P20: Pressure
Covering pressure in solids p=F/A, atmospheric pressure, pressure in liquids, upthrust, floating and sinking, and the relationship between pressure and depth.
Pressure in Solids
Pressure = force / area
p = F / A
p = pressure (Pa or N/m²), F = force (N), A = area (m²)
Pressure is the force per unit area. The SI unit of pressure is the pascal (Pa), where 1 Pa = 1 N/m². A force applied over a smaller area creates a larger pressure.
Worked Example
A box of weight 500 N has a base area of 2 m². Calculate the pressure it exerts on the floor.
p = F / A = 500 / 2 = 250 Pa
Worked Example
A person weighing 700 N stands on one foot with a contact area of 0.02 m². Calculate the pressure.
p = F / A = 700 / 0.02 = 35 000 Pa = 35 kPa
The same force can produce very different pressures depending on the area over which it acts. This is why sharp knives cut better than blunt ones, and why camels have large feet - to reduce pressure on sand.
| Situation | Low pressure (large area) | High pressure (small area) |
|---|---|---|
| Walking on snow | Snowshoes | High heels |
| On the ground | Elephant's foot | Stiletto heel |
| Cutting | Blunt knife | Sharp knife |
| Vehicle on soft ground | Wide tyres | Narrow tyres |
Atmospheric Pressure
The atmosphere is a layer of air above the Earth's surface. Atmospheric pressure is caused by the weight of air molecules in the atmosphere pressing down on surfaces. At sea level, atmospheric pressure is approximately 101 000 Pa (101 kPa).
Key facts about atmospheric pressure:
- Air molecules move randomly and collide with surfaces, creating pressure
- Atmospheric pressure decreases with increasing altitude (height above sea level)
- At higher altitude, there are fewer air molecules per unit volume, so pressure is lower
- The atmosphere gets less dense with increasing altitude
The decrease in atmospheric pressure with altitude explains why:
- Water boils at a lower temperature on a mountain (lower pressure means less energy needed for bubbles to form)
- Climbers at high altitude may experience altitude sickness due to lower oxygen partial pressure
- Aircraft cabins are pressurised to maintain comfortable pressure for passengers
Worked Example
A suction cup has an area of 0.005 m². Atmospheric pressure is 101 000 Pa. Calculate the force holding the cup against a wall.
F = p × A = 101 000 × 0.005 = 505 N
This large force is why suction cups can hold significant weight - atmospheric pressure pushes them against the surface.
Atmospheric pressure questions often involve explaining why a suction cup sticks or why a drinking straw works. The key principle is that removing air from one side creates a pressure difference, and the higher pressure on the other side pushes the object.
Pressure in Liquids
Pressure due to a column of liquid = density × gravitational field strength × height
p = ρgh
p = pressure (Pa), ρ = density (kg/m³), g = gravitational field strength (N/kg), h = depth or height of liquid (m)
Pressure in a liquid increases with depth. This is because the deeper you go, the greater the weight of liquid above. The pressure at any point in a liquid acts equally in all directions.
Worked Example
Calculate the pressure at a depth of 10 m in water (density = 1000 kg/m³, g = 9.8 N/kg).
p = ρgh = 1000 × 9.8 × 10 = 98 000 Pa = 98 kPa
Worked Example
The density of mercury is 13 600 kg/m³. Calculate the pressure at a depth of 0.5 m in mercury.
p = ρgh = 13 600 × 9.8 × 0.5 = 66 640 Pa = 66.6 kPa
Pressure in a liquid depends on three factors: the density of the liquid, the gravitational field strength, and the depth below the surface. It does NOT depend on the volume of liquid, the shape of the container, or the surface area.
The container shape does not affect the pressure at a given depth. A narrow tube and a wide tank with the same liquid at the same depth have the same pressure at the bottom. This is called the hydrostatic paradox.
Pressure and Depth
As depth increases, pressure increases. This has important consequences:
- Dam walls are thicker at the bottom than at the top to withstand greater pressure
- Submarines must have stronger hulls to dive deeper
- Deep-sea creatures are adapted to high pressure environments
The pressure difference between two depths in the same liquid:
Pressure difference = ρgΔh
Δp = pressure difference (Pa), ρ = density (kg/m³), g = 9.8 N/kg, Δh = difference in depth (m)
Worked Example
A swimming pool is 2 m deep. Calculate the pressure difference between the surface and the bottom of the pool (density of water = 1000 kg/m³).
Δp = ρgΔh = 1000 × 9.8 × 2 = 19 600 Pa = 19.6 kPa
At any given depth, the pressure acts equally in all directions. This means the water pushes sideways on the walls of the container as well as downwards on the base. Holes at different depths in a water container show water escaping faster from lower holes (greater pressure).
Upthrust
Upthrust is the upward force exerted on an object immersed in a fluid. It is caused by the pressure difference between the top and bottom surfaces of the object. Since pressure increases with depth, the upward force on the bottom of the object is greater than the downward force on the top.
Upthrust = weight of fluid displaced
Upthrust = ρ₁Vg
ρ₁ = density of fluid (kg/m³), V = volume of fluid displaced (m³), g = gravitational field strength (N/kg)
Worked Example
A solid block with volume 0.003 m³ is fully submerged in water (density = 1000 kg/m³). Calculate the upthrust.
Upthrust = ρVg = 1000 × 0.003 × 9.8 = 29.4 N
Upthrust equals the weight of the fluid displaced by the object. This is known as Archimedes' principle. An object that is partially submerged only displaces a volume of fluid equal to the volume that is below the surface.
Upthrust depends on the density of the fluid, NOT the density of the object. The same object experiences different upthrust in different fluids. For example, upthrust is greater in salt water (higher density) than in fresh water.
Floating and Sinking
An object floats if the upthrust equals its weight. An object sinks if its weight is greater than the upthrust available.
The rules of floating and sinking:
- If the object's density is less than the fluid's density, it floats
- If the object's density is greater than the fluid's density, it sinks
- If the object's density equals the fluid's density, it remains at whatever depth it is placed
A floating object displaces a volume of fluid equal to its own weight. This is why ships float even though steel is denser than water - the ship's hollow shape means the overall density (total mass / total volume) is less than water.
Worked Example
A wooden block of mass 0.6 kg floats in water. Calculate the volume of water it displaces (density of water = 1000 kg/m³).
When floating, upthrust = weight = 0.6 × 9.8 = 5.88 N
Weight of displaced water = 5.88 N
Mass of displaced water = 5.88 / 9.8 = 0.6 kg
Volume of displaced water = mass / density = 0.6 / 1000 = 0.0006 m³ = 600 cm³
Worked Example
An iceberg is 90% submerged. Explain this using density.
When floating: weight of iceberg = weight of displaced water
ρᵢcₑVᵢcₑg = ρwₐₜₑᵣVdᵢₛₚg
ρᵢcₑVᵢcₑ = ρwₐₜₑᵣ × 0.9Vᵢcₑ
ρᵢcₑ = 0.9 × ρwₐₜₑᵣ = 0.9 × 1000 = 900 kg/m³
The density of ice is approximately 900 kg/m³, which is 90% of water's density, so 90% is submerged.
Atmospheric Pressure and Weather (Higher)
Low atmospheric pressure at the Earth's surface is associated with rising air and often brings unsettled weather (clouds, rain). High atmospheric pressure is associated with sinking air and usually brings settled, dry weather.
A barometer is used to measure atmospheric pressure. A simple mercury barometer uses a column of mercury in an inverted tube. The height of the mercury column is proportional to atmospheric pressure:
Atmospheric pressure = ρgh (for the mercury column)
Standard atmospheric pressure supports a mercury column of about 760 mm (0.76 m)
Worked Example
A mercury barometer shows a column height of 760 mm. Verify that this corresponds to approximately 101 000 Pa (density of mercury = 13 600 kg/m³).
p = ρgh = 13 600 × 9.8 × 0.76 = 101 293 Pa ≈ 101 kPa
Compressing Gases (Higher)
Unlike liquids, gases are compressible. When a gas is compressed into a smaller volume, the same number of molecules occupies a smaller space. The molecules collide more frequently with the walls, increasing the pressure.
For a fixed mass of gas at constant temperature:
Pressure is inversely proportional to volume
p₁V₁ = p₂V₂ (Boyle's law)
Worked Example
A gas has a volume of 0.02 m³ at a pressure of 100 000 Pa. The gas is compressed to a volume of 0.005 m³ at constant temperature. Calculate the new pressure.
p₁V₁ = p₂V₂
100 000 × 0.02 = p₂ × 0.005
2000 = 0.005p₂
p₂ = 400 000 Pa
Practice Questions
1. A force of 200 N is applied to an area of 0.04 m². Calculate the pressure.
p = F/A = 200/0.04 = 5000 Pa = 5 kPa.
2. Calculate the pressure at a depth of 50 m in sea water of density 1025 kg/m³.
p = ρgh = 1025 × 9.8 × 50 = 502 250 Pa = 502 kPa.
3. Explain why a dam wall is thicker at the bottom than at the top.
Pressure in the water increases with depth (p = ρgh). At the bottom of the dam, the water pressure is greatest, so the wall must be thicker to withstand the larger force. At the top, the pressure is lower so the wall can be thinner.
4. An object of weight 30 N is placed in water and experiences an upthrust of 30 N. State whether it floats, sinks, or stays where it is and explain why.
It floats. The upthrust equals the weight, so the resultant force is zero. If it is fully submerged, it will rise until partially submerged at a level where the upthrust (now from a smaller displaced volume) equals its weight.
5. A block of wood with volume 0.002 m³ and density 600 kg/m³ is placed in water (density 1000 kg/m³). Calculate the fraction of the block that is submerged.
When floating, weight = upthrust. ρwood × V × g = ρwater × Vsubmerged × g. Vsubmerged/V = ρwood/ρwater = 600/1000 = 0.6. So 60% of the block is submerged.
Maths Skills
Pressure in Solids: p = F/A
Rearrange to find any unknown: F = p × A or A = F/p. Always convert units: force in newtons, area in m², pressure in Pa (N/m²). Common conversions: 1 kPa = 1000 Pa, 1 MPa = 1 000 000 Pa. For irregular shapes, calculate area using appropriate geometry (e.g. A = πr² for a circle).
Worked Example
A drawing pin has a head area of 1 cm² (1 × 10−4 m²) and a point area of 0.05 mm² (5 × 10−8 m²). A force of 10 N is applied to the head. Calculate the pressure at the head and at the point.
Head: p = 10 / 1 × 10−4 = 100 000 Pa = 100 kPa
Point: p = 10 / 5 × 10−8 = 200 000 000 Pa = 200 MPa
The pressure at the point is 2000 times greater — this is why the pin enters the wall.
Pressure in Liquids: p = ρgh
This formula gives the pressure due to a column of liquid of density ρ and height h. Total pressure at a depth includes atmospheric pressure: ptotal = patm + ρgh. Convert all units consistently: ρ in kg/m³, g in N/kg, h in m, p in Pa.
Worked Example
Calculate the total pressure at a depth of 30 m in seawater (ρ = 1025 kg/m³, atmospheric pressure = 101 kPa, g = 9.8 N/kg).
p = 101 000 + (1025 × 9.8 × 30) = 101 000 + 301 350 = 402 350 Pa = 402.4 kPa
Converting Pressure Units
1 atm = 101 325 Pa = 101.3 kPa. 1 bar = 100 000 Pa = 100 kPa. 1 mmHg = 133 Pa. When solving problems, always convert all pressures to the same unit (preferably Pa) before calculating.
Common Misconceptions
Direction of Pressure in Fluids
Pressure in a fluid only acts downwards. Pressure in a fluid acts equally in all directions at any given depth. This is why water pushes sideways on dam walls and upwards on the bottom of a submerged object (creating upthrust). A hole at any depth in a container will leak water sideways, not just downwards, proving the pressure acts in all directions.
Why Objects Float
Objects float because they are light. Objects float when the upthrust (weight of displaced fluid) equals or exceeds their weight. Whether an object floats depends on its overall density relative to the fluid. A massive steel ship floats because its hollow shape gives it a low overall density (total mass / total volume is less than water). A small steel ball sinks because steel is denser than water, regardless of how light the ball is.
6-Mark Extended Question
Explain why an object may float in one liquid but sink in another, in terms of density and upthrust. [6 marks]
Whether an object floats depends on whether the upthrust from the liquid can support its weight (1). Upthrust equals the weight of the fluid displaced by the object, which depends on the density of the fluid (1). If the object is placed in a denser liquid, the weight of the displaced volume of liquid is greater, so the upthrust is greater (1). For example, an egg sinks in fresh water but floats in very salty water because salt water is denser (1). In fresh water, the weight of the displaced water (upthrust) is less than the egg's weight, so it sinks (1). In salt water, the same displaced volume has a greater weight, producing a greater upthrust that can equal the egg's weight, so it floats. The critical factor is the ratio of the object's density to the fluid's density: if the object is less dense, it floats; if more dense, it sinks (1).
AO3: Analyse and Evaluate
A student measures the pressure at different depths in an unknown liquid using a pressure sensor. The atmospheric pressure is 101 000 Pa.
| Depth (m) | Total pressure (Pa) |
|---|---|
| 0 | 101 000 |
| 0.5 | 106 900 |
| 1.0 | 112 800 |
| 1.5 | 118 700 |
| 2.0 | 124 600 |
| 2.5 | 130 500 |
(a) Show that the pressure due to the liquid alone increases linearly with depth. (b) Calculate the density of the liquid. (c) Suggest what the liquid might be.
Evaluation
(a) Subtract atmospheric pressure to find the liquid pressure: at 0.5 m, p = 106 900 − 101 000 = 5900 Pa. At 1.0 m: 11 800 Pa. At 1.5 m: 17 700 Pa. At 2.0 m: 23 600 Pa. At 2.5 m: 29 500 Pa. The pressure increases by 5900 Pa per 0.5 m, which is 11 800 Pa per metre — a constant rate, confirming a linear relationship.
(b) From p = ρgh: ρ = p/(gh) = 11 800 / (9.8 × 1) = 1204 kg/m³.
(c) This density is close to that of salt water (approximately 1025–1200 kg/m³) or possibly glycerine (1260 kg/m³). Given the context, the most likely liquid is dense salt water or a saline solution.
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