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=FAP = \frac{F}{A}
    • where PP is pressure, FF is the normal force, and AA is the contact area.
  • SI unit: the pascal (Pa).
    • 1 Pa = 1 N/m21\ \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∝FP \propto F
    • 2) The area over which the force is applied (smaller area → larger pressure):
    • Pressure varies inversely with area: P∝1AP \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=mg=200 kg×gW = m g = 200 \text{ kg} \times g
    • Use standard near Earth gravity: g≈9.8 m/s2g \approx 9.8\ \text{m/s}^2
    • W=200×9.8=1960 NW = 200 \times 9.8 = 1960\ \text{N}
    • Pressure: P=WA=1960 N9 m2≈218 PaP = \frac{W}{A} = \frac{1960\ \text{N}}{9\ \text{m}^2} \approx 218\ \text{Pa}

  • Example 2: Two bicycle tires have a pressure of 1.5×104 N/m21.5 \times 10^4\ \text{N/m}^2 each. Each tire has contact area A=0.5 m2.A = 0.5\ \text{m}^2. What is the weight of the bicycle?

    • Force per tire: Fi=p×A=(1.5×104)×0.5=7500 NF_i = p \times A = (1.5 \times 10^4) \times 0.5 = 7500\ \text{N}
    • Total weight: W=2×7500=15000 NW = 2 \times 7500 = 15000\ \text{N}
    • Mass of bicycle: m=Wg=150009.8≈1530 kgm = \frac{W}{g} = \frac{15000}{9.8} \approx 1530\ \text{kg}

  • Example 3: A box weighs 100 N and exerts a pressure of 50 N/m250\ \text{N/m}^2. What is its area?

    • Area: A=FP=100 N50 N/m2=2 m2A = \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 m2A = 0.25\ \text{m}^2 and exerts pressure P=800 N/m2P = 800\ \text{N/m}^2. What is the weight?

    • Weight: W=P×A=800×0.25=200 NW = 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:
    • g/cm3\text{g/cm}^3
    • kg/m3\text{kg/m}^3
  • Formula:
    • ρ=mV\rho = \frac{m}{V}
    • where ρ\rho is density, mm is mass, and VV is volume.
  • For a rectangular solid, volume is:
    • V=l×w×hV = 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 kgm = 1\ \text{kg}, but occupy very different volumes due to density differences.

Buoyancy and density in liquids

  • Buoyancy principle (simplified):
    • If density of object ρ<em>object<ρ</em>liquid\rho<em>{object} < \rho</em>{liquid}, the object floats on the liquid.
    • If ρ<em>object>ρ</em>liquid\rho<em>{object} > \rho</em>{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:
    • ρ<em>object<ρ</em>liquid⇒floats\rho<em>{object} < \rho</em>{liquid} \Rightarrow \text{floats}
    • ρ<em>object>ρ</em>liquid⇒sinks\rho<em>{object} > \rho</em>{liquid} \Rightarrow \text{sinks}
    • ρ<em>object=ρ</em>liquid⇒fully submerged (neutrally buoyant)\rho<em>{object} = \rho</em>{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 m×2 m×0.5 m=1 m3V = 1\ \text{m} \times 2\ \text{m} \times 0.5\ \text{m} = 1\ \text{m}^3
    • Density: ρ=mV=10 kg1 m3=10 kg/m3\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: ρbasalt≈3200 kg/m3\rho_{basalt} \approx 3200\ \text{kg/m}^3
    • Required mass to resist wave shifting: m=2000 kgm = 2000\ \text{kg}
    • Required volume: V=mρ=20003200=0.625 m3V = \frac{m}{\rho} = \frac{2000}{3200} = 0.625\ \text{m}^3
    • If the rock were a cube, edge length would be:s=V3=0.6253≈0.85 ms = \sqrt[3]{V} = \sqrt[3]{0.625} \approx 0.85\ \text{m}

  • Try this! Density calculations
    1) Aluminum block: volume V=15 cm3V = 15\ \text{cm}^3, mass m=40.5 gm = 40.5\ \text{g}

    • Density: ρ=mV=40.515=2.7 g/cm3\rho = \frac{m}{V} = \frac{40.5}{15} = 2.7\ \text{g/cm}^3
      2) Mechanical pencil: density ρ=3000 g/cm3\rho = 3000\ \text{g/cm}^3, volume V=15.8 cm3V = 15.8\ \text{cm}^3
    • Mass: m=ρV=3000×15.8=47400 g=47.4 kgm = \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 NW = 20{,}000\ \text{N}
    • Each foot area: A<em>foot=0.05 m2A<em>{foot} = 0.05\ \text{m}^2; 4 feet → total contact area A=4A</em>foot=0.20 m2A = 4A</em>{foot} = 0.20\ \text{m}^2
    • Pressure exerted on ground: P=WA=200000.20=100,000 PaP = \frac{W}{A} = \frac{20000}{0.20} = 100{,}000\ \text{Pa}
      4) Office safe on floor:
    • Weight: W=500 NW = 500\ \text{N}
    • Base area: A=1.25 m2A = 1.25\ \text{m}^2
    • Pressure: P=WA=5001.25=400 PaP = \frac{W}{A} = \frac{500}{1.25} = 400\ \text{Pa}

Quick reference: key formulas

  • Pressure: P=FAP = \frac{F}{A}
  • Force (weight) relation: F=mgF = m g
  • Density: ρ=mV\rho = \frac{m}{V}
  • Rectangular volume: V=lwhV = l w h
  • For a given pressure newton per square meter: 1 Pa = 1 N/m21\ \text{N/m}^2
  • Mass-volume relationship used in density: m=ρVm = \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∝FP \propto F
  • Increasing area while keeping force fixed decreases pressure: P∝1AP \propto \frac{1}{A}
  • Density ties mass and volume: ρ=mV\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: g/cm3,kg/m3\text{g/cm}^3, \text{kg/m}^3
  • When converting between cm^3 and m^3, remember: 1 m3=106 cm31\ \text{m}^3 = 10^6\ \text{cm}^3
  • When converting between Newtons, kilograms, and meters per second squared, use F=mgF = m g with g≈9.8 m/s2g \approx 9.8\ \text{m/s}^2 (Earth).