🚢 Buoyancy & Archimedes' Principle

Part 1 of 7 — Why Things Float

You already know objects less dense than a fluid will float. But how much force does a fluid exert upward? And why does it exert an upward force at all?

The answer is Archimedes' Principle — one of the oldest and most elegant results in physics.

Where Does the Buoyant Force Come From?

Consider a cube submerged in water. Fluid pressure acts on all six faces:

• Left and right faces: Pressure forces cancel (same depth → same pressure)
• Front and back faces: Cancel by the same reasoning
• Top face: Pressure pushes down — $F_{\text{top}} = P_{\text{top}} \times A$
• Bottom face: Pressure pushes up — $F_{\text{bottom}} = P_{\text{bottom}} \times A$

Since the bottom face is deeper than the top face:

$P_{\text{bottom}} > P_{\text{top}} \implies F_{\text{bottom}} > F_{\text{top}}$

The net upward force is the buoyant force:

$F_B = F_{\text{bottom}} - F_{\text{top}} = (P_{\text{bottom}} - P_{\text{top}}) \times A = \rho g h \times A = \rho g V$

This net upward force exists because pressure increases with depth.

Archimedes' Principle

The buoyant force on an object equals the weight of the fluid displaced by the object.

$F_B = \rho_{\text{fluid}} \cdot V_{\text{displaced}} \cdot g$

Where:

• $F_B$ = buoyant force (N)
• $\rho_{\text{fluid}}$ = density of the fluid (kg/m³)
• $V_{\text{displaced}}$ = volume of fluid displaced (= volume of the object that's submerged)
• $g$ = 9.8 m/s² (or 10 for AP estimates)

The Key Insight

The buoyant force depends on:

• ✅ The fluid's density (not the object's density)
• ✅ The volume of fluid displaced
• ❌ NOT the object's mass, weight, or density (those determine whether it sinks or floats, but not the buoyant force formula itself)

Concept Check — Buoyancy Basics

Buoyant Force Drill (use $g = 10$ m/s², $\rho_{\text{water}} = 1000$ kg/m³)

A block with volume $V = 0.005$ m³ is fully submerged in water.

1. Buoyant force on the block (in N)
2. If the block has mass 3 kg, will it sink or float? (type "sink" or "float")
3. If the block has mass 8 kg, its apparent weight when submerged (in N)

Exit Quiz

⚖️ Sink, Float, or Hover?

Part 2 of 7 — The Three Cases

Now that you understand the buoyant force, let's systematically analyze what happens when you place an object in a fluid. There are exactly three possible outcomes.

The Three Cases

Case 1: Sinking ($\rho_{\text{obj}} > \rho_{\text{fluid}}$)

• Weight > buoyant force → net force downward
• Object accelerates to the bottom
• $V_{\text{displaced}} = V_{\text{object}}$ (fully submerged)
• Examples: rock in water, iron in water

Case 2: Floating ($\rho_{\text{obj}} < \rho_{\text{fluid}}$)

• Object rises until buoyant force equals weight
• Only partially submerged
• $V_{\text{displaced}} < V_{\text{object}}$
• Fraction submerged: $V_{\text{sub}}/V_{\text{total}} = \rho_{\text{obj}}/\rho_{\text{fluid}}$
• Examples: wood in water, ice in water, oil on water

Case 3: Neutral Buoyancy ($\rho_{\text{obj}} = \rho_{\text{fluid}}$)

• Weight = buoyant force at any depth
• Object hovers in equilibrium
• Fully submerged but not sinking
• Examples: submarine at operating depth, fish adjusting swim bladder

Floating Objects: The Equilibrium Condition

When an object floats in equilibrium:

$F_B = W$

$\rho_{\text{fluid}} V_{\text{sub}} g = \rho_{\text{obj}} V_{\text{total}} g$

$\frac{V_{\text{sub}}}{V_{\text{total}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}$

This is the fraction submerged equation. Let's use it:

Example: Wooden Block

A wooden block ($\rho = 700$ kg/m³) floats in water ($\rho = 1000$ kg/m³).

$\text{Fraction submerged} = \frac{700}{1000} = 0.70 = 70\%$

70% of the block is underwater, 30% sticks above the surface.

Example: Ice in Seawater

Ice ($\rho = 917$ kg/m³) in seawater ($\rho = 1025$ kg/m³):

$\text{Fraction submerged} = \frac{917}{1025} = 0.895 = 89.5\%$

Only 10.5% of an iceberg is visible above the surface!

Concept Check — Sink vs. Float

Floating Drill (use $g = 10$ m/s²)

A rectangular barge ($6.0 \times 3.0 \times 1.0$ m) has mass 12,000 kg. It floats in freshwater ($\rho = 1000$ kg/m³).

1. Volume of the barge (in m³)
2. Volume of water displaced (in m³)
3. Depth the barge sinks to (draft, in m)

Round all answers to 3 significant figures.

Exit Quiz

🏗️ Apparent Weight & Submerged Objects

Part 3 of 7 — Measuring Buoyancy in the Lab

One of the most common AP Physics 2 lab setups involves weighing objects in air vs. in water. Let's master the analysis.

Apparent Weight

When an object is submerged and hanging from a scale:

$W_{\text{apparent}} = W_{\text{true}} - F_B$

$W_{\text{app}} = mg - \rho_{\text{fluid}} V_{\text{obj}} g$

Free-Body Diagram

For a submerged object hanging from a string:

• Tension $T$ (upward) = what the scale reads = apparent weight
• Buoyant force $F_B$ (upward)
• Weight $W$ (downward)

Equilibrium: $T + F_B = W \implies T = W - F_B$

Measuring Density

You can find an object's density by weighing it in air and in water:

$\rho_{\text{obj}} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}} \times \rho_{\text{water}}$

This works because $W_{\text{air}} - W_{\text{water}} = F_B = \rho_w V g$, and $W_{\text{air}} = \rho_{\text{obj}} V g$.

Lab Analysis Drill (use $g = 10$ m/s², $\rho_{\text{water}} = 1000$ kg/m³)

A metal block weighs 45 N in air and 35 N when fully submerged in water.

1. Buoyant force on the block (in N)
2. Volume of the block (in m³)
3. Density of the block (in kg/m³)

Round all answers to 3 significant figures.

Sinking Objects: Net Force Analysis

For an object that sinks (but hasn't reached the bottom yet), the net downward force is:

$F_{\text{net}} = W - F_B = (\rho_{\text{obj}} - \rho_{\text{fluid}}) V g$

The acceleration is:

$a = \frac{F_{\text{net}}}{m} = \left(1 - \frac{\rho_{\text{fluid}}}{\rho_{\text{obj}}}\right)g$

Example

An iron ball ($\rho = 7800$ kg/m³) sinks in water. Its initial acceleration (ignoring drag):

$a = \left(1 - \frac{1000}{7800}\right)(10) = (1 - 0.128)(10) = 8.72 \text{ m/s}^2$

That's only slightly less than free fall! Iron is so much denser than water that buoyancy barely slows it.

But for an object with $\rho = 1200$ kg/m³:

$a = \left(1 - \frac{1000}{1200}\right)(10) = (0.167)(10) = 1.67 \text{ m/s}^2$

Much slower — buoyancy is providing significant support.

Concept Check

The King's Crown Problem (Archimedes' Original!)

Story: King Hiero II gave a goldsmith pure gold to make a crown. The king suspected the goldsmith mixed in cheaper silver. Archimedes was asked to determine if the crown was pure gold without damaging it.

Solution: Weigh the crown in air: $W_{\text{air}} = 25.0$ N. Weigh it in water: $W_{\text{water}} = 22.6$ N.

$F_B = 25.0 - 22.6 = 2.4$ N

$V = F_B/(\rho_w g) = 2.4/(1000 \times 10) = 2.4 \times 10^{-4}$ m³

$\rho = m/V = 2.5 / (2.4 \times 10^{-4}) = 10{,}417$ kg/m³

Verdict: Pure gold has $\rho = 19{,}300$ kg/m³. This crown has $\rho \approx 10{,}400$ kg/m³ — the goldsmith was a fraud! (The crown was likely a gold-silver alloy.)

Exit Quiz

🧊 Floating Object Problems

Part 4 of 7 — The Problems AP Loves Most

Floating objects are AP exam favorites because they combine buoyancy, density, and equilibrium in satisfying ways. Let's master every variant.

The Floating Equilibrium Equation

For any floating object, exactly two forces balance:

$F_B = W$

$\rho_{\text{fluid}} V_{\text{sub}} g = \rho_{\text{obj}} V_{\text{total}} g$

From this, we can derive:

$\frac{V_{\text{sub}}}{V_{\text{total}}} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}} \quad \text{(fraction submerged)}$

$\frac{V_{\text{above}}}{V_{\text{total}}} = 1 - \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}} \quad \text{(fraction above)}$

Common Floating Scenarios

Adding Weight to Floating Objects

Classic problem: A wooden raft floats in water. How much extra mass can you put on it before it sinks?

Strategy

The raft sinks when it's fully submerged ($V_{\text{sub}} = V_{\text{total}}$). At this point:

$\rho_w V_{\text{raft}} g = (m_{\text{raft}} + m_{\text{extra}}) g$

$m_{\text{extra}} = \rho_w V_{\text{raft}} - m_{\text{raft}}$

Worked Example

A wooden raft has volume 2.0 m³ and mass 1200 kg. Maximum extra mass:

$m_{\text{extra}} = (1000)(2.0) - 1200 = 2000 - 1200 = 800 \text{ kg}$

The raft can hold up to 800 kg before going under!

General formula: $m_{\text{extra}} = (\rho_{\text{fluid}} - \rho_{\text{obj}}) \times V_{\text{obj}}$

Floating Problems Quiz

Raft Problem Drill (use $g = 10$ m/s²)

A styrofoam raft ($\rho = 50$ kg/m³) is $2.0 \times 1.0 \times 0.20$ m.

1. Mass of the raft (in kg)
2. Maximum buoyant force when fully submerged (in N)
3. Maximum extra mass it can carry before sinking (in kg)

Exit Quiz

🔗 Multi-Object & Multi-Fluid Buoyancy

Part 5 of 7 — Complex Buoyancy Scenarios

AP Physics 2 loves problems with objects in layered fluids, objects connected by strings, and objects stacked on top of each other while floating. Let's conquer them all.

Objects in Layered Fluids

When two immiscible fluids form layers (e.g., oil on water), an object can float at the interface, partially submerged in each.

Setup

A block of density $\rho_{\text{obj}}$ where $\rho_{\text{oil}} < \rho_{\text{obj}} < \rho_{\text{water}}$.

The block sinks through the oil but floats on the water.

Equilibrium Equation

$W = F_{B,\text{oil}} + F_{B,\text{water}}$

$\rho_{\text{obj}} V_{\text{total}} g = \rho_{\text{oil}} V_{\text{oil}} g + \rho_{\text{water}} V_{\text{water}} g$

Where $V_{\text{oil}}$ is the volume in the oil layer and $V_{\text{water}}$ is the volume in the water layer.

Since the block is fully submerged: $V_{\text{oil}} + V_{\text{water}} = V_{\text{total}}$

Worked Example

A cube (side 10 cm, $\rho = 850$ kg/m³) in oil ($\rho_{\text{oil}} = 700$ kg/m³) over water ($\rho_w = 1000$ kg/m³):

$850 V = 700 V_{\text{oil}} + 1000 V_{\text{water}}$

Let $V_{\text{water}} = f \cdot V$ (fraction in water), so $V_{\text{oil}} = (1-f) \cdot V$:

$850 = 700(1-f) + 1000f = 700 + 300f$

$f = 150/300 = 0.50$

Half the cube is in water, half in oil.

Layered Fluid Quiz

Connected Objects: The Balloon-on-a-String Problem

Setup: A light object (e.g., a balloon or cork) is tied by a string to the bottom of a tank filled with water. The object is less dense than water and would float, but the string holds it down.

Free-Body Diagram

• Buoyant force $F_B$ (upward)
• Weight $W$ (downward)
• Tension $T$ (downward — the string pulls it down)

$F_B = W + T$

$T = F_B - W = (\rho_{\text{fluid}} - \rho_{\text{obj}}) V g$

Reverse Case: Dense Object on a String from Above

An object denser than water hangs from a string:

$T + F_B = W \implies T = W - F_B$

Key Distinction

Connected Object Problems (use $g = 10$ m/s², $\rho_w = 1000$ kg/m³)

A wooden ball ($\rho = 500$ kg/m³, $V = 0.002$ m³) is tied to the bottom of a pool by a string.

1. Weight of the ball (in N)
2. Buoyant force on the ball (in N)
3. Tension in the string (in N)

Stacked Floating Objects

Problem: A block of wood floats in water. A second, smaller block is placed on top of the first. How does the system float?

Strategy

Treat the stack as one floating object:

$\rho_w V_{\text{sub}} g = (m_1 + m_2) g$

$V_{\text{sub}} = \frac{m_1 + m_2}{\rho_w}$

The first block sinks deeper to support the extra weight.

Example

Block 1: 2.0 kg, volume 0.003 m³. Block 2: 0.5 kg placed on top.

$V_{\text{sub}} = \frac{2.0 + 0.5}{1000} = 0.0025 \text{ m}^3$

Since $V_{\text{sub}} = 0.0025 < V_1 = 0.003$ m³, the stack still floats (part of block 1 above water).

Exit Quiz

🎈 Buoyancy in Gases & Real-World Applications

Part 6 of 7 — Beyond Liquids

Archimedes' Principle applies to all fluids — including gases. Hot air balloons, helium balloons, and even the atmosphere itself rely on the same physics.

Buoyancy in Air

The atmosphere is a fluid! Every object in air experiences a buoyant force:

$F_{B,\text{air}} = \rho_{\text{air}} V_{\text{obj}} g$

With $\rho_{\text{air}} \approx 1.2$ kg/m³ at sea level.

Why We Usually Ignore It

For most solid objects, air buoyancy is negligible:

• 1 kg iron block: $V \approx 1.3 \times 10^{-4}$ m³ → $F_B \approx 0.0016$ N (0.016% of weight)
• 1 kg wood: $V \approx 1.4 \times 10^{-3}$ m³ → $F_B \approx 0.017$ N (0.17% of weight)

But for large, low-density objects, it matters!

Hot Air Balloons

A hot air balloon works by heating air inside the envelope, reducing its density:

$F_B = \rho_{\text{cold air}} V g, \quad W = \rho_{\text{hot air}} V g + W_{\text{basket}}$

The balloon rises when $\rho_{\text{cold}} V > \rho_{\text{hot}} V + m_{\text{basket}}$.

Helium Balloons

$\rho_{\text{He}} \approx 0.16$ kg/m³ vs. $\rho_{\text{air}} \approx 1.2$ kg/m³

Lift per m³: $(1.2 - 0.16)(10) \approx 10.4$ N/m³

A 1 m³ helium balloon can lift about 1 kg!

Gas Buoyancy Quiz

Real-World Buoyancy Applications

🐟 Fish Swim Bladders

Fish control their depth by adjusting their swim bladder volume:

• To rise: Expand the bladder → more volume → lower average density → net upward force
• To dive: Compress the bladder → less volume → higher average density → net downward force
• To hover: Adjust until $\rho_{\text{fish}} = \rho_{\text{water}}$ → neutral buoyancy

🤿 Scuba Diving

Divers use a BCD (buoyancy control device):

• At depth, the wetsuit and BCD compress → volume decreases → density increases → tendency to sink
• Divers add air to the BCD to compensate
• On ascent, they must vent air as the BCD expands

🚢 Plimsoll Lines on Ships

Ships have markings showing the safe waterline in different conditions:

• Freshwater: Ship sits lower (less dense → displaces more volume)
• Saltwater (tropical): Ship sits higher (saltwater is denser than freshwater)
• Saltwater (winter): Ship sits highest (densest water)

⚗️ Hydrometers

A hydrometer is a floating device that measures liquid density:

• It floats higher in denser liquids
• The scale reads density directly where the liquid surface crosses the stem
• Used to check battery acid, wine fermentation, antifreeze

Application Problems (use $g = 10$ m/s²)

A helium balloon has volume 0.50 m³. The envelope + string mass is 0.010 kg. $\rho_{\text{air}} = 1.2$ kg/m³, $\rho_{\text{He}} = 0.16$ kg/m³.

1. Buoyant force on the balloon (in N)
2. Total weight (helium + envelope) (in N)
3. Maximum payload mass it can lift (in kg)

Round all answers to 3 significant figures.

Exit Quiz

🎯 Buoyancy Synthesis & AP Review

Part 7 of 7 — Putting It All Together

This final part consolidates everything: concept maps, common mistakes, mixed problems, and exam-style questions.

Complete Buoyancy Concept Map

$F_B = \rho_{\text{fluid}} \cdot V_{\text{displaced}} \cdot g$

Decision Tree

1. Is the object fully submerged?

   • Yes → $V_{\text{disp}} = V_{\text{obj}}$
   • No (floating) → $V_{\text{disp}} < V_{\text{obj}}$

2. Is the object floating?

   • Yes → $F_B = W$ → use fraction submerged formula
   • No → $F_B < W$ (sinking) or $F_B = W$ (neutrally buoyant while submerged)

3. Are there other forces?

   • String from above: $T + F_B = W$
   • String from below: $F_B = W + T$
   • Normal force (sitting on bottom): $N + F_B = W$ → only if $W > F_B$ (object denser than fluid)

Top 5 AP Mistakes

AP Tricky Questions

Mixed Problem Drill (use $g = 10$ m/s², $\rho_w = 1000$ kg/m³)

An object has mass 2.0 kg and volume $8.0 \times 10^{-4}$ m³.

1. Density of the object (in kg/m³)
2. Will it sink or float in water? (type "sink" or "float")
3. Its apparent weight when fully submerged in water (in N)

AP-Style FRQ Practice

Problem Setup

A cylindrical cup (mass 0.15 kg, outer radius 4.0 cm, height 10.0 cm) floats upright in water. Small lead pellets are gradually added to the cup.

(a) With no pellets, how deep does the cup float?

$V_{\text{sub}} = m/\rho_w = 0.15/1000 = 1.5 \times 10^{-4}$ m³

$V_{\text{sub}} = \pi r^2 d$ → $d = V/(\pi r^2) = 1.5 \times 10^{-4}/(\pi (0.04)^2) = 0.030$ m = 3.0 cm

(b) What mass of pellets makes the cup sink to 8.0 cm?

$V_{\text{sub}} = \pi (0.04)^2 (0.08) = 4.02 \times 10^{-4}$ m³

$m_{\text{total}} = \rho_w V_{\text{sub}} = 1000 \times 4.02 \times 10^{-4} = 0.402$ kg

$m_{\text{pellets}} = 0.402 - 0.15 = 0.252$ kg ≈ 0.25 kg

(c) At what pellet mass does the cup just go under?

$V_{\text{max}} = \pi (0.04)^2 (0.10) = 5.03 \times 10^{-4}$ m³

$m_{\text{max}} = 1000 \times 5.03 \times 10^{-4} = 0.503$ kg

$m_{\text{pellets,max}} = 0.503 - 0.15 = 0.353$ kg ≈ 0.35 kg

Beyond this mass, water floods the top and the cup sinks rapidly!

Final Exit Quiz