Pressure, density, upthrust
Why This Matters
# Pressure, Density, and Upthrust - Summary This lesson covers three fundamental fluid mechanics concepts essential for Cambridge IGCSE Physics. Students learn to calculate pressure using P = F/A, density using ρ = m/V, and understand Archimedes' principle of upthrust, which explains why objects float or sink based on displaced fluid weight. These topics are frequently examined through calculation questions and practical applications including hydraulic systems, floating/sinking scenarios, and pressure variations in liquids and gases.
Key Words to Know
Core Concepts & Theory
Pressure is defined as force per unit area acting perpendicular to a surface. The formula is:
P = F/A
where P = pressure (Pa or N/m²), F = force (N), A = area (m²)
Cambridge Definition: Pressure is the force acting normally (at right angles) per unit area.
Density is mass per unit volume of a substance:
ρ = m/V
where ρ (rho) = density (kg/m³ or g/cm³), m = mass (kg or g), V = volume (m³ or cm³)
Upthrust (or buoyancy) is the upward force exerted by a fluid on an object immersed in it. This force results from pressure differences in the fluid.
Key Principle: Pressure in fluids increases with depth because of the weight of fluid above. The formula is:
P = hρg
where h = depth (m), ρ = fluid density (kg/m³), g = gravitational field strength (N/kg)
Archimedes' Principle states: The upthrust on an object is equal to the weight of fluid displaced.
Mnemonic for Pressure: "Force Across Pushes" (F/A = P)
Mnemonic for Density: "Mary Visited Denmark" (m/V = ρ)
Visual Aid: Imagine pressure as thumbtacks—the sharper (smaller area), the deeper they penetrate for the same force. Density is how tightly packed particles are in a given space. Upthrust is like invisible hands pushing upward when you're underwater.
Detailed Explanation with Real-World Examples
Pressure in Action: When wearing high heels versus trainers, the same body weight creates different pressures. High heels (small area) produce enormous pressure—up to 3 million Pa—damaging wooden floors. Trainers distribute force over larger areas, reducing pressure. This explains why camels have wide feet for desert sand, and snowshoes prevent sinking.
Hydraulic systems use pressure transmission through liquids. Car brakes apply small force over small area (pedal), creating pressure transmitted equally throughout brake fluid, producing large force over large area (brake pads). Formula: P₁ = P₂, so F₁/A₁ = F₂/A₂.
Atmospheric Pressure (≈100,000 Pa at sea level) crushes objects when air is removed. The classic demonstration: evacuating a can makes it implode spectacularly as external pressure overwhelms internal pressure.
Density Applications: Ships float despite being made of steel (density 7,850 kg/m³) because their overall density including air spaces is less than water (1,000 kg/m³). Submarines control density by flooding/emptying ballast tanks. Hot air balloons work because heated air is less dense than surrounding cool air.
Upthrust Explained: Swimming feels easier than walking because water provides upthrust equal to your displaced water's weight. Icebergs float with 90% submerged because ice (920 kg/m³) is slightly less dense than seawater (1,025 kg/m³). The upthrust exactly equals the iceberg's weight when floating.
Analogy: Think of density as "particle crowding"—lead atoms are tightly packed (very crowded = high density), while polystyrene has huge air gaps (sparse = low density).
Worked Examples & Step-by-Step Solutions
Example 1: A block weighing 50 N rests on a table. The contact area is 0.02 m². Calculate the pressure.
Solution:
- Given: F = 50 N, A = 0.02 m²
- Required: P = ?
- Formula: P = F/A
- P = 50/0.02 = 2,500 Pa or 2.5 kPa
Examiner Note: Always include units. State the formula before substituting values.
Example 2: A diver descends to 30 m depth in seawater (density 1,030 kg/m³). Calculate the pressure due to seawater. (g = 10 N/kg)
Solution:
- Given: h = 30 m, ρ = 1,030 kg/m³, g = 10 N/kg
- Formula: P = hρg
- P = 30 × 1,030 × 10 = 309,000 Pa or 309 kPa
- Total pressure = atmospheric + seawater = 100,000 + 309,000 = 409 kPa
Examiner Note: Question may ask for total pressure—don't forget atmospheric pressure!
Example 3: A 2 kg object (volume 0.0005 m³) is fully submerged in water (1,000 kg/m³). Calculate upthrust and determine if it floats.
Solution:
- Volume displaced = 0.0005 m³
- Mass of water displaced = ρV = 1,000 × 0.0005 = 0.5 kg
- Upthrust = weight of water displaced = mg = 0.5 × 10 = 5 N
- Object weight = 2 × 10 = 20 N
- Since weight (20 N) > upthrust (5 N), object sinks
Examiner Note: Compare forces to predict floating/sinking.
Common Exam Mistakes & How to Avoid Them
Mistake 1: Area Confusion
Error: Using total surface area instead of contact area for pressure calculations.
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Cambridge Exam Technique & Mark Scheme Tips
Command Word Mastery:
- Calculate: Show formula, substitution, answer with units (3 marks typical)
- Explain...
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Exam Tips
- 1.Always include units in your answers (e.g., Pressure in Pascals (Pa) or N/m², Density in kg/m³ or g/cm³, Force in Newtons (N)).
- 2.When solving problems, write down the formula first, then substitute the numbers, and finally calculate the answer.
- 3.Remember that pressure in fluids increases with depth and density; this is a common question.
- 4.For upthrust questions, think about how much fluid is being pushed out of the way by the object.
- 5.Practice drawing simple diagrams to show forces and areas for pressure problems; it helps visualize the problem.