define and use pressure

Cambridge A-Level Physics 9702 – Equilibrium of Forces: Define and Use Pressure

Equilibrium of Forces – Defining and Using Pressure

1. What is Pressure?

Pressure is the magnitude of the normal force exerted per unit area on a surface. It quantifies how a force is distributed over an area.

Mathematically, pressure $p$ is defined as

$$p = \frac{F_{\perp}}{A}$$

where $F_{\perp}$ is the component of the force perpendicular to the surface and $A$ is the area over which the force acts.

2. Units of Pressure

The SI unit of pressure is the pascal (Pa), where

$$1\ \text{Pa} = 1\ \frac{\text{N}}{\text{m}^2}$$

Other commonly used units are listed in the table below.

Unit Symbol Conversion to Pa
Pascal Pa 1 Pa
Kilopascal kPa 1 kPa = $10^3$ Pa
Bar bar 1 bar = $10^5$ Pa
Atmosphere atm 1 atm ≈ $1.013\times10^5$ Pa
Torr / mmHg torr 1 torr ≈ 133.3 Pa

3. Pressure in Equilibrium Situations

When an object is in static equilibrium, the sum of forces and the sum of torques are zero. Pressure becomes important when forces are distributed over contact surfaces, such as:

  • Fluid pressure on submerged surfaces.
  • Contact pressure between a block and a rough surface.
  • Air pressure acting on a balloon.

4. Using Pressure to Solve Problems

Typical steps:

  1. Identify the normal force acting on the surface.
  2. Determine the area over which the force is applied.
  3. Calculate pressure using $p = F/A$.
  4. Apply equilibrium conditions ($\sum F = 0$, $\sum \tau = 0$) to relate pressures to other forces.

5. Example Problem

Problem: A rectangular block of mass $5.0\ \text{kg}$ rests on a horizontal table. The contact area between the block and the table is $0.025\ \text{m}^2$. Calculate the pressure exerted by the block on the table.

Solution:

  1. Calculate the weight (normal force): $F = mg = 5.0\ \text{kg} \times 9.81\ \text{m s}^{-2} = 49.05\ \text{N}$.
  2. Use the pressure formula: $$p = \frac{F}{A} = \frac{49.05\ \text{N}}{0.025\ \text{m}^2} = 1962\ \text{Pa}$$
  3. Express the result in kilopascals: $p = 1.96\ \text{kPa}$.

6. Common Misconceptions

  • Pressure vs. Force: Pressure is not a force; it is force per unit area. Two forces of equal magnitude can produce different pressures if their contact areas differ.
  • Direction of Pressure: Pressure acts perpendicular (normal) to the surface. Any component of force parallel to the surface does not contribute to pressure.
  • Uniform Pressure Assumption: In many A‑Level problems, pressure is assumed uniform over the contact area unless otherwise specified (e.g., fluid depth variations).

7. Summary Checklist

  • Remember the definition $p = F_{\perp}/A$.
  • Use SI units: N for force, m² for area, Pa for pressure.
  • Apply equilibrium conditions together with pressure calculations.
  • Check whether pressure is uniform or varies across the surface.
Suggested diagram: A block on a table showing the weight acting downwards, the normal reaction upwards, and the contact area labelled $A$.