---
title: "Unit 8.2 | Buoyancy and Archimedes’ Principle"
description: "Learn buoyancy for AP Physics 1: Archimedes' principle F_B = ρVg, floating fractions, apparent weight, comparing buoyant forces, with practice."
featured_image: "https://nerd-notes.com/wp-content/uploads/2026/10/133453.jpg"
url: "https://nerd-notes.com/unit-8-2-buoyancy-and-archimedes-principle/"
date_modified: "2026-10-31T09:35:27+00:00"
---

# Unit 8.2 | Buoyancy and Archimedes’ Principle

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Why do ships float and rocks feel lighter underwater? The buoyant force, F_B = ρVg, and four problem types that cover every buoyancy question.

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### Unit 8 Breakdown

You are on Lesson 2 of 4

1. [Unit 8.1 | Density and Pressure](https://nerd-notes.com/unit-8-1-density-and-pressure/)
2. [Unit 8.2 | Buoyancy and Archimedes’ Principle](https://nerd-notes.com/unit-8-2-buoyancy-and-archimedes-principle/)
3. [Unit 8.3 | Fluid Flow and the Continuity Equation](https://nerd-notes.com/unit-8-3-fluid-flow-and-the-continuity-equation/)
4. [Unit 8.4 | Bernoulli’s Equation](https://nerd-notes.com/unit-8-4-bernoullis-equation/)

#### In this lesson: 

- Where the buoyant force comes from: the fluid pushes harder on the deeper side
- Archimedes’ principle: [katex] F_B = \rho_f V_{sub}\, g [/katex], the weight of the displaced fluid
- Float or sink: compare densities, and find how much of a floating object sits underwater
- The 4 problem types: floating objects, weighing underwater, comparing buoyant forces, and held, released and lifted

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### Introduction

This lesson covers **buoyancy** and **Archimedes’ principle**: the upward push a fluid gives anything placed in it, and how to use it with Newton’s laws.

In [Unit 8.1](https://nerd-notes.com/unit-8-1-density-and-pressure/) you saw that pressure grows with depth. Put an object in a fluid and its bottom is deeper than its top, so the fluid pushes up harder than it pushes down. That difference is the buoyant force. After that, every problem is a free-body diagram from Unit 2 with one extra force.

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### Where the Buoyant Force Comes From

\(P_{top}\)

\(P_{bottom}\)

\(F_B = P_{bottom}A - P_{top}A\)

\(= \rho_f g\,(\Delta h)\,A\)

\(F_B = \rho_f V g\)

*The side pushes cancel. The bottom face is deeper than the top face, so the upward push wins.*

Take a block of height [katex] \Delta h [/katex] and face area [katex] A [/katex], fully under water:

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1. The fluid pushes down on the top face with [katex] P_{top}A [/katex] and up on the bottom face with [katex] P_{bottom}A [/katex]. The pushes on the sides cancel.
2. The pressure difference between the faces is [katex] P_{bottom} – P_{top} = \rho_f g\,\Delta h [/katex] (Unit 8.1).
3. So the net upward force is [katex] F_B = \rho_f g\,\Delta h\,A = \rho_f V g [/katex], because [katex] \Delta h\,A [/katex] is the block’s volume.

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### Archimedes’ Principle

[katex] F_B = \rho_f V_{sub}\, g [/katex]

The buoyant force equals the weight of the fluid the object pushes out of the way. [katex] \rho_f [/katex] is the density of the **fluid**, not the object, and [katex] V_{sub} [/katex] is the volume of the object that is **under** the surface. For a fully submerged object, [katex] V_{sub} [/katex] is its whole volume.

▶ **Watch:** Archimedes’ principle, worked from scratch (starts at 9:01).

[Embedded content](https://www.youtube-nocookie.com/embed/uubzy_lNCV8?start=541&rel=0)

The buoyant force does not depend on the object’s mass, its material or how deep it is (once it is fully under). An aluminum ball and a lead ball of the same size feel the same buoyant force.

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### Float or Sink?

A fully submerged object of density [katex] \rho_o [/katex] has weight [katex] \rho_o V g [/katex] and buoyant force [katex] \rho_f V g [/katex]. Compare them and only the densities are left:

| **Densities** | **What happens** |
| --- | --- |
| [katex] \rho_o > \rho_f [/katex] | Sinks: the weight beats the largest possible buoyant force |
| [katex] \rho_o = \rho_f [/katex] | Neutrally buoyant: stays at any depth |
| [katex] \rho_o < \rho_f [/katex] | Rises, then floats with only part of it under |

A steel ship floats because its *average* density, steel plus all the air inside the hull, is less than water’s.

Every buoyancy question on the AP exam is one of the four types below. In each one, start with a free-body diagram.

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| **Problem type** | **The clue** | **The setup** |
| --- | --- | --- |
| 1. Floating objects | “floats”, “what fraction is submerged” | [katex] F_B = mg [/katex], so [katex] \frac{V_{sub}}{V} = \frac{\rho_o}{\rho_f} [/katex] |
| 2. Weighing underwater | a spring scale, string tension, “apparent weight” | [katex] T + F_B = mg [/katex] |
| 3. Comparing buoyant forces | two objects or two fluids, “which has the greater buoyant force” | Compare [katex] \rho_f [/katex] and [katex] V_{sub} [/katex] only |
| 4. Held, released and lifted | pushed under, let go, balloons, a scale under the container | [katex] \sum F = ma [/katex] with [katex] F_B [/katex] as one force |

Fluid Statics Part 2 – Archimedes Principle

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### Problem Type 1: Floating objects

**How to spot it:** Something “floats”, “floats at rest”, or the question asks what fraction or how much of it is below the surface.

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**Setup:** Floating at rest means [katex] F_B = mg [/katex], so [katex] \rho_f V_{sub}\,g = \rho_o V g [/katex]. The fraction under water is [katex] \frac{V_{sub}}{V} = \frac{\rho_o}{\rho_f} [/katex].

\(mg\)

\(F_B\)

\(V_{sub}\)

\(F_B = mg\)

\(\rho_f V_{sub}\, g = \rho_o V g\)

\(V_{sub}/V = \rho_o/\rho_f\)

*A floating object sinks until the water it displaces weighs as much as it does.*

A wooden block [katex] 0.40\,\text{m} [/katex] × [katex] 0.40\,\text{m} [/katex] × [katex] 0.20\,\text{m} [/katex] (density [katex] 600\,\text{kg/m}^3 [/katex]) floats flat in water. How deep does its bottom sit, and how much mass can you stack on top before the block goes completely under?

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Fraction under: [katex] \frac{600}{1000} = 0.60 [/katex], so [katex] (0.60)(0.20) = 0.12 \text{ m} [/katex] of its height is under water. The block has [katex] V = 0.032 \text{ m}^3 [/katex] and [katex] m = (600)(0.032) = 19.2 \text{ kg} [/katex]. Fully under, the buoyant force can hold [katex] \rho_f V = (1000)(0.032) = 32 \text{ kg} [/katex], so you can add [katex] 32 – 19.2 = 12.8 \text{ kg} [/katex].

▶ **Watch:** floating and sinking from the forces (starts at 15:13).

[Embedded content](https://www.youtube-nocookie.com/embed/uubzy_lNCV8?start=913&rel=0)

**AP trap:** Using the object’s density in [katex] F_B = \rho V g [/katex], or using the whole volume of a floating object. The buoyant force uses the fluid’s density and only the submerged volume.

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Question 16

A solid plastic cube with uniform density (side length = 0.5 m) of mass 100 kg is placed in a vat of fluid whose density is 1200 kg/m³. What fraction…

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Question 17

Balsa wood with an average density of 130 kg/m³ is floating in pure water. What percentage of the wood is submerged?

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Question 18

A cube of unknown material and uniform density floats in a container of water with 60 % of its volume submerged. If this same cube were placed in a…

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### Problem Type 2: Weighing underwater: tension and apparent weight

**How to spot it:** An object hangs from a spring scale or string while under water: “the scale reads”, “apparent weight”, “tension in the cord”.

**Setup:** Free-body diagram: [katex] T [/katex] and [katex] F_B [/katex] up, [katex] mg [/katex] down. At rest, [katex] T + F_B = mg [/katex], so the scale reads [katex] T = mg – \rho_f V g [/katex]. The drop in the reading is the buoyant force.

\(T\)

\(F_B\)

\(mg\)

\(T + F_B = mg\)

\(T = mg - \rho_f V g\)

*The water holds up part of the weight, so the scale reads less than \(mg\).*

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A [katex] 2.0\,\text{kg} [/katex] rock with a volume of [katex] 8.0 \times 10^{-4} \text{ m}^3 [/katex] hangs from a spring scale, completely under water. What does the scale read?

Weight: [katex] (2.0)(9.8) = 19.6 \text{ N} [/katex]. Buoyant force: [katex] F_B = (1000)(8.0 \times 10^{-4})(9.8) \approx 7.8 \text{ N} [/katex]. Scale: [katex] T = 19.6 – 7.8 \approx 11.8 \text{ N} [/katex]. The rock “weighs” 40% less under water.

Lowering the rock into the water

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*If the rock is a straight-sided block 10 cm tall, the reading drops as more of it goes under, then stays at 11.8 N once it is fully submerged: deeper does not mean more buoyant force.*

**AP trap:** Adding the buoyant force to the weight, or forgetting it. The scale reading is the weight minus the buoyant force.

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Question 19

A student measures the weight of an irregularly shaped metal object in air and finds it to be 15 N. When the object is fully submerged in water, its…

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Question 20

An object with a density of 4000 kg/m³ is weighed in air and then weighed while fully submerged in a fluid with a density of 1000 kg/m³. What is the…

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Question 21

A sphere of mass M and volume V is suspended from a spring scale and completely submerged in a fluid of density ρ_f. What is the reading on the…

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Question 22

A piece of aluminum with a mass of 1.0 kg and density of 2700 kg/m³ is suspended from a light cord and then completely immersed in a container of…

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### Problem Type 3: Comparing buoyant forces

**How to spot it:** Two objects, two fluids, or two depths: “which buoyant force is greater”, “how does the buoyant force change”, ratios.

**Setup:** Ignore what the object is made of. The buoyant force depends only on [katex] \rho_f [/katex] and [katex] V_{sub} [/katex]. If an object floats, its buoyant force equals its weight instead.

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Two balls have the same volume. A wooden one floats with 60% of its volume under water. A steel one rests on the bottom of the pool. Which has the greater buoyant force, and by what factor?

The steel ball is fully under, so it displaces [katex] V [/katex]. The wooden ball displaces only [katex] 0.60V [/katex]. So the steel ball has the greater buoyant force, by [katex] \frac{1}{0.60} \approx 1.7 [/katex] times. It still sinks because its weight is far greater.

**AP trap:** Thinking a floating object has more buoyant force than a sinking one. Floating only means the buoyant force matches the weight. A sunken object often has the larger buoyant force; it just has an even larger weight.

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Question 23

Two identical spheres are held under water. Sphere A is made of aluminum (density 2700 kg/m³) and Sphere B is made of lead (density 11300 kg/m³)…

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Question 24

A cube is fully submerged at a depth of 1 m in a large tank of water. It is then moved to a depth of 3 m. How does the buoyant force on the cube at 3…

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Question 25

Cubes 1 and 2 are made of the same material. Cube 1 has edges of length L and is fully submerged in a fluid of density 0.5 ρ. Cube 2 has edges of…

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Question 26

A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is F₀. The water is then poured out and…

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### Problem Type 4: Held, released and lifted: Newton’s laws with buoyancy

**How to spot it:** An object is pushed under and released, a balloon lifts a load, something sinks or rises with an acceleration, or a scale sits under the container.

**Setup:** Draw the [free-body diagram](https://nerd-notes.com/unit-2-3-drawing-fbds-right/) with [katex] F_B [/katex] as one more force and use [katex] \sum F = ma [/katex]. Newton’s third law: the object pushes down on the fluid with a force equal to [katex] F_B [/katex].

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A [katex] 0.50\,\text{kg} [/katex] ball with a volume of [katex] 2.0 \times 10^{-3} \text{ m}^3 [/katex] is held completely under water. How hard must you push to hold it there, and what is its acceleration just after you let go? (Ignore drag.)

Buoyant force: [katex] (1000)(2.0 \times 10^{-3})(9.8) = 19.6 \text{ N} [/katex]. Weight: [katex] (0.50)(9.8) = 4.9 \text{ N} [/katex]. To hold it: [katex] F_{push} = 19.6 – 4.9 = 14.7 \text{ N} [/katex] down. Let go and that push is gone: [katex] a = \frac{19.6 – 4.9}{0.50} \approx 29 \text{ m/s}^2 [/katex] upward.

▶ **Watch:** the forces framework this type uses, step by step.

[Embedded content](https://www.youtube.com/embed/SIQZyz-cZQs?feature=oembed)

A beaker of water sits on a scale and a hanging object is lowered into it. The water pushes up on the object with [katex] F_B [/katex], so the object pushes down on the water with [katex] F_B [/katex], and the scale reading goes up by exactly [katex] F_B [/katex].

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**AP trap:** Keeping the pushing force after the object is released, or leaving out the third-law pair. The force on the water is equal and opposite to the buoyant force, not to the object’s weight (unless it floats).

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Question 27

In the lab, a student is given a glass beaker filled with water with an ice cube of mass m and volume V_c floating in it. The downward force of…

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Question 28

A block of weight W is floating in water, and one-third of the block is above the surface of the water. Which of the following correctly describes…

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Question 29

A student places a wooden block of mass m in a container of water. The block floats with half of its volume above the surface of the water. The…

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Question 30

A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the…

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Applying Buoyancy in Real Scenarios

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Buoyancy

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### Lesson 8.3 Preview

So far the fluid has been at rest. In the [next lesson](https://nerd-notes.com/unit-8-3-fluid-flow-and-the-continuity-equation/) it starts to flow, and one rule, the continuity equation, explains why water speeds up when you squeeze the end of a hose. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-8-2-buoyancy-and-archimedes-principle/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-8-2-buoyancy-and-archimedes-principle/
