When a small push seems to become a large force
A small piston is pressed with one hand. A much wider piston on the other side of a liquid-filled system supports a substantially greater force. The first impression is that the machine has made something out of nothing. The correct explanation is more useful: pressure can be transmitted through an enclosed fluid, while the areas and distances through which the two pistons move are different. A student who can separate force, pressure, area and distance has understood the centre of a hydraulic problem. A student who merely remembers to multiply two numbers has not yet reached that point.
This guide continues the earlier study of turning effects and equilibrium in Volume 0434. A lever compares moments about a pivot. A hydraulic press compares forces acting over different areas through a common pressure change. The two machines can both give a force advantage, but they do so by different physical mechanisms. Recognising the difference is a useful examination skill: a plausible answer to the wrong physical question is still wrong.
The examples below are invented classroom calculations, not certified equipment ratings or directions for constructing lifting machines. Real hydraulic equipment needs suitable design, inspection, safety protection and supervision. The educational task is to understand an ideal model, explain what it predicts, and know what it cannot prove.
The examination you are preparing for
Singapore’s Singapore–Cambridge Secondary Education Certificate begins in 2027. The G3 Physics syllabus is K323, with the earlier 6091 code appearing only as a reference to the predecessor examination. This is not a new national expectation that students abandon scientific explanation for tricks. The official SEAB 2027 school-candidate syllabus list identifies the current subject code, while the G3 Physics K323 document explains the learning outcomes.
For K323 Physics, Paper 1 is a one-hour, 40-mark multiple-choice paper weighted 30 per cent. Paper 2 is a one-hour-forty-five-minute, 80-mark structured and free-response paper weighted 50 per cent. Paper 3 is a one-hour-fifty-minute, 40-mark practical paper weighted 20 per cent. The pressure topic explicitly includes force per unit area, pressure transmission in hydraulic systems with particular reference to the hydraulic press, density, liquid-column pressure and manometers. These are specific Physics arrangements; a student taking G3 combined science should check the relevant combined-science code and paper structure rather than assuming they are identical.
A pressure question can test recall, application to an unfamiliar diagram, unit conversions or the judgement needed to reject a misleading conclusion. Read the command word before deciding how much to write. Define requires a precise relationship. Calculate requires substituted quantities and units. Explain requires a causal account. Evaluate asks how reliable a claim remains once the assumptions of the model are examined.
Begin with four quantities, not a formula sheet
Force measures a push or pull and is expressed in newtons, N. Area measures the size of a surface and is expressed in square metres, m², when working in SI units. Pressure measures force acting normally per unit area and is expressed in pascals, Pa, where one pascal equals one newton per square metre. Distance measures movement and can be expressed in metres, m. These quantities work together, but they are not interchangeable.
Write pressure = normal force ÷ area, or p = F/A. The phrase normal force matters: it means the component of force perpendicular to the surface. In introductory examination diagrams the pushing force is commonly perpendicular to the piston face, but that condition still has to be noticed. If the same force is spread over twice the area, the average pressure is halved. If the same area experiences twice the normal force, the pressure is doubled. These are statements about ratios, not about a particular machine brand or type.
A frequent confusion arises from the word pressure in everyday language. Someone says that a wider platform creates more pressure because it can carry more weight. In fact, at the same applied force, greater area means less pressure. A wider hydraulic piston can exert a greater force at the same transmitted pressure, but that is a different comparison: the force changes while pressure stays common in the ideal model. Always identify which quantity has been held constant.
Understand pressure before discussing transmission
Imagine a flat block with weight 120 N resting on one of its faces. If the contact area is 0.030 m², its average contact pressure is 120 ÷ 0.030 = 4000 Pa. Turn the block onto a face with area 0.015 m², assuming the same weight is supported and the contact is reasonably uniform. Its contact pressure becomes 8000 Pa. Nothing about its mass or gravitational force has doubled. Only the area has changed. The pressure is now twice as large.
This block problem establishes a simple habit. Before applying p = F/A, make two short annotations: which force is normal to this particular face, and what is the area of that face? The area might be given in square centimetres or derived from dimensions. A rectangle measuring 5 cm by 4 cm has an area of 20 cm², not 20 cm. When converting to SI, 20 cm² = 20 × 0.0001 m² = 0.0020 m². Squared units must be converted with squared conversion factors.
Practice expressing the same answer in two ways: 120 N over 20 cm² is 6 N/cm², which equals 60,000 N/m² or 60,000 Pa. The numerical value changes with the unit, although the physical pressure does not. This is why a bare number such as 6 without a unit makes a poor final answer.
From pressure on a surface to pressure transmitted by a liquid
An ideal hydraulic model consists of two pistons connected by a confined fluid. One piston is smaller, with area A1. The other is larger, with area A2. Applying force F1 to the small piston produces an additional pressure p1 = F1/A1. Under the usual simplified assumptions — a fluid at rest, negligible losses and pistons at the same relevant level — the applied pressure change is transmitted to the other piston. Thus F1/A1 = F2/A2.
The equality refers to a transmitted pressure change in this model. It should not be carelessly expanded into the statement that every point in every liquid always has the same total pressure. In a stationary liquid, pressure can also change with depth. Nor does it follow that every real hydraulic device is perfectly efficient. Seals, friction, hose expansion, fluid compressibility and other practical details can affect performance. The ideal equation teaches the central relationship before those additional complications are considered.
Why use a liquid? In the introductory explanation, a confined liquid transmits the imposed pressure change and changes volume very little under ordinary working pressures. The important idea is not that the fluid manufactures a larger force. It distributes a pressure change over a different area. If the large piston has fifteen times the face area, it may experience fifteen times the output force in the ideal model.
Derive the force ratio in one visible line
Start with p1 = p2. Substitute the expressions F1/A1 and F2/A2. Rearranging gives F2 = F1 × (A2/A1). This single line is more informative than memorising a slogan about big pistons. It explains why the direction of the area ratio matters: the force on the second piston is multiplied by the area of the second piston divided by the area of the first. Reverse the labels and the ratio reverses.
For example, let the small piston area be 6 cm² and the large piston area be 90 cm². The area ratio is 90/6 = 15. An input force of 30 N gives an ideal output force of 30 × 15 = 450 N. The common applied pressure is 30/6 = 5 N/cm², or 50,000 Pa. Dividing the output force, 450 N, by its 90 cm² area gives the same 5 N/cm². This reverse calculation tests both sides of the equation without adding a new formula.
A disciplined answer keeps the word ideal where it belongs. The result 450 N is what the specified model predicts. It does not independently show that a real press is capable of safely supporting 450 N, that its seals are sound or that its frame is rated for that loading. Such claims require evidence that is absent from the question.
Area ratios are not radius ratios
Circular pistons appear often because their radii or diameters can be given instead of their areas. A circle’s area is πr². If one piston has twice the radius of another, its face area is four times as large. Therefore, at equal transmitted pressure, its force can be four times as large in the ideal model. Students who use a ratio of two rather than four have confused a linear dimension with a two-dimensional measure.
Suppose the small piston has radius 1.5 cm and the large piston radius 4.5 cm. Their radius ratio is three, so their area ratio is nine. You do not need to approximate π to find this force ratio, because π cancels. If the input force is 24 N, the ideal output is 216 N. The same approach works with diameters: doubling diameter doubles radius and quadruples area. A common error is to square only one side or to mix the radius from one piston with the diameter from the other.
The mathematical reasoning here connects to the secondary geometry owner. The purpose of that connection is narrow: area scales with the square of a linear factor. It is not a second lesson about every circle theorem.
Force advantage has a movement cost
A good physics answer addresses the distance moved by each piston. Assume the liquid is effectively incompressible and the motion is slow, with no leakage. If the small piston moves through distance d1, it displaces a volume A1 d1. The other piston must move to accommodate the same liquid volume, so A1 d1 = A2 d2. Therefore d2 = d1 × A1/A2. The larger piston moves a shorter distance.
Return to the area ratio of fifteen. If the small piston moves 15 cm, the large piston moves just 1 cm in the ideal model. An output force fifteen times as great is paired with a movement only one-fifteenth as great. This is not an arbitrary penalty invented to balance the mathematics. It follows from conserving the displaced fluid volume. Under ideal conditions the work input F1 d1 and work output F2 d2 are equal.
Using the earlier forces, 30 N through 0.15 m gives 4.5 J of work input. At the output, 450 N through 0.01 m also gives 4.5 J. The matching values remove the apparent mystery. The machine changes the combination of force and movement; it does not provide limitless energy. In a real system, losses can make useful output work lower than input work.
The practical meaning of conservation
When a question gives a dramatic force multiplication, ask what has happened to displacement and energy. Suppose a student writes, “A hydraulic press multiplies the force by twenty, so it produces twenty times the useful work.” That conclusion does not follow. It ignores the shorter output travel. It also assumes that the real machine has no losses. The correct scientific statement is conditional: in an ideal model, a twentyfold increase in force is associated with one-twentieth of the displacement for the same transferred volume.
In an evaluation, examine the system boundary. Which piston is being pushed? Which piston is doing useful work? Is the fluid enclosed? Are there leaks? Is friction ignored? Are the piston movements measured? How is the applied force produced? Not every one of these questions appears in an examination mark scheme, but they train students to identify assumptions before turning a calculation into a physical claim. Good reasoning is not the same as writing a long disclaimer. It is stating the relevant limit at the moment it matters.
This work-and-energy link also builds on the earlier reverse lever calculations in Volume 0435. Both contexts punish a choice of ratio made without first identifying input and output quantities.
Do not confuse hydraulic transmission with liquid-column pressure
Hydraulic pressure transmission and pressure variation with depth are both in the G3 Physics pressure topic, but they are different problems. For a liquid at rest, the additional pressure attributable to the depth of a liquid column is p = hρg, where h is depth, ρ is density and g is gravitational field strength. If water density is taken as 1000 kg/m³ and g as 10 N/kg, the pressure due to a water column 0.8 m deep is 0.8 × 1000 × 10 = 8000 Pa.
This 8000 Pa is a pressure difference due to the depth of the column. It is not automatically the total absolute pressure at that point because atmospheric pressure may also act at a free surface. The result depends on what the question asks and which pressure reference is implied. In a simple textbook situation where both compared locations are at the same liquid level, the pressure-change approach F1/A1 = F2/A2 is appropriate. If the pistons or measuring points are at different elevations, a complete analysis may need to account for hydrostatic differences.
A learner who writes only “liquid pressure is the same everywhere” may get an easy ideal-press question right by accident, then fail a depth question. Better language is to say that an imposed pressure change is transmitted in a confined liquid, with conditions specified.
Density is a relationship, not a memorised label
Density means mass divided by volume, ρ = m/V. To use the liquid-column formula, mass density must be in suitable units, often kg/m³. A given density of 0.8 g/cm³ becomes 800 kg/m³, not 80 or 8000. Because a cubic centimetre equals one millionth of a cubic metre, converting between grams per cubic centimetre and kilograms per cubic metre involves a factor of 1000. This is a common place for a large numerical error.
A liquid with density 800 kg/m³ at depth 0.5 m, using g = 10 N/kg, produces a liquid-column pressure difference of 4000 Pa. Double the depth while keeping density and g unchanged, and that difference doubles. Double the density instead, and it also doubles. If both double, the pressure difference becomes four times as great. These predictions come from the product hρg and are useful multiple-choice checks.
The atmospheric-pressure and manometer portions of the K323 syllabus extend this reasoning into measurement. A manometer can reveal a pressure difference through the difference in liquid levels. It is not necessary to pretend that a hydraulic press and a manometer are the same apparatus merely because both involve fluids and pressure.
A manometer question needs a reference point
Suppose a U-tube manometer has a 0.12 m difference between the levels of an oil of density 850 kg/m³. Using g = 10 N/kg, the magnitude of the pressure difference is Δp = 850 × 10 × 0.12 = 1020 Pa. Whether the gas pressure is greater or less than atmospheric depends on which side has been pushed lower by the pressure. The numerical calculation by itself does not establish the sign or direction.
Before calculating, label the two pressure sources, the height difference and the liquid density. The height difference is not necessarily the total height of liquid visible in one arm. If the question shows the lower liquid level on the gas-connected side, the gas has pushed down more strongly there, indicating pressure greater than the reference atmosphere in the standard open-tube arrangement. If the level is higher on the gas side, the relative pressure is lower.
Students often rush because the formula looks short. The more reliable sequence is to interpret the physical arrangement first, calculate second and describe the result with its reference pressure third. In Physics, a pressure difference is not a complete statement until the comparison is clear.
Force, weight and supported load are not synonyms
In an examination problem, an output piston force may be compared with the weight of a load. Use the distinction deliberately. Mass is in kilograms; weight is a force in newtons, normally W = mg. If a calculated output force is 300 N and the problem gives g = 10 N/kg, the equivalent weight of a 30 kg mass is 300 N. But saying “the press is rated to lift 30 kg” would add an unsupported engineering claim. The ideal force comparison assumes an appropriate configuration and neglects friction and other details.
Sometimes the question asks whether a system can balance a load rather than move it upward. At equilibrium, the relevant forces must balance. To produce upward acceleration, the net force needs to be considered. A simple comparison between the applied hydraulic force and the load weight may be sufficient for an idealised static calculation, but it is not evidence about stability, braking or safe operation.
This discipline links back to the previous display-stand EMS evidence audit in Volume 0437. Both settings show why solving a numerical equality does not provide permission to advertise a safety guarantee.
The hidden role of the diagram
A hydraulic diagram is a relationship map. It may show different widths, connecting tubes, pressure gauges and weights. First find the input piston and output piston. Then mark the normal forces and the areas, and write units beside every number. Do not measure widths with a ruler unless the question explicitly says the diagram is to scale. An attractive sketch may exaggerate the piston sizes to make the idea visible.
If an image says the first piston diameter is 2 cm and the second is 8 cm, the area ratio is sixteen. If the second piston simply looks about four times wider in the drawing, do not rely on appearance to confirm the ratio. Labels are evidence; drawing proportions may not be. Likewise, the symbol p beside a gauge may refer to pressure, while P in a different question could refer to power. Read the key and the context.
When answers disagree, re-read the question to decide whether it gives radii, diameters or piston areas. A small labelling mistake can survive all subsequent correct algebra. Students who annotate before calculating can catch it before it becomes a mark-losing chain.
Worked problem: convert units before you compare
A fictional system has a small piston area of 8 cm² and a large piston area of 0.012 m². An ideal additional pressure of 25,000 Pa is applied through the small piston. Find the force applied to the small piston and the corresponding output force. First convert 8 cm² to 0.0008 m². Then F1 = pA1 = 25,000 × 0.0008 = 20 N. The large piston force is F2 = pA2 = 25,000 × 0.012 = 300 N.
The ratio check is 0.012 / 0.0008 = 15, and 300 / 20 is also 15. The physical story agrees with the arithmetic. A tempting wrong answer is 0.2 N because someone treats 8 cm² as 0.000008 m². Another is 3000 N because the decimal point in 0.012 is moved while copying. A reverse calculation is a small investment that can prevent these errors.
For full working, show the conversion, the pressure equation, the substitution and the unit of the force. The calculator should carry out the arithmetic, not decide what each symbol means. The question may allow direct ratio reasoning, but a transparent conversion makes the answer easier to verify.
Worked problem: start from the required output
Sometimes the examiner reverses the usual story. Suppose the ideal output piston area is 0.025 m² and the target output force is 500 N. The small piston area is 0.0010 m². What input force would give the target under the model? First find the required common pressure, p = 500/0.025 = 20,000 Pa. Then F1 = 20,000 × 0.0010 = 20 N. Alternatively, use F1 = F2 × (A1/A2) = 500 × 0.0010/0.025 = 20 N.
Why is this worth practising? Many candidates have memorised the forward expression F2 = F1 A2/A1. They hesitate when the unknown appears on the other side. A student who begins with pressure equality can rearrange naturally, whichever quantity is missing. Neither piston is magical: each force is simply its area multiplied by the common ideal pressure change.
It also helps to make an estimate first. The input area is one twenty-fifth of the output area. The input force should therefore be one twenty-fifth of the output force under the stated ideal assumptions. If you obtain 12,500 N, your result contradicts that simple proportional picture.
Worked problem: include movement in the answer
An ideal pair of pistons has areas 4 cm² and 60 cm². The smaller piston moves downward by 12 cm. Find how far the larger piston rises. Since the areas have a ratio of fifteen, the output moves 12/15 = 0.8 cm. If the question also gives an input force of 45 N, the ideal output force is 675 N. Check work: the input movement is 0.12 m and the output movement is 0.008 m. Input work is 45 × 0.12 = 5.4 J; output work is 675 × 0.008 = 5.4 J.
This calculation offers two independent checks: transferred fluid volume and equal ideal work. Volume check: 4 × 12 = 48 cm³ displaced by the small piston, while 60 × 0.8 = 48 cm³ received at the large piston. A correct answer should satisfy the pressure relation and the volume relation simultaneously.
Do not claim that the output would travel 12 cm just because both pistons connect to the same liquid. Sharing a pressure change is not the same as sharing a displacement. A large-area output piston requires more fluid volume for every centimetre of its movement.
A multiple-choice question that tests concepts
Consider four claims about an ideal hydraulic press with pistons at the same level. Claim A says the force is the same at both pistons. Claim B says the pressure change transmitted to each piston is the same. Claim C says the larger piston must move farther than the smaller one when equal liquid volumes are displaced. Claim D says the machine creates energy because its output force is greater. Only Claim B is the required general relationship.
Claim A is false unless the two areas happen to be equal or another circumstance makes forces equal. Claim C reverses the volume argument. Claim D confuses force multiplication with useful work. A candidate who identifies the mathematical link among area, pressure and displacement can eliminate three distractors systematically. Guessing from the words greater or smaller is less dependable.
Try a variation: the two piston areas are equal. Then, under ideal conditions, the applied pressure change produces equal forces and equal movement distances. This special case does not contradict the general principle. It illustrates the reason for reading the actual geometry before choosing a memorised story.
A structured answer: explain rather than merely calculate
Imagine a question asking: “Explain why a force of 25 N on a small piston can produce an ideal force of 250 N at a larger piston.” An answer such as “Because the large piston is bigger” is incomplete. It identifies a feature but not the mechanism. A stronger response says that the force on the smaller piston produces an additional pressure equal to its force divided by its area. That pressure change is transmitted through the confined liquid to the larger piston. Since the output piston has ten times the area, the same applied pressure change corresponds to ten times the force.
The explanation should not insist that no energy is lost unless the model explicitly states that losses are neglected. The statement about the force ratio belongs to the ideal conditions. An added sentence can explain that the larger output piston moves a shorter distance for equal displaced volume. This shows deeper understanding when the question asks about efficiency or physical limitations.
Answer the command word and the marks available. If the examiner requests two reasons, do not provide an unrelated paragraph on environmental advantages of hydraulic devices. Be specific, causal and economical.
A calculation with a deliberate trap
A question gives a circular input piston of diameter 3 cm, a circular output piston of radius 6 cm and input force 20 N. What is the ideal output force? The input radius is 1.5 cm. The output radius is 6 cm, four times as great. The area ratio is therefore sixteen. The ideal output force is 320 N. A careless candidate might compare 3 and 6, conclude the ratio is two and give 40 N.
Notice that no numerical value of π is needed. F2/F1 = A2/A1 = (r2/r1)². This is a better route than computing both areas to several decimal places, because it removes an unnecessary source of rounding. Once the ratio is known, a reasonableness check is immediate: the larger piston has much more area, so its output force should be correspondingly greater.
A second trap is to report the common pressure in pascals without converting cm². The input area is π(0.015 m)², not π(1.5 m)². When the radius is supplied in centimetres, keep the units consistent within the ratio or convert both radii into metres before calculating an absolute pressure.
Measurement and practical reasoning
The physics practical assessment tests planning, manipulation, measurement, observation, data presentation, analysis and evaluation. A pressure lesson can develop these abilities even when a particular practical examination task is not known in advance. Picture a supervised classroom demonstration using a manufacturer-approved, low-pressure teaching apparatus. Learners compare applied forces and piston areas, record quantities, and discuss why measured results may not perfectly match an ideal model. The question is not “Can we produce a spectacular load?” but “What observation supports the relationship, and what might distort the result?”
A fair comparison would identify the variables that change and those that should remain controlled. In a piston-area comparison, a student needs to know whether the input pressure is held constant, whether the same apparatus and fluid are used, and whether friction or trapped air makes observations less reliable. Repeat readings can reveal random variation; a calibration offset can produce a systematic discrepancy. Graphing force against area at a fixed applied pressure should give an ideal proportional relationship through the origin, but real data may not sit perfectly on that line.
Never infer a safe working load from a single classroom trial. An educational measurement and an engineering certification answer different questions. The secondary science experimental-skills route develops this distinction in more detail.
How to build a table and a conclusion
Suppose a worksheet reports three pairs of ideal input conditions: at 10,000 Pa, an output area of 0.005 m² gives 50 N; an area of 0.010 m² gives 100 N; an area of 0.015 m² gives 150 N. The learner may conclude that, for the same pressure and within this model, output force is directly proportional to output piston area. The claim is bounded by the conditions, not by every hydraulic machine in existence.
The same table does not show that the largest piston gives the greatest distance moved. It does not establish the power required over time, the efficiency of the apparatus, the strength of the frame, or the maximum permitted load. If a poster claims all these things from three force-area rows, it has stepped beyond its evidence. The correct response is to narrow the claim to what was actually varied and measured or modelled.
When writing a scientific conclusion, connect the evidence to the mathematical relationship: F = pA. State the held-constant pressure and the changed area. If measurements are involved, use the actual trend and identify anomalies before announcing a perfect rule. Distinguish results, inference and limitation.
Reading a difficult data-based stem
Longer questions often include a description, diagram, table and additional information. Before solving, divide the stem into four parts. First locate the task: are you being asked to calculate an input force, compare pressures, choose a piston size or challenge a safety claim? Second list the relevant givens with units. Third identify which model is permitted: ideal pressure transmission, p = hρg, density or an energy relation. Fourth note any stated limitations such as negligible friction, a fixed fluid level or a particular value of g.
A narrative about a workshop, farm or school exhibition is context rather than a request to invent engineering specifications. If a task asks what the model predicts, remain within it. If it asks whether a public claim is justified, inspect evidence that the model has not provided. This deliberate separation reduces errors made by importing assumptions from the story rather than following the physics.
An effective way to manage a long stem is to draw two columns headed Given and Required. For example, “input radius 2 cm; output diameter 12 cm; input force 35 N” belongs under Given. “Ideal output force” belongs under Required. Converting the output diameter to radius 6 cm is the next defensible step; guessing from the picture is not.
Practice set one: four foundation checks
Question 1. A force of 75 N acts normally over an area of 0.025 m². Calculate pressure in pascals. Question 2. A force of 100 N acts normally over 200 cm². Express the pressure in pascals. Question 3. In an ideal hydraulic press, piston areas are 5 cm² and 25 cm². If 18 N acts on the small piston, calculate the force on the large piston. Question 4. Two circular pistons have radii of 2 cm and 6 cm. State their area ratio, small to large, and explain why simply comparing radii is insufficient.
Work without a calculator for the ratio questions first. Include an estimate beside each numerical answer. For Question 1, 75 divided by roughly one fortieth should yield a few thousand pascals. For Question 2, remember that 200 cm² is 0.020 m². For Question 3, identify that the second area is five times the first. For Question 4, the larger radius is three times the smaller radius, so its area must be nine times as great.
These questions test different weak links. A wrong unit in Question 2 suggests a conversion problem. A reversed ratio in Question 3 suggests confusion about which piston is output. A ratio of three rather than nine in Question 4 indicates that the meaning of area has not been applied.
Practice set two: transfer, movement and depth
Question 5. An ideal hydraulic system has input area 0.0015 m² and output area 0.018 m². The desired output force is 360 N. Find the input force and the common additional pressure. Question 6. The same piston pair displaces 90 cm³ of liquid from the input side. How far does the output piston move? Convert its output area to cm² before calculating. Question 7. Calculate the pressure due to an oil column 0.75 m deep if density is 800 kg/m³ and g is 10 N/kg. Is this necessarily the absolute pressure at the point? Question 8. A U-tube liquid manometer has a level difference of 0.10 m and fluid density 1200 kg/m³. Calculate the pressure-difference magnitude using g = 10 N/kg.
These are deliberately mixed. Not every question about a liquid uses the same equation. Question 5 is pressure transmission. Question 6 is a volume or displacement question. Question 7 is pressure due to liquid depth. Question 8 is a manometer reading. The examination skill is to decide which model the language and diagram require before manipulating numbers.
For Question 6, 0.018 m² equals 180 cm², so the displacement in centimetres can be found without rounding a large decimal. In Questions 7 and 8, compare the pressure to an appropriate reference and avoid claiming a pressure direction when the diagram has not supplied enough information.
Worked answers to the foundation and transfer sets
Answer 1. p = F/A = 75/0.025 = 3000 Pa. Answer 2. 200 cm² = 0.020 m²; p = 100/0.020 = 5000 Pa. Answer 3. Equal ideal pressure change gives F2 = 18 × 25/5 = 90 N. Answer 4. The radii ratio is 1:3, and the areas ratio is 1:9 because area is proportional to the square of radius. The output force ratio follows the areas, not the radii.
Answer 5. Pressure = 360/0.018 = 20,000 Pa. Input force = 20,000 × 0.0015 = 30 N. A ratio check gives output area twelve times input area and output force twelve times input force. Answer 6. The output area is 180 cm². Displaced volume divided by area gives 90/180 = 0.5 cm of output travel. The input side, whose area is 15 cm², would move 90/15 = 6 cm if that entire volume is displaced through its movement.
Answer 7. p = hρg = 0.75 × 800 × 10 = 6000 Pa. It is the pressure contribution associated with that depth; without a pressure reference it should not be labelled the absolute pressure. Answer 8. Δp = 0.10 × 1200 × 10 = 1200 Pa. The magnitude alone does not say which side is higher in pressure. A correct explanation of direction needs the arrangement and relative heights of the two arms.
Practice set three: a written judgement question
A fictional brochure says: “Our educational hydraulic model has a piston-area ratio of ten. A student applied 12 N at the small piston. Therefore the large piston will always deliver 120 N in every setting, travel the same distance as the small piston, and support any 120 N object safely.” Identify two defensible statements and three conclusions that require correction or further evidence.
A strong answer begins by identifying the stated ideal conditions rather than assuming every word of the brochure is true. If the apparatus is treated as an ideal hydraulic system and the areas ratio is ten, a 12 N input corresponds to an ideal 120 N output. The large piston need not travel the same distance. With conserved displaced volume it travels one tenth as far as the input piston. Whether the real output reaches that force depends on practical losses. Whether an object is safely supported depends on equipment ratings, geometry, operating conditions and other evidence.
The brochure also omits the distinction between a force calculation and lifting a moving load. If a question adds the weight of a particular object, compare forces under stated assumptions. Avoid claiming a machine is suitable merely because one numerical threshold is met. In an evaluation question, the most convincing answer names what the evidence supports and what measurement or specification remains missing.
Why students lose marks in an otherwise short chapter
The pressure chapter is not difficult because its equations are long. It is difficult because short equations can be applied in the wrong setting. Students confuse pressure and force, turn a radius ratio into an area ratio, convert cm² as though it were cm, assume every point in a liquid has equal pressure, forget the difference between gauge and absolute pressure, and treat an ideal numerical output as a practical safety claim.
The best repair is to classify each error. If the student writes the wrong equation, practise identifying the physical situation before any arithmetic. If the student writes the correct equation but substitutes the wrong area, practise diagram labelling. If the arithmetic is wrong, practise conversions and estimation. If the calculation is right but the explanation overreaches, practise adding one necessary condition to a factual claim. Different mistakes need different fixes.
A simple check sequence is: draw or annotate the situation; identify the force and area; convert units if an SI pressure is required; select the relationship; calculate; reverse-check the ratio or units; answer in the language of the question. This process is slower during early practice and faster after it becomes familiar.
A focused revision sequence across four sessions
In the first study session, spend time on the meaning of pressure using blocks, feet, supports and labelled contact areas. Work five straightforward p = F/A problems and explain aloud why an unchanged force distributed over more area gives less pressure. Do not move to hydraulic presses until this statement can be used without prompting. Conceptual fluency is a better foundation than the number of pages completed.
In the second session, derive the ideal hydraulic equality from pressure transmission. Work problems with area values in the same units, then with cm² and m² mixed. Add problems where the unknown is the input force, not always the output. In the third session, introduce circular pistons, fluid volume and movement distance. Verify results using area ratios and conservation of ideal work. In the fourth session, mix hydraulic questions with liquid depth, density and manometer comparisons, then write two short explanations correcting exaggerated public claims.
The sessions are suggested practice, not a claim about the official SEC timetable or the time every student must spend. If a student is still misreading area units, repeat that foundation before progressing. Reviewing errors at intervals is more useful than repeating only the questions already answered correctly.
A short note for parents and tutors
An adult can help without turning every homework problem into an explanation the learner only copies. Ask the student to point to the small and large piston, name the unit of area, and tell you what remains unchanged in an ideal pressure-transmission comparison. Then ask which quantity becomes larger and what happens to displacement. These questions reveal the mental model more clearly than a quick declaration that the topic is understood.
If an answer is wrong, find the first step where the meaning changed. It may be an unnecessary square, a centimetre conversion, a reversed input label or an unjustified safety conclusion. Let the learner correct that step and carry the rest forward. The same style of help is useful in the PSLE-to-secondary Science handover: identify cause, evidence and the difference between a statement and a justified explanation.
For structured secondary support, the experimental-skills hub extends the method to controlled investigations and interpretation. The learner should become independent of the prompt, not dependent on more prompts.
What changes in the next guides
The next guide takes the same physics into Mathematics: reverse calculations with piston areas, circular dimensions, feasible intervals and checks that prevent an impossible design answer. A subsequent English guide treats the way visual and numerical claims are presented to readers, and an EMS case puts all three disciplines in conversation. This sequence builds from a physical relationship to calculation, explanation and evidence-based judgement. It is not an invitation to build or operate an actual press.
The essential understanding from this volume is compact. Pressure is normal force divided by area. In the stated ideal hydraulic model, an imposed pressure change is transmitted through the confined liquid, so piston forces scale with their areas. With conserved transferred volume, greater output force is paired with shorter output movement. Liquid pressure also varies with depth, which is why pressure transmission and hydrostatic calculations must be distinguished. Real equipment performance and safety require evidence beyond these ideal equations.
Official reference for current coverage and assessment: SEAB 2027 G3 syllabuses and SEAB’s SEC overview. Read the document for the examination year and subject level actually entered. The worked values in this guide are teaching examples, not past-paper reproductions or measured machinery specifications.
