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ESO Y4B · Physics and Chemistry
Interaction · Forces and pressure in fluids

Archimedes' principle and upthrust

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 on your body, and it is provided by the liquid itself in which you are submerged.

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

Upthrust: the force exerted by the liquid

If you push a beach ball down with your hand to get it under the water in a swimming pool, you feel a strong vertical force resisting you, and it gets bigger the more volume you submerge: what exactly produces that upward force, and what does its value depend on?

Force diagram of a block completely submerged in a liquid: the upthrust E acts upwards, the weight acts downwards, and the pressure is greater on the lower face than on the upper face. liquid body E (upthrust) weight lower pressure higher pressure
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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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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
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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

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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 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 upthrust takes away weight

Two newton meters hold the same stone: in air they read 50 newtons, its real weight; with the stone submerged in water they read 38 newtons, because the 12-newton upthrust acting upwards takes away weight. 01020 304050 50 N newton meter P = 50 N 01020 304050 38 N water P = 50 N E = 12 N in air submerged in water P apparent = P real − E = 50 N − 12 N = 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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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
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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

The submerged fraction at equilibrium

If you place a cork stopper and a block of beech wood in a bucket of water, both float, but the cork barely gets its base wet while the wood sits almost entirely below the surface: what does the exact part of the body that is hidden depend on?

Body floating at rest in a liquid: the free surface separates the emerged volume, V(out), from the submerged volume, V(sub). V(out) V(sub) free surface
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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 upthrust it receives? 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 litres.
Solve
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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: $d_{hielo}\cdot V\cdot g = d_{agua}\cdot V_{sum}\cdot g$. 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
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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

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
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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
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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
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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

Back to the start: In the swimming pool you notice that you weigh less and can easily lift another person because what you are holding is their apparent weight: the upward upthrust ($E = d_{\text{agua}} \cdot V \cdot g$) counteracts almost $100\,\%$ of their real weight, since the human body has an average density practically identical to that of water.

Summary

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