Back to homePhysics · High School Year 3 · Physics-Math · Change subject
Start for free
Prep 3 Tech · Physics
Interaction · Forces and pressure in fluids

Archimedes' Principle: Buoyant Force

Before you start, make sure you know: Density (Secondary Year 2)
Before you start, make sure you know: Balanced forces (Secondary Year 2)

You get into the pool and notice that you weigh less: the water seems to hold you up. In a cenote or on a trajinera in Xochimilco the same thing happens: lifting a friend while in the water is easy, something impossible outside of it. A new force is acting on your body, and it is applied by the very liquid you are submerged in.

Buoyant force: the force the liquid exerts

If you use your hand to push a beach ball under the water, you feel a vertical force pushing back, and it grows the more volume you submerge. What produces that upward force and what does its value depend on?

Force diagram on a block fully submerged in a liquid: the buoyant force E acts upward, the weight acts downward, and the pressure is greater on the bottom face than on the top face. liquid body E (buoyant force) weight lower pressure higher pressure

It is a contact force: it is exerted by the fluid touching the body. Before using it in calculations, it helps to know its value.

Practice★☆☆Not started
When you get into a swimming pool you feel lighter, as if the water were supporting you. What force appears when you submerge yourself? What exerts it and in which…

When you get into a swimming pool you feel lighter, as if the water were supporting you. What force appears when you submerge yourself? What exerts it and in which direction does it push?

Solve
Practice★☆☆Not started
A cube fully submerged in water displaces a volume of $0.002\ \text{m}^3$ (about $2\ \text{L}$), as shown in the figure. Calculate the upthrust it receives using…

A cube fully submerged in water displaces a volume of $0.002\ \text{m}^3$ (about $2\ \text{L}$), as shown in the figure.

d = 1 000 kg/m³ V = 0.002 m³ E = ?

Calculate the upthrust it receives using $E = d\times V\times g$. Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

Solve
Practice★☆☆Not started
Two identical bottles filled with sand are fully submerged in water, one at a depth of $1\ \text{m}$ and the other at $5\ \text{m}$, as shown in the figure. Does one…

Two identical bottles filled with sand are fully submerged in water, one at a depth of $1\ \text{m}$ and the other at $5\ \text{m}$, as shown in the figure.

d = 1 000 kg/m³ 1 m E = ? 5 m E = ?

Does one receive more upthrust than the other? Justify your answer by looking at the quantities on which upthrust depends.

Solve

Does it float or sink? Comparing densities

Practice★☆☆Not started
On the table you have a cork stopper ($d = 240\ \text{kg/m}^3$) and an iron nut ($d = 7\,870\ \text{kg/m}^3$). Without carrying out calculations, predict which one will…

On the table you have a cork stopper ($d = 240\ \text{kg/m}^3$) and an iron nut ($d = 7\,870\ \text{kg/m}^3$). Without carrying out calculations, predict which one will float and which will sink in water ($d = 1\,000\ \text{kg/m}^3$) and explain the criterion you used.

Solve
Practice★☆☆Not started
Ice cubes float in a glass of water. Knowing that the density of ice is about $900\ \text{kg/m}^3$, explain why they float and state the condition that the density of a…

Ice cubes float in a glass of water. Knowing that the density of ice is about $900\ \text{kg/m}^3$, explain why they float and state the condition that the density of a body must meet to float in a liquid.

Solve
Practice★★☆Not started
A block of $3\ \text{kg}$ has a volume of $0.004\ \text{m}^3$ and is fully submerged in water ($d = 1\,000\ \text{kg/m}^3$), as shown in the figure. Take…

A block of $3\ \text{kg}$ has a volume of $0.004\ \text{m}^3$ and is fully submerged in water ($d = 1\,000\ \text{kg/m}^3$), as shown in the figure. Take $g = 10\ \text{m/s}^2$.

d = 1 000 kg/m³ P = ? E = ? m = 3 kg V = 0.004 m³
  1. a) Calculate its weight.
  2. b) Calculate the upthrust it receives.
  3. c) Does it float or sink? State what net force acts on it and in which direction.
Solve
Practice★★☆Not started
A metal part of $2\ \text{kg}$ has a volume of $0.0005\ \text{m}^3$ ($0.5\ \text{L}$). Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$. a)…

A metal part of $2\ \text{kg}$ has a volume of $0.0005\ \text{m}^3$ ($0.5\ \text{L}$). Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

  1. a) Calculate its weight.
  2. b) Calculate the upthrust it receives if fully submerged in water.
  3. c) Does it float or sink? Also calculate its density and compare it with that of water to confirm your answer.
Solve

Apparent weight: when buoyant force takes away weight

Two spring scales hold the same stone: in the air they read 50 newtons, its real weight; with the stone submerged in water they read 38 newtons, because the 12-newton upward buoyant force takes away weight. 01020 304050 50 N P = 50 N 01020 304050 38 N water P = 50 N E = 12 N in the air submerged in water P apparent = P real − E = 50 N − 12 N = 38 N
Practice★★☆Not started
A stone of $5\ \text{kg}$ and a volume of $0.001\ \text{m}^3$ hangs from a force meter and is fully submerged in water, as shown in the figure. Data:…

A stone of $5\ \text{kg}$ and a volume of $0.001\ \text{m}^3$ hangs from a force meter and is fully submerged in water, as shown in the figure. Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 9.8\ \text{m/s}^2$.

? N d = 1 000 kg/m³ E = ? m = 5 kg V = 0.001 m³
  1. a) Calculate its true weight.
  2. b) Calculate the upthrust from the water.
  3. c) What reading does the force meter show with the stone submerged? Explain why this "apparent weight" is smaller.
Solve
Practice★★☆Not started
An object weighs $12\ \text{N}$ in air and, when fully submerged in water, the force meter reads $8\ \text{N}$. Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take…

An object weighs $12\ \text{N}$ in air and, when fully submerged in water, the force meter reads $8\ \text{N}$. Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

  1. a) What is the upthrust it receives?
  2. b) Calculate the volume of the object.
  3. c) Using its true weight and volume, calculate its density in $\text{kg/m}^3$.
Solve
Practice★★☆Not started
An egg of $60\ \text{g}$ has a volume of $55\ \text{cm}^3$. Take $g = 10\ \text{m/s}^2$. a) Calculate its weight in newtons (take care with the units). b) Calculate the…

An egg of $60\ \text{g}$ has a volume of $55\ \text{cm}^3$. Take $g = 10\ \text{m/s}^2$.

  1. a) Calculate its weight in newtons (take care with the units).
  2. b) Calculate the upthrust it receives when fully submerged in tap water ($d = 1\,000\ \text{kg/m}^3$) and deduce whether it floats or sinks.
  3. c) On dissolving a large amount of salt, the water reaches a density of $1\,200\ \text{kg/m}^3$: calculate the new upthrust and explain why the egg now floats.
Solve
Practice★★★Not started
A crown of $1\ \text{kg}$ should be made of pure gold ($d_{oro} = 19\,300\ \text{kg/m}^3$). You weigh it submerged in water and the newton meter reads $9.2\ \text{N}$.…

A crown of $1\ \text{kg}$ should be made of pure gold ($d_{oro} = 19\,300\ \text{kg/m}^3$). You weigh it submerged in water and the newton meter reads $9.2\ \text{N}$. Given data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

  1. a) What volume would the crown occupy if it were pure gold? Express it in $\text{cm}^3$.
  2. b) From the newton meter reading, calculate the actual upthrust and the actual volume of the crown.
  3. c) Calculate the actual density of the crown and determine whether it is pure gold or alloyed with a less dense metal, as Archimedes suspected.
Solve

Same density, another liquid: lake, sea, and salt water

The buoyant force depends on the density of the liquid. The same body receives more buoyant force in a denser liquid, which is why you float higher in the sea than in a pool, and in the Dead Sea it is almost impossible to sink.

Liquid Approximate density (kg/m³)
Fresh water 1,000
Seawater 1,025
Dead Sea 1,240
Air 1.2
Practice★☆☆Not started
A steel ship weighing thousands of metric tons floats, but a small steel ball sinks to the bottom. If both are made of the same material, explain what enables the ship…

A steel ship weighing thousands of metric tons floats, but a small steel ball sinks to the bottom. If both are made of the same material, explain what enables the ship to float (think about its hollow shape and average density).

Solve
Practice★★☆Not started
A log of $0.01\ \text{m}^3$ is fully submerged, first in a lake and then in the sea. Take $g = 10\ \text{m/s}^2$. a) Calculate the upthrust it receives in the lake (…

A log of $0.01\ \text{m}^3$ is fully submerged, first in a lake and then in the sea. Take $g = 10\ \text{m/s}^2$.

  1. a) Calculate the upthrust it receives in the lake ($d = 1\,000\ \text{kg/m}^3$).
  2. b) Calculate the upthrust it receives in the sea ($d = 1\,025\ \text{kg/m}^3$).
  3. c) Using these results, explain why swimmers float "higher" in the sea than in a swimming pool.
Solve

The submerged fraction at equilibrium

If you put a cork stopper and a block of wood in a bucket of water, both float, but the cork barely gets wet while the wood is almost entirely submerged. What determines how much stays under the water?

Body floating at rest in a liquid: the free surface separates the emerged volume, V outside, from the submerged volume, V sub. V outside V sub free surface
Practice★★☆Not started
A wooden block of $6\ \text{kg}$ floats in equilibrium on the water of a swimming pool ($d = 1\,000\ \text{kg/m}^3$), as shown in the figure. Take…

A wooden block of $6\ \text{kg}$ floats in equilibrium on the water of a swimming pool ($d = 1\,000\ \text{kg/m}^3$), as shown in the figure. Take $g = 10\ \text{m/s}^2$.

d = 1.000 kg/m³ m = 6 kg P = ? E = ? V(sum) = ?
  1. a) If it floats in equilibrium, what is the buoyant force it experiences? Justify your answer.
  2. b) Calculate the volume of water it displaces, that is, its submerged volume. Express it in $\text{m}^3$ and in liters.
Solve
Practice★★☆Not started
An ice cube ($d_{hielo} = 900\ \text{kg/m}^3$) floats in a glass of water ($d_{agua} = 1\,000\ \text{kg/m}^3$), as shown in the figure. a) In equilibrium, the weight of…

An ice cube ($d_{hielo} = 900\ \text{kg/m}^3$) floats in a glass of water ($d_{agua} = 1\,000\ \text{kg/m}^3$), as shown in the figure.

V(sum) / V = ? ice: d = 900 kg/m³ water: d = 1 000 kg/m³
  1. a) In equilibrium, the weight of the ice equals the upthrust: $d_{hielo}\times V\times g = d_{agua}\times V_{sum}\times g$. Simplify the expression and solve for the submerged fraction $\frac{V_{sum}}{V}$.
  2. b) Calculate what percentage of the ice cube remains under water and what percentage sticks out above the water.
  3. c) Relate the result to the phrase "this is just the tip of the iceberg".
Solve
Practice★★★Not started
A yacht of $4\,000\ \text{kg}$ floats in the sea ($d = 1\,025\ \text{kg/m}^3$). Take $g = 10\ \text{m/s}^2$. a) Calculate its weight and reason what upthrust it receives…

A yacht of $4\,000\ \text{kg}$ floats in the sea ($d = 1\,025\ \text{kg/m}^3$). Take $g = 10\ \text{m/s}^2$.

  1. a) Calculate its weight and reason what upthrust it receives when floating in equilibrium.
  2. b) Calculate the submerged volume of the hull.
  3. c) The yacht enters a freshwater river ($d = 1\,000\ \text{kg/m}^3$): calculate the new submerged volume and explain whether the boat floats higher or lower in the water.
Solve
Practice★★★Not started
An iceberg of $10\,000\ \text{m}^3$ floats in the sea. Data: $d_{hielo} = 900\ \text{kg/m}^3$, $d_{mar} = 1\,025\ \text{kg/m}^3$. a) Calculate what fraction of its…

An iceberg of $10\,000\ \text{m}^3$ floats in the sea. Data: $d_{hielo} = 900\ \text{kg/m}^3$, $d_{mar} = 1\,025\ \text{kg/m}^3$.

  1. a) Calculate what fraction of its volume remains submerged (remember: in equilibrium, $\frac{V_{sum}}{V} = \frac{d_{hielo}}{d_{mar}}$).
  2. b) What volume of the iceberg sticks out above the water?
  3. c) Using these results, explain why icebergs are so dangerous to shipping even though the visible part seems small.
Solve
Practice★★★Not started
A cork raft ($d = 240\ \text{kg/m}^3$) has a volume of $0.5\ \text{m}^3$ and floats in water ($d = 1\,000\ \text{kg/m}^3$). Take $g = 10\ \text{m/s}^2$. a) What fraction…

A cork raft ($d = 240\ \text{kg/m}^3$) has a volume of $0.5\ \text{m}^3$ and floats in water ($d = 1\,000\ \text{kg/m}^3$). Take $g = 10\ \text{m/s}^2$.

  1. a) What fraction of the raft is submerged when it floats on its own?
  2. b) Calculate the maximum upthrust that the raft can provide, i.e. when fully submerged.
  3. c) Calculate the weight of the raft itself and deduce the maximum mass it can carry on top without sinking completely.
Solve
Practice★★★Not started
A person of $70\ \text{kg}$ floats in equilibrium, first in a swimming pool ($d = 1\,000\ \text{kg/m}^3$) and then in the Dead Sea ($d = 1\,240\ \text{kg/m}^3$). a)…

A person of $70\ \text{kg}$ floats in equilibrium, first in a swimming pool ($d = 1\,000\ \text{kg/m}^3$) and then in the Dead Sea ($d = 1\,240\ \text{kg/m}^3$).

  1. a) Write down the equilibrium condition $P = E$ using the person's mass and the submerged volume, and verify that simplifying gives $V_{sum} = \frac{m}{d}$: you do not even need to know $g$.
  2. b) Calculate the submerged volume in the swimming pool and express it in liters.
  3. c) Calculate the submerged volume in the Dead Sea and explain, by comparing both results, why people float "like a cork" there and it is almost impossible to sink.
Solve

Ships, submarines, and balloons: average density

Cross section of a hollow steel ship floating on water thanks to the air inside it, next to a solid steel block that sinks to the bottom. air steel hull water solid block of steel air inside the hull: low average density → floats solid steel: denser than water → sinks

A hot-air balloon, like the ones that fly over Teotihuacán, works the same way: when the air inside is heated, it expands and its density drops. The weight of the air it contains decreases and the buoyant force from the cold outside air ($E = d_{\text{aire}}\, V g$) ends up lifting it. The force available to lift the envelope, the basket, and the passengers is the difference between the buoyant force and the weight of the hot air. It stops rising because the air is less dense at greater altitude and the buoyant force decreases.

Practice★☆☆Not started
A hot-air balloon rises when the burner heats the air inside it. Explain what happens to the density of the air when heated and why the upthrust from the cool outside…

A hot-air balloon rises when the burner heats the air inside it. Explain what happens to the density of the air when heated and why the upthrust from the cool outside air ends up lifting the balloon.

Solve
Practice★★★Not started
A submarine of $800\ \text{t}$ ($800\,000\ \text{kg}$) travels fully submerged, in equilibrium, in the sea ($d = 1\,025\ \text{kg/m}^3$), as shown in the figure. Take…

A submarine of $800\ \text{t}$ ($800\,000\ \text{kg}$) travels fully submerged, in equilibrium, in the sea ($d = 1\,025\ \text{kg/m}^3$), as shown in the figure. Take $g = 10\ \text{m/s}^2$.

m = 800 000 kg sea: d = 1 025 kg/m³ ballast tanks E = ? P = ?
  1. a) What is the upthrust it receives? Justify your answer using the equilibrium condition.
  2. b) Calculate the volume of water it displaces, i.e. the volume of the submarine.
  3. c) To surface, the submarine expels sea water from its ballast tanks and replaces it with air: explain which quantity changes, whether weight or upthrust, and why this causes it to rise.
Solve
Practice★★★Not started
A hot-air balloon has a volume of $2\,000\ \text{m}^3$, as shown in the figure. The outside air has a density of $1.2\ \text{kg/m}^3$ and the hot air inside has a…

A hot-air balloon has a volume of $2\,000\ \text{m}^3$, as shown in the figure. The outside air has a density of $1.2\ \text{kg/m}^3$ and the hot air inside has a density of $1.0\ \text{kg/m}^3$. Take $g = 10\ \text{m/s}^2$.

V = 2 000 m³ d(int) = 1.0 kg/m³ m = ? E = ? P = ? d(ext) = 1.2 kg/m³
  1. a) Calculate the upthrust exerted by the outside air on the balloon.
  2. b) Calculate the weight of the hot air filling the balloon.
  3. c) The difference between the two is the force available to lift the envelope, basket, and passengers: what is the maximum combined mass they can have for the balloon to ascend?
Solve
Practice★★★Not started
A party balloon contains $5\ \text{L}$ of helium ($0.005\ \text{m}^3$). Given data: $d_{aire} = 1.2\ \text{kg/m}^3$, $d_{helio} = 0.18\ \text{kg/m}^3$; take…

A party balloon contains $5\ \text{L}$ of helium ($0.005\ \text{m}^3$). Given data: $d_{aire} = 1.2\ \text{kg/m}^3$, $d_{helio} = 0.18\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

  1. a) Calculate the upthrust exerted by the air on the balloon.
  2. b) Calculate the weight of the helium filling it.
  3. c) The balloon's envelope has a mass of $3\ \text{g}$: what maximum extra mass (ribbon, decorations) can it carry to keep ascending?
  4. d) Explain why the balloon stops rising once it reaches a certain height (think about how the density of air changes with altitude).
Solve

Back to the beginning: in the pool you weigh less because the water pushes you upward with $E = d_{\text{agua}}\, V\, g$. Your body has an average density very close to that of water, so the buoyant force offsets almost all of your weight, and what you hold up of another person is only their apparent weight.

Summary

See all the exercises in this section