Density and pressure
Section: Forces, Density and Pressure | Syllabus: Cambridge AS Level Physics 9702
Pressure In everyday life, 'force' and 'pressure' are often used interchangeably, but in physics they are distinct. Pressure is a measure of how concentrated a force is. p = fraction Where: p = Pressure (Pascal, Pa or N m^-2) F = Force perpendicular to the area (N) A = Area (m^2) Tip: Area Conversions Common mistake: forgetting to square the conversion factor.
1 cm = 10^-2 m 1 cm^2 = (10^-2)^2 = 10^-4 m^2 1 mm = 10^-3 m 1 mm^2 = (10^-3)^2 = 10^-6 m^2 Worked Example: Drawing Pin Scenario: A student pushes a drawing pin into a wall with a force of 1 N. Explain why it goes into the wall but not their thumb.
Area of flat head = 1.2 cm^2 = 1.2 × 10^-4 m^2 Area of point = 0.01 mm^2 = 1 × 10^-8 m^2 Calculation Pressure at Thumb: p = 1 / (1.2 × 10^-4) 8.3 kPa Pressure at Wall: p = 1 / (1 × 10^-8) = 100 MPa Conclusion: The pressure at the point is 12,000 times greater!
This high concentration of force allows it to penetrate the wall. Comparison: Car vs Tractor Car: Mass 1500 kg. 4 Tyres. Contact patch per tyre: 195 mm × 85 mm. Tractor: Mass 3200 kg. Total contact area 0.80 m^2.
Which sinks in mud? Car Area: 4 × (0.195 × 0.085) = 0.0663 m^2. Car Pressure: p = (1500 × 9.81) / 0.0663 222 kPa. Tractor Pressure: p = (3200 × 9.81) / 0.80 39 kPa. Even though the tractor is heavier, its huge tyres spread the weight, exerting much less pressure.
The car is more likely to sink. Pressure in a Fluid (Hydrostatic Pressure) Pressure increases with depth because of the weight of the fluid column above you. For every 10m you dive, pressure increases by about 1 atmosphere.
Figure 4.18: Deriving Hydrostatic Pressure (Show a column/cube of water of height h and base area A. Highlight that the Volume V = hA and Mass m = ρ V. Weight acting down is W = mg = ρ A h g.) p = ρ g h Derivation: Force (Weight) = m g = (ρ V) g = ρ (A h) g Pressure = Force / Area = (ρ A h g) / A p = ρ g h This strictly applies to incompressible fluids (liquids).
It is not accurate for large heights of gases (like the atmosphere) because density ρ changes with height. Upthrust and Archimedes' Principle Objects feel lighter in water because of upthrust . This force arises because pressure at the bottom of an object is higher than at the top.
Figure 4.19: Origin of Upthrust (Show a submerged rectangular block. Top face at depth h_1, Bottom face at depth h_2. Smaller arrow down F_1 on top. Larger arrow up F_2 on bottom. Resultant is Upthrust.) Deriving the Formula Total Upward Force = Force on Bottom - Force on Top Upthrust = p_2 A - p_1 A = (ρ g h_2)A - (ρ g h_1)A = ρ g A (h_2 - h_1) = ρ g V (since A × height = Volume) Archimedes' Principle The upthrust on a body is equal to the weight of the fluid displaced by the body.
Worked Example: Submarine Atlantis XII Scenario: Submarine (approx cylinder, length 20 m, width 4 m). Depth 30 m. Seawater density 1024 kg m^-3. Atmospheric pressure 101 kPa. 1. Total Pressure p_water = ρ g h = 1024 × 9.81 × 30 = 301,363 Pa 301 kPa.
p_total = 301 + 101 = 402 kPa. 2. Upthrust & Mass Volume (cylinder V = π r^2 l) = π × 2^2 × 20 251.3 m^3. Upthrust F = ρ g V = 1024 × 9.81 × 251.3 2.52 MN. If hovering (equilibrium), Weight = Upthrust.
Mass = 2.52 × 10^6 / 9.81 257 tonnes. Deep Thinking: Balloons A helium balloon rises because the upthrust (weight of air displaced) is greater than its weight. It stops rising when the air becomes too thin (low density), so the weight of displaced air decreases until it exactly balances the balloon's weight.
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