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What is pressure?

Key idea: Understand pressure as force per unit area (p=F/A), including area unit conversions, exam tips, and worked examples for O Level Physics.

  • SEC G3 Physics 2027
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Learning objectives

  • Define pressure as force per unit area
  • Apply pressure = force ÷ area
  • Explain pressure transmission in a hydraulic press
  • Apply density = mass ÷ volume
  • Apply liquid-column pressure = height × density × gravitational field strength
  • Explain how liquid-column height measures atmospheric pressure
  • Explain how a manometer measures pressure difference

1. Definition

A. Pressure

A drawing pin can be pushed into wood easily, but your finger cannot. The pin tip has a much smaller contact area, so it produces a much larger pressure.

Pressure and contact area comparisonTwo cases with equal downward force. A wide shoe has a larger contact area and lower pressure, while a sharp tip has a smaller area and higher pressure.Large contact areaSmall contact areaA is largeF (same)p = F / ALow pA is smallF (same)p = F / AHigh p
Same force on a smaller contact area produces greater pressure.

Pressure, p, is the force per unit area, where the force acts perpendicular to the surface.

p = F/A

  • p = pressure (Pa)
  • F = perpendicular force (N)
  • A = area (m²)
  • SI unit: pascal (Pa), where 1 Pa = 1 N m⁻²
  • Pressure is a scalar quantity.

You may also see:

  • 1 kPa = 10³ Pa
  • 1 MPa = 10⁶ Pa

2. Key Ideas

  • Pressure increases when:
    • the force increases, or
    • the contact area decreases.
ChangeEffect on pressure pReason
F increases (area same)p increasesp = F/A
A increases (force same)p decreasesp = F/A
  • In many “object on a surface” questions, the force is the weight: F = W = mg (in newtons).
  • Convert area units carefully:
    • 1 cm² = 1 × 10⁻⁴ m²
    • 1 mm² = 1 × 10⁻⁶ m²

3. Detailed Explanations

A. Why the drawing pin works

With the same force:

  • large area (finger) → low pressure → wood does not yield easily
  • small area (pin tip) → high pressure → wood yields more easily

B. What force should you use in p = F/A?

Use the perpendicular force on the surface.

In most O Level questions:

  • object resting on a surface → F is the weight (mg)
  • object pressing on a surface → F is the force pushing on the surface
Where this goes next

4. Common Mistakes

  • Using A in cm² or mm² without converting to m².
  • Forgetting that force must be in newtons (use F = mg, not just mass).
  • Forgetting that area conversions are squared (e.g. 1 cm² ≠ 10⁻² m²).
  • Using the wrong face area when finding maximum/minimum pressure.

5. Exam Tips

  • Write p = F/A first, then:
    1. find F in newtons (often F = mg),
    2. convert A to m²,
    3. substitute and calculate,
    4. state the final unit (Pa).
  • For “maximum pressure”, use the smallest contact area; for “minimum pressure”, use the largest contact area.

6. Worked Examples

Modelled example 1

Pressure from weight and area

Core

Problem

Plastic blocks of total mass 42 kg rest on a person. The contact area is 840 cm². Take g = 10 N kg⁻¹.

  1. Find the pressure on the person.
  2. State what happens to the pressure if the area decreases while the mass stays the same.
Study the worked solution

  1. Find the perpendicular force

    Method

    Convert the blocks’ mass to their weight.

    Reason

    Pressure uses force in newtons; the downward force on the person is the blocks’ weight.

    Working

    F = mg = (42)(10) = 420 N

  2. Convert the contact area

    Reason

    Pressure in pascals requires area in square metres, and 1 cm² = 10⁻⁴ m².

    Working

    840 cm² = 840 × 10⁻⁴ = 0.084 m²

  3. Calculate the pressure

    Reason

    The full perpendicular force is spread over the stated contact area.

    Working

    p = F/A = 420/0.084 = 5.0 × 10³ Pa

  4. Predict the effect of reducing area

    Working

    With the same force, decreasing A increases p = F/A.

Guided practice 2

Same force, different area

About 6 min

Problem

A girl has mass 50 kg. Find the pressure she exerts on a trampoline:

  1. standing upright on area 500 cm²;
  2. lying down on area 6000 cm².

Take g = 10 N kg⁻¹.

Calculate and compare both pressures

Unit: Pa
Unit: Pa

Hints

Hint 1: identify what stays constant
Her weight is the same in both positions; calculate it once using F = mg.
Hint 2: convert each area
Multiply each area in cm² by 10⁻⁴ before using p = F/A.
View solution step by step

  1. Find the common force

    Method

    Use the girl’s weight for both contact arrangements.

    Reason

    Changing position changes contact area but not her mass or gravitational field strength.

    Working

    F = mg = (50)(10) = 500 N

  2. Calculate the standing pressure

    Reason

    The standing contact area is 500 × 10⁻⁴ = 0.050 m².

    Working

    p_standing = 500/0.050 = 1.0 × 10⁴ Pa

  3. Calculate the lying pressure

    Reason

    The lying contact area is 6000 × 10⁻⁴ = 0.60 m².

    Working

    p_lying = 500/0.60 = 8.33 × 10² Pa

  4. Compare the results

    Working

    Lying down spreads the same force over twelve times the area, so the pressure is one twelfth as large.

Common misconception 3

Roller blades vs sneakers

Find and correct the mistake

Learner response

The pressure on the ground is greater when the same person wears roller blades than when they wear flat-soled sneakers. A student writes:

Roller blades produce more pressure because the person becomes heavier.

Locate the first error and give the correct explanation.

Diagnose the comparison

What is the first error?

View solution step by step

  1. Locate the first error

    Method

    Keep the person’s weight approximately constant.

    Reason

    Changing footwear does not meaningfully change the person’s mass, so it does not explain the pressure difference.

    Working

    The perpendicular force F is about the same in both cases.

  2. Correct the area comparison

    Reason

    Roller-blade wheels contact the ground over a smaller total area than flat sneaker soles.

    Working

    For the same F, a smaller A gives a larger p = F/A.

Examiner practice 4

Maximum vs minimum pressure (different faces)

4 marks

Examination question

A wooden plank is 100 cm × 20 cm × 2 cm and has weight 40 N. Calculate the maximum and minimum pressure it can exert on a table. [4 marks]

Show the face choices, conversions and pressures

View solution step by step

  1. Choose the face for maximum pressure

    1 mark

    Method

    Select the smallest face and convert its area.

    Reason

    For a fixed weight, pressure is greatest when contact area is smallest.

    Working

    Aₘᵢₙ = (20)(2) = 40 cm² = 4.0 × 10⁻³ m²

  2. Calculate the maximum pressure

    1 mark

    Reason

    The plank’s full 40 N weight acts over the smallest face.

    Working

    pₘₐₓ = 40/(4.0 × 10⁻³) = 1.0 × 10⁴ Pa = 10 kPa

  3. Choose the face for minimum pressure

    1 mark

    Method

    Select the largest face and convert its area.

    Reason

    For the same weight, pressure is least when contact area is largest.

    Working

    Aₘₐₓ = (100)(20) = 2000 cm² = 0.20 m²

  4. Calculate the minimum pressure

    1 mark

    Reason

    The force remains 40 N; only the supporting face changes.

    Working

    pₘᵢₙ = 40/0.20 = 200 Pa

Challenge 5

Knife edge pressure (unit conversions)

Minimal support

Estimation transfer

A force of 20 N is applied to a knife edge. The contact region is approximately 3.0 cm long and 0.10 mm thick. Estimate the pressure in Pa and MPa.

Derive the contact area before calculating pressure

Unit: m²
Unit: Pa

Hints

Hint 1: convert both contact dimensions
Use 1 cm = 10⁻² m and 1 mm = 10⁻³ m.
Hint 2: build the area before using pressure
Multiply the converted edge length by its converted thickness, then use p = F/A.
View solution step by step

  1. Convert the two lengths

    Method

    Convert each contact dimension to metres before multiplying.

    Reason

    The two dimensions start in different units, while the pressure calculation requires area in m².

    Working

    3.0 cm = 0.030 m, 0.10 mm = 1.0 × 10⁻⁴ m

  2. Estimate the contact area

    Reason

    The narrow contact region is approximated as a rectangle.

    Working

    A = (0.030)(1.0 × 10⁻⁴) = 3.0 × 10⁻⁶ m²

  3. Calculate and convert the pressure

    Reason

    The applied force is perpendicular to the estimated contact area.

    Working

    p = 20/(3.0 × 10⁻⁶) = 6.7 × 10⁶ Pa = 6.7 MPa

7. Mind Stretchers

Mind stretcher 1: SnowshoesExtension

Explain why people wear snowshoes when walking on snow.

Show Answer

Snowshoes increase the contact area A with the snow. For the same weight F, a larger A gives a smaller pressure p = F/A, so you sink less.

Mind stretcher 2: High heels vs flat shoesExtension

A 60 kg person stands on one high heel with contact area 2.0 cm². Estimate the pressure on the ground (take g = 10 N kg⁻¹). Comment on why high heels can damage floors.

Show Answer

Force (weight): F = mg = (60)(10) = 600 N

Convert area: 2.0 cm² = 2.0 × 10⁻⁴ m²

Pressure: p = 600/(2.0 × 10⁻⁴) = 3.0 × 10⁶ Pa = 3.0 MPa

Very large pressure over a small area can mark or damage some floors.

8. Practice and next step

Complete the Pressure quiz, then continue to Hydrostatic pressure to apply pressure reasoning to liquid columns.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027