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ESO Y4A · Physics and Chemistry
Interaction · Forces and pressure in fluids

Archimedes' principle and buoyant force

You get into the swimming pool and notice that you weigh less: the water seems to hold you up. You can easily lift another person when they are in the water, something impossible outside it. A new force is acting, and it is provided by the liquid itself.

Before you start, make sure you know: Density (2nd ESO)
Before you start, make sure you know: Equilibrium of forces (2nd ESO)

Buoyant force: the force exerted by the liquid

If you push a ball down towards the bottom of a swimming pool, you feel a clear force pushing your hands upwards and, when you let go, it shoots up to the surface: where does this force come from and what quantities does it depend on?

Body immersed in a liquid subject to two opposite vertical forces: the buoyant force E upwards, applied at the centre of the body, and the weight downwards. liquid body E (buoyant force) weight
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State Archimedes' principle in your own words: what upthrust is, in what direction and sense it acts, and what its magnitude is equal to.

State Archimedes' principle in your own words: what upthrust is, in what direction and sense it acts, and what its magnitude is equal to.

Solve
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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
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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 $E = d\cdot V\cdot g$. Data: $d_{agua} = 1\,000\ \text{kg/m}^3$; take $g = 10\ \text{m/s}^2$.

Solve

Does it float or sink? Comparing densities

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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
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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
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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

Apparent weight: when the buoyant force reduces weight

A stone hanging from a newton meter: in air it reads its real weight, 50 N; submerged in water it receives an upward buoyant force and the newton meter reads less, 38 N, which is the apparent weight. 50 N P newton meter stone 38 N water P E in air: reads P = 50 N submerged: reads P − E = 38 N
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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
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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

Ships, submarines and balloons: average density

Cross-section of a floating ship: the steel hull, very dense, encloses a lot of air, so the average density of the ship is lower than that of seawater and the ship floats with part of the hull submerged below the waterline. air inside the hull steel hull ρ ≈ 7 850 kg/m³ waterline seawater ρ ≈ 1 025 kg/m³ average density of the ship (steel + air) lower than that of water → the ship floats
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A steel ship weighing thousands of tonnes floats, but a small steel ball sinks to the bottom. If both are made of the same material, explain what enables the ship to…

A steel ship weighing thousands of tonnes 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
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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
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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

Back to the start: in the swimming pool you can lift another person because you only support their apparent weight ($P_{\text{aparente}} = P_{\text{real}} - E$); the buoyant force of the water compensates for almost all of their real weight.

Summary

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