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

1.8 Pressure

Learning Objective

Define pressure as force per unit area and use the equation p = F ⁄ A to solve problems, including situations involving fluids.

Key Statement (Cambridge Syllabus 1.8)

“Pressure varies with force and area (p = F/A).”

“In a fluid at rest the pressure increases with depth and with the density of the fluid (Δp = ρ g Δh).”

1. Definition of Pressure

Pressure (p) is the magnitude of a force (F) acting perpendicularly (normal) to a surface, divided by the area (A) over which the force is distributed.

Mathematical form

p = \frac{F}{A}

where

  • p – pressure (SI unit: pascal, Pa)

  • F – force (newton, N)

  • A – area (square metre, m²)

2. Units and Conversions

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

3. Rearranging the Formula

Any of the three variables can be isolated:

  • Force: F = p × A

  • Area:  A = F ⁄ p

  • Pressure: p = F ⁄ A (original form)

4. Everyday Illustrations of “Force ÷ Area”

  • Knife cutting – The blade’s edge has a very small contact area; therefore even a modest hand force produces a high pressure that can shear food.

  • Snowshoes – By spreading a person’s weight over a large area, the pressure on the snow is reduced, preventing sinking.

  • Tyre pressure – Inflating a tyre raises the internal pressure, allowing the tyre to support the vehicle’s weight over a relatively small contact patch with the road.

  • Blood pressure – The heart exerts a force on the blood; the resulting pressure (≈120 mm Hg) is the force per unit area exerted on the arterial walls.

5. Pressure in Fluids (Hydrostatic Pressure)

In a fluid at rest the pressure increases with depth because the weight of the fluid above adds to the pressure. This is expressed by the syllabus wording:

“In a fluid at rest the pressure increases with depth and with the density of the fluid (Δp = ρ g Δh).”

\Delta p = \rho \, g \, \Delta h

  • Δp – increase in pressure (Pa)

  • ρ – density of the fluid (kg m⁻³)

  • g – acceleration due to gravity (≈9.81 m s⁻²)

  • Δh – vertical depth measured from the free surface (m)

Key points

  • The pressure at a given depth is the same in all directions (isotropic).

  • Atmospheric pressure acts on the free surface; the absolute pressure at depth is p = p_atm + ρ g h.

  • Denser fluids (e.g., water vs. oil) produce a larger pressure increase for the same depth.

6. Key Points to Remember

  • Pressure rises when the same force acts on a smaller area.

  • The force must be perpendicular (normal) to the surface for the simple formula p = F/A to apply directly.

  • In fluids, pressure varies with depth and density according to Δp = ρ g Δh.

  • Always use SI units (N, m², Pa) in calculations; convert kPa, mm Hg, etc., only after the algebraic step.

  • When converting units, keep track of the conversion factor (e.g., 1 kPa = 1 000 Pa, 1 mm Hg ≈ 133.3 Pa).

7. Worked Example – Solid Contact

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

Solution:

  1. Identify the known quantities: F = 250 N, A = 0.05 m².

  2. Apply the pressure formula: p = F / A.

  3. Substitute the numbers: p = 250 N ÷ 0.05 m² = 5 000 Pa.

  4. Convert to kilopascals if required: 5 000 Pa = 5 kPa.

Therefore, the pressure is 5 000 Pa (5 kPa).

8. Worked Example – Hydrostatic Pressure

Question: What is the gauge pressure 3.0 m below the surface of a freshwater lake? (Take ρwater = 1 000 kg m⁻³, g = 9.81 m s⁻².)

Solution:

  1. Use the hydrostatic equation: Δp = ρ g h.

  2. Insert the values: Δp = (1 000 kg m⁻³)(9.81 m s⁻²)(3.0 m).

  3. Calculate: Δp = 29 430 Pa ≈ 29.4 kPa.

The gauge pressure at 3 m depth is ≈ 29 kPa (above atmospheric pressure).

9. 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.

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

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

  • “Pressure in a fluid is the same at all depths.” – Pressure increases linearly with depth according to Δp = ρ g Δh.

10. Practice Questions

  1. A nail with a tip area of 2.0 × 10⁻⁶ m² 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 m² on which a force of 200 N is applied. The large piston has an area of 0.5 m². What force is exerted by the large piston? (Assume the fluid transmits pressure equally.)

  3. Convert a pressure of 120 kPa to mm Hg. (Use 1 mm Hg ≈ 133.3 Pa.)

  4. What is the gauge pressure 5 m below the surface of oil with density ρ = 800 kg m⁻³? (Take g = 9.81 m s⁻².)

11. Suggested Diagram

Block pressing on a horizontal surface showing force F normal to the surface, contact area A, and resulting pressure p = F/A

Block pressing on a horizontal surface: the force F acts normal to the surface, the contact area is A, and the resulting pressure is p = F/A.

12. Summary

Pressure quantifies how concentrated a force is over an area. The fundamental relationship p = F/A applies to solid contacts, while fluids obey the additional rule Δp = ρ g Δh. Mastery of unit conversion, careful identification of the contact area, and awareness of how depth and density affect fluid pressure are essential for solving all IGCSE 0625 pressure problems.