Q: What does IP Physics Notes (Upper Secondary, Year 3-4): 5) Pressure cover? A: Force-per-area basics, liquid columns, hydraulic systems, and Boyle's law for IP Y3-Y4 applied pressure questions.
Quick recap -- Pressure concentrates force over area. In fluids it scales with depth, density, and gravity; in gas systems it trades off with volume when temperature stays fixed.
The core idea is simple: Pressure is force spread over area.
Use it as a working check: Smaller area gives higher pressure for the same force. In liquids, pressure increases with depth, density, and gravity. In sealed gases, pressure and volume trade off at constant temperature.
Then go one layer deeper: Use the pool wall and hydraulic jack examples to practise choosing between P=F/A, P=ρgh, Pascal's principle, and Boyle's law.
For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: solid and liquid pressure, hydraulic systems, atmospheric pressure, manometers, and Boyle's law follow Marcus Pang's current Chapter 5 route.
Eclat blended link: density is introduced in Chapter 3 and reused here in liquid-column pressure. Review ρ=m/V before applying P=ρgh.
Eclat extension: Boyle's law belongs to Marcus's IP chapter but is not a named outcome in K323 Topic 5. Keep it in your assessed scope only when your school has introduced the fixed-mass, constant-temperature model.
2027 national comparison: K323 Topic 5 overlaps with pressure, hydraulics, density, liquid-column pressure, atmospheric pressure, and manometer use.
Check your school: follow the gauge-pressure and atmospheric-pressure convention used in the current task.
Before substituting numbers, identify what is physically changing in the question.
Question cue
Start with
Check before calculating
A force acts on a stated contact area
P=AF
Use the force perpendicular to the surface and convert area to m2.
A point is deeper in a liquid
P=ρgh
Use vertical depth below the surface, not slanted distance or container width.
Two pistons are linked by trapped liquid
A1F1=A2F2
A U-tube column height is different on two sides
ΔP=ρgΔh
Decide whether the question wants gauge pressure or comparison with atmospheric pressure.
A sealed gas changes volume at constant temperature
P1V1=P2V2
Common trap: do not mix the models in one step. For example, a hydraulic jack first uses equal pressure in the liquid, then uses piston areas to find different forces.
Defining Pressure
Pressure is force acting perpendicular to a surface per unit area.
Formula:
P=AF
P in Pa=N⋅m−2
F in N
A in m2
Increase pressure by raising the normal force or shrinking the contact area. Decrease it by spreading the force over a wider area.
Pressure in Solids: Force vs Area
Scenario
Contact area
Applied force
Resulting pressure
Drawing pin point
Very small
Thumb force
Very high -- tip bites into board
Snowshoe
Wide
Body weight
Low -- spreads load so you do not sink
Knife edge
Sharpened
Cutting push
High -- blade slices material
When the contact is angled, resolve the force to the component perpendicular to the surface before using P=AF.
Engineers tweak both force and area (e.g., tyre width, footprints of machine supports) to keep ground pressure within safe limits.
Contact-area pressure checkpoint
For solid contact questions, pause before substituting into P=AF. Most wrong answers come from using the wrong area unit or the wrong force component.
Question clue
First check
Calculation move
Common trap
Area is given in cm2 or mm2
Pressure in pascals needs m2.
Convert area before division.
Treating 4.0cm2 as 4.0m2.
A slanted push acts on a surface
Only the perpendicular component produces pressure on that surface.
Use the normal component of the force.
Using the full angled force without resolving.
A shoe, snowshoe, tyre, or support has a wider base
Compare cases with the same weight first.
Larger area gives lower pressure when force is unchanged.
Saying "larger area always lowers pressure" even when the force also changes.
A sharp point or blade is used
Contact area is deliberately tiny.
Same force over smaller area gives higher pressure.
Saying sharp objects work because they use a bigger force.
Worked check: a small rubber foot is pushed against the floor with 120N over a contact area of 4.0cm2.
A=4.0cm2=4.0⋅10−4m2,P=4.0e−4120=3.0⋅105Pa.
Misconception check: smaller area raises pressure only when the force being compared is the same. If both force and area change, calculate F/A for each case instead of judging from area alone.
Hydrostatic Pressure
Fluids transmit pressure equally in all directions; deeper layers carry the weight of fluid above.
Density is mass per unit volume, ρ=m/V. This chapter reuses the density work introduced in Chapter 3.
Pressure at depth h in a fluid of density ρ:
P=ρgh
Derivation: take a column of cross-sectional area A and height h. Its weight is W=ρAhg. Using P=AF
Hydrostatic pressure depends only on depth, density, and gravity -- not on container shape.
Applied cases: dam walls thicken near the base, submarines need hulls that withstand higher ρgh at depth.
Worked Example: Swimming Pool Wall
At 3.0m depth, water exerts pressure
P=ρgh=1000kg⋅m−3×9.81m⋅s−2×3.0m=2.94×104Pa
Each square metre of wall must therefore resist about 29.4kN pushing sideways.
Pascal's Principle & Hydraulic Systems
Pascal's principle: a pressure change applied to an enclosed, incompressible fluid transmits undiminished to every part of the fluid.
In a hydraulic press with pistons of area A1 and A2:
P=A1F1=A2F2
Force multiplication:
F2=F1×A1A2
Trade-off: the larger output force moves a shorter distance so overall energy is conserved. Hydraulic brakes, car jacks, and presses all rely on this ratio.
Worked Example: Service Garage Jack
A mechanic pushes with 180N on a 5.0cm2 input piston. The output piston area is 150cm2.
F2=180×5.0150=5400N
The car experiences a lifting force of about 5.4kN.
Measuring Pressure
Barometer: a sealed mercury column. Atmospheric pressure equals the hydrostatic pressure of the mercury column: Patm=ρHggh. Standard sea-level pressure 1.013×105Pa supports 760mm of mercury.
Manometer: U-tube with one side connected to the gas source.
Closed-end: pressure equals ρgh of the column difference.
Open-end: difference between gas and atmospheric pressure given by the column height difference.
Digital sensors convert diaphragm deformation into electrical signals -- but conceptually they still report force per area.
Manometer direction checkpoint
For an open-end manometer, first decide which side has pushed the liquid level lower. That side has the higher pressure.
If the gas side liquid level is lower than the open side, then Pgas>Patm, so Pgas=Patm+ρgΔh.
If the gas side liquid level is higher than the open side, then Pgas<Patm, so Pgas=Patm−ρgΔh
If both liquid levels are equal, then Pgas=Patm.
Worked check: in an open water manometer, the gas side is 4.0cm lower than the open side. The gas pressure is higher than atmospheric pressure by ρgΔh=1000×9.81×0.040=392Pa. If Patm=1.01⋅105Pa, then Pgas≈1.014⋅105Pa.
Misconception check: do not automatically add ρgΔh. Add it only when the gas pushes its side lower; subtract it when the gas side is higher.
Gas Pressure & Boyle's Law
For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume:
P1V1=P2V2
Syringe: pulling the plunger increases volume, lowering pressure so fluid/air enters.
Breathing: diaphragmatic movement changes lung volume, producing pressure gradients with the atmosphere.
Scuba ascent: decreasing external pressure lets lung volume expand; divers ascend slowly so the gas escapes instead of over-expanding tissues.
Boyle's law setup checkpoint
Before using P1V1=P2V2, check that the question is describing the same trapped gas at constant temperature. Then keep one pressure unit and one volume unit throughout the equation.
Question clue
What to check first
Setup move
Common trap
"Fixed mass of gas" or "sealed syringe"
The same gas particles stay inside.
Use P1V1=P2V2.
Applying Boyle's law when gas is leaking or being added.
"Temperature remains constant"
Average kinetic energy is unchanged.
Pressure changes only because volume changes.
Using Boyle's law during heating or cooling without checking temperature.
Volume is halved
Smaller volume means more frequent wall collisions.
Pressure doubles if the gas mass and temperature stay fixed.
Saying pressure halves because volume halves.
Mixed units such as cm3 and m3
Both volumes must use the same unit.
Convert before substituting, or keep both volumes in cm3
Worked check: a sealed syringe contains gas at 100kPa and 60cm3. The plunger is pushed in slowly at constant temperature until the volume is 40cm3.
P2=V2P1V1=40100×60=150kPa.
Misconception check: Boyle's law is inverse, not direct. A smaller volume gives a larger pressure only when the amount of gas and temperature stay constant.
Practice Quiz
Run through hydrostatic calculations, hydraulic ratios, and Boyle's law quick-fire checks to consolidate your pressure toolkit.
Key Takeaways
Start every problem by identifying the relevant area, depth, or volume change, then apply P=AF or P=ρgh.
Pascal's principle enables hydraulic multipliers but keeps the pressure equal throughout the fluid.
Pressure measurement tools compare unknown pressure to a known fluid column height.
Gas pressure trades with volume when temperature is fixed -- keep PV constant and track unit conversions carefully.
Equal pressure does not mean equal force; the larger piston has the larger force.
Keep the same pressure units on both sides and convert volume units consistently.
