The next question after turning effects
A force can push, pull and turn an object. The preceding volumes explored how the position of a force changes its turning effect about a pivot. Now we ask a different question: how concentrated is that force on a surface? A person standing on a floor, a tyre pressing against a road and water pressing against a wall all involve force acting over area. Pressure is the quantity that lets us compare these situations. Once you understand it, you can connect contact surfaces, liquid depth, density, hydraulic systems, a barometer and a manometer without memorising six disconnected tricks.
Why this chapter follows the previous cycle
The previous EMS Lever Demonstration Audit, Volume 0678 finished a model-evidence-language cycle. Here we begin another sequence with a fresh Physics foundation. In earlier beam questions the key geometric idea was the perpendicular distance to a pivot, so the relationship involved force multiplied by distance. In pressure, the key geometric idea is the contact area, so the relationship involves force divided by area. Mixing those formulas because the diagrams both show forces is a common error. The next Mathematics chapter will take responsibility for proportions, area conversion, tables and bounds, while this chapter teaches what the physical quantities mean.
Check the 2027 G3 Physics rules, not old assumptions
The first Singapore–Cambridge Secondary Education Certificate examination is in 2027. The G3 Physics code is K323 and the earlier reference O-Level code is 6091. The official 2027 K323 syllabus explicitly includes pressure as force per unit area; density as mass per unit volume; liquid-column pressure as height × density × gravitational field strength; pressure transmission in a hydraulic press; measuring atmospheric pressure with a liquid column; and manometers for pressure differences. K323 Paper 1 is a 40-mark one-hour multiple-choice paper worth 30%; Paper 2 is an 80-mark one-hour-45-minute structured/free-response paper worth 50%; and Paper 3 is a 40-mark one-hour-50-minute practical worth 20%. Verify the current code using the SEAB 2027 G3 index. The questions below are our own teaching examples, not official questions.
First principles: pressure is force distributed over an area
Pressure is force acting normally per unit area. Write p = F/A. A force F measured in newtons divided by an area A measured in square metres gives pressure p in pascals, written Pa. One pascal is one newton per square metre. An important phrase is ‘normal to the surface’: the force component perpendicular to the area matters in the simplest pressure relation. In ordinary school word problems a downward load on a horizontal surface often supplies that normal force. In a tilted or constrained system, you must first identify what actually acts perpendicular to the contact surface.
Do not mistake force for pressure
Two objects may each exert a downward force of 100 N yet create very different pressures when their contact areas differ. A broad, flat base spreads the force over a larger area and gives lower average pressure; a narrow point concentrates the force and gives higher average pressure. The force has not changed. Only the distribution has. This distinction explains why some practical designs use wide supports and others use narrow tips, though real materials and safety must be considered. If the question asks which exerts the greater pressure, identify both force and contact area before deciding.
The conversion that appears before almost every calculation
Area conversion is not the same as length conversion. One metre equals 100 centimetres, but one square metre equals 10,000 square centimetres. To convert 50 cm² into m², divide by 10,000 to obtain 0.0050 m². If a 150 N perpendicular force acts over 0.0050 m², the average pressure is 30,000 Pa. Many pupils write 150/50 and label the result pascals, silently keeping the area in cm². The number 3 N/cm² describes the same pressure but must be converted to 30,000 N/m² to be written as 30,000 Pa.
A worked example with ordinary contact area
A rectangular block presses vertically on a table with a perpendicular force of 240 N. Its contact face measures 0.20 m by 0.15 m, so the contact area is 0.030 m². The average pressure is 240/0.030 = 8,000 Pa. Turn the same block onto a 0.20 m by 0.10 m face, keeping the total force at 240 N. The area becomes 0.020 m² and the pressure rises to 12,000 Pa. The block’s weight has not mysteriously increased. A correct explanation states that the same force is distributed over a smaller contact area.
How to make a sensible calculation check
Before reaching for a calculator, predict the direction of the change. With force unchanged, doubling area halves pressure. With area unchanged, doubling force doubles pressure. If your computed answer goes the wrong way, check the division and the area units. This qualitative test is especially valuable when a problem has several steps. Write p = F/A, substitute values with units, then inspect the result. A plausible number in the wrong unit is still wrong. A correct numerical result without a clear meaning is fragile when the question changes.
Pressure in liquids is a different physical setting
At a point below the surface of a liquid, pressure is not produced merely by a single solid object resting on a face. In a stationary liquid of approximately uniform density, the additional pressure due to a liquid column is given by p = hρg, where h is vertical depth or height of column, ρ (rho) is density and g is gravitational field strength. Each symbol has a role: height in metres, density in kilograms per cubic metre and g in newtons per kilogram or metres per second squared. Their product has units of pascals.
Why depth matters more than the shape of the container
Consider two differently shaped open containers filled with the same liquid to the same vertical depth. At points the same distance below the surface, the hydrostatic pressure due to the liquid is the same under the standard stationary-liquid model, even if one vessel is wide and the other narrow. The important variables are depth, density and gravitational field strength. This result may seem surprising because one container holds more liquid. But pressure at a point is not simply the total liquid mass divided by the bottom area in an arbitrary shape. Distinguish local pressure from the overall forces acting on an entire container.
A careful depth calculation
Use water of density 1,000 kg/m³ and take g = 10 N/kg as stated for this illustrative question. At a depth of 0.60 m, the pressure due to the water column is 0.60 × 1,000 × 10 = 6,000 Pa. At 1.20 m it is 12,000 Pa. The deeper point has twice the hydrostatic pressure contribution in this simple model. If atmospheric pressure acts above an open liquid, total absolute pressure at depth additionally includes the atmospheric pressure at the surface. Unless a question requests the absolute pressure, be clear whether you are reporting the extra pressure due to the liquid.
Gauge pressure and absolute pressure
An open container’s liquid surface is exposed to atmospheric pressure. The simple p = hρg formula gives the pressure increase between the surface and the depth considered. For an idealised open vessel, absolute pressure at depth equals atmospheric pressure plus this increase. A pressure gauge often displays a difference relative to atmospheric pressure rather than an absolute value. Examination tasks may describe the intended reference explicitly. Never insert 101,000 Pa from memory into an answer that asks only for pressure due to a liquid column; equally, do not omit the surface pressure when the problem clearly asks for absolute pressure and supplies the necessary information.
Density is not pressure
Density describes mass per unit volume: ρ = m/V. If a sample has mass 1.8 kg and volume 0.002 m³, its density is 900 kg/m³. That number is not a pressure, since its unit is not pascals. To work out liquid-column pressure you additionally need height and gravitational field strength. Greater density gives greater liquid pressure at the same depth under comparable conditions. But a liquid’s density alone tells you nothing about the depth at which a measurement was taken. In a multi-step question, calculate density before inserting it into hρg.
Two liquids can be compared without a calculator
At the same depth with the same gravitational field strength, pressure due to the liquid column is proportional to density. If liquid A has density 1,000 kg/m³ and liquid B has density 800 kg/m³, A produces 1.25 times the hydrostatic pressure contribution of B. At different depths, compare the products ρh. For example, 1,000 × 0.40 and 800 × 0.50 are both 400, so the pressure contributions match at the same g. Notice the structure: both density and vertical depth matter. A correct comparison often needs only the ratio, not two bulky calculations.
What happens to pressure sideways?
At a given point in a static liquid, pressure is transmitted in all directions and acts normal to a surface. A horizontal wall can experience a force from the liquid; pressure is not a force that acts only downward. The magnitude of force over a chosen small surface can be estimated using F = pA when the pressure is effectively uniform on that surface. A large vertical wall may experience pressure that increases with depth, so treating the entire wall as being at one pressure can be an inappropriate simplification unless the question gives an average or a suitable model. Read the physical geometry before substituting.
The hydraulic principle
In a simplified hydraulic system, an applied pressure change is transmitted through an enclosed liquid. If two connected pistons have areas A₁ and A₂ and the relevant pressure difference is the same, then F₁/A₁ = F₂/A₂. This does not say the two forces are equal. The larger piston can experience a larger force because the same pressure acts over a larger area. The relationship assumes an appropriate idealised system. Real devices have losses, seals, friction, mechanical constraints and rated limits; do not treat a school calculation as a licence to build or modify lifting equipment.
A worked hydraulic calculation
Suppose a small piston has area 0.025 m² and receives a force of 200 N. Its added pressure is 200/0.025 = 8,000 Pa. A connected large piston has area 0.20 m². In the idealised model its force is 8,000 × 0.20 = 1,600 N. This is eight times the input force because the area is eight times larger. The hydraulic device has not created energy. The large piston moves a smaller distance when the small piston moves, assuming incompressible liquid and ignoring losses. Physics calculations must respect both the force relation and the displacement trade-off.
The hidden distance trade-off
In an ideal incompressible hydraulic system, displaced volume is conserved. If the small piston moves 0.40 m with area 0.025 m², it displaces 0.010 m³. To displace the same volume at an area of 0.20 m², the large piston moves 0.050 m. Ideal input work is 200 N × 0.40 m = 80 J. Ideal output work is 1,600 N × 0.050 m = 80 J. There is a gain in force, not a gain in energy. This calculation is an extension that helps explain why a machine’s force advantage should not be mistaken for free work.
The atmosphere exerts pressure too
Atmospheric pressure results from the surrounding air. One classic method of measuring it uses the vertical height of a liquid column in a barometer, related to the liquid’s density and gravitational field strength. This is a conceptual and examination explanation; pupils should not attempt to construct devices containing hazardous liquids. A question may give a height and a suitable liquid density and ask you to calculate a corresponding pressure. Use the vertical height of the column and check the stated assumptions. Atmospheric pressure can vary; no single numerical value should be inserted when the problem provides a different condition.
A manometer measures a difference
A manometer uses liquid-level differences to indicate a pressure difference. Under the relevant idealised conditions, the difference can be related to Δp = ρgΔh. The height difference is the vertical separation between appropriate liquid surfaces, not the entire curved length of a tube. A pupil who traces a longer path along the glass will overstate the height. Check which sides are connected to which pressures, what density applies and whether the question is asking for a difference or an absolute pressure. A pressure difference is not the same as the pressure on one side.
A worked manometer example
A simple manometer contains a liquid of density 1,200 kg/m³, with a vertical level difference of 0.15 m. Take g = 10 N/kg. The corresponding pressure difference is 1,200 × 10 × 0.15 = 1,800 Pa. This does not tell you the absolute pressure of the unknown gas unless another reference pressure is supplied. If the open side is exposed to atmospheric pressure, the direction of the level difference may help identify whether the gas pressure is above or below atmospheric pressure. Use the diagram’s orientation before assigning the sign.
Common answers that lose their meaning
‘Pressure increases because mass increases’ may be irrelevant when the question is about moving deeper into the same liquid. ‘Pressure is larger because force is larger’ can be wrong when contact area grows proportionally. ‘Water pressure at the bottom depends on the container’s volume’ overlooks vertical depth. ‘A hydraulic press makes more energy’ confuses force and work. ‘The manometer reads 15 cm, therefore the pressure is 15 Pa’ treats height as pressure. The cure is not a larger formula sheet. It is to name the physical quantity and its unit before calculating.
A reliable four-stage answer
Stage one: describe the situation with a word and a sketch if one is helpful. Stage two: name the relationship that fits the quantity being asked about, such as p = F/A or p = hρg. Stage three: convert units and substitute values, carrying the units into the calculation. Stage four: interpret direction, magnitude and limitations. If the question compares two arrangements, explain which variable changed while others were held fixed. If the question requests an evaluation, say whether the stated data are sufficient. The answer is complete when a reader can follow the physical meaning without guessing what your numbers represent.
Twenty pressure problems with worked reasoning
These are original practice cases arranged from direct substitution to interpretation and model limits. Cover each answer first, write an equation and one sentence about physical meaning, then compare with the commentary. Where g is needed and not otherwise specified in a case, use the illustrative value 10 N/kg stated in the case. The focus is choosing the right quantity and explaining it, not merely matching a calculator display.
Pressure practice 1: A box exerts a perpendicular force of 180 N on 0.030 m². Find average pressure.
Use p = F/A, so p = 180/0.030 = 6,000 Pa. The answer describes average normal force per unit contact area. If the box is rotated but its force remains the same, a new pressure requires the new contact area. Writing ‘6,000 N’ would report a force, not a pressure. Always finish with both numerical magnitude and unit.
Pressure practice 2: A 90 N load acts on 300 cm². Calculate pressure in pascals.
Convert first: 300 cm² = 300/10,000 = 0.030 m². Then p = 90/0.030 = 3,000 Pa. The usual trap is to divide by 300 and label the result Pa. In that case the numerical quotient would be in N/cm². A sound response makes the area conversion visible so the reader knows exactly what was divided.
Pressure practice 3: Two supports carry the same normal force, but one has twice the area.
The support with twice the area has half the average pressure if the normal force is unchanged. Let the original pressure be F/A. The second is F/(2A) = 0.5F/A. This reasoning is more important than any invented measurement. Be careful not to assume that contact area and total force must change together merely because the object looks larger.
Pressure practice 4: A sharp tool and a broad pad each transmit 50 N over very different areas.
For a fixed perpendicular force, the smaller contact area yields the greater average pressure. The equation predicts the direction of change but does not alone determine penetration, damage or safe use. Those outcomes also depend on material behaviour and conditions. In an examination explanation, connect force concentration to area without turning a simple model into an unsupported safety statement.
Pressure practice 5: A water column is 0.75 m deep with density 1,000 kg/m³ and g = 10 N/kg.
The pressure due to the column is 0.75 × 1,000 × 10 = 7,500 Pa. This is a pressure increase relative to the surface reference for the standard open-liquid model. If the question asks for absolute pressure, additional surface pressure information is needed. Identify the requested reference before you add any atmospheric value.
Pressure practice 6: At depths 0.4 m and 1.6 m in the same stationary liquid, compare hydrostatic pressure contributions.
The deeper pressure contribution is four times the shallower one: 1.6/0.4 = 4, because density and g remain unchanged. You do not need the actual density to form the ratio. This direct proportionality is often faster and clearer than two separate substitutions. Do not claim that the absolute pressures have a 4:1 ratio unless the surface pressure is negligible or otherwise handled.
Pressure practice 7: Liquid A has density 900 kg/m³ at depth 0.8 m; liquid B has density 1,200 kg/m³ at 0.6 m.
Compare ρh: A gives 900 × 0.8 = 720; B gives 1,200 × 0.6 = 720. Their liquid-column pressure contributions are equal if g is the same. This is a useful check on the tendency to choose the denser liquid automatically. Density matters, but so does depth. The quantities balance through multiplication, not by looking for the largest single number.
Pressure practice 8: A sample mass is 2.4 kg and its volume is 0.003 m³. Find density.
ρ = m/V = 2.4/0.003 = 800 kg/m³. That is a density result, not a pressure. To calculate a liquid-column pressure, you must also know the relevant depth and g. A multi-step question may offer more data than needed at one stage. Label the intermediate quantity so that you do not confuse it with the final requested quantity.
Pressure practice 9: A liquid has density 800 kg/m³, depth 0.5 m and g = 10 N/kg. Find column pressure.
The liquid-column contribution is 800 × 0.5 × 10 = 4,000 Pa. The multiplication gives a pressure only because each term carries the appropriate unit. A student who uses volume in place of depth or uses the total mass of the container has answered a different problem. The formula refers to vertical height, density and g.
Pressure practice 10: Water fills a tall narrow tank and a short wide tank to the same height.
At the same depth and with the same stationary water, the hydrostatic pressure contribution is equal. The total forces on the bottom or sides may differ because the relevant areas and shapes differ, but that does not change the pressure at matching depths in the idealised model. State the distinction between pressure at a point and total force on a surface.
Pressure practice 11: An open vessel has 6,000 Pa of liquid-column pressure at a point and atmospheric pressure of 100,000 Pa.
The absolute pressure is 100,000 + 6,000 = 106,000 Pa if those are the given conditions. If asked instead for gauge pressure relative to the atmosphere, the answer would be 6,000 Pa. The difference is not a rounding issue; it is a different reference. Read whether the question says ‘due to water’, ‘above atmosphere’ or ‘total’.
Pressure practice 12: A small piston area is 0.010 m² and input force is 120 N; a large piston area is 0.040 m².
The added fluid pressure is 120/0.010 = 12,000 Pa. At the larger area, ideal output force is 12,000 × 0.040 = 480 N. The force is multiplied by four because area is multiplied by four. The larger piston moves proportionally less for the same displaced volume. This idealised calculation does not establish any machine’s safe capacity.
Pressure practice 13: A hydraulic press has an area ratio of 1:6. Find the ideal force ratio.
When the same pressure change acts at both pistons, the force ratio follows the area ratio: output force can be six times the input force if the output piston area is six times greater. Do not confuse this with an energy ratio. For an incompressible ideal system, the output distance is one-sixth of the input distance. Ratios capture the pattern before numbers are supplied.
Pressure practice 14: A small piston of 0.020 m² moves 0.30 m; the larger piston area is 0.12 m².
The displaced volume is 0.020 × 0.30 = 0.0060 m³. The larger piston’s ideal displacement is 0.0060/0.12 = 0.050 m. The sixfold area increase produces a sixfold decrease in displacement. This is the companion to force amplification; it prevents the false inference that hydraulics produce more energy than they receive.
Pressure practice 15: A manometer liquid has density 1,100 kg/m³ and vertical level difference 0.20 m; g = 10 N/kg.
The pressure difference is 1,100 × 10 × 0.20 = 2,200 Pa. Report it as a difference. Whether the gas side is higher or lower than atmosphere requires interpreting which liquid surface is lower in the actual diagram. A numeric magnitude without the directional interpretation can be incomplete when the question asks which side has greater pressure.
Pressure practice 16: A U-tube is curved, and a student measures 45 cm along its wall instead of 18 cm vertically.
The required hydrostatic height difference is the vertical 18 cm, or 0.18 m, not the distance along the glass. Pressure difference depends on vertical separation, density and g. Replacing height with path length introduces an error even if the arithmetic is flawless. The diagram should be read geometrically before any value is placed into Δp = ρgΔh.
Pressure practice 17: A student claims that doubling the density always doubles absolute pressure at depth.
Doubling density doubles the liquid-column contribution hρg if h and g stay fixed. But absolute pressure also includes any surface pressure. If the latter is unchanged and non-zero, absolute pressure does not generally double. This is an example where leaving out the reference gives a wrong general statement. A good response names which component of pressure is proportional to density.
Pressure practice 18: A piston pressure is 9,000 Pa over 0.050 m². Find normal force.
Rearrange p = F/A to F = pA. Then F = 9,000 × 0.050 = 450 N. The rearrangement should make sense dimensionally: Pa × m² gives newtons. If your answer is in newtons per square metre after solving for force, the units reveal the error. Algebra is useful when it preserves the physical meaning of each symbol.
Pressure practice 19: A force of 60 N produces 15,000 Pa on a pad. Find its contact area.
Use A = F/p = 60/15,000 = 0.0040 m², equivalent to 40 cm². The answer is an area, not a length. Do not take a square root unless a shape’s side length is also requested and the shape is specified. The units provide a quick test: N divided by N/m² is m².
Pressure practice 20: A public caption says, ‘Water always presses harder at the bottom because there is more water in a wide tank’.
Rewrite it as: ‘For the same stationary liquid, the pressure contribution increases with vertical depth; a wider tank does not by itself increase the pressure at a matching depth.’ The original mixes a valid depth trend with a misleading appeal to total water volume. The corrected statement identifies the relevant variable and avoids a false generalisation about container width.
Pressure practice 21: A hydraulic piston creates a larger force, so a student claims it makes unlimited useful work.
The claim is wrong. Ideal hydraulic force increase comes with reduced displacement, so input and output work can match without creating energy. Real machines have losses. Explain the trade-off using F × distance, not just F/A. This is a powerful way to connect pressure to conservation of energy while respecting the limits of the simplified model.
Pressure practice 22: A pressure reading is printed to six decimal places after measuring an area to two significant figures.
Calculator precision is not measurement precision. Show the intermediate value accurately enough to avoid rounding error, then report the final answer to the accuracy required by the question and supported by data. The Physics syllabus expects measurements, calculations and evaluation; an excessively long decimal is not inherently more scientific. State the limitations where they matter.
Pressure practice 23: A graph of pressure against depth is straight through the origin for pressure due to a liquid column.
For constant density and g, p = ρgh is a direct proportional relationship with gradient ρg. The phrase ‘due to the liquid column’ matters: absolute pressure against depth would generally have a non-zero intercept equal to the pressure at the surface. Do not describe the graph’s intercept without checking the reference. This links the model to a mathematical interpretation.
Pressure practice 24: A student’s final answer says ‘the liquid was under 3,000 N of pressure’.
The unit is wrong. Pressure is expressed in pascals, equivalent to N/m²; newtons are force. If the answer actually calculated 3,000 Pa, revise the unit and check how that pressure was determined. If the student intended force on a surface, a corresponding area is needed for F = pA. A clean conclusion names exactly the quantity found.
A revision route from easier to harder
First complete contact pressure calculations with areas already in square metres. Then repeat with cm² and mm² so the unit conversions become fluent. Next compare hydrostatic pressure in identical and different liquids; add density calculations only after the depth idea is stable. Move to hydraulic relations when the pressure interpretation is secure, and finish with pressure differences and graph questions. Do not start with mixed questions if you still cannot explain what one pascal means. A learner progresses faster when the first unstable relationship is repaired before complexity is added.
An original longer response with full reasoning
Question: A tank holds water to a depth of 1.5 m. A pressure sensor is placed 0.4 m below the surface. Explain whether moving the sensor to 0.9 m below the surface raises the liquid-column pressure, and calculate each value for density 1,000 kg/m³ and g = 10 N/kg. Response: At the initial depth, p = 0.4 × 1,000 × 10 = 4,000 Pa. At the deeper position, p = 0.9 × 1,000 × 10 = 9,000 Pa. The increase is 5,000 Pa because the vertical height of liquid above the sensor has increased by 0.5 m while density and g remain unchanged. This does not state the absolute pressure unless the surface pressure is accounted for. Notice that the explanation and the calculation support the same conclusion.
The last check before an examination answer is finished
Circle the requested quantity. Underline the supplied units. Write a relationship that produces the unit you need. Substitute the numbers with converted units and test the direction of change. Then decide whether your conclusion refers to an ideal model, a measured apparatus, a liquid-column contribution or an absolute pressure. This distinction is central to correct reasoning. When the context is unfamiliar, reconstruct it from the basic quantities rather than searching your memory for a worked example that looks similar.
Continue with the Secondary learning route, the PSLE foundation route and the Secondary Mathematics owner. The authoritative scope and papers are listed in SEAB K323 G3 Physics (2027) and the SEC G3 syllabus index. No unsupervised practical or load-bearing experiment is recommended here.
