is not a force: multiply by g = 9.8 N/kg to get the weight before using p = F ÷ A. 1 mm² = 10−6 m² and 1 cm² = 10−4 m².
Pressure & Depth
- The pressure a liquid exerts grows with depth and with the liquid's density: Δp = ρgΔh.
- It does not depend on the shape of the container or the amount of liquid: a thin tube and a wide tank of the same depth give the same pressure at the bottom.
- The pressure at a point acts equally in all directions, at right angles to any surface. This is why a dam is built thicker at the bottom.
Pascal's Principle
- Pascal's principle: a pressure applied to a liquid in a closed system is passed on equally to every part of it.
- In a hydraulic system a small force on a small piston makes a pressure p = F1 ÷ A1; the same pressure on a large piston gives a large force F2 = pA2. Car brakes and car jacks work this way.
- It works because liquids are almost incompressible. An air bubble in brake fluid squashes, so the brakes feel soft. The small piston must move further than the large one: the force is bigger, but no energy is gained.
Key terms
- Pressure
- force per unit area acting at right angles to a surface, in Pa
- Pascal (Pa)
- a pressure of one newton per square metre
- Hydraulic system
- a closed system of liquid that passes on a pressure to multiply a force
- Incompressible
- keeps the same volume when squeezed, as liquids almost do
Equations
| Equation | What it says, symbols and units |
|---|---|
p = FA | Pressure is force per unit area: the same force on a bigger area gives a smaller pressure.
|
Δp = ρgΔh | Pressure in a liquid rises with depth and with the liquid's density.
Edexcel writes the pressure at depth h as p = hρg. |
F1A1 = F2A2 | In a hydraulic system both pistons (at the same height) feel the same pressure.
|
Diagram to know
A hydraulic jack: a narrow cylinder and a wide cylinder joined at the bottom and filled with oil, a small piston pushed down with F₁ and a large piston pushing up a load with F₂. Label both areas and mark that the pressure is the same throughout.
Worked example
Water stands 8.0 m deep behind a lock gate. A hatch of area 0.40 m² sits at the bottom of the gate. Find the pressure the water exerts on the hatch and the force on it. (Density of water 1000 kg/m³.)
- p = ρgh = 1000 × 9.8 × 8.0 = 78 400 Pa (78.4 kPa).
- F = pA = 78 400 × 0.40 = 31 360 N.
- That is about the weight of a 3.2 tonne mass, on a door less than half a square metre in size.
Answer: 78 400 Pa; 31 360 N (about 31 kN)
Common mistakes
- Putting a mass in kg into p = F ÷ A. Pressure needs the force in newtons.
- Converting cm² to m² by dividing by 100. Divide by 10 000 (10⁴); for mm² divide by 10⁶.
- Thinking a wider container gives more pressure at the bottom because it holds more water. Only depth and density matter.
Exam tip
For a "why is a dam thicker at the bottom" question, link the steps: pressure increases with depth, so the force on each square metre of the wall is greatest at the bottom. One-word answers like "pressure" score nothing.
AP Physics and NGSS: AP Physics 1 (Unit 8) uses the same p = F/A and p = ρgh, with Pascal's principle for hydraulic lifts. Pascal's principle is not on Cambridge 0625 or the GCSE specifications in this form.
4 The Air on Top
What you need to know
Total Pressure
- The air presses on the surface of every liquid. The total (absolute) pressure at depth h is the air's pressure plus the liquid's: p = p0 + ρgh. At sea level p0 is about 101 kPa.
- Every 10 m of water adds about 98 kPa, roughly one more atmosphere, so the total pressure 10 m down is about double that at the surface.
- A pressure gauge (such as a tyre gauge) reads 0 in open air: it shows only the pressure above the air's, the gauge pressure ρgh. The pressure difference between two depths, ρgΔh, does not depend on the air.
A Column of Air
- Atmospheric pressure is caused by the weight of the air above. A column of liquid in a barometer is held up by it: p0 = ρgh, about 0.76 m of mercury or about 10.3 m of water.
- Only the vertical height of the column matters, not the width or tilt of the tube. A suction pump cannot lift water more than about 10 m, because the air's push is what raises it.
- Higher up there is less air above you, so the atmospheric pressure is lower. A falling barometer reading often means a storm is coming.
Key terms
- Atmospheric pressure
- the pressure caused by the weight of the air above, about 101 kPa at sea level
- Absolute pressure
- the total pressure, including the air's
- Gauge pressure
- the pressure above atmospheric pressure, as a gauge reads it
- Barometer
- an instrument that measures atmospheric pressure, for example with a column of mercury
Equations
| Equation | What it says, symbols and units |
|---|---|
p = p0 + ρgh | Absolute pressure at depth h: the air's pressure on the surface plus the liquid's.
|
Δp = ρgΔh | Pressure in a liquid rises with depth and with the liquid's density.
This is the gauge pressure, and the difference between any two depths. |
Diagram to know
A mercury barometer: a long tube, closed at the top, upside down in a dish of mercury, with a vacuum above the column. Mark the vertical height h from the surface in the dish to the top of the column.
Worked example
A diver works 15 m below the surface of the sea (density 1030 kg/m³). Atmospheric pressure is 101 kPa. Find (a) the pressure due to the water, (b) the total pressure on her, and (c) the height of mercury (13 600 kg/m³) in a barometer at the surface.
- (a) ρgh = 1030 × 9.8 × 15 = 151 410 Pa, about 151 kPa. A depth gauge would read this.
- (b) Total pressure = 101 + 151 = 252 kPa: about two and a half atmospheres.
- (c) h = p0 ÷ (ρg) = 101 000 ÷ (13 600 × 9.8) = 0.76 m.
Answer: (a) 151 kPa (b) 252 kPa (c) 0.76 m
Common mistakes
- Forgetting atmospheric pressure when a question asks for the total pressure, or adding it when the question asks for the pressure due to the water.
- Mixing kPa and Pa in one sum. Convert both to the same unit before adding.
- Thinking a wider barometer tube gives a shorter column. Only the vertical height balances the air.
Exam tip
Look for the words "total" or "absolute" (add the atmosphere) and "due to the water" or "gauge" (leave it out). Say which you are giving in your answer line.
AP Physics and NGSS: AP Physics 1 (Unit 8) uses P = P0 + ρgh and the difference between absolute and gauge pressure directly. Cambridge 0625 (Supplement) and GCSE (Physics only) need the pressure change with depth, ρgΔh, and what causes atmospheric pressure.
5 Upthrust and Newton's Laws
What you need to know
Where Upthrust Comes From
- Upthrust comes from pressure. The bottom of an object is deeper than its top, so the fluid pushes up on the bottom harder than it pushes down on the top.
- For a block of height h with top and bottom faces of area A: Fb = (pbottom − ptop)A = ρghA = ρgV, which is Archimedes' principle again.
- Going deeper adds the same pressure to both faces, so the difference, and the upthrust, stays the same once the object is fully under. The air's pressure and the side pushes cancel.
Apparent Weight
- An object hanging still in a fluid is in equilibrium: the string's pull T and the upthrust together hold up its weight, so T = W − Fb. The spring balance reads this smaller apparent weight; the real weight mg has not changed.
- The object pushes down on the fluid as hard as the fluid pushes up on it, so a scale under the beaker reads more by exactly the upthrust. The two readings together still add up to the full weight.
Released Under Water
- Released in a fluid, an object has its weight down and the upthrust up. Taking up as positive, its acceleration at the moment of release is a = (Fb − W) ÷ m.
- Two objects of the same volume get the same upthrust; the lighter one accelerates upwards faster. Once it moves, drag grows with speed until the forces balance and it reaches a terminal speed.
Key terms
- Apparent weight
- what a spring balance reads when an object hangs in a fluid: its weight minus the upthrust
- Pressure difference
- the bottom face's pressure minus the top face's; times the area, the upthrust
- Equilibrium
- zero resultant force, so no acceleration
- Terminal speed
- the steady speed reached when drag balances the other forces
Equations
| Equation | What it says, symbols and units |
|---|---|
Fb = (pbottom − ptop)A = ρgV | The upthrust is the pressure difference between the bottom and top faces, times their area.
|
T = W − Fb | Apparent weight: the pull a balance needs to hold an object still in a fluid.
|
a = Fb − Wm | Newton's second law for an object let go in a fluid, before drag builds up; up is positive.
|
Worked example
A rock of mass 1.5 kg and volume 0.00060 m³ hangs from a spring balance, fully under fresh water (1000 kg/m³). Find the balance reading. The string is then cut: find the rock's acceleration at that moment.
- Weight W = 1.5 × 9.8 = 14.7 N.
- Upthrust Fb = ρVg = 1000 × 0.00060 × 9.8 = 5.88 N.
- Balance reading T = 14.7 − 5.88 = 8.82 N, about 8.8 N.
- String cut, up positive: a = (5.88 − 14.7) ÷ 1.5 = −5.88 m/s². The rock accelerates downwards at about 5.9 m/s², less than g because the upthrust still acts.
Answer: 8.8 N; 5.9 m/s² downwards
Common mistakes
- Saying the rock's weight gets smaller under water. Its weight is still mg; the balance reads less because the upthrust holds up part of it.
- Using the depth of the object, not its height, when finding the upthrust from the pressure difference.
- Forgetting the sign: if the upthrust is less than the weight, the acceleration is downwards.
Exam tip
When two readings are involved (a balance holding the object and a scale under the beaker), write the third-law pair in words: "the object pushes down on the water with a force equal to the upthrust". That sentence explains why the scale's reading rises.
AP Physics and NGSS: AP Physics 1 (Unit 8) puts the buoyant force into free-body diagrams and ΣF = ma exactly like this. GCSE asks for the pressure-difference explanation of upthrust at Higher tier, Physics only.
6 Bernoulli & Flow
What you need to know
Continuity
- Continuity: in a pipe with no leaks, the volume of liquid passing each point every second is the same everywhere, so A1v1 = A2v2.
- Where the pipe or river is narrower, the fluid flows faster: a third of the area means three times the speed.
Bernoulli
- Bernoulli's principle: along a level pipe, where a fluid flows faster its pressure is lower. The fluid speeds up into a narrow part because the pressure behind it is higher.
- For an ideal fluid (no friction) in a level pipe, the pressure drop is ΔP = ½ρ(vfast² − vslow²), so doubling the flow speed gives four times the drop.
- Two ships sailing close side by side are pushed together: the water between them flows faster, so its pressure is lower than on their outer sides.
Lift & Stall
- A wing makes the air flow faster over its top than under it. The lower pressure above and the higher pressure below give an upward force, lift: L = ΔP × A.
- Increasing the angle of the wing increases the lift up to about 15°; beyond that the airflow leaves the top surface and the lift collapses, a stall. A racing car's upside-down wing gives downforce instead.
Torricelli
- Torricelli's law: water leaves a hole at depth h below the surface with the speed it would gain falling h: v = √(2gh). The speed does not depend on the liquid's density.
Key terms
- Continuity
- the volume flow rate Av is the same at every point along a pipe
- Volume flow rate
- the volume of fluid passing a point each second, in m³/s
- Bernoulli's principle
- in a flowing fluid, faster flow means lower pressure (at the same height)
- Stall
- the sudden loss of lift when a wing's angle is too steep
Equations
| Equation | What it says, symbols and units |
|---|---|
A1v1 = A2v2 | Continuity: the volume flow rate is the same all along a pipe.
|
ΔP = ½ρ(v22 − v12) | Bernoulli for a level pipe: the pressure drop from the slow part (v₁) to the fast part (v₂).
|
v = √(2gh) | Torricelli: the speed of water leaving a hole at depth h below the surface.
|
Diagram to know
A level pipe that narrows in the middle, with a vertical open tube above the wide part and above the narrow part. The water column stands lower above the narrow part, where the flow is faster and the pressure lower.
Worked example
Water (1000 kg/m³) flows at 1.0 m/s along a level pipe of cross-section 0.0060 m². The pipe narrows to 0.0020 m². Find the speed in the narrow part and the drop in pressure there.
- Continuity: v2 = A1v1 ÷ A2 = 0.0060 × 1.0 ÷ 0.0020 = 3.0 m/s.
- v2² − v1² = 9.0 − 1.0 = 8.0 m²/s².
- ΔP = ½ × 1000 × 8.0 = 4000 Pa: the pressure in the narrow part is 4.0 kPa lower.
Answer: 3.0 m/s; a pressure drop of 4000 Pa
Common mistakes
- Thinking the pressure is higher where the fluid flows faster. It is lower.
- Subtracting the speeds and then squaring. Square each speed first, then subtract.
- Using the height of the hole above the ground in Torricelli's law instead of its depth below the water surface.
Exam tip
In a Bernoulli question, start by writing which part is faster and why (continuity: narrower, so faster). The direction of the pressure change usually earns the first mark, before any number.
AP Physics and NGSS: Continuity, Bernoulli and Torricelli are AP Physics 1 Unit 8 content, with NGSS HS-PS3-1 for the energy view. They are not on Cambridge 0625 or the GCSE specifications.
7 Bernoulli in Full
What you need to know
Uphill Pipes
- The full Bernoulli equation for an ideal flow: P + ½ρv² + ρgy is the same at every point. Each term is energy per cubic metre.
- Going up a pipe at a steady speed, the pressure falls by ρgΔy: about 9.8 kPa for every metre of water. A water main at 300 kPa (gauge) can lift water only about 30 m, which is why the top floor's tap is weaker.
Energy per Litre
- Bernoulli's equation is conservation of energy. The pressure difference does work on each parcel of fluid, W = ΔP × V, and that work becomes kinetic energy or height energy.
- The mass flow rate ρAv is the same at every section of the pipe. Only a pump adds energy to the flow; friction takes it away.
The Siphon
- A siphon carries liquid up over a crest and down to an outlet below the liquid's surface, with no pump. The air pressing on the surface drives the liquid up.
- The outlet speed is v = √(2gΔh), where Δh is the drop from the surface to the outlet; the crest's height does not change it. The flow stops if the outlet is raised above the surface.
- The pressure is lowest at the crest. The crest can be at most about 10 m above the liquid, where that pressure would fall to zero.
Key terms
- Bernoulli equation
- P + ½ρv² + ρgy is constant along an ideal flow
- Mass flow rate
- the mass of fluid passing a point each second, ρAv, in kg/s
- Ideal fluid
- a fluid with no friction (viscosity) that cannot be compressed
- Siphon
- a full tube that carries liquid over a crest to an outlet lower than the liquid's surface
Equations
| Equation | What it says, symbols and units |
|---|---|
P + ½ρv2 + ρgy = constant | Bernoulli's equation: pressure energy, kinetic energy and height energy per cubic metre add to the same total.
|
W = ΔP V, mass flow rate = ρAv | Work done on a parcel of fluid by a pressure difference, and the mass flow rate that stays the same along a pipe.
|
Worked example
A water main supplies water at a gauge pressure of 300 kPa at street level. The water flows at a steady speed up a pipe of the same width to a tap 12 m higher. Find the gauge pressure at the tap and the greatest height the main could lift water to. (Density of water 1000 kg/m³.)
- Same speed, so only height and pressure change: the drop in pressure is ρgΔy = 1000 × 9.8 × 12 = 117 600 Pa, about 118 kPa.
- Pressure at the tap = 300 − 117.6 = 182 kPa (to 3 s.f.).
- Greatest height: the pressure falls to zero when ρgh = 300 000 Pa, so h = 300 000 ÷ (1000 × 9.8) = 31 m.
Answer: 182 kPa; about 31 m
Common mistakes
- Leaving out the ρgy term when the pipe changes height.
- Thinking a siphon's crest height sets the outlet speed. Only the drop from the surface to the outlet does.
- Saying a siphon works by suction. The atmosphere pushes the liquid up; the low pressure at the crest only allows it.
Exam tip
Write Bernoulli's equation for two named points (1 and 2) with every term, then cross out the terms that are the same at both points. It shows the examiner your reasoning and stops you dropping a term.
AP Physics and NGSS: This lesson is AP Physics 1 Unit 8 (with NGSS HS-PS3-1 for the energy view). IGCSE and GCSE do not cover fluids in motion.
8 The Full Picture
What you need to know
Air Pressure
- Air molecules hit every surface, so the air exerts atmospheric pressure on everything: about 101 kPa at sea level.
- It is caused by the weight of the air above, so it falls as you climb. It falls fastest near the ground, where the air is squashed by the air above and is densest.
- A sealed packet swells on a mountain: the air inside is still at the pressure it was sealed at, and the thinner air outside pushes less. The resultant force on a lid is the pressure difference times its area.
Hot Air
- Heated air expands, so it becomes less dense than the cooler air around it, and rises.
- A hot-air balloon lifts off when the weight of the cold air it pushes aside is more than its total weight, including the hot air inside it. The load it can carry is m = (ρoutside − ρinside)V.
- The same idea explains convection currents and a lava lamp: a warmer, less dense part of a fluid rises through the cooler part.
The Finale
- The whole topic in one task: a diver's lift bag filled with air at depth raises an object from the seabed when the upthrust on the bag and object is more than their total weight.
- The steps are always the same: density and volume give the upthrust; mass and g give the weight; pressure depends on depth; the resultant force gives the acceleration.
Key terms
- Atmospheric pressure
- the pressure of the air, caused by the weight of the air above
- Pressure difference
- the difference between the pressures on two sides of a surface; times the area, the resultant force
- Convection
- the rising of a warmer, less dense part of a fluid through the cooler part
- Lift (of a balloon)
- the mass a balloon can carry: the mass of cold air displaced minus the mass of hot air inside
Equations
| Equation | What it says, symbols and units |
|---|---|
p = FA | Pressure is force per unit area: the same force on a bigger area gives a smaller pressure.
|
ρ = mV | Density is the mass of each cubic metre of a material; a solid less dense than a liquid floats in it.
|
mlift = (ρout − ρin)V | The greatest mass a hot-air balloon can carry, envelope, basket and people included.
|
Worked example
A hot-air balloon's envelope holds 2500 m³. The air outside has a density of 1.20 kg/m³ and the hot air inside 0.96 kg/m³. What is the greatest mass of balloon, basket and passengers it can lift?
- Mass of cold air pushed aside = 1.20 × 2500 = 3000 kg. Its weight is the upthrust.
- Mass of hot air inside = 0.96 × 2500 = 2400 kg, which the balloon must also hold up.
- Greatest load = 3000 − 2400 = 600 kg.
Answer: 600 kg
Common mistakes
- Saying hot air rises because heat rises. Heat is energy, not a substance; the hot air rises because it is less dense.
- Forgetting the mass of the hot air inside the envelope when finding what a balloon can lift.
- Using one pressure, not the pressure difference, to find the force on a sealed lid or window.
Exam tip
In a long fluids question, write each quantity on its own line with its unit (volume, density, mass, weight, upthrust, resultant) before combining them. Marks are given for each step, even if the final answer is wrong.
AP Physics and NGSS: AP Physics 1 (Unit 8) brings the same ideas together in multi-step buoyancy and pressure problems. Warm fluids rising because they are less dense is also NGSS MS-PS1-4 and Cambridge 0625 2.2.1 (thermal expansion).
Written in Physiworld's own words. The course tags come from the same mapping as the syllabus maps; equations are on the IGCSE, GCSE, AP Physics 1 and NGSS formula sheets. Values use g = 9.8 N/kg.
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