Physics · Secondary School · Hydrostatics

Hydrostatics

A steel ship that floats, the ache in our ears at the bottom of a swimming pool, juice rising up a straw and the brakes of a car all depend on a handful of ideas about liquids and gases at rest. We start with density and pressure, see how pressure grows with depth and arrive at buoyancy, which explains why some things float and others sink. The calculations of the lesson use \(g = 10\ \text{m/s}^2\).

1Density
2Pressure
3Stevin's law
4Atmospheric pressure
5Communicating vessels
6Pascal's principle
7Buoyancy
✓Challenges

STEP 1

Why does ice float and iron sink?

The density of a body is its mass divided by the volume it takes up. A kilogram of lead and a kilogram of cotton wool have the same mass, and the lead fits into a volume many times smaller because it is much denser.

When the body is solid and made of a single material, its density coincides with the specific mass of that material. For a hollow body we divide the mass by the total volume, air included, and so a hollow steel ball can have a density well below that of steel.

The volume of an irregular stone can be measured by lowering it into a graduated container and seeing how far the water level rises. A body floats in water when it is less dense than water, and that is the case of ice, which takes up more room than the water it formed from.

\(d = \dfrac{m}{V}\)\(1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3\)\(1\ \text{L} = 1000\ \text{cm}^3\)Typical values, in g/cm³: wood 0.6 · oil 0.9 · ice 0.92 · water 1.0 · aluminium 2.7 · steel 7.8 · lead 11.3 · gold 19.3. A litre of water has a mass of 1 kg, and a cubic metre, a tonne.

Let's discuss

Choose ice and lower the block into the beaker. How far does the level rise, and does that number seem to be the volume of the whole block?
Now push the ice down with the stick until it is fully under water, and see what changes in the beaker reading when the top part also goes in.
Keep the material and double the edge of the cube. Our first intuition may say that the density changes too, so check what happens to the mass, the volume and the density.
Choose steel and increase the hollow part until the block starts to float. What might that fraction have to do with a steel ship?

Edge of the cube

Hollow part

Mass

Volume

Density

In kg/m³

Beaker reading

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A glass block has a mass of 200 g and a volume of 80 cm³. What is its density, and does it sink or float in water, whose density is \(1.0\ \text{g/cm}^3\)?

Show solution
The density is the mass divided by the volume, \(d = \dfrac{m}{V} = \dfrac{200}{80} = 2.5\ \text{g/cm}^3\).
Since \(2.5 > 1.0\), the glass is denser than water and sinks.
\(d = 2.5\ \text{g/cm}^3\), and the block sinks.
basic
Alcohol has a density of \(0.8\ \text{g/cm}^3\). How much is that in \(\text{kg/m}^3\)?

Show solution
A cubic metre has \(10^6\ \text{cm}^3\) and a kilogram has 1000 g, and so \(1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3\).
We just multiply by a thousand, \(0.8 \cdot 1000 = 800\ \text{kg/m}^3\).
\(800\ \text{kg/m}^3\)
basic
Ice floats in a glass of water. What does that tell us about the density of ice, and what happens to the volume of water when it freezes?

Show solution
A body floats in a liquid when it is less dense than the liquid.
Ice has a density of about \(0.92\ \text{g/cm}^3\), lower than that of liquid water, \(1.0\ \text{g/cm}^3\).
The mass does not change when water freezes, and since \(d = m/V\) goes down, the volume has to go up.
Ice is less dense than water, and water expands when it freezes.
basic
What is the mass of 2 L of cooking oil, whose density is \(0.9\ \text{g/cm}^3\)? Remember that \(1\ \text{L} = 1000\ \text{cm}^3\).

Show solution
The volume is \(V = 2\ \text{L} = 2000\ \text{cm}^3\).
Solving \(d = m/V\) for the mass, we get \(m = d \cdot V = 0.9 \cdot 2000 = 1800\ \text{g}\).
\(m = 1800\ \text{g} = 1.8\ \text{kg}\)
intermediate
A measuring cylinder holds 50 mL of water. When we drop a 54 g stone into it, the level rises to 70 mL. What is the density of the stone?

Show solution
The volume of the stone is the volume of water it displaces, \(V = 70 - 50 = 20\ \text{mL} = 20\ \text{cm}^3\).
With the mass measured on the balance, \(d = \dfrac{54}{20} = 2.7\ \text{g/cm}^3\).
\(d = 2.7\ \text{g/cm}^3\)
intermediate
A hollow steel ball has a mass of 780 g and an external volume of 200 cm³, and the specific mass of steel is \(7.8\ \text{g/cm}^3\). What is the density of the ball, and what is the volume of the hollow part?

Show solution
The density of the body uses the total volume, air included, \(d = \dfrac{780}{200} = 3.9\ \text{g/cm}^3\).
The steel alone takes up \(V_{\text{steel}} = \dfrac{780}{7.8} = 100\ \text{cm}^3\).
What is left is the hollow part, \(200 - 100 = 100\ \text{cm}^3\).
\(d = 3.9\ \text{g/cm}^3\), half the specific mass of steel, and the hollow part is \(100\ \text{cm}^3\).
intermediate
We often hear that ‘a kilo of lead weighs more than a kilo of cotton wool’. What is wrong with the sentence, and what is the real difference between the two?

Show solution
The masses are equal, 1 kg, and so are the weights, \(W = m\,g = 10\ \text{N}\) with \(g = 10\ \text{m/s}^2\), if we neglect the buoyancy of the air (which we will see in step 7).
The difference lies in the density, since lead, at about \(11.3\ \text{g/cm}^3\), takes up a much smaller volume than cotton wool.
Both weigh the same, and the lead, being denser, takes up less volume.
intermediate
We mix 300 cm³ of water (\(1.0\ \text{g/cm}^3\)) with 200 cm³ of sugar syrup (\(1.3\ \text{g/cm}^3\)). Assuming the volumes simply add up, what is the density of the mixture?

Show solution
The total mass is \(300 \cdot 1.0 + 200 \cdot 1.3 = 300 + 260 = 560\ \text{g}\).
With the volumes added, \(V = 500\ \text{cm}^3\), and \(d = \dfrac{560}{500} = 1.12\ \text{g/cm}^3\).
\(d = 1.12\ \text{g/cm}^3\)
challenge
A 1930 g crown takes up 120 cm³. Gold has \(19.3\ \text{g/cm}^3\) and silver, \(10.5\ \text{g/cm}^3\). Is the crown pure gold? If it is an alloy of gold and silver, how much silver does it contain?

Show solution
The density of the crown is \(d = \dfrac{1930}{120} \approx 16.1\ \text{g/cm}^3\), lower than that of gold, and so it cannot be pure gold.
Calling the mass of silver \(x\), the volumes of the two parts add up to 120, \(\dfrac{1930 - x}{19.3} + \dfrac{x}{10.5} = 120\).
This gives \(100 - \dfrac{x}{19.3} + \dfrac{x}{10.5} = 120\), that is, \(x\left(\dfrac{1}{10.5} - \dfrac{1}{19.3}\right) = 20\).
Hence \(x \approx \dfrac{20}{0.0434} \approx 461\ \text{g}\).
The crown is not pure gold and contains about 461 g of silver (and 1469 g of gold).
challenge
A glass bottle with 1 L of water goes into the freezer, and the water turns into ice of density \(0.92\ \text{g/cm}^3\). What is the volume of the ice, and why can full bottles burst in the freezer?

Show solution
The mass of water is \(m = 1.0 \cdot 1000 = 1000\ \text{g}\), and it is conserved on freezing.
The volume of the ice is \(V = \dfrac{1000}{0.92} \approx 1087\ \text{cm}^3\).
The volume grows by about \(87\ \text{cm}^3\), almost 9%, and the ice pushes on the walls of the bottle, which have nowhere to give.
\(V \approx 1087\ \text{cm}^3\), and this expansion of almost 9% can break the glass.

STEP 2

The same force, different areas

Pressure measures how a force is spread over a surface. It is the perpendicular force divided by the area on which it acts, so that the same force produces a large pressure on a small area and a small pressure on a large area.

A sharp knife cuts better than a blunt one for this reason. On a drawing pin, the force we barely feel on the wide head becomes, at the fine point, a pressure able to pierce wood, while snowshoes do the opposite and spread the weight of the walker over a large area.

In the International System we measure pressure in pascals, a small unit, and in everyday life we also meet the atmosphere (atm) and the millimetre of mercury (mmHg), which we will understand better in step 4.

\(p = \dfrac{F}{A}\)\(1\ \text{Pa} = 1\ \text{N/m}^2\)\(1\ \text{atm} \approx 1.0 \cdot 10^5\ \text{Pa}\)\(1\ \text{atm} = 760\ \text{mmHg}\)\(1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\). In the simulation we use \(g = 10\ \text{m/s}^2\) and assume a very simple snow, in which the foot sinks 1 cm for every 1 kPa of pressure until it reaches firm ground. Real snow tends to be far less regular.

Let's discuss

With the same person, swap the trainers for snowshoes and compare how many times the area grows with how many times the pressure drops.
With ‘On one foot’ switched on, the weight stays the same and the area halves. What do you expect to happen to the pressure, before you check?
Choose the stiletto heel and try to explain how a person so much lighter than an elephant can press harder on the ground than the elephant does.
Double the mass without changing the footwear. The pressure doubles too, and does the sinking in our snow model follow?

Mass of the person

Weight

Contact area

Pressure

In atm

Sinking

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A box exerts a force of 600 N on an area of \(0.03\ \text{m}^2\) of the floor. What is the pressure?

Show solution
The pressure is the force divided by the area, \(p = \dfrac{F}{A} = \dfrac{600}{0.03} = 20\,000\ \text{Pa}\).
\(p = 2.0 \cdot 10^4\ \text{Pa}\)
basic
Why does a sharp knife cut better than a blunt one, if the force of the hand is the same?

Show solution
The pressure is the force divided by the area, \(p = F/A\).
On a sharp edge, the area in contact with the food is very small, and the same force produces a much greater pressure.
With a smaller area the pressure is greater, and it is the pressure that breaks the material.
basic
A bicycle tyre is pumped to 3 atm. How much is that in pascals? And a pressure of 380 mmHg, how much is it in atm? Use \(1\ \text{atm} \approx 1.0 \cdot 10^5\ \text{Pa} = 760\ \text{mmHg}\).

Show solution
In pascals, \(3 \cdot 1.0 \cdot 10^5 = 3.0 \cdot 10^5\ \text{Pa}\).
In atmospheres, \(\dfrac{380}{760} = 0.5\ \text{atm}\).
\(3.0 \cdot 10^5\ \text{Pa}\) and \(0.5\ \text{atm}\)
basic
A 2 kg brick measures 20 cm × 10 cm × 5 cm and lies on its largest face. What pressure does it exert on the table? Use \(g = 10\ \text{m/s}^2\).

Show solution
The weight of the brick is \(W = m\,g = 2 \cdot 10 = 20\ \text{N}\).
The largest face has \(A = 0.20 \cdot 0.10 = 0.02\ \text{m}^2\).
The pressure is \(p = \dfrac{20}{0.02} = 1000\ \text{Pa}\).
\(p = 1000\ \text{Pa}\)
intermediate
The same brick as in the previous exercise (2 kg, 20 cm × 10 cm × 5 cm) is stood on end, resting on its smallest face. What is the pressure now, and how many times greater is it than on the largest face? Use \(g = 10\ \text{m/s}^2\).

Show solution
The smallest face has \(A = 0.10 \cdot 0.05 = 0.005\ \text{m}^2\).
With the same weight, \(p = \dfrac{20}{0.005} = 4000\ \text{Pa}\).
The ratio \(\dfrac{4000}{1000} = 4\) is the same as the ratio between the areas, \(0.02 / 0.005\).
\(p = 4000\ \text{Pa}\), four times greater, because the area became four times smaller.
intermediate
A 60 kg person rests for a moment on the heel of a single shoe, with an area of \(1\ \text{cm}^2\). Compare with the pressure when she wears trainers and stands on one foot, with an area of \(150\ \text{cm}^2\). Use \(g = 10\ \text{m/s}^2\).

Show solution
The force is the weight, \(F = 60 \cdot 10 = 600\ \text{N}\), and \(1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\).
On the heel, \(p = \dfrac{600}{10^{-4}} = 6.0 \cdot 10^6\ \text{Pa}\), about 60 atm.
In trainers, \(p = \dfrac{600}{0.015} = 4.0 \cdot 10^4\ \text{Pa}\).
The ratio between the two is \(\dfrac{6.0 \cdot 10^6}{4.0 \cdot 10^4} = 150\).
\(6.0 \cdot 10^6\ \text{Pa}\) on the heel and \(4.0 \cdot 10^4\ \text{Pa}\) in trainers, 150 times less.
intermediate
We push a drawing pin with 20 N. The head has \(1\ \text{cm}^2\) and the point, \(0.1\ \text{mm}^2\). Neglecting the mass of the pin, what is the pressure on the finger and what is the pressure on the wall?

Show solution
With negligible mass, the pin passes the same 20 N force to both ends.
On the finger, \(A = 10^{-4}\ \text{m}^2\) and \(p = \dfrac{20}{10^{-4}} = 2.0 \cdot 10^5\ \text{Pa}\).
At the point, \(0.1\ \text{mm}^2 = 10^{-7}\ \text{m}^2\) and \(p = \dfrac{20}{10^{-7}} = 2.0 \cdot 10^8\ \text{Pa}\).
\(2.0 \cdot 10^5\ \text{Pa}\) on the finger and \(2.0 \cdot 10^8\ \text{Pa}\) on the wall, a thousand times more.
intermediate
A 70 kg person walks on snow in boots, with a total contact area of \(0.05\ \text{m}^2\). With snowshoes, the area becomes \(0.35\ \text{m}^2\). Work out both pressures. Use \(g = 10\ \text{m/s}^2\).

Show solution
The weight is \(F = 70 \cdot 10 = 700\ \text{N}\).
In boots, \(p = \dfrac{700}{0.05} = 14\,000\ \text{Pa}\).
On snowshoes, \(p = \dfrac{700}{0.35} = 2000\ \text{Pa}\).
14,000 Pa in boots and 2000 Pa on snowshoes, seven times less, and so the person sinks less.
challenge
A 4000 kg elephant stands with its four feet on the ground, each with \(0.1\ \text{m}^2\). A 50 kg ballerina stands on the tip of one shoe, with \(5\ \text{cm}^2\) of contact. Who exerts more pressure on the ground? Use \(g = 10\ \text{m/s}^2\).

Show solution
The elephant weighs 40,000 N, spread over \(4 \cdot 0.1 = 0.4\ \text{m}^2\), which gives \(p = \dfrac{40\,000}{0.4} = 1.0 \cdot 10^5\ \text{Pa}\).
The ballerina weighs 500 N over \(5 \cdot 10^{-4}\ \text{m}^2\), and \(p = \dfrac{500}{5 \cdot 10^{-4}} = 1.0 \cdot 10^6\ \text{Pa}\).
The ballerina, with \(1.0 \cdot 10^6\ \text{Pa}\), ten times the pressure of the elephant.
challenge
An 80 kg person lies on a bed of nails, each with a point of \(1\ \text{mm}^2\). Suppose the skin can take up to \(5 \cdot 10^5\ \text{Pa}\) without injury and the weight is shared equally among the nails. What is the minimum number of nails in contact with the body? Use \(g = 10\ \text{m/s}^2\).

Show solution
The weight is \(F = 800\ \text{N}\), and the minimum contact area is \(A = \dfrac{F}{p_{\text{max}}} = \dfrac{800}{5 \cdot 10^5} = 1.6 \cdot 10^{-3}\ \text{m}^2\).
Each point has \(1\ \text{mm}^2 = 10^{-6}\ \text{m}^2\).
The number of nails is \(N = \dfrac{1.6 \cdot 10^{-3}}{10^{-6}} = 1600\).
At least 1600 nails.

STEP 3

The deeper, the greater the pressure

A point inside a liquid at rest holds up the weight of the column of liquid above it, as well as the pressure of the air on the surface. From this comes Stevin's law, the law of hydrostatic pressure, according to which pressure grows in a straight line with depth.

The pressure depends only on the depth and on the liquid. Vessels of very different shapes, filled with the same liquid to the same height, have the same pressure at the bottom, a result that tends to surprise and that we call the hydrostatic paradox.

\(p = p_0 + d\,g\,h\)\(\Delta p = d\,g\,\Delta h\)\(p_0\) is the pressure at the surface (atmospheric pressure, \(\approx 1.0 \cdot 10^5\ \text{Pa}\)), \(d\) in kg/m³ and \(h\) in metres. With \(d = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\), 10 m of water add \(10^5\ \text{Pa}\), about 1 atm. We take the liquid to be incompressible, a very good approximation for water.

In water, every 10 m of depth adds about 1 atm, and a diver at 20 m is already under about 3 atm. By the same law, the tap on the ground floor gets the water with more force than the one on the top floor, since it sits further below the level of the water tank.

Tyre gauges and many pressure gauges show only what the gas or liquid adds to the pressure of the air, which is why a flat tyre, full of air at atmospheric pressure, reads zero.

Let's discuss

Drag the gauge downwards and watch the graph. Does the pressure seem to grow by the same amount with every metre?
Take the gauge to 10 m in water and read the total pressure. How many atmospheres are there, and where does each of them come from?
Swap water for mercury without moving the gauge and see by how many times the \(d\,g\,h\) part increases. Does the number remind you of the ratio between the densities?
Look at the four vessels, which hold very different amounts of liquid. Intuition tends to say that the fullest one presses harder on its bottom, and what does the simulation show?
Switch off the addition of \(p_0\) and see what the gauge now shows. What happens to the line on the graph, and why does a tyre gauge work this way?

Depth of the gauge

Depth

d·g·h part

Total pressure

Total in atm

Per metre

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What is the pressure 5 m deep in a swimming pool? Take \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\), \(d_{\text{water}} = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
The column of water adds \(d\,g\,h = 1000 \cdot 10 \cdot 5 = 5.0 \cdot 10^4\ \text{Pa}\).
Adding the atmosphere, \(p = p_0 + d\,g\,h\) \(= 1.0 \cdot 10^5 + 0.5 \cdot 10^5\) \(= 1.5 \cdot 10^5\ \text{Pa}\).
\(p = 1.5 \cdot 10^5\ \text{Pa}\), about 1.5 atm.
basic
Three containers of different shapes (a cylinder, a funnel that widens upwards and a vase that narrows upwards) have the same base area and are filled with water to the same height. In which of them is the pressure at the bottom greatest?

Show solution
By Stevin's law, \(p = p_0 + d\,g\,h\) depends only on the liquid and the depth.
The shape of the container and the amount of water do not appear in the formula.
The pressure at the bottom is the same in all three, the so-called hydrostatic paradox.
basic
A diver goes down to 30 m in the sea. Using the rule that every 10 m of water adds about 1 atm, what is the total pressure on the diver?

Show solution
The column of water adds about \(\dfrac{30}{10} = 3\ \text{atm}\).
With the atmosphere above the surface, \(p \approx 1 + 3 = 4\ \text{atm}\).
\(p \approx 4\ \text{atm}\)
basic
Two points in a lake are 2 m and 6 m deep. What is the pressure difference between them? Use \(d = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
The difference depends only on the difference in height, \(\Delta p = d\,g\,\Delta h\) \(= 1000 \cdot 10 \cdot 4\) \(= 4.0 \cdot 10^4\ \text{Pa}\).
Atmospheric pressure appears at both points and cancels in the subtraction.
\(\Delta p = 4.0 \cdot 10^4\ \text{Pa}\)
intermediate
The water level in the tank of a block of flats is 8 m above the ground-floor tap and 3 m above the top-floor tap. What is the extra pressure, above atmospheric, at each tap? Use \(g = 10\ \text{m/s}^2\).

Show solution
On the ground floor, \(d\,g\,h = 1000 \cdot 10 \cdot 8 = 8.0 \cdot 10^4\ \text{Pa}\).
On the top floor, \(1000 \cdot 10 \cdot 3 = 3.0 \cdot 10^4\ \text{Pa}\).
That is why the water comes out with more force on the ground floor.
\(8.0 \cdot 10^4\ \text{Pa}\) on the ground floor and \(3.0 \cdot 10^4\ \text{Pa}\) on the top floor.
intermediate
In a tank of oil (\(d = 800\ \text{kg/m}^3\)), what is the total pressure 2 m down? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\) and \(g = 10\ \text{m/s}^2\).

Show solution
The oil adds \(d\,g\,h = 800 \cdot 10 \cdot 2 = 16\,000\ \text{Pa}\).
Adding the atmosphere, \(p = 100\,000 + 16\,000 = 116\,000\ \text{Pa}\).
\(p = 1.16 \cdot 10^5\ \text{Pa}\)
intermediate
Why are dams much thicker at the base than at the top?

Show solution
The pressure of the water grows with depth, \(p = p_0 + d\,g\,h\).
Near the bottom, each square metre of the wall takes a much greater force than near the surface.
The base has to withstand the greater pressure, which is at the bottom.
intermediate
A tank holds 2 m of water and, on top of it, 1 m of oil (\(d = 800\ \text{kg/m}^3\)). What is the pressure at the bottom? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\) and \(g = 10\ \text{m/s}^2\).

Show solution
Each layer adds its share to the pressure of the air.
The oil contributes \(800 \cdot 10 \cdot 1 = 8000\ \text{Pa}\) and the water, \(1000 \cdot 10 \cdot 2 = 20\,000\ \text{Pa}\).
At the bottom, \(p = 100\,000 + 8000 + 20\,000 = 128\,000\ \text{Pa}\).
\(p = 1.28 \cdot 10^5\ \text{Pa}\)
challenge
An aquarium has a base of \(0.5\ \text{m} \times 0.3\ \text{m}\) and 40 cm of water. Work out the force of the water on the bottom (not counting the atmosphere) and compare it with the weight of the water. Now imagine a vase with the same base and the same height of water, narrowing upwards. Does the force on the bottom change? Use \(g = 10\ \text{m/s}^2\).

Show solution
The pressure of the water on the bottom is \(p = d\,g\,h = 1000 \cdot 10 \cdot 0.4 = 4000\ \text{Pa}\).
With \(A = 0.5 \cdot 0.3 = 0.15\ \text{m}^2\), the force is \(F = p\,A = 4000 \cdot 0.15 = 600\ \text{N}\).
The water takes up \(V = 0.15 \cdot 0.4 = 0.06\ \text{m}^3\), has a mass of 60 kg and weighs 600 N, the same value.
In the narrowing vase, \(p\) and \(A\) do not change, and the force is still 600 N, even with less water. The sloping walls push the water down and make up the difference.
\(F = 600\ \text{N}\) in both cases.
challenge
A submarine is 200 m deep, with air at 1 atm inside. What is the resultant force of the pressure on a \(0.5\ \text{m}^2\) hatch? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\), \(d = 1000\ \text{kg/m}^3\) (neglecting the difference between fresh and salt water) and \(g = 10\ \text{m/s}^2\).

Show solution
Outside, \(p = 1.0 \cdot 10^5 + 1000 \cdot 10 \cdot 200 = 2.1 \cdot 10^6\ \text{Pa}\).
Inside there is \(1.0 \cdot 10^5\ \text{Pa}\), and the difference between the two sides is \(2.0 \cdot 10^6\ \text{Pa}\).
The resultant force is \(F = 2.0 \cdot 10^6 \cdot 0.5 = 1.0 \cdot 10^6\ \text{N}\).
\(F = 1.0 \cdot 10^6\ \text{N}\), the weight of about 100 tonnes.

STEP 4

We live at the bottom of an ocean of air

Air has weight too, and the layer of atmosphere above us exerts, at sea level, a pressure of about \(10^5\ \text{Pa}\). In 1643, Torricelli filled a one-metre glass tube with mercury, closed the end and turned the tube upside down in a dish that also held mercury.

The column dropped until it was some 76 cm tall, leaving an almost perfect vacuum at the top of the tube. The weight of the air on the dish balances the column, and that is why we say that 1 atm equals 76 cmHg, or 760 mmHg. With water, which is 13.6 times less dense, the column would need some 10 m.

The higher we climb, the less air is left above us and the lower the pressure, and in La Paz the mercury column is close to two thirds of what it measures on the coast. The drinking straw and the suction cup depend on atmospheric pressure, since in both cases we lower the pressure inside and it is the air outside that pushes.

\(p_{\text{atm}} = d_{\text{Hg}}\,g\,h\)\(1\ \text{atm} = 76\ \text{cmHg} \approx 1.0 \cdot 10^5\ \text{Pa}\)\(1\ \text{atm} \approx 10\ \text{m}\) of waterThe simulation uses the measured values (1 atm = 1.013·10⁵ Pa and the real g, 9.8 m/s²), which give 76.0 cm of mercury and 10.3 m of water; in the calculations, with \(g = 10\ \text{m/s}^2\), the numbers change little. For altitude, we assume an atmosphere at uniform temperature, in which the pressure halves roughly every 5.8 km.

Let's discuss

Swap mercury for water and watch the column grow. Why does it end up 13.6 times taller?
Tilt the tube little by little and follow the length of the column and its vertical height. What happens when the tube tilts too far?
Take the mercury barometer to La Paz and then to the top of Everest, and note how much the column drops in each case.
With water in the tube, what seems to be the greatest depth of a well from which a suction pump, installed at the top, can draw water?

Altitude

Tilt of the tube

Pressure

In kPa

In mmHg

Column (vertical)

Length in the tube

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
When we drink juice through a straw, what pushes the juice up?

Show solution
When we suck, we lower the pressure of the air inside the mouth and the straw.
The atmospheric pressure on the surface of the juice in the glass stays the same and is now greater than the pressure inside the straw.
It is the atmosphere that pushes the juice, and all we do is lower the pressure inside the straw.
basic
In La Paz, Bolivia, at an altitude of about 3600 m, a mercury barometer reads 49 cmHg. How much is that in atm? Use \(1\ \text{atm} = 76\ \text{cmHg}\).

Show solution
We only need to compare with the column at sea level, \(p = \dfrac{49}{76} \approx 0.64\ \text{atm}\).
\(p \approx 0.64\ \text{atm}\), about two thirds of the pressure at sea level.
basic
Had Torricelli used water instead of mercury, how tall would the column have been at sea level? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\), \(d = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
In equilibrium, the pressure of the column equals that of the atmosphere, \(d\,g\,h = p_0\).
Solving for the height, \(h = \dfrac{1.0 \cdot 10^5}{1000 \cdot 10} = 10\ \text{m}\).
\(h \approx 10\ \text{m}\), which would make the barometer rather impractical.
basic
How does a suction cup stay stuck to a smooth tile?

Show solution
When we press the suction cup, we push out part of the air that was between it and the tile.
The pressure in that space drops below atmospheric, and the air outside pushes the cup against the wall.
The difference between the pressure outside and the pressure inside holds the cup in place.
intermediate
Show that a 76 cm column of mercury (\(d = 13\,600\ \text{kg/m}^3\)) exerts about \(1 \cdot 10^5\ \text{Pa}\). Use \(g = 10\ \text{m/s}^2\).

Show solution
The pressure of the column is \(p = d\,g\,h = 13\,600 \cdot 10 \cdot 0.76\).
The calculation gives \(p = 103\,360\ \text{Pa} \approx 1.03 \cdot 10^5\ \text{Pa}\).
\(p \approx 1.0 \cdot 10^5\ \text{Pa}\), the atmospheric pressure at sea level.
intermediate
How tall would a barometer made with oil (\(d = 800\ \text{kg/m}^3\)) be at sea level? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\) and \(g = 10\ \text{m/s}^2\).

Show solution
As before, \(h = \dfrac{p_0}{d\,g} = \dfrac{1.0 \cdot 10^5}{800 \cdot 10} = 12.5\ \text{m}\).
\(h = 12.5\ \text{m}\), since the less dense the liquid, the taller the column.
intermediate
What force does the atmosphere exert on a \(1\ \text{m}^2\) table top? Why does the table not break? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\).

Show solution
The force is \(F = p_0\,A = 1.0 \cdot 10^5 \cdot 1 = 1.0 \cdot 10^5\ \text{N}\), the weight of some 10,000 kg with \(g = 10\ \text{m/s}^2\).
The air is also under the table top and pushes upwards with a force that is practically equal.
\(F = 1.0 \cdot 10^5\ \text{N}\), and the forces from above and below balance.
intermediate
In a mountain town, the mercury column of the barometer is 70 cm. What is the atmospheric pressure in pascals? Use \(d_{\text{Hg}} = 13\,600\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
The pressure of the air is that of the column it holds up, \(p = 13\,600 \cdot 10 \cdot 0.70 = 95\,200\ \text{Pa}\).
\(p \approx 9.5 \cdot 10^4\ \text{Pa}\), a little less than at sea level.
challenge
In 1654, Otto von Guericke put two metal hemispheres together and pumped the air out from inside them. Suppose a radius of 0.25 m and a perfect vacuum. What force is needed to pull them apart? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\) and \(\pi \approx 3.14\).

Show solution
The force that matters is the pressure times the area of the circle of contact, \(A = \pi r^2\) \(= 3.14 \cdot 0.0625 \approx 0.196\ \text{m}^2\).
Hence \(F = p_0\,A\) \(= 1.0 \cdot 10^5 \cdot 0.196 \approx 1.96 \cdot 10^4\ \text{N}\).
With \(g = 10\ \text{m/s}^2\), this is equivalent to the weight of about 2000 kg.
\(F \approx 2.0 \cdot 10^4\ \text{N}\), like lifting a large car.
challenge
A suction pump installed at the top of a well draws water by lowering the pressure in the pipe. What is the greatest depth from which it can draw water at sea level? And in a town where the pressure is 0.7 atm? Use \(1\ \text{atm} \approx 1.0 \cdot 10^5\ \text{Pa}\), \(d = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
Even with a perfect vacuum in the pipe, it is the atmosphere that pushes the water up, and the limit is \(d\,g\,h = p_0\).
At sea level, \(h = \dfrac{1.0 \cdot 10^5}{10^4} = 10\ \text{m}\).
With 0.7 atm, \(h = \dfrac{0.7 \cdot 10^5}{10^4} = 7\ \text{m}\).
10 m at sea level and 7 m in the high town. Deeper wells need a pump down at the bottom, which pushes the water up.

STEP 5

Liquids that communicate

When containers are joined at the base and hold the same liquid at rest, the free surface sits at the same height in all of them, whatever their shape. We call these containers communicating vessels, and the explanation comes from Stevin's law, because points of the same liquid at the same height have the same pressure.

The spout of a teapot and the builder's water level use this principle, and the water from the tank reaches the taps of a house for the same reason.

With two liquids that do not mix, such as water and oil in a U-tube, the levels no longer match. At the height of the interface, the pressures in the two arms are equal, which gives \(d_1\,h_1 = d_2\,h_2\), and the less dense liquid stands higher.

If we measure the two heights, this relation gives us the density of an unknown liquid from the density of water.

\(p_A = p_B\) (same liquid, same height)\(d_1\,h_1 = d_2\,h_2\)The heights \(h_1\) and \(h_2\) are measured from the level of the interface. Since \(g\) and \(p_0\) appear on both sides, they cancel, and we can use densities in g/cm³ and heights in cm.

Let's discuss

Before pouring anything, compare the levels in the two arms. What happens to them when oil is poured into the left arm?
Pour in 10 cm of paraffin and check, with the numbers from the simulation, whether the rule \(d_1\,h_1 = d_2\,h_2\) holds.
Switch to mercury and water. Why does a tall column of water seem to shift the mercury so little?
If you only knew the density of water, how could you use this tube to measure the density of an oil?

Poured column h₁

h₁ (light liquid)

h₂ (base liquid)

d₁·h₁

d₂·h₂

Level difference

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Why does the spout of a teapot (or of a watering can) need to reach the height of the rim of the container?

Show solution
The teapot and the spout form communicating vessels, and with the same liquid the level is the same in both.
With the spout lower than the rim, the water would pour out of the spout before the teapot was full.
Since the levels even out, a low spout would limit how much the teapot can hold.
basic
Builders use a clear hose full of water to mark two points at the same height on distant walls. Why does this work?

Show solution
The two ends of the hose are communicating vessels with the same liquid, open to the same atmosphere.
In equilibrium, the surface of the water sits at the same height at both ends, whatever path the hose takes.
The water level at the two ends is always the same, and so it marks the same height.
basic
In a U-tube with water, we pour oil (\(0.8\ \text{g/cm}^3\)) into one arm until it forms a column 10 cm above the interface. How high is the water, in the other arm, above that level?

Show solution
The points at the height of the interface are connected through the water and have the same pressure, which gives \(d_1\,h_1 = d_2\,h_2\).
With the numbers, \(0.8 \cdot 10 = 1.0 \cdot h_2\), and \(h_2 = 8\ \text{cm}\).
\(h_2 = 8\ \text{cm}\)
basic
In a U-tube with mercury (\(13.6\ \text{g/cm}^3\)), a 27.2 cm column of water balances how many centimetres of mercury above the interface?

Show solution
The U-tube rule is \(d_{\text{water}}\,h_{\text{water}} = d_{\text{Hg}}\,h_{\text{Hg}}\).
So \(1.0 \cdot 27.2 = 13.6 \cdot h\), and \(h = 2\ \text{cm}\).
\(h = 2\ \text{cm}\) of mercury
intermediate
In a U-tube with water, a 12 cm column of an unknown oil balances 9 cm of water above the interface. What is the density of the oil?

Show solution
Equal pressures give \(d_{\text{oil}} \cdot 12 = 1.0 \cdot 9\).
Solving for the density, \(d_{\text{oil}} = \dfrac{9}{12} = 0.75\ \text{g/cm}^3\).
\(d = 0.75\ \text{g/cm}^3\)
intermediate
In the U-tube with water and oil, the free surface of the oil stands higher than that of the water. Explain why.

Show solution
At the height of the interface, the pressures in the two arms are equal, \(d_1\,h_1 = d_2\,h_2\).
Since the oil is less dense, it needs a taller column to produce the same pressure.
The less dense liquid stands higher, because it needs more height to weigh the same per unit area.
intermediate
In a U-tube with water, we pour oil (\(0.9\ \text{g/cm}^3\)) until it forms a 20 cm column above the interface. What is the difference in level between the free surfaces of the oil and the water?

Show solution
The water above the interface has \(h_2\) given by \(0.9 \cdot 20 = 1.0 \cdot h_2\), that is, \(h_2 = 18\ \text{cm}\).
The difference in level is \(20 - 18 = 2\ \text{cm}\), with the oil higher.
A difference of 2 cm
intermediate
In the U-tube of exercise 3 (10 cm of oil of \(800\ \text{kg/m}^3\) on top of the water, balancing 8 cm of water), what is the pressure at the height of the interface? Use \(p_0 = 1.0 \cdot 10^5\ \text{Pa}\) and \(g = 10\ \text{m/s}^2\).

Show solution
Through the oil arm, \(p = p_0 + d\,g\,h\) \(= 100\,000 + 800 \cdot 10 \cdot 0.10\) \(= 100\,800\ \text{Pa}\).
Through the water arm, \(100\,000 + 1000 \cdot 10 \cdot 0.08 = 100\,800\ \text{Pa}\), the same value.
\(p = 100\,800\ \text{Pa}\) on both sides
challenge
A U-tube of constant cross-section holds mercury, at the same level in both arms. We pour 13.6 cm of water into one arm. How far does the mercury drop in that arm, and how far does it rise in the other?

Show solution
The water balances a difference in mercury level \(\Delta h\) such that \(1.0 \cdot 13.6 = 13.6 \cdot \Delta h\), and \(\Delta h = 1\ \text{cm}\).
Since the cross-section is constant, the mercury that drops on one side rises on the other by the same amount \(x\), and the difference in level is \(2x\).
Hence \(2x = 1\ \text{cm}\) and \(x = 0.5\ \text{cm}\).
The mercury drops 0.5 cm on one side and rises 0.5 cm on the other.
challenge
In a U-tube with mercury, we pour 27.2 cm of water into arm A. How much oil (\(0.8\ \text{g/cm}^3\)) do we need to pour into arm B for the mercury levels to be equal again?

Show solution
With the mercury level, the pressures on its surface are equal only if the two columns above have the same \(d\,h\).
So \(1.0 \cdot 27.2 = 0.8 \cdot h\), and \(h = 34\ \text{cm}\).
34 cm of oil

STEP 6

How to lift a car with one hand

A liquid hardly changes volume when we compress it, and so it transmits pressure. According to Pascal's principle, an increase in pressure at one point of a liquid in equilibrium reaches every other point and the walls of the container in full.

In the hydraulic press, we push a small-area piston and the same pressure reaches a large-area piston. Since \(F = p\,A\), the force on the large piston is multiplied by the ratio between the areas, and this is what happens in the car lift at the garage and in hydraulic brakes, where the foot on the pedal squeezes the pads of all four wheels.

\(\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}\)\(A_1\,d_1 = A_2\,d_2\)\(F_1\,d_1 = F_2\,d_2\)We assume the pistons are at the same level, frictionless and of negligible mass, and \(g = 10\ \text{m/s}^2\). In the simulation, the displacements of the two pistons are drawn to the same scale; the car is just an icon.

Nobody gets energy for free, though. The volume of liquid that leaves one cylinder enters the other, and the small piston has to go down much further than the large one goes up. With friction neglected, the work \(F\,d\) is the same on both sides.

In a hydraulic jack, a valve lets the oil through in one direction only. We can then pump several times, and each time the small piston rises it fills up again with oil from the reservoir.

Let's discuss

Drag the small piston down to the end of its stroke and compare how far it went down with how far the car went up. Does the ratio between the two distances remind you of the ratio between the areas?
Click ‘Pump’ several times and follow the work on both sides, which ought to be the same if energy is conserved.
Reduce the area of the small piston. What do we gain from this, and what do we lose?
Why can an air bubble in the brake fluid be dangerous? Think about what air does when it is compressed.

Area of small piston A₁

Area of large piston A₂

Mass of the car

Force on small piston

Weight of the car

Pressure in the oil

Total travel of small piston

Rise of the car

Work on each side

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
In a hydraulic press, the smaller piston has \(10\ \text{cm}^2\) and the larger, \(500\ \text{cm}^2\). We apply 100 N to the smaller one. What is the force on the larger one?

Show solution
The pressure is the same on both pistons, \(\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}\), and so \(F_2 = 100 \cdot \dfrac{500}{10}\).
The calculation gives \(F_2 = 5000\ \text{N}\).
\(F_2 = 5000\ \text{N}\), 50 times greater
basic
State Pascal's principle and give an everyday example.

Show solution
An increase in pressure applied at one point of a liquid in equilibrium is transmitted in full to every point of the liquid and to the walls of the container.
Hydraulic car brakes, the car lift at the garage and the dentist's chair use this property.
The increase in pressure is transmitted in full, as in hydraulic brakes.
basic
A hydraulic lift raises a 1200 kg car on a piston whose area is 60 times that of the piston on which we apply the force. What force do we need to apply? Use \(g = 10\ \text{m/s}^2\) and neglect the mass of the pistons.

Show solution
The large piston holds up the weight of the car, \(F_2 = 1200 \cdot 10 = 12\,000\ \text{N}\).
On the small piston, \(F_1 = F_2 \cdot \dfrac{A_1}{A_2} = \dfrac{12\,000}{60} = 200\ \text{N}\).
\(F_1 = 200\ \text{N}\)
basic
A classmate says that the hydraulic press ‘multiplies energy’, since we put in 100 N and get out 5000 N. Is she right?

Show solution
The force is multiplied, and the displacement is divided by the same factor.
Neglecting friction, the work \(F\,d\) done on the small piston equals the work done by the large piston.
The classmate is wrong, because energy is conserved, and we gain force at the cost of displacement.
intermediate
In the press of exercise 1 (\(10\ \text{cm}^2\) and \(500\ \text{cm}^2\)), the smaller piston goes down 25 cm. How far does the larger one go up?

Show solution
The volume of liquid that leaves one cylinder enters the other, \(A_1\,d_1 = A_2\,d_2\).
With the numbers, \(d_2 = \dfrac{10 \cdot 25}{500} = 0.5\ \text{cm}\).
\(d_2 = 0.5\ \text{cm}\)
intermediate
Still on the same press, work out the work done on the smaller piston (100 N over 25 cm) and the work done by the larger piston (5000 N over 0.5 cm). What can we conclude?

Show solution
On the smaller piston, \(W_1 = 100 \cdot 0.25 = 25\ \text{J}\).
On the larger one, \(W_2 = 5000 \cdot 0.005 = 25\ \text{J}\).
\(W_1 = W_2 = 25\ \text{J}\), and energy is conserved.
intermediate
The pistons of a press have radii of 2 cm and 10 cm. What force on the smaller piston balances 10,000 N on the larger one?

Show solution
The area grows with the square of the radius, \(\dfrac{A_2}{A_1} = \left(\dfrac{10}{2}\right)^2 = 25\).
Hence \(F_1 = \dfrac{10\,000}{25} = 400\ \text{N}\).
\(F_1 = 400\ \text{N}\)
intermediate
In a car's brakes, the pedal turns the 50 N of the foot into 200 N on the master cylinder, of area \(2\ \text{cm}^2\). The piston at each wheel has \(8\ \text{cm}^2\). What is the pressure in the fluid, and what is the force on each wheel piston?

Show solution
The pressure in the fluid is \(p = \dfrac{200}{2 \cdot 10^{-4}} = 1.0 \cdot 10^6\ \text{Pa}\).
It reaches every wheel equally, and at each one \(F = p\,A = 1.0 \cdot 10^6 \cdot 8 \cdot 10^{-4} = 800\ \text{N}\).
\(p = 1.0 \cdot 10^6\ \text{Pa}\) and \(F = 800\ \text{N}\) at each wheel.
challenge
A hydraulic lift has pistons of \(15\ \text{cm}^2\) and \(0.3\ \text{m}^2\) and must raise a 1500 kg car to a height of 1.8 m. What is the force on the smaller piston? How far would it have to go down in total? If each pump stroke moves the smaller piston 20 cm, how many strokes are needed? Use \(g = 10\ \text{m/s}^2\).

Show solution
The ratio between the areas is \(\dfrac{A_2}{A_1} = \dfrac{0.3}{0.0015} = 200\), and \(F_1 = \dfrac{15\,000}{200} = 75\ \text{N}\).
The displacement is divided by the same factor, and the smaller piston needs to go down \(d_1 = 200 \cdot 1.8 = 360\ \text{m}\) in total.
That makes \(N = \dfrac{360}{0.2} = 1800\) strokes.
Checking with energy, \(W = 75 \cdot 360 = 27\,000\ \text{J} = 15\,000 \cdot 1.8\).
\(F_1 = 75\ \text{N}\), 360 m of travel in total and 1800 strokes.
challenge
In a press filled with oil (\(d = 800\ \text{kg/m}^3\)), the smaller piston (\(20\ \text{cm}^2\)) holds up a 2 kg body, and the larger piston (\(0.2\ \text{m}^2\)) is 1 m below it. What mass on the larger piston is in equilibrium? And if the two pistons were at the same level? Neglect the mass of the pistons and use \(g = 10\ \text{m/s}^2\).

Show solution
Under the smaller piston, the pressure above atmospheric is \(p_1 = \dfrac{20}{0.002} = 10\,000\ \text{Pa}\).
One metre lower, Stevin's law adds \(d\,g\,h = 800 \cdot 10 \cdot 1 = 8000\ \text{Pa}\), and \(p_2 = 18\,000\ \text{Pa}\).
The force on the larger piston is \(F_2 = 18\,000 \cdot 0.2 = 3600\ \text{N}\), the weight of 360 kg.
At the same level, we would have \(F_2 = 10\,000 \cdot 0.2 = 2000\ \text{N}\), or 200 kg.
360 kg with the difference in level and 200 kg at the same level.

STEP 7

Why do some things float?

On a submerged body, the liquid pushes from all sides, and it pushes harder from below, where it is deeper, than from above. The resultant of these forces points upwards and is called the buoyant force, or upthrust.

Archimedes' principle tells us how large it is. The buoyant force equals the weight of the liquid displaced by the body, \(B = d_{\text{liq}}\,V_{\text{sub}}\,g\), and depends on the submerged volume, not on the material the body is made of.

\(B = d_{\text{liq}}\,V_{\text{sub}}\,g\)\(W_{\text{app}} = W - B\)\(\dfrac{V_{\text{sub}}}{V} = \dfrac{d_{\text{body}}}{d_{\text{liq}}}\) (floating)In the simulation, \(g = 10\ \text{m/s}^2\), the body is a cube (the egg is only drawn differently) and we assume the tank is so wide that the level hardly rises when the body goes in. The motion after release is shown in slow motion.

Hanging from a spring balance, a submerged body seems lighter, and the balance reads the apparent weight \(W - B\). Once released, it sinks when it is denser than the liquid. When it is less dense, it rises and floats with a submerged fraction equal to \(d_{\text{body}}/d_{\text{liq}}\).

This explains the steel ship, which is hollow and has an average density lower than that of water, and the egg, which sinks in fresh water and floats in salt water. The hydrometer uses the same idea, because it sinks deeper in less dense liquids.

Let's discuss

With the body hanging, lower it little by little and follow the reading on the spring balance. From what point does it stop changing, and why?
Choose the egg and release it first in fresh water and then in very salty water. What changed from one case to the other, the weight of the egg or the maximum buoyant force?
Release the wood and wait for the body to stop bobbing. Does the submerged fraction match the ratio \(d_{\text{body}}/d_{\text{liq}}\)?
Choose iron and mercury. What do you expect to see before releasing it, knowing that iron sinks in water without hesitation?
Increase the volume while keeping the density. Does the submerged fraction change, or does it seem to depend only on the ratio between the densities?

Density of the body

Volume of the body

Immersion (base below the surface)

Weight W

Buoyant force B

Spring balance (W − B)

Submerged fraction

d_body / d_liq

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A \(500\ \text{cm}^3\) body is fully submerged in water. What is the buoyant force? Use \(d = 1000\ \text{kg/m}^3\) and \(g = 10\ \text{m/s}^2\).

Show solution
In cubic metres, \(V = 500\ \text{cm}^3 = 5 \cdot 10^{-4}\ \text{m}^3\).
The buoyant force is \(B = d\,V\,g = 1000 \cdot 5 \cdot 10^{-4} \cdot 10 = 5\ \text{N}\).
\(B = 5\ \text{N}\)
basic
A stone weighs 30 N in air. Fully submerged in water, it receives a buoyant force of 10 N. What does the spring balance holding it read?

Show solution
The spring balance measures the apparent weight, \(W - B = 30 - 10 = 20\ \text{N}\).
20 N
basic
Steel is almost 8 times denser than water. How does a steel ship manage to float?

Show solution
What decides is the average density of the ship, which includes the huge volume of air inside the hull.
The hull displaces a volume of water whose weight equals the weight of the ship before it is fully submerged.
Being hollow, the ship has an average density lower than that of water.
basic
An egg sinks in a glass of fresh water and floats when we dissolve plenty of salt in the water. Why?

Show solution
Salt increases the density of the water, and with it the buoyant force \(B = d_{\text{liq}}\,V\,g\).
When the salt water becomes denser than the egg, the buoyant force with the egg fully submerged exceeds its weight, and the egg rises.
The salt water becomes denser than the egg.
intermediate
A block of wood (\(0.6\ \text{g/cm}^3\)) floats in water. What fraction of its volume is submerged?

Show solution
When floating, \(B = W\), that is, \(d_{\text{liq}}\,V_{\text{sub}}\,g = d_{\text{body}}\,V\,g\).
The submerged fraction is \(\dfrac{V_{\text{sub}}}{V} = \dfrac{0.6}{1.0} = 0.6\).
60% of the volume is submerged.
intermediate
Ice has a density of \(0.92\ \text{g/cm}^3\) and sea water, \(1.03\ \text{g/cm}^3\). What fraction of an iceberg is above the water?

Show solution
The submerged fraction is \(\dfrac{V_{\text{sub}}}{V} = \dfrac{0.92}{1.03} \approx 0.89\).
What sticks out is \(1 - 0.89 = 0.11\).
About 11% of the iceberg is above the water, the famous tip of the iceberg.
intermediate
A spring balance reads 12 N with a stone in air and 8 N with the stone fully submerged in water. Work out the buoyant force, the volume and the density of the stone. Use \(g = 10\ \text{m/s}^2\).

Show solution
The buoyant force is the difference between the readings, \(B = 12 - 8 = 4\ \text{N}\).
The volume comes from \(B = d\,V\,g\), \(V = \dfrac{4}{1000 \cdot 10} = 4 \cdot 10^{-4}\ \text{m}^3 = 400\ \text{cm}^3\).
The mass is 1.2 kg, and \(d = \dfrac{1.2}{4 \cdot 10^{-4}} = 3000\ \text{kg/m}^3\).
\(B = 4\ \text{N}\), \(V = 400\ \text{cm}^3\) and \(d = 3.0\ \text{g/cm}^3\).
intermediate
A block floats in water with 72% of its volume submerged. In oil, the same block floats with 90% submerged. What is the density of the block, and what is that of the oil?

Show solution
In water, the submerged fraction is the ratio between the densities, and \(d_{\text{block}} = 0.72 \cdot 1.0 = 0.72\ \text{g/cm}^3\).
In oil, \(0.90 \cdot d_{\text{oil}} = 0.72\), and \(d_{\text{oil}} = 0.8\ \text{g/cm}^3\).
\(0.72\ \text{g/cm}^3\) and \(0.8\ \text{g/cm}^3\), the same principle as the hydrometer.
challenge
A polystyrene slab of \(0.2\ \text{m}^3\) and density \(50\ \text{kg/m}^3\) serves as a raft on water. What is the largest mass it can carry without sinking? Use \(g = 10\ \text{m/s}^2\).

Show solution
The maximum buoyant force occurs with the slab fully submerged, \(B_{\text{max}} = 1000 \cdot 0.2 \cdot 10 = 2000\ \text{N}\).
The slab itself weighs \(50 \cdot 0.2 \cdot 10 = 100\ \text{N}\).
The load can weigh up to \(2000 - 100 = 1900\ \text{N}\), or 190 kg.
Up to 190 kg
challenge
A block of aluminium of \(100\ \text{cm}^3\) (\(2.7\ \text{g/cm}^3\)), hanging from a spring balance, is fully lowered into a glass of water standing on a set of scales, without touching the bottom. What does the spring balance read, and by how much does the reading on the scales increase? Use \(g = 10\ \text{m/s}^2\).

Show solution
The block has 270 g, and \(W = 0.27 \cdot 10 = 2.7\ \text{N}\). The buoyant force is \(B = 1000 \cdot 10^{-4} \cdot 10 = 1\ \text{N}\).
The spring balance reads the apparent weight, \(2.7 - 1 = 1.7\ \text{N}\).
By Newton's third law, the block pushes the water down with 1 N, and the scales now read 1 N more, the equivalent of 100 g.
The spring balance reads 1.7 N, and the scales go up by 1 N (100 g).

WRAP-UP

Challenges

Density and buoyancy

A 400 g block takes up 500 cm³. Does it sink or float in water? If it floats, what fraction of its volume is submerged?

Show solution

The density is \(d = \dfrac{400}{500} = 0.8\ \text{g/cm}^3\), lower than that of water, and the block floats.
Floating, \(\dfrac{V_{\text{sub}}}{V} = \dfrac{0.8}{1.0} = 0.8\), that is, 80% of the volume is submerged.

Pressure

A 70 kg person stands on one foot, with 140 cm² of contact. What is the pressure on the ground, in pascals and in atmospheres? (\(g = 10\ \text{m/s}^2\), \(1\ \text{atm} \approx 1.0 \cdot 10^5\ \text{Pa}\))

Show solution

The weight is 700 N and the area, \(0.014\ \text{m}^2\), and so \(p = \dfrac{700}{0.014} = 50\,000\ \text{Pa}\).
That gives \(5.0 \cdot 10^4\ \text{Pa}\), or 0.5 atm.

Stevin's law

What is the total pressure 25 m deep in a lake? (\(p_0 = 1.0 \cdot 10^5\ \text{Pa}\), \(d = 1000\ \text{kg/m}^3\), \(g = 10\ \text{m/s}^2\))

Show solution

The water adds \(d\,g\,h = 1000 \cdot 10 \cdot 25 = 2.5 \cdot 10^5\ \text{Pa}\).
With the atmosphere, \(p = 3.5 \cdot 10^5\ \text{Pa}\), about 3.5 atm.

Atmospheric pressure

In a town where the mercury barometer reads 68 cm, how tall would a water barometer be? (\(d_{\text{Hg}} = 13.6\ \text{g/cm}^3\))

Show solution

The two columns exert the same pressure, \(d_{\text{Hg}}\,h_{\text{Hg}} = d_{\text{water}}\,h_{\text{water}}\).
Hence \(h_{\text{water}} = 13.6 \cdot 68 \approx 925\ \text{cm}\), about 9.2 m.

Communicating vessels

In a U-tube with water, we pour 15 cm of oil (\(0.8\ \text{g/cm}^3\)) into one arm. How high is the water above the interface, and what is the difference in level between the free surfaces?

Show solution

At the height of the interface, \(0.8 \cdot 15 = 1.0 \cdot h_2\), and \(h_2 = 12\ \text{cm}\).
The oil stands \(15 - 12 = 3\ \text{cm}\) above the surface of the water.

Pascal's principle

A hydraulic jack has pistons of 4 cm² and 400 cm² and lifts a 1000 kg car. What force do we apply to the smaller piston? If it goes down 10 cm, how far does the car go up? (\(g = 10\ \text{m/s}^2\))

Show solution

The ratio between the areas is 100, and \(F_1 = \dfrac{10\,000}{100} = 100\ \text{N}\).
The displacement is divided by the same factor, \(d_2 = \dfrac{10}{100} = 0.1\ \text{cm} = 1\ \text{mm}\).

Buoyancy

A spring balance reads 5 N with an object in air and 3 N with the object fully submerged in water. What is the volume of the object, and what is its density? (\(g = 10\ \text{m/s}^2\))

Show solution

The buoyant force is \(B = 5 - 3 = 2\ \text{N}\), and the volume comes from \(V = \dfrac{2}{1000 \cdot 10} = 2 \cdot 10^{-4}\ \text{m}^3\), or \(200\ \text{cm}^3\).
The mass is 0.5 kg, or 500 g, and \(d = \dfrac{500}{200} = 2.5\ \text{g/cm}^3\).

Thinking, no sums

A glass is filled with water to the brim, with an ice cube floating in it. When the ice melts, does the water overflow?

Show solution

When floating, the ice displaces a volume of water whose weight equals its own weight.
On melting, it turns into exactly that amount of water, which fills the space the submerged part used to take up, and the level does not change.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on the same topics, which may well be the best practice once these are done.

Density · Step 1
Joana works at a recycling cooperative and needs to sort a solid metal part that arrived with no markings at all. On the balance, the part has a mass of 540 g. She takes a measuring cylinder with 300 mL of water, lowers the part in completely and sees the level rise to 500 mL. To decide, the cooperative has a table with the specific mass of the metals it usually receives.

Specific mass
Metal d (g/cm³)

Magnesium 1.7

Aluminium 2.7

Zinc 7.1

Iron 7.9

Copper 8.9

What is the density of the part, and what can Joana conclude about the material?
1. 0.37 g/cm³, and the part is not made of any metal in the table.
2. 1.1 g/cm³, the value closest to magnesium.
3. 1.8 g/cm³, the value closest to magnesium.
4. 2.7 g/cm³, that of aluminium.
5. 2.7 g/cm³, but the part could be any metal, because density changes with the size of the part.
Show solution

Answer: D.
The volume of the part is the volume of water it displaces, and so we look at how far the level rose.
\(V = 500 - 300 = 200\ \text{cm}^3\), since 1 mL is 1 cm³
\(d = \dfrac{m}{V} = \dfrac{540}{200} = 2.7\ \text{g/cm}^3\)
The value matches that of aluminium, and since the part is solid, this density is the specific mass of the material itself.
Dividing the mass by the final reading of the cylinder gives 1.1 g/cm³, and dividing by the initial reading gives 1.8 g/cm³. The 0.37 g/cm³ comes from dividing the volume by the mass. Option E forgets that the density of a material does not depend on the size of the part.
Stevin's law · Step 3
In a lab lesson, Davi and Marina lowered a pressure sensor, little by little, into an open tank of a liquid they did not know. Every half metre they noted the total pressure, and the points fell on the straight line of the graph below. Take \(g = 10\ \text{m/s}^2\) and assume the liquid is incompressible.

Based on the graph, what is the density of the liquid?
1. 1.2 kg/m³
2. 1000 kg/m³
3. 1200 kg/m³
4. 2400 kg/m³
5. 6200 kg/m³
Show solution

Answer: C.
The line starts at 100 kPa because the air already presses on the surface, and what matters is how much the pressure grows with depth.
\(\Delta p = 124 - 100 = 24\ \text{kPa}\) over \(\Delta h = 2\ \text{m}\)
With \(\Delta p = d\,g\,\Delta h\) and 24 kPa = 24,000 Pa, the density is
\(d = \dfrac{24\,000}{10 \cdot 2} = 1200\ \text{kg/m}^3\)
The 1000 appears when we assume, without looking at the graph, that the liquid is water, and the 1.2 when we forget to convert kPa to Pa. Forgetting the depth leads to 2400, and using the total pressure of 124 kPa, atmosphere included, leads to 6200.
Real measured points tend to scatter a little around the line, and so it pays to take the slope from two points far apart, as we did.
Communicating vessels · Step 5
Otávio set up a glass U-tube for the school science fair. He put water in the tube and then poured soya oil into one arm, until it formed a column 25 cm above the interface between the two liquids. The density of soya oil is \(0.92\ \text{g/cm}^3\) and that of water, \(1.0\ \text{g/cm}^3\), and the two do not mix.

Once everything has stopped moving, how high is the water, in the other arm, above the interface, and how do the free surfaces end up?
1. 25 cm, and the two free surfaces are at the same level.
2. 23 cm, and the surface of the oil is 2 cm above the surface of the water.
3. 23 cm, and the surface of the water is 2 cm above the surface of the oil.
4. 27 cm, and the surface of the water is 2 cm above the surface of the oil.
5. 23 cm, and the surface of the oil is 25 cm above the surface of the water.
Show solution

Answer: B.
At the height of the interface, the two arms are connected through the water and have the same pressure, which gives us the U-tube rule.
With \(d_1\,h_1 = d_2\,h_2\), we have \(0.92 \cdot 25 = 1.0 \cdot h_2\), and \(h_2 = 23\ \text{cm}\).
The oil, being less dense, needs a taller column, and its surface stands \(25 - 23 = 2\ \text{cm}\) above the surface of the water.
Option A applies the rule of communicating vessels, which only holds for a single liquid, to two different liquids. C puts the denser liquid on top, D inverts the ratio between the densities and E confuses the oil column with the difference in level.
In a real tube, the curve of the surface against the glass probably shifts the reading by a millimetre or two.
Pascal's principle · Step 6
Antônio, a motorbike mechanic, is explaining to an apprentice how the front disc brake works. When the rider squeezes the lever, the master-cylinder piston, of \(1.5\ \text{cm}^2\), pushes the brake fluid with a force of 60 N and moves 12 mm. The fluid carries the pressure to the calliper piston, of \(9\ \text{cm}^2\), which presses the pad against the disc. Assume the fluid is incompressible, with no air bubbles, and neglect friction.

What is the force of the calliper piston on the pad, and how far does that piston move?
1. 10 N and 72 mm
2. 60 N and 2 mm
3. 360 N and 12 mm
4. 360 N and 72 mm
5. 360 N and 2 mm
Show solution

Answer: E.
The fluid passes the same pressure to both pistons, and the force is multiplied by the ratio between the areas.
\(p = \dfrac{60}{1.5 \cdot 10^{-4}} = 4.0 \cdot 10^5\ \text{Pa}\), and at the calliper
\(F_2 = 4.0 \cdot 10^5 \cdot 9 \cdot 10^{-4} = 360\ \text{N}\)
The volume of fluid that leaves the master cylinder enters the calliper, \(A_1\,d_1 = A_2\,d_2\), and so \(d_2 = \dfrac{1.5 \cdot 12}{9} = 2\ \text{mm}\).
Checking with energy, \(60 \cdot 12 = 360 \cdot 2 = 720\ \text{N}\cdot\text{mm}\), the same work on both sides.
Option A inverts the ratio between the areas, and B confuses equal pressure with equal force. C and D forget that the gain in force costs displacement, and D goes as far as multiplying both, as if the press created energy.
Buoyancy · Step 7
A steel ferry carries cars across a river. The hull is shaped like a flat-bottomed box, 12 m long, 5 m wide and 1.2 m high, and the empty ferry has a mass of 18 t. For safety, the harbour authority only allows the hull to sink to 0.8 m. Take the density of the river water as \(1000\ \text{kg/m}^3\).

What is the largest load, in tonnes, that the ferry can carry within this limit?
1. 30 t
2. 48 t
3. 54 t
4. 66 t
5. None, because steel is denser than water and the ferry only floats when empty.
Show solution

Answer: A.
When floating, the buoyant force equals the weight of the ferry plus its load, and the buoyant force is the weight of the water displaced by the submerged part of the hull.
\(V_{\text{sub}} = 12 \cdot 5 \cdot 0.8 = 48\ \text{m}^3\), which corresponds to \(48\,000\ \text{kg} = 48\ \text{t}\) of displaced water
\(m_{\text{load}} = 48 - 18 = 30\ \text{t}\)
The 48 appears when we forget the mass of the ferry itself, and the 66 when we add it instead of subtracting it. The 54 comes from using the full height of the hull instead of the 0.8 m limit. Option E looks at the density of steel when what decides is the average density of the hull, which is hollow.
The 0.4 m margin between the limit and the edge of the hull seems large. With wind, waves and badly placed cars, however, the ferry may tilt, and it is this margin that keeps the water from coming in.

Specific mass
Metal	d (g/cm³)
Magnesium	1.7
Aluminium	2.7
Zinc	7.1
Iron	7.9
Copper	8.9

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

ENEM 2010, Day 1, blue booklet, question 81. A solid iron sculpture at the bottom of an empty swimming pool resists the workers trying to lift it, and the question asks why filling the pool with water would make the job easier, which leads us to buoyancy.
ENEM 2012, Day 1, blue booklet, question 67. The manual of a bidet shower requires a minimum water pressure, and we need to recognise, in the drawing of the plumbing, which height between the water tank and the shower decides that pressure.
ENEM 2013, Day 1, blue booklet, question 57. A plastic bottle pierced at three heights does not leak while it is capped and leaks once it is uncapped, and the question asks for the role of atmospheric pressure in each case.
ENEM 2013, Day 1, blue booklet, question 61. A hydraulic accessibility lift raises a person in a wheelchair, and from the ratio between the areas we work out the force the pump has to exert on the fluid.
ENEM 2018, Day 2, blue booklet, question 134. Anyone who tries to drink juice through two straws, one in the glass and one outside it, tends to fail, and the question asks for the explanation in terms of the pressure inside the mouth.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Density\(d = m/V\)

Units of density\(1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3\)

Pressure\(p = F/A\)

Atmosphere\(1\ \text{atm} = 76\ \text{cmHg} \approx 10^5\ \text{Pa}\)

Stevin's law\(p = p_0 + d\,g\,h\)

Diving\(10\ \text{m}\) of water \(\approx 1\ \text{atm}\)

U-tube\(d_1\,h_1 = d_2\,h_2\)

Hydraulic press\(F_1/A_1 = F_2/A_2\)

Displacements\(A_1\,d_1 = A_2\,d_2\)

Buoyant force\(B = d_{\text{liq}}\,V_{\text{sub}}\,g\)

Apparent weight\(W_{\text{app}} = W - B\)

Floating\(V_{\text{sub}}/V = d_{\text{body}}/d_{\text{liq}}\)