Define pressure as force per unit area; recall and use the equation p = F / A

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – Topic 1.8 Pressure

1.8 Pressure

Learning Objective

Define pressure as force per unit area and use the equation \$p = \frac{F}{A}\$ to solve problems.

Definition of Pressure

Pressure (\$p\$) is the amount of force (\$F\$) applied perpendicular to a surface divided by the area (\$A\$) over which the force acts.

Mathematically,

\$p = \frac{F}{A}\$

where:

  • \$p\$ = pressure (in pascals, Pa)

  • \$F\$ = force (in newtons, N)

  • \$A\$ = area (in square metres, m²)

Units of Pressure

QuantitySI UnitSymbolEquivalent
ForcenewtonN1 N = 1 kg·m s⁻²
Areasquare metrem²–
PressurepascalPa1 Pa = 1 N m⁻²
Pressure (alternative)kilopascalkPa1 kPa = 1 000 Pa
Pressure (alternative)millimetre of mercurymm Hg1 mm Hg ≈ 133.3 Pa

Rearranging the Formula

The basic equation can be rearranged to find any of the three variables:

  • Force: \$F = p \times A\$

  • Area: \$A = \frac{F}{p}\$

  • Pressure: \$p = \frac{F}{A}\$ (original form)

Key Points to Remember

  • Pressure increases when the same force is applied to a smaller area.

  • The direction of the force must be perpendicular (normal) to the surface for the formula to apply directly.

  • SI unit of pressure is the pascal (Pa); in everyday contexts, kilopascals (kPa) or millimetres of mercury (mm Hg) are often used.

  • When converting between units, keep track of the conversion factor (e.g., 1 kPa = 1 000 Pa).

Example Problem

Question: A rectangular block exerts a force of 250 N on a surface. The contact area is 0.05 m². Calculate the pressure exerted by the block.

Solution:

  1. Write down the known values: \$F = 250\ \text{N}\$, \$A = 0.05\ \text{m}^2\$.

  2. Use the pressure formula: \$p = \frac{F}{A}\$

  3. Substitute the numbers: \$p = \frac{250\ \text{N}}{0.05\ \text{m}^2} = 5\,000\ \text{Pa}\$

  4. If required, convert to kilopascals: \$5\,000\ \text{Pa} = 5\ \text{kPa}\$

Therefore, the pressure exerted is \$5\,000\ \text{Pa}\$ (or \$5\ \text{kPa}\$).

Common Misconceptions

  • “Pressure is the same as force.” – Pressure is force divided by area; a large force over a large area can give a low pressure.

  • “The shape of the object matters for pressure.” – Only the area of contact matters, not the overall shape.

  • “Units can be mixed arbitrarily.” – Always convert forces to newtons and areas to square metres before using the formula.

Practice Questions

  1. A nail with a tip area of \$2.0 \times 10^{-6}\ \text{m}^2\$ is driven into wood with a force of 30 N. Calculate the pressure at the tip of the nail.

  2. A hydraulic press has a small piston of area \$0.01\ \text{m}^2\$ on which a force of 200 N is applied. If the large piston has an area of \$0.5\ \text{m}^2\$, what is the force exerted by the large piston? (Assume the fluid transmits pressure equally.)

  3. Convert a pressure of \$120\ \text{kPa}\$ to mm Hg (use \$1\ \text{mm Hg} \approx 133.3\ \text{Pa}\$).

Suggested Diagram

Suggested diagram: A block pressing on a surface showing force \$F\$, contact area \$A\$, and resulting pressure \$p\$.

Summary

Pressure quantifies how concentrated a force is over an area. The fundamental relationship \$p = F/A\$ allows us to calculate any one of the three variables when the other two are known. Mastery of unit conversion and careful identification of the contact area are essential for solving IGCSE physics problems involving pressure.