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Interaction · Forces and pressure in fluids

Archimedes' principle and upthrust

You wade into the sea at Salthill in Galway and notice that you weigh less: the water seems to hold you up. You can easily lift a friend when they are in the water, something impossible on the beach. A new force is acting on your body, and it is provided by the liquid itself in which you are submerged.

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 in Cork, 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 U acts upwards, the weight acts downwards, and the pressure is greater on the lower face than on the upper face. liquid body U (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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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 $E = d\times V\times g$. 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
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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

Does it float or sink? Comparing densities

Floating in equilibrium

When a body floats at rest, it is in equilibrium: its weight and the upthrust are equal in size, $W = U$. The upthrust comes only from the part of the body that is below the surface, so $U = \rho_{\text{liquid}} \cdot V_{\text{submerged}} \cdot g$. This is the same idea as Newton's first law: no net force, so no acceleration.

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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
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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
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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

Apparent weight: when upthrust takes away weight

Two newton balances 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 balance W = 50 N 01020 304050 38 N water W = 50 N U = 12 N in air submerged in water W apparent = W − U = 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 oak 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?

This section builds on the floating equilibrium $W = U$ from the previous section.

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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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
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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
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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
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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 litres.
  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
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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 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 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
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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 sea at Salthill you notice that you weigh less and can easily lift a friend because what you are holding is their apparent weight: the upthrust ($U = \rho_{\text{water}} \cdot V \cdot g$) counteracts most of their real weight, since the human body has an average density only slightly less than that of water.

Summary

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