Use pressure, force and contact area
Use the supplied pressure formula to find pressure, force or area, convert area units and compare contact-face choices under an average-pressure model.
Section 4.4 · Measures
Exam scope and assessment notes
Foundation and Higher share the seven published F:4.4A–G Measures rows. The Higher table says See Foundation Tier; there are no additional H4.4 rows. The chosen eight-lesson split covers instrument scales, time intervals, estimates, nearest-degree angle measurement, three-figure bearings, average speed, density and pressure. Eight lessons and thirty-two original worked/staged cases are StudyForA’s design, not a syllabus-prescribed count. Each structural source link requires the corresponding complete teaching record carrying that source ID, a worked explanation and all Guided, Independent and Transfer stages. F4.4F requires the complete speed lesson; broad F4.4G requires all three speed, density and pressure lessons, not a speed-only partial link. Source binding and substantive structure must belong to the same record; a separate untagged duplicate cannot supply missing evidence. These are structural links, not exhaustive subskill or specialist approval. Preserve the actual calibrated instrument ticks, pointer/meniscus positions, ruler endpoints, units and protractor alignment action. Read labelled scales rather than assuming CSS pixels are millimetres. Use only diagrams explicitly supplied as accurate for measurement; ordinary examination diagrams need not be to scale and must not be measured without instructions. Numerical reference readings, protractor openings and the measured bearing remain hidden before a calibrated learner attempt; revealing a reference is not proof of instrument operation. Keep ideal model values separate from approximate human readings and nearest-degree rounding; no universal automatic tolerance or measurement competence is established. Bearings start at the named origin and run clockwise from declared north, with parallel north assumptions stated and leading zeros retained. Distinguish reverse rays, interior route angles and forward heading changes; no magnetic or globe-navigation correction is inferred. Clock intervals retain am/pm, noon/midnight and day order. Average speed uses total route distance over the full named elapsed interval, including stated stops; displacement and instantaneous speed are different. Calculating time by division requires positive speed: at constant zero speed a positive distance is impossible, while zero distance alone does not determine elapsed time. Keep density mass/volume units compatible and volume positive; liquid displacement is final minus initial level. Combined volumes are added only under the stated no-void/no-volume-change assumption. The pressure formula is supplied in every pressure stage; do not infer that the density or average-speed formula is also supplied. Use a provided force in newtons, positive contact area, correct squared-area conversion and an explicitly limited average-pressure model. No real-world safety or route-feasibility certification follows. The source assessment context uses SI units and may require mathematical instruments. These lessons do not complete4.5 nearest-mm line drawing, formal shape/compass constructions or scale-drawing problems,5.2 transformations,4.8 bearing trigonometry,4.9/4.10 full mensuration or4.11 map/similarity work. No blanket formal-proof waiver is imported. Written practice is self-checked, not automatically marked; acknowledgements and attempts do not award marks or establish mastery. Specialist approval and formal assessment validation remain unestablished. Publisher navigation is separate evidence: literal F4.4D appears in Book2 Shape and Space6, map physical10/pages128–160, and Hub36 Bearings and Scale Drawings, co-tagged F4.5C/F4.11B; those co-tags confer no construction/map credit here. Literal F4.4G appears in Book1 Number5, map physical5/pages255–268, and Hub13 Compound Measures. A/B/C/E/F have no literal4.4 chapter or Hub tag in that map; do not invent one from nearby time or scale-drawing topics. Book1 public uncorrected-proof contents start Number5/Compound measures at256 and end its summary at269, conflicting with map255–268; preserve both references. Book2 chapter start128 and next start161 agree with its map range, but no explicit bearings contents heading locates every skill. Neither relevant chapter body is present in the public samples. Full final-book/Teaching Hub content and pagination alignment remain unverified; chapter links are not full-book or modular teaching credit. The current2026/2027 manual lists linear4MA1 F/H in November2026 and June2027; FR/HR are June2027 only, based on centre eligibility. Vietnam is on the regional-centre list; learner residence alone does not choose an option. The table points to page2 although the list is on physical/printed3. Modular4XMAF/4XMAH cash-ins and4WM units remain separate entries for international centres. This entry context does not create a2027 revision of the Issue2 specification.
Before you begin
- Use a supplied formula and rearrange multiplication or division.
- Multiply two labelled dimensions to form a rectangular area, using 1 m²=10000 cm².
- Convert 1000 Pa to 1 kPa and distinguish a given force from mass.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Given P=F/A, which area in m² equals 160 cm²?
What you will learn
- Use the given P=F/A relationship with force in newtons and area in square metres.
- Rearrange for force or contact area and convert cm² to m² correctly.
- Compare positive contact areas with a stated pressure limit and explain the model’s limits.
Given P=F/A, a force of 720 N acts normally over a contact area of 0.12 m². Find the average pressure in Pa and kPa. Then find the pressure if the area doubles to 0.24 m² while the force remains 720 N. Explain the direction of the change.
Read, draw and calculate before comparing. Every requested part and explanation needs your own attempt. The teal protractor and orange rays show your actual placements, including wrong ones. Comparisons do not award marks or establish practical instrument competence.
Coordinates use the original worksheet grid. The entire drawing and your tools rotate together; clockwise bearings retain their direction. Case or orientation changes clear all responses, placements and references. Named given points can be placed exactly; other drawing clicks use the nearest half-unit grid point and never snap to a solution.
Given panel 1
- Contact force (N)
- 720
- Contact area (m²)
- 0.12
- New contact area (m²)
- 0.24
Given relationship: P=F/A
The board retains the last submitted valid placement. Editing a draft invalidates its submission without moving the last drawing; submit again before comparison.
Read the given formula and its units
For these tasks the supplied relationship is P=F/A, where P is average pressure, F is force normal to the contact and A is contact area. Use force in newtons and area in square metres to obtain N/m², called pascals, Pa. One kilopascal is 1000 Pa. State which symbol is unknown before choosing the operation.
Force in newtons is not a mass in kilograms. All forces needed here are provided, so no unstated gravitational conversion is required. The source says that a pressure formula will be given; it does not make the same promise for density or average speed. This simple contact-pressure model does not establish a general fluid-pressure law or a safety assessment of a real surface.
Rearrange and convert the contact area
From P=F/A, multiply by A to get F=P×A, or rearrange to A=F/P when the pressure is positive. Ordinary contact areas in these models must be positive; division by zero is undefined and a negative contact area does not describe the stated situation. Check which quantity is required rather than always dividing the first two given numbers.
Convert area units by the square of the length conversion. Since 1 m=100 cm, 1 m²=10000 cm². An area in cm² is therefore divided by 10000 to write it in m² before using newtons per square metre. Converting by only 100 treats an area like a length and gives a pressure wrong by a factor of 100.
Use the largest permitted contact area
For a fixed positive force, a larger contact area gives a smaller average pressure because that force is divided across more area. To meet a maximum pressure, the area must be at least force divided by that maximum. Equality is included when the instruction says “no greater than”; a slightly smaller positive area would give a larger pressure.
For a rectangular object, work out the possible contact-face areas from the labelled dimensions and calculate each pressure before choosing an orientation. Assume the stated full, flat contact and the given force; do not infer an unmentioned mass or extra support. The calculation is a model comparison, so a numerical pass against its limit does not certify real materials or loading conditions.
Sources for this lesson
Pause and explain
Given P=F/A with a fixed positive force, which contact choice minimises average pressure?
Worked example
Given P=F/A, a force of 720 N acts normally over a contact area of 0.12 m². Find the average pressure in Pa and kPa. Then find the pressure if the area doubles to 0.24 m² while the force remains 720 N. Explain the direction of the change.
Show the worked solution
- The force and area already have compatible units for pascals. Substitute in the given formula: P=720÷0.12=6000 N/m²=6000 Pa.
- Convert pascals to kilopascals: 6000÷1000=6 kPa. This conversion changes the unit label and numerical value together.
- For the larger area, use the unchanged force: P=720÷0.24=3000 Pa, or 3 kPa.
- Doubling the area with fixed force halves the pressure. Both areas are positive, so the quotient and this comparison are valid in the stated average-contact model.
Answer Initial pressure: 6000 Pa=6 kPa. At 0.24 m²: 3000 Pa=3 kPa, half the original pressure.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Given P=F/A, a force of 630 N acts on an area of 210 cm². Find pressure in Pa and kPa. Then find the force required for the same pressure on an area of 0.035 m². Show the area conversion and the rearranged formula.
Give me a hint
Divide cm² by 10000 to get m². For the second question use F=P×A.
Compare my reasoning
- Convert contact area: 210 cm²÷10000=0.021 m². Keep the force as the given 630 N.
- Calculate P=630÷0.021=30000 Pa. Converting gives 30000÷1000=30 kPa.
- For the new force at that pressure, use F=P×A with pressure in Pa: 30000×0.035=1050 N.
- Check that 1050÷0.035=30000 Pa. The force is larger because the new area is larger while the required pressure is unchanged.
210 cm²=0.021 m². Pressure: 30000 Pa=30 kPa. Force for 0.035 m² at that pressure: 1050 N.
Look for these in your work
- Convert the square-centimetre area using 10000.
- Distinguish the 30000 Pa value from the equivalent 30 kPa value.
- Use F=P×A with pressure in Pa to find and check the force of 1050 N.
Given P=F/A, a force of 800 N acts on 0.025 m². Independently find the pressure in Pa and kPa. To keep pressure no greater than 20 kPa with that same force, find the minimum contact area in m² and cm². Explain whether equality at the limit is allowed.
Give me a hint
Convert the maximum pressure to Pa. Rearrange to find the minimum positive area, then check it by substitution.
Compare my reasoning
- For the original contact, P=800÷0.025=32000 Pa=32 kPa. That exceeds the later 20 kPa limit.
- Convert the limit: 20 kPa=20000 Pa. Rearranging gives the threshold area A=F/P=800÷20000=0.04 m².
- Convert back to square centimetres: 0.04×10000=400 cm². This is the minimum area for the fixed-force model.
- At A=0.04 m² the pressure is exactly 20000 Pa, which is allowed by “no greater than”. Larger positive areas also meet the limit; smaller positive areas give higher pressure, so A≥0.04 m².
Original pressure: 32000 Pa=32 kPa. Minimum area: 0.04 m²=400 cm²; equality is allowed, so A≥0.04 m².
Look for these in your work
- Calculate the original pressure with correct Pa and kPa units.
- Rearrange with 20000 Pa to find the threshold area.
- Convert to 400 cm² and explain both equality and the direction of the area requirement.
Given P=F/A, a rectangular crate presses down with a force of 1200 N. Its labelled dimensions are 0.2 m, 0.3 m and 0.5 m. Use the three possible full rectangular contact faces in the diagram. Find the average pressure on each face, then choose every orientation meeting a maximum pressure of 10 kPa. Assume full flat contact and the stated downward force.
Give me a hint
Multiply each pair of dimensions for a face area. The largest face should give the lowest pressure.
Compare my reasoning
- The three possible face areas are 0.2×0.3=0.06 m², 0.2×0.5=0.10 m² and 0.3×0.5=0.15 m². Keep each face paired with its area.
- Using the fixed force, the pressures are 1200÷0.06=20000 Pa, 1200÷0.10=12000 Pa and 1200÷0.15=8000 Pa.
- These are 20 kPa, 12 kPa and 8 kPa. Only 8 kPa is no greater than 10 kPa, so the 0.3 m by 0.5 m face, area 0.15 m², is admissible.
- The largest face minimises the model pressure. This conclusion uses full flat contact and a provided force, not an unstated mass conversion or a certification that a real support is safe.
Face pressures: 0.06 m²→20000 Pa, 0.10 m²→12000 Pa, 0.15 m²→8000 Pa. Only the 0.15 m² face meets the 10 kPa maximum.
Look for these in your work
- Show all three labelled face-area calculations.
- Calculate each pressure and compare in the same unit as the 10 kPa limit.
- Choose only the 0.3 m by 0.5 m contact face and state the average-contact assumptions.
Common mistakes
- Using kilograms as newtons without a stated force conversion.
- Converting cm² to m² by dividing by 100 rather than 10000, or mixing Pa and kPa.
- Assuming a smaller area lowers pressure at fixed force, allowing zero area, or interpreting a model limit as safety certification.
What can you explain now?
Given P=F/A and a fixed positive force, what happens to pressure when a positive area is halved?
Compare with the explanation
The pressure doubles.
The denominator is halved while the numerator stays fixed, so the quotient doubles. The comparison assumes positive areas; a zero area would make the formula undefined.
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Make it stick
Starting from the given P=F/A, recall the forms for force and area. State the cm²-to-m² conversion and explain why a larger positive contact area lowers average pressure at fixed force.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.