Edexcel iGCSE Physics -5.5 Pressure, Force, and Area- Study Notes- New Syllabus
Edexcel iGCSE Physics -5.5 Pressure, Force, and Area- Study Notes- New syllabus
Edexcel iGCSE Physics -5.5 Pressure, Force, and Area- Study Notes -Edexcel iGCSE Physics – per latest Syllabus.
Key Concepts:
update
Pressure, Force and Area
Pressure describes how concentrated a force is over a surface. A large force acting on a small area produces a high pressure, while the same force acting over a larger area produces a lower pressure.
Relationship Between Pressure, Force and Area
The relationship is:
\( \mathrm{pressure = \dfrac{force}{area}} \)
\( \mathrm{P = \dfrac{F}{A}} \)
- \( \mathrm{P} \) = pressure (pascal, Pa)
- \( \mathrm{F} \) = force (newton, N)
- \( \mathrm{A} \) = area (square metre, m²)
1 pascal = 1 newton per square metre (N/m²).
Understanding the Relationship
- For a constant force, decreasing the area increases pressure.
- For a constant area, increasing the force increases pressure.
- Pressure depends on both force and contact area.
Everyday examples:
- Sharp knives cut better because they have a small contact area.
- High-heeled shoes exert more pressure than flat shoes.
- Snowshoes reduce pressure to prevent sinking into snow.
Rearranging the Formula

- Force: \( \mathrm{F = P \times A} \)
- Area: \( \mathrm{A = \dfrac{F}{P}} \)
Key Idea
- Pressure is force per unit area.

- Small area → high pressure.
- Large area → low pressure.
Important Points to Remember
- Area must be in m².
- Force must be in newtons.
- Pressure is measured in pascals.
Example
A force of \( \mathrm{600\ N} \) acts on an area of \( \mathrm{0.30\ m^2} \).
Calculate the pressure exerted.
▶️ Answer / Explanation
Use:
\( \mathrm{P = \dfrac{F}{A}} \)
\( \mathrm{P = \dfrac{600}{0.30}} \)
\( \mathrm{P = 2000\ Pa} \)
Example
A pressure of \( \mathrm{5.0 \times 10^4\ Pa} \) acts on an area of \( \mathrm{0.20\ m^2} \).
Calculate the force producing this pressure.
▶️ Answer / Explanation
Rearrange:
\( \mathrm{F = P \times A} \)
\( \mathrm{F = 5.0 \times 10^4 \times 0.20} \)
\( \mathrm{F = 1.0 \times 10^4\ N} \)
