Unit 8.2 | Buoyancy and Archimedes’ Principle

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.

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].

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?

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).

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.

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.

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.

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

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.

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.

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 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].

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.

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].

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).

Lesson 8.3 Preview

So far the fluid has been at rest. In the next lesson 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/