Archimedes’ principle

General physics

🎯 Learning Objectives

By the end of this lesson, you will be able to:
• Understand that upthrust acting on an object in a fluid is due to differences in hydrostatic pressure
• Calculate the upthrust acting on an object using Archimedes’ principle: F = ρgV
• Explain why objects float, sink, or remain suspended in fluids
• Apply Archimedes’ principle to solve buoyancy problems in engineering and everyday situations
• Analyze the relationship between object density, fluid density, and buoyant behavior

🗣️ Language Objectives

Students will develop their physics communication skills by:
• Using precise scientific terminology when describing buoyancy and upthrust concepts
• Explaining Archimedes’ principle using appropriate technical vocabulary
• Reading and interpreting buoyancy problems written in English with confidence
• Communicating mathematical solutions involving fluid forces clearly in written English
• Understanding and using buoyancy-related terms (upthrust, displacement, immersion) accurately

📚 Key Terms and Translations

English Term	Russian Translation	Kazakh Translation
Upthrust	Выталкивающая сила	Итеруші күш
Buoyancy	Плавучесть	Жүзу қабілеті
Archimedes’ principle	Принцип Архимеда	Архимед принципі
Displacement	Вытеснение	Ығыстыру
Immersion	Погружение	Батыру
Float	Плавать	Жүзу
Sink	Тонуть	Батып кету
Submerged	Погруженный	Батқан

🃏 Vocabulary Study Cards

Upthrust (Buoyant Force)

Definition: Upward force exerted by a fluid on an immersed object

Formula: F = ρ_fluidgV_displaced

Direction: Always vertically upward

Example: Force that makes objects feel lighter underwater

Archimedes’ Principle

Statement: Upthrust equals weight of displaced fluid

Formula: F_upthrust = m_{displaced fluid} × g

Key Point: Depends on displaced volume, not object’s total volume

Application: Ship design, hot air balloons, submarines

Floating Condition

Equilibrium: Weight = Upthrust

Density relation: ρ_object < ρ_fluid

Partial submersion: Object floats at surface

Example: Ice floating on water (ρ_ice = 917 kg/m³)

Sinking Condition

Imbalance: Weight > Upthrust

Density relation: ρ_object > ρ_fluid

Complete submersion: Object sinks to bottom

Example: Stone in water (ρ_stone ≈ 2500 kg/m³)

📖 Glossary of Terms

Upthrust (Buoyant Force)

The upward force exerted by a fluid on any object placed in it. This force arises due to the pressure difference between the top and bottom surfaces of the submerged object, with higher pressure at greater depths pushing upward more strongly than the lower pressure at shallow depths pushing downward.

Translation

Russian: Выталкивающая сила — это направленная вверх сила, оказываемая жидкостью на любой объект, помещенный в нее. Эта сила возникает из-за разности давлений между верхней и нижней поверхностями погруженного объекта, при этом более высокое давление на больших глубинах толкает вверх сильнее, чем более низкое давление на малых глубинах толкает вниз.

Kazakh: Итеруші күш — бұл сұйықтықтың оған орналастырылған кез келген затқа жоғары бағытта көрсететін күші. Бұл күш суға батырылған заттың жоғарғы және төменгі беттері арасындағы қысым айырмашылығынан туындайды, мұнда үлкен тереңдіктегі жоғары қысым кіші тереңдіктегі төмен қысымның төмен итеруіне қарағанда жоғары қарай күштірек итереді.

Archimedes’ Principle

A fundamental law of fluid statics stating that the upthrust acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body. This principle explains why objects float or sink and is essential for understanding buoyancy in engineering applications.

Translation

Russian: Принцип Архимеда — это фундаментальный закон статики жидкостей, утверждающий, что выталкивающая сила, действующая на тело, погруженное в жидкость, равна весу жидкости, вытесненной телом. Этот принцип объясняет, почему объекты плавают или тонут, и является основополагающим для понимания плавучести в инженерных приложениях.

Kazakh: Архимед принципі — бұл сұйықтық статикасының іргелі заңы, ол сұйықтыққа батырылған денеге әсер ететін итеруші күш дене ығыстырған сұйықтықтың салмағына тең екенін айтады. Бұл принцип заттардың неге жүзетінін немесе батып кететінін түсіндіреді және инженерлік қолданбаларда жүзу қабілетін түсіну үшін маңызды.

Displacement

The volume of fluid that is moved out of the way when an object is placed in it. According to Archimedes’ principle, the upthrust depends only on this displaced volume, not on the object’s shape, material, or total volume if partially submerged.

Translation

Russian: Вытеснение — это объем жидкости, который перемещается в сторону при помещении в нее объекта. Согласно принципу Архимеда, выталкивающая сила зависит только от этого вытесненного объема, а не от формы объекта, материала или общего объема, если он частично погружен.

Kazakh: Ығыстыру — бұл затты орналастырған кезде жанға жылжитылатын сұйықтықтың көлемі. Архимед принципіне сәйкес, итеруші күш тек осы ығыстырылған көлемге байланысты, заттың пішініне, материалына немесе жартылай батырылған жағдайда жалпы көлеміне байланысты емес.

Neutral Buoyancy

The condition when an object’s weight exactly equals the upthrust acting on it, causing the object to remain suspended at a constant depth in the fluid without rising or sinking. This occurs when the object’s average density equals the fluid’s density.

Translation

Russian: Нейтральная плавучесть — это состояние, когда вес объекта точно равен действующей на него выталкивающей силе, заставляя объект оставаться взвешенным на постоянной глубине в жидкости без всплытия или погружения. Это происходит, когда средняя плотность объекта равна плотности жидкости.

Kazakh: Бейтарап жүзу қабілеті — бұл заттың салмағы оған әсер ететін итеруші күшке дәл тең болатын жағдай, бұл заттың сұйықтықта тұрақты тереңдікте көтерілместен немесе батпастан ілулі тұруына себеп болады. Бұл заттың орташа тығындығы сұйықтықтың тығындығына тең болған кезде пайда болады.

🔬 Theory: Understanding Archimedes' Principle

Origin of Upthrust — Pressure Differences

The upthrust (or buoyant force) acting on an object in a fluid arises from the pressure difference between the top and bottom surfaces of the object.

In a fluid, pressure increases with depth according to the hydrostatic pressure equation: ∆p = ρg∆h

Consider a cubic object of height h submerged in a fluid:

Pressure at top surface: P_top
Pressure at bottom surface: P_bottom = P_top + ρ_fluidgh
Net upward force = (P_bottom — P_top) × Area = ρ_fluidgh × A

Since Volume = Area × height: V = A × h

Upthrust = ρ_fluidgh × A = ρ_fluidg × V
F = ρgV
This is Archimedes’ principle

Translation

Russian: Выталкивающая сила, действующая на объект в жидкости, возникает из-за разности давлений между верхней и нижней поверхностями объекта. В жидкости давление увеличивается с глубиной согласно уравнению гидростатического давления.

Kazakh: Сұйықтықтағы затқа әсер ететін итеруші күш заттың жоғарғы және төменгі беттері арасындағы қысым айырмашылығынан туындайды. Сұйықтықта қысым гидростатикалық қысым теңдеуіне сәйкес тереңдікпен өседі.

Archimedes’ Principle Statement

states that: «The acting on a body in a fluid is equal to the of the fluid by the body.»

F_upthrust = Weight of displaced fluid
F = m_displaced × g = ρ_fluid × V_displaced × g
F = ρgV

Key points about the displaced volume:

For fully submerged objects: V_displaced = V_object
For floating objects: V_displaced < V_object
The shape of the object doesn’t matter, only the displaced volume

Translation

Russian: Принцип Архимеда утверждает: «Выталкивающая сила, действующая на тело, погруженное в жидкость, равна весу жидкости, вытесненной телом.» Ключевые моменты о вытесненном объеме включают различия для полностью погруженных и плавающих объектов.

Kazakh: Архимед принципі былай дейді: «Сұйықтыққа батырылған денеге әсер ететін итеруші күш дене ығыстырған сұйықтықтың салмағына тең.» Ығыстырылған көлем туралы негізгі мәселелер толық батырылған және жүзетін заттар үшін айырмашылықтарды қамтиды.

Conditions for Floating, Sinking, and Neutral Buoyancy

The behavior of an object in a fluid depends on the comparison between its weight and the upthrust:

Floating (ρ_object < ρ_fluid)

Condition: Weight < Maximum possible upthrust

Result: Object partially submerged at surface

Equilibrium: Weight = Upthrust from displaced volume

Neutral Buoyancy (ρ_object = ρ_fluid)

Condition: Weight = Upthrust when fully submerged

Result: Object suspended at any depth

Application: Submarines, underwater vehicles

Sinking (ρ_object > ρ_fluid)

Condition: Weight > Maximum possible upthrust

Result: Object sinks to bottom

Net force: Downward force = Weight — Upthrust

Translation

Russian: Поведение объекта в жидкости зависит от сравнения между его весом и выталкивающей силой. Существуют три основных условия: плавание (когда плотность объекта меньше плотности жидкости), нейтральная плавучесть (равные плотности) и погружение (плотность объекта больше плотности жидкости).

Kazakh: Сұйықтықтағы заттың мінез-құлығы оның салмағы мен итеруші күш арасындағы салыстыруға байланысты. Үш негізгі жағдай бар: жүзу (заттың тығындығы сұйықтықтың тығындығынан аз), бейтарап жүзу қабілеті (тең тығындықтар) және батып кету (заттың тығындығы сұйықтықтың тығындығынан көп).

Theory Questions

Easy Question: A wooden block with volume 0.2 m³ floats in water. If the density of wood is 600 kg/m³, what volume of the block is submerged?

Answer

For floating equilibrium: Weight of block = Upthrust
ρ_wood × V_total × g = ρ_water × V_submerged × g
600 × 0.2 = 1000 × V_submerged
V_submerged = (600 × 0.2)/1000 = 0.12 m³
Percentage submerged = (0.12/0.2) × 100% = 60%

Medium Question: A steel sphere (density 7800 kg/m³) with radius 0.1m is completely submerged in water. Calculate the upthrust acting on it and determine the net force.

Answer

Volume of sphere: V = (4/3)πr³ = (4/3)π(0.1)³ = 4.19 × 10⁻³ m³
Upthrust: F = ρ_watergV = 1000 × 9.8 × 4.19 × 10⁻³ = 41.1 N
Weight of sphere: W = ρ_steelgV = 7800 × 9.8 × 4.19 × 10⁻³ = 320.5 N
Net downward force = W — F = 320.5 — 41.1 = 279.4 N
The sphere will sink as the net force is downward.

Medium Question: An iceberg floats in seawater. If 90% of the iceberg is submerged, calculate the density of ice. (Density of seawater = 1025 kg/m³)

Answer

For floating equilibrium: Weight of ice = Upthrust from displaced seawater
ρ_ice × V_total × g = ρ_seawater × V_submerged × g
ρ_ice × V_total = ρ_seawater × 0.9V_total
ρ_ice = 0.9 × 1025 = 922.5 kg/m³
This explains why 90% of an iceberg is underwater — ice is slightly less dense than seawater.

Hard Question (Critical Thinking): A submarine needs to achieve neutral buoyancy at a depth of 100m. The submarine’s hull has a volume of 200 m³ and mass of 180,000 kg (including crew and equipment). Calculate the volume of water that must be taken into the ballast tanks. At this depth, seawater density is 1027 kg/m³. Analyze why submarines use ballast tanks rather than fixed weights, and discuss the safety implications of this design choice.

Answer

Calculation of ballast water needed:

For neutral buoyancy: Total weight = Total upthrust
(Mass of submarine + Mass of ballast water) × g = ρ_seawater × V_total × g

Let V_ballast = volume of ballast water needed
Mass of ballast water = ρ_seawater × V_ballast = 1027 × V_ballast

180,000 + 1027V_ballast = 1027 × 200
180,000 + 1027V_ballast = 205,400
1027V_ballast = 25,400
V_ballast = 24.73 m³

Analysis of ballast tank advantages:

1. **Controllable buoyancy:** Can adjust from positive (surface) to neutral (submerged) to negative (emergency descent)
2. **Operational flexibility:** Can maintain precise depth without propulsion
3. **Energy efficiency:** No constant power needed to maintain depth
4. **Emergency capabilities:** Can surface rapidly by blowing ballast tanks

Safety implications:

**Positive aspects:**
— Emergency blow system provides rapid surface capability
— Multiple independent ballast tanks prevent total failure
— Can compensate for hull compression at depth
— Allows precise depth control for safety margins

**Risk factors:**
— Ballast system failure could trap submarine underwater
— Requires backup compressed air systems
— Complex valving creates potential failure points
— Must account for pressure changes affecting buoyancy

**Comparison with fixed weights:**
Fixed weights would make depth control impossible and eliminate emergency surface capability, making submarine operation extremely dangerous.

The ballast tank system is essential for safe submarine operation, providing both operational capability and critical safety features.

💪 Memorization Exercises for Key Terms

Complete the Definitions

1. _______ is the upward force exerted by a fluid on an immersed object.

Answer

Upthrust (or buoyant force)

2. Archimedes’ principle states that upthrust equals the _______ of the _______ displaced.

Answer

weight of the fluid displaced

3. The formula for upthrust is: F = _______

Answer

F = ρgV (where ρ is fluid density, g is gravitational acceleration, V is displaced volume)

4. An object will float if its density is _______ than the fluid’s density.

Answer

less than (lower than)

5. _______ buoyancy occurs when an object’s weight exactly equals the upthrust.

Answer

Neutral buoyancy

🎥 Educational Video Resource

Additional Video Resources:

• Khan Academy: Archimedes’ Principle and Buoyancy

• Crash Course Physics: Buoyancy

• Veritasium: Why Don’t We All Float?

📐 Worked Problem Examples

Example 1: Floating Object Analysis

Problem: A cork with density 240 kg/m³ and volume 50 cm³ is placed in water. Calculate: (a) the upthrust when the cork is floating, (b) the volume of cork submerged, (c) what would happen if the cork were held completely underwater.

Cork floating in water

Step-by-Step Solution

Given:
— Density of cork: ρ_cork = 240 kg/m³
— Volume of cork: V_cork = 50 cm³ = 50 × 10⁻⁶ m³
— Density of water: ρ_water = 1000 kg/m³

Part (a): Upthrust when floating
When floating, upthrust = weight of cork
Weight = ρ_cork × V_cork × g = 240 × 50 × 10⁻⁶ × 9.8 = 0.118 N
Therefore, upthrust = 0.118 N

Part (b): Volume submerged
For equilibrium: Weight of cork = Upthrust from displaced water
ρ_cork × V_cork × g = ρ_water × V_submerged × g
240 × 50 × 10⁻⁶ = 1000 × V_submerged
V_submerged = (240 × 50 × 10⁻⁶)/1000 = 12 × 10⁻⁶ m³ = 12 cm³

Percentage submerged = (12/50) × 100% = 24%

Part (c): Completely underwater
If held completely underwater:
Upthrust = ρ_water × V_cork × g = 1000 × 50 × 10⁻⁶ × 9.8 = 0.49 N
Weight = 0.118 N
Net upward force = 0.49 — 0.118 = 0.37 N

The cork would experience a strong upward force and rise rapidly to the surface when released.

Example 2: Ship Design Problem

Problem: A cargo ship has a mass of 50,000 tonnes when empty. The ship’s hull displaces 60,000 m³ of seawater when fully loaded. Calculate: (a) the maximum cargo mass the ship can carry, (b) the draft (depth submerged) when carrying 30,000 tonnes of cargo, given that the ship’s cross-sectional area at the waterline is 5000 m². (Density of seawater = 1025 kg/m³)

Complete Solution with Ship Design Analysis

Given:
— Empty ship mass = 50,000 tonnes = 50,000,000 kg
— Maximum displacement volume = 60,000 m³
— Cross-sectional area = 5000 m²
— Density of seawater = 1025 kg/m³
— Cargo mass for part (b) = 30,000 tonnes = 30,000,000 kg

Part (a): Maximum cargo capacity
At maximum load, ship displaces 60,000 m³ of seawater
Maximum total weight = Maximum upthrust
(m_ship + m_cargo) × g = ρ_seawater × V_displaced × g

m_ship + m_cargo = 1025 × 60,000 = 61,500,000 kg
m_cargo = 61,500,000 — 50,000,000 = 11,500,000 kg = 11,500 tonnes

Part (b): Draft with 30,000 tonnes cargo
Total mass = 50,000,000 + 30,000,000 = 80,000,000 kg
For equilibrium: Total weight = Upthrust
80,000,000 × g = 1025 × V_displaced × g
V_displaced = 80,000,000/1025 = 78,048.8 m³

But wait! This exceeds the maximum displacement of 60,000 m³.
This means the ship would be overloaded and would sink!

Let’s recalculate the maximum safe cargo:
Maximum safe cargo = 11,500 tonnes (from part a)

For 30,000 tonnes cargo, the ship cannot float safely. If we assume the ship design allows this displacement:
Draft = V_displaced / Cross-sectional area = 78,048.8 / 5000 = 15.6 m

Engineering Analysis:
This problem illustrates the critical importance of load line regulations in shipping. The Plimsoll line on ships marks safe loading limits to prevent overloading and ensure adequate freeboard (height above water) for safe operation in various sea conditions.

Safety considerations:
— Overloading reduces stability and increases capsize risk
— Insufficient freeboard allows water to enter in rough seas
— Load distribution affects ship’s center of gravity and stability

Example 3: Submarine Ballast Analysis

Problem: A research submarine has a total volume of 150 m³ and a hull mass of 120,000 kg. To achieve neutral buoyancy at the surface, determine: (a) the mass of ballast water needed, (b) the percentage of internal volume occupied by ballast water, (c) analyze what happens if the submarine descends to 200m depth where seawater density increases to 1027 kg/m³.

Advanced Submarine Analysis

Given:
— Submarine volume = 150 m³
— Hull mass = 120,000 kg
— Surface seawater density = 1025 kg/m³
— Deep seawater density (200m) = 1027 kg/m³

Part (a): Ballast water mass for neutral buoyancy
For neutral buoyancy: Total weight = Total upthrust
(m_hull + m_ballast) × g = ρ_seawater × V_total × g

At surface:
120,000 + m_ballast = 1025 × 150
120,000 + m_ballast = 153,750
m_ballast = 33,750 kg

Part (b): Percentage volume of ballast water
Volume of ballast water = m_ballast / ρ_water = 33,750 / 1000 = 33.75 m³
Percentage = (33.75 / 150) × 100% = 22.5%

Part (c): Effect of depth change
At 200m depth, if no ballast adjustment is made:
Total weight = 120,000 + 33,750 = 153,750 kg
Upthrust at depth = 1027 × 150 × g = 154,050g N
Weight = 153,750g N

Net upward force = (154,050 — 153,750) × g = 300g = 2,940 N

Analysis:
The submarine becomes slightly positively buoyant due to increased water density at depth. This means:

1. **Automatic ascent tendency:** Without depth control, the submarine would rise
2. **Ballast adjustment needed:** Must take on additional 300 kg of water for neutral buoyancy
3. **Control systems:** Submarines use trim tanks for fine depth control
4. **Safety margins:** Real submarines maintain slight positive buoyancy for safety

Practical considerations:
— Hull compression at depth slightly reduces internal volume
— Temperature changes affect ballast water density
— Emergency blow systems must overcome maximum depth pressure
— Multiple ballast tank systems provide redundancy for safety

This explains why submarines need sophisticated ballast control systems for precise depth maintenance and safety.

🧪 Interactive Investigation - PhET Simulation

Explore buoyancy and Archimedes’ principle using this interactive simulation:

Investigation Tasks:

Task 1: Test different materials (wood, aluminum, lead) in water. Record their floating/sinking behavior and measure the upthrust in each case.

Task 2: For a floating object, measure the submerged volume and verify Archimedes’ principle by calculating the expected upthrust.

Task 3: Change the fluid from water to honey or gasoline. Observe how the same object behaves differently and explain why.

Investigation Answers and Analysis

Task 1 Analysis:
Students should observe:
— Wood (density ~600 kg/m³): Floats, upthrust = weight of wood
— Aluminum (density ~2700 kg/m³): Sinks, upthrust < weight
— Lead (density ~11,340 kg/m³): Sinks rapidly, upthrust much < weight

The upthrust depends on displaced volume and fluid density, not object material.

Task 2 Verification:
For floating objects, students can verify:
— Measure submerged volume from simulation
— Calculate expected upthrust = ρ_water × g × V_submerged
— Compare with weight of object
— They should be equal for floating equilibrium

Task 3 Fluid Comparison:
Same object in different fluids:
— Honey (higher density): Object floats higher, less submerged
— Gasoline (lower density): Object floats lower, more submerged
— Some objects that sink in water may float in honey

This demonstrates that buoyancy depends on fluid density, not just object properties.

👥 Collaborative Group Activity

Work with your team to complete this interactive buoyancy challenge:

Group Design Challenge:

Design a Density-Testing Device

Challenge: Your team must design a simple device to determine the density of irregular objects using Archimedes’ principle.

Requirements:

Measure object weight in air and in water
Calculate density using buoyancy measurements
Test accuracy with known materials
Design works for objects denser than water

Deliverables:

Detailed procedure with step-by-step instructions
Mathematical derivation of density formula from Archimedes’ principle
Error analysis and accuracy testing results
Demonstration with unknown sample (5 minutes maximum)

Alternative Group Activities:

• Buoyancy Investigation: Test the floating behavior of various materials in different liquids

• Ship Design Challenge: Design a clay boat that can carry maximum cargo without sinking

• Submarine Simulation: Model ballast tank operation for different depths and conditions

📝 Individual Assessment - Structured Questions

Question 1: Analysis and Application

A hot air balloon has a total volume of 2500 m³. The balloon fabric and basket have a combined mass of 150 kg. Calculate the mass of hot air needed inside the balloon to achieve neutral buoyancy in cold air at 15°C. Given: density of cold air = 1.225 kg/m³, density of hot air at 100°C = 0.946 kg/m³. Analyze what happens if the air inside cools down and determine the minimum temperature needed to maintain lift with a 70 kg passenger.

Answer

Part 1: Mass of hot air for neutral buoyancy
For neutral buoyancy: Total weight = Upthrust from displaced cold air
(m_fabric + m_{hot air}) × g = ρ_{cold air} × V_balloon × g
150 + m_{hot air} = 1.225 × 2500 = 3062.5 kg
m_{hot air} = 3062.5 — 150 = 2912.5 kg

Verification with hot air density:
Expected hot air mass = ρ_{hot air} × V = 0.946 × 2500 = 2365 kg
This is less than calculated (2912.5 kg), so balloon will have positive buoyancy.
Net lift = (2912.5 — 2365) × g = 547.5g = 5365 N

Part 2: Effect of cooling
If hot air cools and density increases toward cold air density:
At ρ = 1.225 kg/m³: m_air = 3062.5 kg (neutral buoyancy)
Current hot air mass = 2365 kg, so balloon loses lift as air cools.

Part 3: Minimum temperature with 70 kg passenger
Total mass to lift = 150 + 70 = 220 kg
Required hot air mass = 3062.5 — 220 = 2842.5 kg
Required hot air density = 2842.5/2500 = 1.137 kg/m³

Using ideal gas law approximation: ρ ∝ 1/T
T_hot/T_cold = ρ_cold/ρ_hot
T_hot = 288K × (1.225/1.137) = 310K = 37°C

Analysis: Hot air balloons require continuous heating to maintain lift. As air cools, density increases and lift decreases. The minimum temperature of 37°C is much lower than typical operating temperatures (60-80°C), providing safety margin for varying conditions.

Question 2: Synthesis and Critical Thinking

An oil spill cleanup team uses containment booms that must float on the ocean surface while containing oil underneath. A boom section is 10m long, has a rectangular cross-section 0.8m wide × 0.6m high, and is made of foam with density 200 kg/m³. Calculate: (a) the equilibrium floating position, (b) the maximum weight of oil that can be contained before the boom sinks, (c) analyze the design requirements for effective oil containment considering wave action and varying oil densities.

Answer

Given data:
— Boom dimensions: 10m × 0.8m × 0.6m
— Boom volume: V = 10 × 0.8 × 0.6 = 4.8 m³
— Foam density: ρ_foam = 200 kg/m³
— Seawater density: ρ_seawater = 1025 kg/m³

Part (a): Equilibrium floating position
Boom mass: m_boom = 200 × 4.8 = 960 kg
For floating equilibrium: Weight = Upthrust
m_boom × g = ρ_seawater × V_submerged × g
960 = 1025 × V_submerged
V_submerged = 960/1025 = 0.937 m³

Submerged depth = V_submerged/(length × width) = 0.937/(10 × 0.8) = 0.117 m
Freeboard (height above water) = 0.6 — 0.117 = 0.483 m

Part (b): Maximum oil containment capacity
For boom to just reach full submersion:
Total weight when fully submerged = Maximum upthrust
(m_boom + m_oil) × g = ρ_seawater × V_total × g
960 + m_oil = 1025 × 4.8 = 4920 kg
m_oil = 4920 — 960 = 3960 kg

Assuming oil density ≈ 850 kg/m³:
V_oil = 3960/850 = 4.66 m³

Part (c): Design analysis for effective containment

Wave action considerations:
1. **Dynamic loading:** Waves create additional forces requiring safety margins
2. **Freeboard requirements:** Need substantial height above water for wave tolerance
3. **Flexibility:** Boom must flex with waves without breaking containment
4. **Anchor points:** Must resist wave-induced lateral forces

Variable oil density effects:**
— Light crude oil (800 kg/m³): Can contain more volume
— Heavy crude oil (950 kg/m³): Less volume but similar mass capacity
— Weathered oil increases density over time

Improved design recommendations:**
1. **Increased freeboard:** Design for 20-30% safety margin above calculated equilibrium
2. **Compartmentalization:** Multiple sealed sections prevent total failure
3. **Ballast system:** Adjustable ballast for different oil types and sea conditions
4. **Drainage system:** Automatic oil collection to prevent overloading
5. **Material selection:** Chemical resistance to prevent degradation

Engineering trade-offs:**
— Higher freeboard increases wind resistance
— Heavier construction improves stability but reduces portability
— Larger cross-section improves capacity but increases deployment complexity

The design must balance buoyancy, structural integrity, environmental resistance, and operational requirements for effective oil spill response.

Question 3: Complex Analysis

A deep-sea research vessel uses syntactic foam for buoyancy control. The foam has density 400 kg/m³ and is enclosed in titanium spheres. Each sphere has internal volume 0.5 m³ and wall thickness 2cm. At 4000m depth, the external pressure is 40 MPa, compressing the foam by 5%. Calculate: (a) the buoyancy force per sphere at this depth, (b) the change in buoyancy due to compression, (c) analyze how this affects vehicle trim and depth control systems.

Answer

Given data:
— Foam density: ρ_foam = 400 kg/m³
— Internal volume: V_internal = 0.5 m³
— Wall thickness: t = 0.02 m
— Depth: 4000m, Pressure: 40 MPa
— Foam compression: 5%
— Seawater density at depth: ρ_seawater ≈ 1030 kg/m³ (accounting for compression)
Step 1: Calculate sphere dimensions
Internal radius: r_i = (3V/4π)^(1/3) = (3×0.5/4π)^(1/3) = 0.492 m
External radius: r_e = r_i + t = 0.492 + 0.02 = 0.512 m
External volume: V_external = (4/3)π(0.512)³ = 0.563 m³
Part (a): Buoyancy force at 4000m depth
Upthrust = ρ_seawater × g × V_external
F_buoyancy = 1030 × 9.8 × 0.563 = 5684 N
Sphere weight calculation:
Foam mass: m_foam = 400 × 0.5 = 200 kg
Titanium shell volume: V_shell = V_external — V_internal = 0.563 — 0.5 = 0.063 m³
Titanium mass: m_Ti = 4500 × 0.063 = 283.5 kg (ρ_Ti = 4500 kg/m³)
Total weight: W = (200 + 283.5) × 9.8 = 4742 N
Net buoyancy: F_net = 5684 — 4742 = 942 N (upward)
Part (b): Effect of foam compression
Compressed foam volume: V_compressed = 0.5 × 0.95 = 0.475 m³
Mass remains constant, so new foam density: ρ_new = 200/0.475 = 421 kg/m³
External volume unchanged (rigid titanium shell)
New upthrust = 5684 N (same as before)
New weight = (200 + 283.5) × 9.8 = 4742 N (mass unchanged)
Net buoyancy remains: 942 N
However, if we consider the water entering compressed space:
Water volume entering: 0.5 — 0.475 = 0.025 m³
Additional water mass: 1030 × 0.025 = 25.75 kg
New total weight: (200 + 283.5 + 25.75) × 9.8 = 4994 N
Revised net buoyancy: 5684 — 4994 = 690 N
Change in buoyancy: 942 — 690 = 252 N reduction
Part (c): Impact on vehicle trim and depth control
Trim effects:
1. **Reduced positive buoyancy:** Vehicle becomes less buoyant, tendency to sink deeper
2. **Uneven compression:** If spheres compress differently, creates trim imbalance
3. **Progressive effect:** Compression increases with depth, affecting stability
Depth control challenges:**strong>
1. **Depth hunting:** Vehicle may oscillate around target depth due to changing buoyancy
2. **Control system adaptation:** Must compensate for depth-dependent buoyancy changes
3. **Ballast requirements:** May need active ballast adjustment during deep dives
Engineering solutions:
1. **Pressure-compensated design:** Use incompressible syntactic foam formulations
2. **Active buoyancy control:** Variable ballast tanks to compensate for compression
3. **Distributed buoyancy:** Multiple smaller spheres to minimize individual failure impact
4. **Predictive control:** Software compensation based on depth/pressure measurements
Safety considerations:**strong>
— Buoyancy reduction affects emergency ascent capability
— Must maintain positive buoyancy margin even with maximum compression
— Backup buoyancy systems essential for deep-sea operations
This analysis shows why deep-sea vehicles require sophisticated buoyancy management systems to maintain precise depth control despite pressure-induced changes in buoyancy characteristics.

Question 4: Engineering Application and Synthesis

Design a floating wind turbine platform for offshore installation. The turbine tower and nacelle have a combined mass of 800 tonnes and center of gravity 90m above the platform. The platform must remain stable in 3m waves while maintaining the turbine vertical within ±5°. Calculate the required platform dimensions and ballast configuration, considering both static equilibrium and dynamic stability. Analyze the engineering trade-offs between stability, cost, and performance.

Answer

Design Requirements Analysis:
Given specifications:
— Turbine mass: 800 tonnes = 800,000 kg
— Center of gravity height: 90m above platform
— Wave height: 3m (significant wave height)
— Tilt tolerance: ±5°
— Seawater density: 1025 kg/m³
Step 1: Basic stability calculations
Platform mass estimation:
For stability, platform mass should be 2-3 times turbine mass
Platform mass ≈ 2000 tonnes = 2,000,000 kg
Total system mass = 2,800,000 kg
Required displacement:
Displaced volume = Total mass / seawater density = 2,800,000 / 1025 = 2732 m³
Step 2: Platform geometry design
Circular platform approach:
Assume cylindrical platform with draft d = 20m (deep draft for stability)
Platform area = 2732 / 20 = 136.6 m²
Platform radius = √(136.6/π) = 6.6m → Use 7m radius
Platform volume = π × 7² × 20 = 3078 m³
Displacement = 3078 × 1025 = 3,155,000 kg
Ballast calculation:
Available buoyancy = 3,155,000 kg
Required mass = 2,800,000 kg
Available for ballast = 355,000 kg ballast water
Step 3: Stability analysis
Metacentric height calculation:
For a floating cylinder:
I = π × r⁴ / 4 = π × 7⁴ / 4 = 5890 m⁴
BM = I / V_submerged = 5890 / 3078 = 1.91 m
Center of buoyancy (B) = draft/2 = 10m below waterline
Center of gravity (G) calculation:
— Platform CG: 10m below waterline
— Turbine CG: 90m above waterline
— Combined CG = (2000×(-10) + 800×90) / 2800 = 18.6m above waterline
BG = 10 — 18.6 = -28.6m (G above B)
GM = BM — BG = 1.91 — (-28.6) = 30.5m
This is negative metacentric height — UNSTABLE!
Step 4: Redesign for stability
Approach 1: Increase platform size
Try radius = 15m, draft = 10m
Platform volume = π × 15² × 10 = 7069 m³
New metacentric calculations:
I = π × 15⁴ / 4 = 99,532 m⁴
BM = 99,532 / 7069 = 14.1m
New combined CG ≈ 10m above waterline (with larger platform mass)
GM = 14.1 — 10 = 4.1m (STABLE)
Approach 2: Semi-submersible design
Multiple columns with submerged pontoons:
— 3 columns, each 4m diameter
— Connected by submerged pontoons at 25m depth
— Lower center of gravity, higher stability
Step 5: Dynamic stability in waves
Wave response analysis:
Natural roll period: T = 2π√(I_mass/ρgV_disp×GM)
For GM = 4.1m: T ≈ 15-20 seconds
Wave periods typically 6-12 seconds, so good separation
Angular displacement in 3m waves:
θ_max = Wave slope × Response factor
For well-designed platform: θ_max < 3° (within ±5° requirement)
Step 6: Engineering trade-offs analysis
Stability vs Cost:
— Larger platforms: Higher stability but much higher material costs
— Semi-submersible: Excellent stability but complex construction
— Spar buoy: Good stability, moderate cost, but requires deep water
Stability vs Performance:
— Higher stability reduces turbine fatigue loads
— Excessive stability can increase acceleration loads
— Optimal GM = 2-5m for floating wind turbines
Alternative configurations:
1. **Spar buoy:** 6m diameter, 100m draft, ballasted bottom
2. **TLP (Tension Leg Platform):** Moored with tension cables
3. **Semi-submersible:** Multiple hulls with cross-bracing
Recommended design:
Semi-submersible with 3 columns (5m diameter each), connected by pontoons at 20m depth:
— Total displacement: 3000 m³
— GM ≈ 3m (stable but not over-stable)
— Natural period > 20 seconds
— Meets tilt requirements in design wave conditions
Critical considerations:
— Mooring system design for station keeping
— Power cable management through platform motion
— Maintenance access and crew transfer systems
— Extreme weather survival capabilities
This design balances stability requirements with practical engineering and economic constraints for offshore wind energy applications.

Question 5: Advanced Critical Analysis

A marine biologist claims that «fish can control their buoyancy by adjusting their swim bladder volume, effectively changing their average density to match the surrounding water.» Evaluate this statement by analyzing the physics of swim bladder operation, calculating the volume changes required for depth adjustments, and discussing the biological and physical limitations. Consider how this mechanism compares to artificial buoyancy control systems and explain why some deep-sea fish lack swim bladders entirely.

Answer

Evaluation: The statement is CORRECT but requires detailed analysis of the mechanisms and limitations.
Physics of swim bladder buoyancy control:
Basic principle:
Fish achieve neutral buoyancy when: ρ_fish = ρ_water
Average fish density: ρ_fish = (m_body + m_air)/(V_body + V_bladder)
By controlling V_bladder, fish can adjust their average density.
Quantitative analysis of volume changes:
Example calculation:
Consider a 1kg fish with 50cm³ swim bladder at 10m depth:
— Fish body density ≈ 1050 kg/m³ (slightly denser than water due to bones, organs)
— Fish body volume ≈ 950 cm³
— Water density = 1000 kg/m³
At 10m depth (2 atmospheres):
Swim bladder pressure = 2 atm = 200 kPa
For neutral buoyancy: (1000g)/(950 + 50)cm³ = 1000 kg/m³ ✓
If fish swims to 20m depth (3 atmospheres):
Without adjustment, bladder compresses to: 50 × (2/3) = 33.3 cm³
New density = 1000/(950 + 33.3) = 1.018 kg/m³
Fish becomes negatively buoyant and sinks.
Required volume adjustment:
For neutral buoyancy at 20m: 1000/(950 + V_new) = 1000
V_new = 50 cm³ (must maintain original volume)
Fish must add gas to restore 16.7 cm³ (50 — 33.3) against 3 atm pressure.
Biological mechanisms:
Physoclistous fish (gas gland system):
1. **Gas secretion:** Specialized gas gland secretes O₂ into bladder
2. **Resorption:** Oval organ reabsorbs excess gas
3. **Active process:** Requires energy and time (hours for large depth changes)
Physostomous fish (pneumatic duct):
1. **Direct connection:** Bladder connected to esophagus
2. **Surface gulping:** Must surface to gulp air for buoyancy adjustment
3. **Limited depth range:** Cannot operate at great depths
Physical and biological limitations:
Pressure limitations:
— Gas becomes increasingly compressed with depth
— At 1000m (100 atm), gas volume is 1% of surface volume
— Energetic cost of gas secretion increases exponentially with pressure
Time constraints:
— Gas secretion/resorption takes hours to days
— Rapid depth changes cause swim bladder trauma
— Fish cannot make quick vertical migrations
Metabolic costs:
— Gas gland operation requires significant energy
— Oxygen concentration in bladder can reach 85% (vs 21% in air)
— Continuous maintenance energy required
Comparison with artificial systems:
Similarities to submarine ballast systems:
— Both use variable volume/density for buoyancy control
— Both require energy input for active control
— Both have limitations in operational depth range
Key differences:
— **Speed:** Submarines can change buoyancy in minutes vs hours for fish
— **Precision:** Fish achieve near-perfect neutral buoyancy
— **Energy source:** Fish use metabolic energy vs mechanical pumps
— **Operational range:** Submarines can operate at any depth with adequate ballast capacity
Why deep-sea fish lack swim bladders:
Physical impossibility at extreme depths:
1. **Pressure effects:** At 6000m (600 atm), maintaining gas-filled bladder requires enormous energy
2. **Structural limitations:** Bladder walls cannot withstand extreme pressure differentials
3. **Gas solubility:** High pressure forces gases into solution
Alternative adaptations:
1. **Lipid-filled livers:** Low-density oils provide buoyancy (sharks, deep-sea fish)
2. **Gelatinous tissues:** Water-filled tissues with lower density
3. **Reduced bone density:** Cartilaginous skeletons reduce overall density
4. **Behavioral adaptations:** Continuous swimming to maintain depth
Evolutionary trade-offs:
— **Energy efficiency:** Lipids provide buoyancy without energy cost
— **Depth flexibility:** No pressure-related limitations
— **Trade-off:** Less precise buoyancy control, requires other adaptations
Engineering insights:
Fish swim bladders demonstrate elegant biological engineering solutions:
— **Graduated depth control:** Some fish have multiple bladder compartments
— **Pressure regulation:** Sophisticated gas gland chemistry
— **Material science:** Bladder walls with optimal strength-to-weight ratios
Conclusion:
The biologist’s statement is accurate but represents only part of the story. Swim bladders are sophisticated buoyancy control systems that work excellently within their operational limits. However, physical laws impose fundamental constraints that explain both their capabilities and limitations, and why evolution has produced alternative solutions for extreme environments.
This analysis demonstrates how physics principles apply universally, from engineered systems to biological adaptations, while highlighting the ingenuity of both natural and artificial solutions to buoyancy control challenges.

🔗 Additional Learning Resources

📚 Comprehensive Study Resources:

🤔 Lesson Reflection and Self-Assessment

💭 Knowledge Self-Check

Conceptual Understanding (Rate 1-5):

□ I understand that upthrust arises from pressure differences in fluids

□ I can apply Archimedes’ principle using F = ρgV correctly

□ I can predict whether objects will float, sink, or achieve neutral buoyancy

□ I understand the relationship between density differences and buoyant behavior

□ I can explain how buoyancy control systems work in ships and submarines

Problem-Solving Skills Assessment:

Which problem-solving strategies worked best for you today?

Identifying the displaced volume correctly for different scenarios
Applying equilibrium conditions (weight = upthrust) for floating objects
Using density comparisons to predict object behavior
Breaking complex buoyancy problems into simpler force analysis
Connecting theoretical principles to real-world applications

What challenges did you encounter?

Distinguishing between total volume and displaced volume
Understanding partially submerged vs. fully submerged scenarios
Working with variable fluid densities at different depths
Visualizing three-dimensional buoyancy effects

Real-World Connections:

How can you apply Archimedes’ principle in everyday life?

Understanding why ice floats and how much is submerged
Appreciating ship design and loading principles
Understanding how hot air balloons and weather balloons work
Recognizing buoyancy effects when swimming or diving
Understanding how submarines control their depth

Language Development Reflection:

New physics vocabulary mastered:

□ Can use «upthrust,» «buoyancy,» and «displacement» correctly in explanations

□ Understand floating conditions and can explain them clearly

□ Can describe Archimedes’ principle in written English

□ Comfortable reading and interpreting buoyancy problems in English

Communication goals for next lesson:

• Practice explaining buoyancy principles using everyday examples

• Use more precise scientific language when describing fluid forces

• Develop confidence in presenting buoyancy solutions step-by-step

Future Learning Goals:

What aspects of buoyancy would you like to explore further?

Fluid dynamics and moving objects through fluids
Advanced applications in marine and aerospace engineering
Biological applications of buoyancy in marine life
Atmospheric buoyancy and weather phenomena
Materials science applications for buoyancy control

How will this knowledge help in future physics topics?

Understanding fluid flow and Bernoulli’s principle
Studying atmospheric physics and meteorology
Analyzing complex fluid-structure interactions
Connecting to thermal physics through hot air balloons

🎯 Action Plan for Continued Learning:

This week I will:

□ Observe floating objects and estimate their density ratios

□ Practice buoyancy calculations with different scenarios

□ Research real applications of Archimedes’ principle in engineering

□ Explore the additional simulations and online resources

□ Prepare questions about fluid applications for next class

Upthrust (Buoyant Force)

Archimedes’ Principle

Floating Condition

Sinking Condition

Upthrust (Buoyant Force)

Archimedes’ Principle

Displacement

Neutral Buoyancy

Origin of Upthrust — Pressure Differences

Archimedes’ Principle Statement

Conditions for Floating, Sinking, and Neutral Buoyancy

Floating (ρobject < ρfluid)

Neutral Buoyancy (ρobject = ρfluid)

Sinking (ρobject > ρfluid)

Theory Questions

Complete the Definitions

Additional Video Resources:

Example 1: Floating Object Analysis

Example 2: Ship Design Problem

Example 3: Submarine Ballast Analysis

Investigation Tasks:

Group Design Challenge:

Design a Density-Testing Device

Alternative Group Activities:

Question 1: Analysis and Application

Question 2: Synthesis and Critical Thinking

Question 3: Complex Analysis

Question 4: Engineering Application and Synthesis

Question 5: Advanced Critical Analysis

📚 Comprehensive Study Resources:

📖 Theory Resources

🎥 Video Tutorials

🧪 Interactive Tools

📝 Practice Problems

💭 Knowledge Self-Check

Conceptual Understanding (Rate 1-5):

Problem-Solving Skills Assessment:

Real-World Connections:

Language Development Reflection:

Future Learning Goals:

🎯 Action Plan for Continued Learning:

Floating (ρ_object < ρ_fluid)

Neutral Buoyancy (ρ_object = ρ_fluid)

Sinking (ρ_object > ρ_fluid)