The key idea
Two equations and a lot of everyday explanations: snowshoes, knife edges, dams thicker at the bottom. The explain questions all reduce to the same sentence — same force, different area, different pressure.
Understand it · 01
Keep force and pressure separate
The same force can produce very different effects depending on its contact area. Pressure is the normal force divided by that area. A person standing on one foot puts approximately the same total force on the ground as when standing on two feet, but the smaller total contact area increases the average pressure. The weight has not increased simply because the pressure is larger.
For a calculation, identify the actual area in contact, not the whole surface area of the object. A rectangular block resting on one face uses that face's area in p = F/A. If the pressure is not uniform, this formula gives an average over the stated area. Express force in newtons and area in square metres to obtain pascals, with one pascal equal to one newton per square metre.
Understand it · 02
Square the length conversion before dividing
A common pressure mistake is converting square centimetres as though they were centimetres. Since 1 cm is 0.01 m, one square centimetre is 0.01 × 0.01 = 0.0001 m². Thus 200 cm² is 0.0200 m². Writing the area conversion as a separate line makes the size of the denominator visible before dividing the force by it.
Check the result with a comparison. Halving the area at fixed force doubles the pressure; doubling the force at fixed area also doubles it. Broad feet, skis and foundations spread forces to reduce pressure. A cutting edge concentrates force over a small area to raise pressure. These explanations need the full chain from force and area to pressure rather than claiming the surface somehow reduces the object's mass.
Understand it · 03
Use vertical depth below the free surface
In a stationary liquid, a deeper layer supports more liquid weight above it. For uniform density, the pressure increase is Δp = ρgΔh. Here Δh is a vertical separation, not distance along a sloping wall or pipe. At the same depth in the same connected liquid at rest, the pressure is the same even if the container has different widths or shapes.
Liquid pressure acts in every direction at a point; a wall feels a force perpendicular to its surface. Pressure increases down a dam wall, helping explain why the lower part needs greater strength. The equation assumes the liquid is at rest, its density is approximately constant and g is effectively constant over the height range. A moving fluid or a large change in density needs a more detailed model.
Understand it · 04
Decide whether to include atmospheric pressure
The expression ρgh gives the increase in pressure below the liquid's surface. If the surface is open to air, the absolute pressure at depth is atmospheric pressure plus this increase. A question asking for pressure due to the liquid wants only ρgh. Read the wording carefully before adding an atmospheric value, and never mix kilopascals with pascals in the same addition.
Atmospheric pressure comes from the atmosphere and generally decreases with altitude. Differences in pressure create net forces: a suction cup is held because the pressure outside exceeds the pressure inside, rather than because an empty space pulls by itself. For fluid-pressure calculations, identify the two pressures acting on opposite sides of a surface; the net force depends on their difference multiplied by area when the pressure difference is uniform.
Key definitions
- Pressure
- The force acting per unit area, at right angles to the surface: p = F ÷ A. Measured in pascals (1 Pa = 1 N/m²).
Equations, units & conditions
Pressure
p = F ÷ A
Units: Pa
Liquid pressure
Δp = ρ × g × Δh
Units: Pa
Worked examples, step by step
Example 1
Calculate the pressure due to sea water (density 1030 kg/m³) at a depth of 25 m. (g = 9.8 m/s²)
Follow the method
- Δp = ρgΔh = 1030 × 9.8 × 25.
Δp ≈ 252,000 Pa ≈ 250 kPa
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 4 suggested marks
1. Find pressure on two feet
A student weighs 600 N. Each shoe has a ground contact area of 150 cm². Calculate the average pressure when standing on both feet, then on one foot, assuming the contact area of each shoe stays unchanged.
Show solution and self-check: question 1
- Total area for two feet = 300 cm² = 0.0300 m².
- Pressure on two feet = 600/0.0300 = 20 000 Pa.
- Area for one foot = 0.0150 m², so pressure = 600/0.0150 = 40 000 Pa.
Both feet: 20 kPa. One foot: 40 kPa.
Self-check · one mark per point
- Adds the two contact areas.
- Converts square centimetres using a factor of 10 000.
- Finds 20 000 Pa on two feet.
- Finds 40 000 Pa on one foot.
Build · 4 suggested marks
2. Liquid pressure and total pressure
A sensor is 3.0 m below the surface of oil of density 800 kg/m³. The oil is at rest and open to air at 100 kPa. Use g = 10 N/kg. Find the pressure due to the oil and the absolute pressure at the sensor.
Show solution and self-check: question 2
- Pressure increase = ρgh = 800 × 10 × 3.0 = 24 000 Pa = 24 kPa.
- Absolute pressure = surface pressure + increase = 100 + 24 = 124 kPa.
Pressure due to oil = 24 kPa; absolute pressure = 124 kPa.
Self-check · one mark per point
- Uses vertical depth in ρgh.
- Finds 24 000 Pa.
- Converts pressure units consistently.
- Adds atmospheric pressure to obtain 124 kPa.
Common mistakes & how to avoid them
- Confusing force and pressure in explanations. A camel's wide feet do not reduce the force on the sand — the weight is unchanged. They increase the area, which reduces the pressure.
- Using the wrong area units: cm² must become m² (divide by 10,000, not 100) before p = F/A gives pascals.
- Confusing the pressure increase due to a liquid with absolute pressure. For a stationary liquid of uniform density, Δp = ρgΔh does not depend on container shape; absolute pressure also includes the pressure at the surface.
- Missing that liquid pressure acts in all directions at a given depth, not just downwards.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why are dam walls thicker at the bottom?
Water pressure increases with vertical depth because deeper levels support more water above them. The lower part of a dam therefore experiences greater pressure and needs greater strength. For water of approximately uniform density, the pressure increase is Δp = ρgΔh.
Compare equal small wall areas at different depths. Greater pressure means a greater perpendicular force on the deeper area because F = pA. The effect is not caused by the water moving faster at the bottom, and the pressure force on a vertical dam wall acts sideways rather than simply following gravity downwards.
The liquid-pressure profile grows approximately linearly with depth under the constant-density model. Actual dam design also depends on geometry, foundations, materials and other loads, so the school explanation identifies the pressure trend rather than supplying a complete engineering design rule. Do not substitute the wall's sloping length for the vertical depth below the water surface.
Why does a sharp knife cut better than a blunt one?
For the same applied force, a sharp edge concentrates that force over a smaller contact area, producing greater pressure. A blunt edge spreads the force over a wider area. The pressure increase helps the edge deform or separate the material being cut.
Hold the comparison variables clear: same force, smaller area, greater pressure. It is not necessary to claim that sharpening makes the hand apply a larger force. This is the same relationship used to explain why a drawing pin's point penetrates more easily than its broad head under comparable forces.
The simple p = F/A calculation represents an average pressure over a stated contact area. A real cutting contact has uneven stresses and changes as material deforms, so it is not usually possible to infer an exact cutting force from edge width alone. The explanation remains useful because it identifies how concentrating a load changes its local effect.
Does the shape of a container affect liquid pressure?
For a stationary liquid of uniform density, pressure at a given vertical depth depends on surface pressure, density and g, not container width or shape. Points at the same depth have equal pressure when those conditions match. The total force can still differ if the area differs.
Connected vessels can have narrow and wide sections while sharing one liquid level. A point in the narrow section does not have lower pressure simply because there is less water directly above that particular cross-section. The complete fluid and its walls transmit forces consistently with the same-depth pressure condition.
Distinguish pressure from total downward force on a base. At equal depth the pressures can be equal, but a wider base experiences a larger force because more area is exposed. Different surface pressures, different liquids or fluid motion can invalidate a simple equal-depth comparison, so check these conditions before interpreting a complicated vessel drawing.
Why must I divide cm² by 10,000 when calculating pressure in pascals?
A square centimetre has two length dimensions to convert. Since 1 cm = 0.01 m, 1 cm² = 0.01 × 0.01 m² = 0.0001 m². Dividing an area in cm² by 10,000 gives m², which combines with newtons to produce pascals.
A hundred centimetres fit along one metre, so a square metre contains a hundred rows of a hundred square centimetres. This geometric picture explains why the area conversion differs from the length conversion. Dividing by only a hundred would make the area in square metres too large and the calculated pressure too small.
Write the area conversion on its own line before substituting into p = F/A. If an object's contact area consists of several separate patches, add their areas before dividing the total normal force when finding average pressure over all of them. Do not use a side face's area if the object is resting on a different face.
Convert before calculating pressure
A 450 N force acts normally over 18 cm². Find the average pressure in pascals.
- Area = 18/10 000 = 0.0018 m².
- Pressure = 450/0.0018 = 250 000 Pa.
Average pressure = 2.5 × 10⁵ Pa = 250 kPa.
When should I add atmospheric pressure to ρgh?
Add the surface pressure when the question asks for absolute pressure at depth. For a liquid open to air, that surface pressure is atmospheric pressure. If the question asks only for pressure due to the liquid or the increase below the surface, use ρgh alone.
The term ρgh measures a difference: how much more pressure there is at the chosen depth than at the reference surface. It does not specify the pressure already acting on that surface. A sealed vessel can have a surface pressure different from atmospheric pressure, so use the actual given value rather than automatically adding a memorised atmospheric number.
Convert all pressures into the same units before adding. A calculation may produce pascals while atmospheric pressure is supplied in kilopascals. Also read a gauge's reference carefully: a gauge reading relative to the atmosphere can be zero while the absolute pressure is about one atmosphere. The two readings use different zero points, not different fluids.
Distinguish added and absolute pressure
Water of density 1000 kg/m³ is open to air at 101 kPa. Find the pressure due to water and absolute pressure 5.0 m below the surface. Use g = 10 N/kg.
- Liquid pressure increase = 1000 × 10 × 5.0 = 50 000 Pa = 50 kPa.
- Absolute pressure = 101 + 50 = 151 kPa.
Pressure due to the water = 50 kPa; absolute pressure = 151 kPa.
How do I find the net force when pressure acts on both sides of a surface?
For uniform pressures over the same flat area, multiply the pressure difference by that area: net pressure force = (p₁ − p₂)A. The forces act in opposite directions, so using only the larger absolute pressure overestimates the net force when the other side also exerts pressure.
Begin by drawing the force from each side separately. A fluid pushes perpendicular to the surface; two fluids on opposite sides therefore produce opposing forces. Equal pressures cancel even if each individual pressure is large. The difference, rather than either absolute value alone, determines the net pressure contribution.
This shortcut assumes each stated pressure is approximately uniform over the area. On a tall submerged panel, pressure varies with depth, so use the appropriate average pressure or the method given in the question. Supports and other forces can also act on the panel; the calculated pressure force is not automatically the resultant of every force in the complete situation.
Subtract before multiplying by area
A flat panel of area 0.15 m² has uniform pressure 108 kPa on its inside and 100 kPa outside. Find the net outward force due to pressure.
- Pressure difference = 108 − 100 = 8 kPa = 8000 Pa.
- Net outward pressure force = 8000 × 0.15 = 1200 N.
Net outward force due to pressure = 1200 N.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define pressure and calculate it with p = F ÷ A
- Explain everyday high- and low-pressure applications
- Calculate the pressure due to a liquid column with Δp = ρgΔh
- Describe how pressure in a liquid varies with depth and density
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Pressure. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/pressure
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