Physics 0625 Study Notes: Pressure (1.8)

Hello future physicist! Welcome to the chapter on Pressure. You've already learned about forces. Pressure is simply the idea that a force doesn't just push; it pushes over an area. How concentrated that push is matters a lot in the real world—from why nails are sharp to how submarines manage to stay intact deep underwater!

Pressure is a key concept that links forces directly to motion and structure, making it vital in understanding mechanics. Don't worry if the formulas look scary; we'll break them down step-by-step.

1. Defining Pressure on Solid Surfaces

Core Concept: Force Distributed Over Area

Imagine trying to hammer a screw into a piece of wood. It won't work easily because the force is spread out over the blunt head of the screw. Now, try a nail—the force is concentrated on the tiny, sharp point. That difference is pressure.

Definition of Pressure (\(p\))

Pressure is defined as the force acting perpendicularly (at a right angle) to a surface, divided by the area over which that force is spread.

Key Definition:
$$ \text{Pressure} = \frac{\text{Force}}{\text{Area}} $$

We recall and use the equation:

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

  • \(p\) is Pressure.
  • \(F\) is the Force applied (measured in Newtons, N).
  • \(A\) is the Area over which the force acts (measured in metres squared, m²).
Units of Pressure

Since force is in Newtons (N) and area is in metres squared (m²), the SI unit for pressure is the Newton per square metre (N/m²).

This unit has a special name: the Pascal (Pa).

1 Pa = 1 N/m²


Did you know? Because the Pascal is a very small unit, we often use kiloPascals (kPa) or megaPascals (MPa) in physics problems!

How Pressure Varies (Everyday Examples)

The core principle is simple:

  1. If you increase the force (F), the pressure (p) increases. (Directly proportional)
  2. If you increase the area (A), the pressure (p) decreases. (Inversely proportional)
Case 1: Increasing Pressure (Small Area)

To exert high pressure, you need a large force acting on a very small area.

  • A sharp knife or pin: The force applied to cut or pierce is concentrated onto the tiny area of the tip or blade edge, creating huge pressure to overcome material resistance.
  • High-heeled shoes: When a person stands, all their weight (force) is concentrated onto the tiny area of the heel, which can cause damage to soft floors.
Case 2: Decreasing Pressure (Large Area)

To exert low pressure, you need to spread the force over a very large area.

  • Tractors and Tanks: These heavy vehicles use wide, large tracks (treads) rather than thin wheels. This large area lowers the pressure exerted on the ground, preventing them from sinking into soft mud or soil.
  • Snowshoes: These are designed to be wide and long. When you walk on snow, the snowshoes spread your weight over a large area, reducing the pressure so you don't sink.
Quick Review: The Squeeze Factor

Think of pressure as the "squeeze." To get a big squeeze:

  • Use a BIG Force.
  • Use a SMALL Area.

Common Mistake Alert! Remember that Force (N) is applied by the object, while Pressure (Pa) is the result of that force being distributed. They are not the same!

Key Takeaway for Section 1

Pressure is concentrated force. High pressure results from focusing force onto a small area (like a nail), while low pressure results from spreading force over a large area (like snowshoes).

2. Pressure in Liquids (Fluid Pressure)

Liquids (and gases, which are both fluids) exert pressure, but unlike a solid pushing down on a single point, fluid pressure acts in all directions at any given depth. This is why a hole punched in the side of a bucket causes water to spray out horizontally.

Core Concept: Factors Affecting Liquid Pressure

The pressure at a point beneath the surface of a liquid is caused by the weight of the fluid column directly above that point.

Qualitatively, the pressure beneath the surface of a liquid changes based on two main factors:

1. Depth (h)

The deeper you go, the greater the pressure.

Explanation: As you dive deeper (increase depth, \(h\)), there is a taller column of water resting on top of you. Since the weight of the water is proportional to the amount of water, the force increases, causing the pressure to rise.

  • Real-world example: Divers feel their ears pop when they descend because of the rapid increase in pressure.
2. Density (\(\rho\))

The denser the liquid, the greater the pressure at any given depth.

Explanation: If you replaced water with mercury (which is far denser), the same height (depth) of the liquid would weigh much more. Since the fluid's weight increases, the pressure increases.

  • Analogy: Comparing a swim in a swimming pool (low density, clean water) versus a swim in the extremely salty Dead Sea (high density, brine). The pressure difference is noticeable, even if slight.

Supplement Content: Calculating Change in Pressure in a Liquid

For extended candidates, we can calculate the change in pressure, \(\Delta p\), as you move from one depth to another within a uniform liquid. If you measure pressure change from the surface, \(\Delta h\) is simply the depth \(h\).

The relationship is given by:

$$ \Delta p = \rho g \Delta h $$

Let's break down these terms (you must recall and use this equation!):

  • \(\Delta p\) (Change in Pressure) measured in Pascals (Pa).
  • \(\rho\) (Rho) is the density of the liquid (measured in kg/m³).
  • \(g\) is the gravitational field strength (or acceleration of free fall) (measured in N/kg or m/s²). Use $9.8$ N/kg or $10$ N/kg, depending on the question instructions.
  • \(\Delta h\) (Delta h) is the change in depth (measured in metres, m).
Step-by-Step Calculation Tip

The most common mistake here is forgetting to use consistent units, especially making sure the density is in kg/m³ and the depth is in metres.

  1. Check Units: Ensure \(\rho\) is kg/m³ and \(\Delta h\) is m.
  2. Use g: Use the gravitational constant provided in the exam data (usually 9.8 N/kg or 10 N/kg).
  3. Multiply: Multiply the three terms: \(\rho \times g \times \Delta h\). The result is the pressure due to the liquid column.


Encouragement: Don't worry if this seems tricky at first! This formula simply mathematically represents what we already know: deep water (large \(h\)) is heavy (large \(\rho\)) because of gravity (\(g\)).

Liquid Pressure Applications

  • Dams: The bottom of a dam wall must be built much thicker than the top. This is because the pressure exerted by the water increases with depth (\(\Delta h\)), meaning the greatest force pushes against the dam at the base.
  • Water Pipes: If a house receives water from a reservoir high up on a hill, the height difference (\(\Delta h\)) creates high water pressure in the pipes, allowing water to reach upper floors without additional pumps.
Key Takeaway for Section 2

Liquid pressure depends on the weight of the liquid above the point being measured. Therefore, pressure increases with depth (\(\Delta h\)) and density (\(\rho\)) of the liquid. The increase in pressure is calculated by \(\Delta p = \rho g \Delta h\).