Buoyancy — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Archimedes' principle, buoyant force calculation, floating and sinking equilibrium, fully/partially submerged objects, force diagram analysis for buoyancy problems, and core problem-solving techniques for all exam question types.

You should already know: Density definition and calculation. Hydrostatic pressure variation with depth. Newton's first and second laws for static and dynamic equilibrium.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Buoyancy?

Buoyancy describes the net upward force exerted on any object immersed in a static fluid. It arises from the well-known property that hydrostatic pressure increases with depth: the pressure pushing upward on the bottom surface of an immersed object is always larger than the pressure pushing downward on the object's top surface, resulting in a net upward force called the buoyant force (standard notation $F_b$, also called upthrust in some textbooks, a perfect synonym).

Buoyancy is a core topic in AP Physics 2 Unit 1 (Fluids), accounting for approximately 1-2% of total exam score, and it appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. It is almost always paired with equilibrium force analysis, a foundational skill from AP Physics 1 that is tested repeatedly across the AP Physics 2 curriculum, making it a high-yield topic to master for quick, consistent points on exam day.

2. Archimedes' Principle

Archimedes' principle is the fundamental physical law that lets us calculate the magnitude of the buoyant force on any immersed object. It states that the magnitude of the buoyant force equals the weight of the fluid displaced by the object. Displaced fluid is the fluid that would occupy the space taken up by the submerged part of the object if the object were removed.

Mathematically, Archimedes' principle is written as: $$F_b={\rho }_f{V}_{d}g$$ Where ${\rho }_f$ is the density of the fluid (not the object, a common point of confusion), $V_d$ is the volume of fluid displaced (equal to the volume of the submerged part of the object), and $g$ is acceleration due to gravity ($9.8{\text{m/s}}^2$, or $10{\text{m/s}}^2$ for approximation on the AP exam).

For a fully submerged object (entire object under the fluid surface), $V_d=V_o$, the total volume of the object. For a partially submerged object (floating on the surface), $V_d$ is only the volume of the object below the fluid line, so it is always less than $V_o$.

Worked Example

A solid aluminum cube with side length $0.10\text{m}$ is fully submerged in pure water. Aluminum has density ${\rho }_{Al}=2700{\text{kg/m}}^3$, and pure water has density ${\rho }_w=1000{\text{kg/m}}^3$. Calculate the magnitude of the buoyant force on the cube.

1. The object is fully submerged, so displaced volume equals the total volume of the cube: $V_d=V_o=(0.10\text{m}{)}^3=0.0010{\text{m}}^3$.
2. Confirm the fluid density is ${\rho }_f={\rho }_w=1000{\text{kg/m}}^3$ (we do not use aluminum density for $F_b$).
3. Substitute into Archimedes' principle: $F_b={\rho }_f{V}_{d}g$.
4. Plug in values: $F_b=(1000{\text{kg/m}}^3)(0.0010{\text{m}}^3)(9.8{\text{m/s}}^2)=9.8\text{N}$.

Exam tip: Always label ${\rho }_f$ (fluid) and ${\rho }_o$ (object) at the start of every problem. ~30% of student errors on basic buoyancy questions come from accidentally swapping these two values when plugging into the formula.

3. Floating and Sinking Equilibrium

Once we can calculate buoyant force, we use Newton's first law of static equilibrium to predict whether an object will float, sink, or stay suspended in a fluid. Any object in a fluid has two primary vertical forces (horizontal forces cancel out, so we can ignore them): the downward weight of the object $W=m_{o}g={\rho }_o{V}_{o}g$, and the upward buoyant force $F_b$.

We compare the two forces to get the outcome:

• If $F_b>W$: Net upward force, the object rises to the surface and becomes a floating object.
• If $F_b<W$: Net downward force, the object sinks to the bottom of the container.
• If $F_b=W$: Net force is zero, the object is neutrally buoyant and stays suspended at any depth.

For a stationary floating object on the surface, equilibrium always gives $F_b=W$. Substituting the formulas for $F_b$ and $W$ gives ${\rho }_f{V}_{d}g={\rho }_o{V}_{o}g$, so $g$ cancels out, leaving the very useful relation: $$\frac{V_d}{V_o}=\frac{{\rho }_o}{{\rho }_f}$$ This means the fraction of the floating object that is submerged is exactly equal to the ratio of the object density to the fluid density.

Worked Example

A solid oak block of total volume $0.02{\text{m}}^3$ floats at rest in fresh water (${\rho }_f=1000{\text{kg/m}}^3$). Oak has density $750{\text{kg/m}}^3$. What is the volume of the block that sits above the water line?

1. The block is in static equilibrium floating, so net vertical force is zero: $F_b=W$.
2. Substitute Archimedes' principle and the weight formula: ${\rho }_f{V}_{d}g={\rho }_o{V}_{o}g$. Cancel $g$ from both sides.
3. Solve for submerged volume: $V_d=V_o\frac{{\rho }_o}{{\rho }_f}=(0.02{\text{m}}^3)\left(\frac{750}{1000}\right)=0.015{\text{m}}^3$.
4. Calculate volume above water by subtracting submerged volume from total volume: $V_{\text{above}}=V_o-V_d=0.02-0.015=0.005{\text{m}}^3$.

Exam tip: If a question asks for volume above the water line, do not stop after calculating submerged volume. Always double-check what the question is asking for before moving on to the next problem.

4. Buoyancy with Tension or Normal Force

Most AP exam questions do not involve freely floating or sinking objects; they include an additional force from a string, scale, or the bottom of the container, so you need to draw a full force diagram and solve for the unknown force. This is one of the most common FRQ contexts for buoyancy, so it is important to master the force balance approach.

Three common scenarios:

1. Dense object fully submerged, hanging from a string above the water: Forces are downward weight $W$, upward buoyant force $F_b$, upward tension $T$. Equilibrium: $F_b+T=W⟹T=W-F_b$. Tension here equals the apparent weight of the object, the force measured by a scale holding the object.
2. Dense object resting on the bottom of a fluid container: The normal force $N$ from the bottom replaces tension: $F_b+N=W⟹N=W-F_b$.
3. Low-density object fully submerged, held down by a string tied to the container bottom: Forces are upward buoyant force $F_b$, downward weight $W$, downward tension $T$. Equilibrium: $F_b=W+T⟹T=F_b-W$.

The key is always to draw the force diagram first to check the direction of the additional force.

Worked Example

A solid plastic sphere of mass $2.0\text{kg}$ and total volume $3.0\times 10^{-3}{\text{m}}^3$ is held fully submerged under water (${\rho }_f=1000{\text{kg/m}}^3$) by a string tied to the bottom of the container. What is the tension in the string?

1. Draw the force diagram: upward buoyant force from water, downward weight of the sphere, downward tension from the string holding it under the surface.
2. Write the equilibrium condition (upward forces = downward forces): $F_b=W+T$.
3. Calculate values for $W$ and $F_b$: $W=mg=(2.0\text{kg})(9.8{\text{m/s}}^2)=19.6\text{N}$; $F_b={\rho }_f{V}_{d}g=(1000{\text{kg/m}}^3)(3.0\times 10^{-3}{\text{m}}^3)(9.8{\text{m/s}}^2)=29.4\text{N}$.
4. Solve for tension: $T=F_b-W=29.4-19.6=9.8\text{N}$.

Exam tip: Never assume tension is always upward. The direction of tension depends on whether the object is trying to sink or rise; the force of tension always opposes the motion the object would have if the string were cut.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Uses the object's density instead of the fluid density in Archimedes' principle $F_b=\rho V_{d}g$. Why: Students default to the density of the object that is the focus of the question, since it is usually highlighted in the problem statement. Correct move: Always label ${\rho }_f$ (fluid) and ${\rho }_o$ (object) at the start of the problem, and confirm you use ${\rho }_f$ for the buoyant force calculation.
• Wrong move: For a floating object, uses the full object volume as $V_d$ instead of only the submerged volume. Why: Students practice mostly with fully submerged problems, so they assume $V_d=V_o$ for all cases. Correct move: For any floating object, immediately write $V_d<V_o$ on your paper, and use the equilibrium relation $V_d=V_o({\rho }_o/{\rho }_f)$ to find the correct displaced volume.
• Wrong move: Writes $W=F_b+T$ for a low-density object held fully submerged under water, when tension should be downward. Why: Students are used to dense objects hanging from strings above water, so they assume tension is always upward. Correct move: Draw a quick 10-second sketch of the scenario to confirm tension direction before writing the equilibrium equation.
• Wrong move: When asked for the apparent weight of a submerged object, reports only the buoyant force instead of $W-F_b$. Why: Confusion between buoyant force (the upward force from the fluid) and apparent weight (the net downward force the object exerts on a scale). Correct move: Remember that apparent weight of a submerged object is always $|W-F_b|$, the net force after accounting for buoyancy.
• Wrong move: Ignores the buoyant force from air when solving problems involving balloons or low-density objects in air. Why: Students associate buoyancy only with water, so they forget air is a fluid that also exerts a buoyant force. Correct move: If the problem involves an object in air, especially a balloon or low-density object, always add the buoyant force from air to your force balance.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A solid wooden block floats in a glass of salt water. Over time, some of the water evaporates, so the density of the remaining salt water decreases. How does the fraction of the block submerged in the water change after evaporation, and why? (A) The submerged fraction increases, because the buoyant force on the block decreases (B) The submerged fraction increases, because the buoyant force remains equal to the weight of the block (C) The submerged fraction decreases, because the buoyant force on the block decreases (D) The submerged fraction decreases, because the buoyant force remains equal to the weight of the block

Worked Solution: The wooden block remains in static floating equilibrium after evaporation, so the buoyant force must always equal the constant weight of the block. This eliminates options A and C, which incorrectly claim the buoyant force changes. We use the floating object relation $\frac{V_d}{V_o}=\frac{{\rho }_o}{{\rho }_f}$. The density of the block ${\rho }_o$ is constant, so as the fluid density ${\rho }_f$ decreases, the submerged fraction $\frac{V_d}{V_o}$ increases. The correct answer is B.

Question 2 (Free Response)

A spherical hot air balloon has a total empty mass (fabric + basket + passengers) of $280\text{kg}$. The balloon is inflated with hot air to a total volume of $1200{\text{m}}^3$. The density of the surrounding cold air is ${\rho }_{cold}=1.20{\text{kg/m}}^3$, and the density of the hot air inside the balloon is ${\rho }_{hot}=0.95{\text{kg/m}}^3$. (a) Calculate the total weight of the fully inflated balloon (including the hot air inside). (b) Calculate the buoyant force on the balloon from the surrounding cold air. (c) Calculate the net upward acceleration of the balloon just after it is released.

Worked Solution: (a) First calculate the mass of the hot air inside the balloon: $m_{hot}={\rho }_{hot}V=(0.95{\text{kg/m}}^3)(1200{\text{m}}^3)=1140\text{kg}$. Total mass of the inflated balloon is $m_{total}=280\text{kg}+1140\text{kg}=1420\text{kg}$. Total weight is $W=m_{total}g=(1420\text{kg})(9.8{\text{m/s}}^2)=13916\text{N}\approx 1.39\times 10^4\text{N}$.

(b) By Archimedes' principle, buoyant force equals the weight of displaced cold air: $F_b={\rho }_{cold}Vg=(1.20{\text{kg/m}}^3)(1200{\text{m}}^3)(9.8{\text{m/s}}^2)=14112\text{N}\approx 1.41\times 10^4\text{N}$.

(c) Use Newton's second law $F_{net}=m_{total}a$. Net upward force is $F_{net}=F_b-W=14112-13916=196\text{N}$. Solve for acceleration: $a=\frac{F_{net}}{m_{total}}=\frac{196}{1420}\approx 0.14{\text{m/s}}^2$ upward.

Question 3 (Application / Real-World Style)

A marine biologist is designing a neutrally buoyant float to carry an underwater sensor, so the float stays suspended at any depth. The float has a rigid hollow plastic shell with total volume $2.5\times 10^{-4}{\text{m}}^3$ and mass $0.20\text{kg}$. What mass of lead must be added to the float to make it neutrally buoyant in salt water (density ${\rho }_{salt}=1025{\text{kg/m}}^3$)? Lead has density ${\rho }_{lead}=11300{\text{kg/m}}^3$, and all added lead will be fully submerged.

Worked Solution: Let $m_l$ = mass of added lead, so volume of lead is $V_l=\frac{m_l}{{\rho }_{lead}}$. Neutral buoyancy requires $F_b=W_{total}$, so ${\rho }_{salt}(V_{shell}+V_l)g=(m_{shell}+m_l)g$. Cancel $g$ from both sides: $$1025\left(2.5\times 10^{-4}+\frac{m_l}{11300}\right)=0.20+m_l$$ Expand the left side: $0.25625+0.0907m_l=0.20+m_l$. Rearrange to solve for $m_l$: $0.05625=0.9093m_l⟹m_l\approx 0.062\text{kg}=62\text{g}$.

In context, adding 62 grams of lead to the float makes it neutrally buoyant, so it will not rise or sink when placed in salt water, which keeps the sensor at a fixed depth for consistent data collection.

7. Quick Reference Cheatsheet

8. What's Next

Buoyancy is a core application of hydrostatic pressure, the foundation of all AP Physics 2 Unit 1 Fluids. Immediately after mastering buoyancy, you will move on to fluid flow, the continuity equation, and Bernoulli's principle, where you will need the same density, volume, and force balance skills you practiced here. Buoyancy also connects to broader concepts across the course, including hydrostatic pressure in gases, and thermal expansion of fluids that changes buoyancy for problems like climate-driven sea level rise from melting ice. Without mastering the force balance and Archimedes' principle skills from this chapter, you will struggle with any fluid problem involving immersed objects, which make up a large fraction of Unit 1 exam points.

Hydrostatic Pressure Fluid Continuity Equation Bernoulli's Principle