Pressure and Density - Comprehensive Notes

Pressure: Fundamentals

Pressure is the push or normal force exerted per unit area on a surface.
In physics, the formal definition is:
- P = rac{F}{A}
- where P is pressure, F is the normal force, and A is the contact area.
SI unit: the pascal (Pa).
- 1 Pa = 1\ \text{N}/\text{m}^2
- Pa is named after Blaise Pascal.
Layman view: pressure is the effect the force has on an object (the push on a surface).

Factors that determine pressure

Pressure depends on two factors:
- 1) Magnitude of the force applied (larger force → larger pressure):
- Pressure varies directly with force: P \propto F
- 2) The area over which the force is applied (smaller area → larger pressure):
- Pressure varies inversely with area: P \propto \frac{1}{A}
In short:
- Greater force with the same area increases pressure.
- The same force spread over a smaller area increases pressure; spreading the same force over a larger area decreases pressure.

Small area vs large area

SMALL AREA → greater pressure (for the same force).
LARGE AREA → smaller pressure (for the same force).

Example problems (Pressure calculations)

Example 1: A cube has edges of 3 m; mass = 200 kg.
- Contact area with floor: base area = $3\text{ m} \times 3\text{ m} = 9\text{ m}^2$.
- Weight: W = m g = 200 \text{ kg} \times g
- Use standard near Earth gravity: g \approx 9.8\ \text{m/s}^2
- W = 200 \times 9.8 = 1960\ \text{N}
- Pressure: P = \frac{W}{A} = \frac{1960\ \text{N}}{9\ \text{m}^2} \approx 218\ \text{Pa}
Example 2: Two bicycle tires have a pressure of 1.5 \times 10^4\ \text{N/m}^2 each. Each tire has contact area A = 0.5\ \text{m}^2. What is the weight of the bicycle?
- Force per tire: F_i = p \times A = (1.5 \times 10^4) \times 0.5 = 7500\ \text{N}
- Total weight: W = 2 \times 7500 = 15000\ \text{N}
- Mass of bicycle: m = \frac{W}{g} = \frac{15000}{9.8} \approx 1530\ \text{kg}
Example 3: A box weighs 100 N and exerts a pressure of 50\ \text{N/m}^2. What is its area?
- Area: A = \frac{F}{P} = \frac{100\ \text{N}}{50\ \text{N/m}^2} = 2\ \text{m}^2
Try this out: A wooden crate has contact area A = 0.25\ \text{m}^2 and exerts pressure P = 800\ \text{N/m}^2. What is the weight?
- Weight: W = P \times A = 800 \times 0.25 = 200\ \text{N}

Density: definition, units, and formula

Density is defined as mass per unit volume.
Common units:
- \text{g/cm}^3
- \text{kg/m}^3
Formula:
- \rho = \frac{m}{V}
- where \rho is density, m is mass, and V is volume.
For a rectangular solid, volume is:
- V = l \times w \times h
Visual examples (different objects with same mass but different density):
- 1 kg of iron nails vs 1 kg of cotton both have mass m = 1\ \text{kg}, but occupy very different volumes due to density differences.

Buoyancy and density in liquids

Buoyancy principle (simplified):
- If density of object \rho{object} < \rho{liquid}, the object floats on the liquid.
- If \rho{object} > \rho{liquid}, the object sinks.
- If densities are equal, the object is neutrally buoyant and can become fully submerged.
Note: In diagrams, these are shown as:
- \rho{object} < \rho{liquid} \Rightarrow \text{floats}
- \rho{object} > \rho{liquid} \Rightarrow \text{sinks}
- \rho{object} = \rho{liquid} \Rightarrow \text{fully submerged (neutrally buoyant)}

Density practice and concepts

Example 1: Wooden box
- Mass = 10 kg; dimensions = 1 m × 2 m × 0.5 m
- Volume: V = 1\ \text{m} \times 2\ \text{m} \times 0.5\ \text{m} = 1\ \text{m}^3
- Density: \rho = \frac{m}{V} = \frac{10\ \text{kg}}{1\ \text{m}^3} = 10\ \text{kg/m}^3
- Will it float or sink in water? Since the wooden box density (10 kg/m^3) is far less than the density of common liquids like water (roughly 1000 kg/m^3), the box will float because its density is smaller than the liquid’s density.
Example 2: Erosion-control rock size
- Basalt density: \rho_{basalt} \approx 3200\ \text{kg/m}^3
- Required mass to resist wave shifting: m = 2000\ \text{kg}
- Required volume: V = \frac{m}{\rho} = \frac{2000}{3200} = 0.625\ \text{m}^3
- If the rock were a cube, edge length would be:s = \sqrt[3]{V} = \sqrt[3]{0.625} \approx 0.85\ \text{m}
Try this! Density calculations
1) Aluminum block: volume V = 15\ \text{cm}^3, mass m = 40.5\ \text{g}
- Density: \rho = \frac{m}{V} = \frac{40.5}{15} = 2.7\ \text{g/cm}^3
  2) Mechanical pencil: density \rho = 3000\ \text{g/cm}^3, volume V = 15.8\ \text{cm}^3
- Mass: m = \rho V = 3000 \times 15.8 = 47400\ \text{g} = 47.4\ \text{kg}
Try this! More density problems
3) Elephant and feet pressure:
- Elephant weight: W = 20{,}000\ \text{N}
- Each foot area: A{foot} = 0.05\ \text{m}^2; 4 feet → total contact area A = 4A{foot} = 0.20\ \text{m}^2
- Pressure exerted on ground: P = \frac{W}{A} = \frac{20000}{0.20} = 100{,}000\ \text{Pa}
  4) Office safe on floor:
- Weight: W = 500\ \text{N}
- Base area: A = 1.25\ \text{m}^2
- Pressure: P = \frac{W}{A} = \frac{500}{1.25} = 400\ \text{Pa}

Quick reference: key formulas

Pressure: P = \frac{F}{A}
Force (weight) relation: F = m g
Density: \rho = \frac{m}{V}
Rectangular volume: V = l w h
For a given pressure newton per square meter: 1 Pa = 1\ \text{N/m}^2
Mass-volume relationship used in density: m = \rho V

Quick glossary

Pressure (Pa): normal force per unit area; depends on force and contact area.
Density (kg/m^3 or g/cm^3): mass per unit volume; determines buoyancy in a liquid.
Buoyancy: upward force from a liquid that can cause objects to float or sink depending on relative densities.
Volume (m^3 or cm^3): space that an object occupies; used to compute density.

Summary of the relationships

Increasing force while keeping area fixed increases pressure: P \propto F
Increasing area while keeping force fixed decreases pressure: P \propto \frac{1}{A}
Density ties mass and volume: \rho = \frac{m}{V}; equal masses can have very different densities depending on volume.
Buoyancy depends on density comparison between object and liquid: ρobject vs ρliquid.

Notes on units and conversions

Mass density units: \text{g/cm}^3, \text{kg/m}^3
When converting between cm^3 and m^3, remember: 1\ \text{m}^3 = 10^6\ \text{cm}^3
When converting between Newtons, kilograms, and meters per second squared, use F = m g with g \approx 9.8\ \text{m/s}^2 (Earth).