# Buoyancy and Archimedes' Principle

> AP Physics 1 · Fluids and Thermal Physics
> Source: https://www.owlsprep.com/study/ap-physics-1-u8-buoyancy-and-archimedes-principle/

This module covers Archimedes' Principle, buoyant force calculation, static equilibrium of floating and submerged objects, and apparent weight analysis, aligned to AP Physics 1 exam expectations.

**Prerequisites:** [Density definition and calculation](https://www.owlsprep.com/study/ap-physics-1-u8-density/); [Static force equilibrium and Newton's first law](https://www.owlsprep.com/study/ap-physics-1-u8-overview/)

## Learning objectives

- Define buoyant force and state Archimedes' Principle
- Calculate buoyant force for fully and partially submerged objects
- Solve static equilibrium problems for floating and submerged objects
- Calculate apparent weight and use it to find object density
- Avoid common exam pitfalls on buoyancy questions

## Core Concepts of Buoyancy

Buoyancy is the net upward force exerted by a static fluid on any object immersed partially or fully in it. It arises because hydrostatic pressure increases with depth: upward pressure on the object's bottom surface is greater than downward pressure on the top surface, creating the net upward force called *buoyant force* ($F_b$).

**Archimedes' Principle** — Any object immersed in a static fluid experiences a buoyant force equal in magnitude to the weight of the fluid displaced by the object.

*Notation:* F_b

**Derivation:** Prove Archimedes' Principle for static fluids

*Starting from:* Static equilibrium of a fluid volume matching the displaced volume of an object

1. Replace the immersed object with an identical volume of the fluid itself. This fluid volume is at rest, so net force on it is zero.
2. The only forces acting on the fluid volume are its own weight (downward) and the net upward force from the surrounding fluid (buoyant force).
3. Force balance gives:
4. $$F_b = W_{displaced}$$
5. Replacing the fluid volume with the original object does not change the force exerted by the surrounding fluid, so the relationship holds.

*Conclusion:* Archimedes' Principle is valid for any object immersed in a static fluid.

> **Exam tip:** Buoyancy is tested conceptually as often as numerically on the AP exam. Be prepared to explain why buoyant force does not change with depth for fully submerged objects.

## Buoyant Force Calculations

From Archimedes' Principle, the core formula for buoyant force is:

$$F_b = \rho_f V_d g$$

Where $\rho_f$ = fluid density, $V_d$ = volume of displaced fluid, and $g$ = acceleration due to gravity. For fully submerged objects, $V_d$ equals the total volume of the object $V_o$. For partially submerged objects, $V_d$ is only the volume of the object below the fluid surface.

**Worked example:** A solid rectangular iron block with total volume $2.5 \times 10^{-3} \text{ m}^3$ is fully submerged in fresh water of density $1000 \text{ kg/m}^3$. Calculate the magnitude of the buoyant force on the block.

1. Confirm the object is fully submerged, so displaced volume equals total object volume:
2. $$V_d = V_o = 2.5 \times 10^{-3} \text{ m}^3$$
3. Write Archimedes' Principle formula:
4. $$F_b = \rho_f V_d g$$
5. Substitute known values ($g = 9.8 \text{ m/s}^2$):
6. $$F_b = (1000 \text{ kg/m}^3)(2.5 \times 10^{-3} \text{ m}^3)(9.8 \text{ m/s}^2)$$
7. Calculate the final result:
8. $$F_b = 24.5 \approx 25 \text{ N}$$
9. Note that the density of iron is not required here, because buoyant force only depends on fluid properties and displaced volume.

> **Exam tip:** Always label $\rho_f$ (fluid density) and $\rho_o$ (object density) at the start of every problem to avoid mixing the two up.

*Calculator:* allowed

## Equilibrium of Floating Objects

A floating object at rest is in static equilibrium, so net vertical force is zero. The only vertical forces are the downward weight of the object $W_o$ and upward buoyant force $F_b$, so force balance gives:

$$F_b = W_o$$

Substituting the formulas for $F_b$ and $W_o$ and canceling $g$ from both sides gives the useful relationship for floating objects:

$$\frac{V_d}{V_o} = \frac{\rho_o}{\rho_f}$$

**Neutral Buoyancy** — Condition when an object's average density equals the fluid density, so the object floats fully submerged at rest at any depth.

*Example:* A scuba diver adjusting their buoyancy to hover at a fixed depth.

**Worked example:** A piece of pine wood has a density of $500 \text{ kg/m}^3$, and floats in pure ethanol of density $789 \text{ kg/m}^3$. What fraction of the wood's volume is above the surface of the ethanol?

1. For a floating object at equilibrium, $\frac{V_d}{V_o} = \frac{\rho_o}{\rho_f}$ applies. Substitute the given densities:
2. $$\frac{V_d}{V_o} = \frac{500}{789} \approx 0.634$$
3. The fraction of volume above the surface is total fraction minus submerged fraction:
4. $$1 - \frac{V_d}{V_o} = 1 - 0.634 = 0.366 \approx 0.37$$
5. Approximately 37% of the wood's volume is above the ethanol surface.

> **Exam tip:** AP questions almost always ask for the fraction of volume *above* the fluid surface, not the submerged fraction. Double-check which quantity the question asks for.

*Calculator:* allowed

## Apparent Weight of Submerged Objects

When an object is suspended and held at rest fully submerged in a fluid, its apparent weight (the force required to support it) is less than its actual weight, because the upward buoyant force counteracts part of the object's weight. For static equilibrium:

$$W_{app} + F_b = W_o \implies W_{app} = W_o - F_b$$

For a fully submerged object, this can be rewritten in terms of densities as:

$$W_{app} = W_o \left(1 - \frac{\rho_f}{\rho_o}\right)$$

This relationship is commonly used to find the density of irregular solid objects by measuring weight in air and apparent weight in a fluid of known density, a common lab-based AP exam problem.

**Worked example:** A solid irregular brass object is weighed in air and found to have a weight of 32 N. When fully submerged in water of density $1000 \text{ kg/m}^3$, its apparent weight is 28 N. What is the density of brass?

1. Rearrange the apparent weight formula to solve for buoyant force:
2. $$F_b = W_o - W_{app} = 32 - 28 = 4 \text{ N}$$
3. For full submersion, $F_b = \rho_f V_o g$, so solve for object volume:
4. $$V_o = \frac{F_b}{\rho_f g} = \frac{4}{(1000)(9.8)} \approx 4.08 \times 10^{-4} \text{ m}^3$$
5. Use actual weight to solve for object density:
6. $$\rho_o = \frac{W_o}{V_o g} = \frac{32}{(4.08 \times 10^{-4})(9.8)} \approx 8000 \text{ kg/m}^3$$
7. This matches the accepted density of brass, as expected.

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. A solid wooden block floats in a beaker of pure fresh water. The beaker is then drained and refilled with ethyl alcohol, which has a lower density than fresh water. The block still floats in the alcohol. What happens to the buoyant force on the block, and what happens to the volume of the block submerged below the fluid surface, compared to fresh water?

   - A) Buoyant force increases, submerged volume increases
   - B) Buoyant force stays the same, submerged volume decreases
   - C) Buoyant force stays the same, submerged volume increases
   - D) Buoyant force decreases, submerged volume decreases

   *Why:* Since the block remains floating, equilibrium requires buoyant force always equals the constant weight of the block. From $V_d = \frac{\rho_o V_o}{\rho_f}$, lower fluid density means larger submerged volume, so C is correct.

> **Exam tip:** If a problem gives you the tension in a string holding a submerged object, that tension equals $W_{app}$ — use it directly in your force balance.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the object's density instead of the fluid's density in the buoyant force formula
  - Why it fails: Students often mix up which density is relevant, since object density determines whether an object sinks or floats
  - Correct: Always explicitly label each variable and write $F_b = \rho_{\text{fluid}} V_{\text{displaced}} g$ at the start of every problem
- **Wrong:** Assuming $V_d = V_o$ for all floating objects
  - Why it fails: Students forget only neutral buoyancy (fully submerged floating) has $V_d = V_o$, not surface floating
  - Correct: Always start with $F_b = W_o$ for floating objects to relate $V_d$ and $V_o$, never assume $V_d = V_o$ unless the object is explicitly fully submerged
- **Wrong:** Claiming buoyant force increases as a fully submerged object moves deeper into the fluid
  - Why it fails: Students confuse increasing pressure with depth with increasing net pressure difference, which does not change for a fixed volume
  - Correct: For incompressible objects and fluids (all AP Physics 1 cases), $V_d$ is constant, so $F_b$ is constant regardless of depth
- **Wrong:** Flipping the density ratio to get $\frac{V_d}{V_o} = \frac{\rho_f}{\rho_o}$ for floating objects
  - Why it fails: Students memorize the ratio instead of deriving it from force balance
  - Correct: Always start from first principles $F_b = W_o$ and cancel terms step-by-step to get the ratio, rather than relying on memory
- **Wrong:** Adding buoyant force to the object's weight when calculating apparent weight
  - Why it fails: Students forget buoyant force acts upward, opposite to weight
  - Correct: Always draw a free-body diagram of the object to confirm force directions before writing the force balance equation

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Buoyant Force (Archimedes' Principle) | $F_b = \rho_f V_d g$ | $\rho_f$ = fluid density, $V_d$ = displaced fluid volume |
| Fully Submerged Buoyant Force | $F_b = \rho_f V_o g$ | $V_o$ = total object volume |
| Floating Object Equilibrium | $F_b = W_o = m_o g$ | Only applies to static floating objects at rest |
| Floating Submerged Fraction | $\frac{V_d}{V_o} = \frac{\rho_o}{\rho_f}$ | Gives fraction of volume below fluid surface |
| Apparent Weight (Submerged) | $W_{app} = W_o - F_b$ | $W_{app}$ equals tension supporting the object |
| Apparent Weight (Density Form) | $W_{app} = W_o \left(1 - \frac{\rho_f}{\rho_o}\right)$ | Only valid for fully submerged objects |
| Neutral Buoyancy Condition | $\rho_o = \rho_f, \; W_o = F_b$ | Object floats fully submerged at any depth |

## What's next

Buoyancy and Archimedes' Principle is the core foundation for all fluid mechanics topics in AP Physics 1. Mastery of this concept is required for multi-concept free-response questions that combine force analysis with fluid properties, and it is regularly tested in conceptual multiple-choice questions. Next, you will build on this understanding to study hydrostatic pressure, then move to fluid flow with the continuity equation and Bernoulli's principle. Buoyancy also connects to core course concepts like static force equilibrium, which is tested throughout the AP Physics 1 exam. A solid grasp of Archimedes' Principle will help you avoid common pitfalls on fluid questions and earn maximum points on exam day.

- [Unit 8 Fluids Overview](https://www.owlsprep.com/study/ap-physics-1-u8-overview/)
- [Fluid Continuity Equation](https://www.owlsprep.com/study/ap-physics-1-u8-fluid-continuity-equation/)
- [Bernoulli's Principle](https://www.owlsprep.com/study/ap-physics-1-u8-bernoulli-s-principle/)

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