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Chapter 4: Pressure in Fluids and its Transmission

On 19 September 1648, a small group of men climbed the Puy de Dôme, an old volcanic peak in central France. They were carrying an odd piece of luggage — a long glass tube, a dish, and a heavy supply of mercury.

The climb was an experiment. A young French thinker, Blaise Pascal, had written to his brother-in-law Florin Périer with an idea he could not test himself. Fill the tube with mercury, invert it into a dish, and the mercury does not all run out. A column of it stays standing, held up by the air pressing down on the dish below. Pascal’s question was simple: if the air is really doing the holding, then higher up a mountain, where there is less air above you, the column should stand lower.

Périer measured the column in a garden at the foot of the mountain. Then his party carried the apparatus to the summit, more than a kilometre higher, and measured again. The mercury had dropped by about three inches. On the way down it climbed back, step by step, exactly as Pascal had predicted.

A column of liquid, it turned out, could be used to weigh the invisible ocean of air we live in. The same idea — that a fluid presses, and that the deeper you go the harder it presses — runs through this whole chapter.

Diagram

Fig. 4.0-1 — Measuring The Mercury Column At Two Heights Two-panel illustration side by side against the sloping outline of a mountain. Left panel, at the foot of the mountain: a dish of mercury with an inverted glass tube standing in it, the mercury inside the tube standing high; label ‘at the foot of the mountain’, ‘mercury column taller’, and mark the space above the mercury in the tube as ‘empty space’. Right panel, at the summit: the same apparatus with the mercury inside the tube standing visibly lower; label ‘at the summit’, ‘mercury column shorter’. Show downward arrows above each dish labelled ‘air pressing on the surface’, with fewer and shorter arrows in the summit panel. Add a vertical double-headed arrow beside each tube labelled ‘height of mercury column’.

NoteCuriosity Corner

Q1. A heavy elephant walks across soft ground and leaves a shallow print, while a woman in narrow heels can dent the same ground deeply. How can the lighter one press harder? Q2. Look at any large dam. Its wall is thin at the top and thick near the base. Why is the extra material needed only at the bottom? Q3. If you punch several holes down the side of a tall can of water, all the jets shoot out — but the lowest one lands farthest away. What is different about the water down there? Q4. A bubble of gas leaves the muddy bed of a lake as a tiny speck, yet it is noticeably bigger by the time it reaches the surface. Nothing has been added to it. So what grew? Q5. A driver presses one small pedal with one foot, and a car weighing over a tonne comes to a stop. How can such a small push be turned into such a large force?

4.1 Thrust And Pressure

4.1.1 Thrust

A force can be applied on a surface in any direction. You can push a wall straight on, or you can drag your palm sideways along it. The two pushes do very different things to the wall.

Physics gives a special name to the first kind. When a force acts on a surface in a direction normal to it — that is, at right angles to the surface — the force is called thrust.

NoteDefinition

Thrust Thrust is the force acting normally on a surface. It is measured in the same units as force; its SI unit is the newton (N).

When a body simply rests on a surface, the force it presses down with is its own weight. That weight acts vertically downwards, at right angles to the horizontal surface below. So

\[\text{thrust exerted by a body on a surface} = \text{weight of the body}\]

This thrust is the same whichever way round the body is placed on the surface. A brick lying flat and the same brick standing on end press down with exactly the same thrust, because the brick’s weight has not changed.

Q. A brick is first laid flat on a table and then stood on its smallest face. Does the thrust on the table change?

A. No. The thrust equals the weight of the brick, and the weight is the same in every position.

Thrust is a force, so it has both a size and a direction. It is therefore a vector quantity, and its direction is always along the normal to the surface.

4.1.2 Units Of Thrust

Thrust is measured in the units of force. In the SI system the unit is the newton (N), and in the CGS system it is the dyne, where

\[1\ \text{N} = 10^{5}\ \text{dyne}\]

There is a second family of force units, called the gravitational units, based on the weight of a chosen mass. In the MKS system this unit is the kilogramme-force (kgf), and in the CGS system it is the gramme-force (gf). One kilogramme-force is simply the weight of a mass of one kilogramme. Since the weight of a mass \(m\) is \(mg\), these units are related to the newton and the dyne through the value of \(g\):

\[1\ \text{kgf} = 9.8\ \text{N} \qquad \text{and} \qquad 1\ \text{gf} = 980\ \text{dyne}\]

4.1.3 Pressure

Two people can press on the ground with the same thrust and yet leave completely different marks. What decides the effect is not the thrust alone, but how the thrust is shared out over the surface.

Stand upright on loose sand and your feet sink in. Lie down on the same sand and you do not sink at all. Your weight has not changed, so the thrust on the sand is the same in both cases. What changed is the area over which that thrust acts — small when you stand, large when you lie down.

The effect of a thrust is less on a large area, and more on a small area. To capture this, we divide the thrust by the area it acts on. This quantity is called pressure.

NoteDefinition

Pressure Pressure is the thrust per unit area of a surface. Its SI unit is the pascal (Pa), which is one newton per square metre.

If a thrust \(F\) acts normally on an area \(A\), then

\[P = \frac{F}{A}\]

where \(F\) is the thrust in N, \(A\) is the area on which it acts in m², and \(P\) is the pressure in Pa. The area used here is the area actually in contact — the base of the brick, not its whole outer surface.

Pressure has a size but no direction of its own, so pressure is a scalar quantity. This is worth pausing on, because thrust and pressure are so easily muddled. Their differences are set out in Table 4.1.

Table 4.1 — Thrust Compared With Pressure | Property | Thrust | Pressure | |—|—|—| | What it is | The force acting normally on a surface | The thrust acting on unit area of the surface | | Depends on area? | No — it is the whole force, however large the surface | Yes — the same thrust gives more pressure on a smaller area | | Type of quantity | Vector (it has a direction, along the normal) | Scalar (it has size only) | | SI unit | newton (N) | pascal (Pa) | | CGS unit | dyne | dyne cm⁻² | | Relation | \(F = P \times A\) | \(P = F/A\) |

P = F/A Pressure tells you how much of the pushing force each square metre of the surface has to carry. Squeeze the same force onto half the area and each part carries twice as much, so the pressure doubles. The force must act at right angles to the surface for this to apply.

4.1.4 Units Of Pressure

Since pressure is thrust divided by area, its unit must be the unit of thrust divided by the unit of area.

In the SI system, the unit of thrust is the newton and that of area is the square metre. So the SI unit of pressure is newton per square metre, written N m⁻². This unit is given the name pascal, symbol Pa, after the French scientist Blaise Pascal.

\[1\ \text{Pa} = 1\ \text{N m}^{-2}\]

NoteDefinition

One Pascal One pascal is the pressure exerted on a surface of area \(1\ \text{m}^2\) by a force of \(1\ \text{N}\) acting normally on it.

If the thrust is measured in kgf and the area in m², the unit of pressure is kgf m⁻².

In the CGS system, the unit of pressure is dyne cm⁻². Since one dyne is \(10^{-5}\ \text{N}\) and one square centimetre is \(10^{-4}\ \text{m}^2\),

\[1\ \text{dyne cm}^{-2} = 0.1\ \text{N m}^{-2} \qquad \text{or} \qquad 1\ \text{N m}^{-2} = 10\ \text{dyne cm}^{-2}\]

If the thrust is measured in gf and the area in cm², the unit of pressure is gf cm⁻².

Two more units are in common use, especially in weather reports — the bar and the millibar, where

\[1\ \text{bar} = 10^{5}\ \text{N m}^{-2} \qquad \text{and} \qquad 1\ \text{millibar} = 10^{-3}\ \text{bar} = 10^{2}\ \text{N m}^{-2}\]

Atmospheric pressure — the pressure exerted by the air above us — is usually stated as the height of the mercury column it can support in a barometer. At normal temperature and pressure this barometric height at sea level is \(0.76\ \text{m}\) of mercury, that is \(76\ \text{cm}\) of mercury or \(760\ \text{mm}\) of mercury. This value is taken as one atmosphere (symbol atm), where

\[1\ \text{atm} = 0.76\ \text{m of Hg} = 1.013 \times 10^{5}\ \text{Pa}\]

Atmospheric pressure is sometimes given in a smaller unit called the torr, named after the scientist Evangelista Torricelli, where

\[1\ \text{torr} = 1\ \text{mm of Hg} \qquad \text{and} \qquad 1\ \text{atm} = 760\ \text{torr}\]

Thrust and pressure Thrust: SI unit newton (N); a vector, always directed along the normal to the surface. Pressure: SI unit pascal (Pa), equal to N m⁻²; a scalar, with no direction of its own. A pressure written without its unit says nothing at all.

4.1.5 Factors Affecting The Pressure On A Surface

From \(P = F/A\) you can read off exactly what pressure depends on. There are two factors, and no others: the magnitude of the thrust applied, and the area on which that thrust acts.

Take a brick of weight \(4\ \text{kgf}\) whose dimensions are \(20\ \text{cm} \times 10\ \text{cm} \times 5\ \text{cm}\). It can be rested on the ground in different ways, as shown in Fig. 4.1-1, and each way puts a different face in contact with the ground.

Diagram

Fig. 4.1-1 — The Same Brick Resting On Two Different Faces Two labelled sketches of the same rectangular brick side by side, drawn in simple perspective with the ground shaded. (a) Brick standing with its longest side (20 cm) vertical, so that the base in contact with the ground is the 5 cm × 10 cm face; mark the edges 20 cm, 10 cm and 5 cm, shade the base, and label ‘thrust = 4 kgf’, ‘area of base = 50 cm²’, ‘pressure = 0.08 kgf cm⁻²’. (b) The same brick lying with its shortest side (5 cm) vertical, so that the base is the 10 cm × 20 cm face; mark the edges, shade the base, and label ‘thrust = 4 kgf’, ‘area of base = 200 cm²’, ‘pressure = 0.02 kgf cm⁻²’. Caption both panels (a) longest side vertical and (b) shortest side vertical.

In Fig. 4.1-1(a) the brick stands with its longest side vertical. The thrust is \(4\ \text{kgf}\) and the base is the \(5\ \text{cm} \times 10\ \text{cm}\) face, so

\[\text{area of base} = 50\ \text{cm}^2, \qquad P = \frac{4\ \text{kgf}}{50\ \text{cm}^2} = 0.08\ \text{kgf cm}^{-2}\]

In Fig. 4.1-1(b) the same brick lies with its shortest side vertical. The thrust is still \(4\ \text{kgf}\), but the base is now the \(10\ \text{cm} \times 20\ \text{cm}\) face, so

\[\text{area of base} = 200\ \text{cm}^2, \qquad P = \frac{4\ \text{kgf}}{200\ \text{cm}^2} = 0.02\ \text{kgf cm}^{-2}\]

The pressure in Fig. 4.1-1(b) is one-fourth of the pressure in Fig. 4.1-1(a), although the brick and its weight are unchanged. So for a given thrust, the larger the area on which it acts, the smaller the pressure it exerts.

Now keep the area fixed and change the thrust instead. Place a second identical brick on top of the first in Fig. 4.1-1(b). The thrust becomes \(8\ \text{kgf}\) while the base is still \(200\ \text{cm}^2\), so

\[P = \frac{8\ \text{kgf}}{200\ \text{cm}^2} = 0.04\ \text{kgf cm}^{-2}\]

The pressure has doubled. So for a given area, the larger the thrust acting on it, the greater the pressure exerted on it.

Q1. A heavy elephant walks across soft ground and leaves a shallow print, while a woman in narrow heels can dent the same ground deeply. How can the lighter one press harder?

A. It is pressure, not thrust, that sinks into the ground. The elephant’s weight is huge, but it is spread over broad feet of large area, so the thrust per unit area stays modest. A narrow heel carries a much smaller weight, but on a contact area of a few square centimetres — so \(F/A\) comes out far larger, and the ground gives way.

Thrust And Pressure Under A Stone Slab

1. Given: weight of the slab \(= 250\ \text{N}\); the face resting on the floor measures \(0.50\ \text{m} \times 0.20\ \text{m}\). To find: the thrust on the floor and the pressure on the floor.

  1. The slab rests on a horizontal floor, so the thrust equals the weight of the slab. The pressure then follows from \(P = F/A\), using the area of the face actually in contact.

  2. Thrust \(F = 250\ \text{N}\). Area \(A = 0.50\ \text{m} \times 0.20\ \text{m} = 0.10\ \text{m}^2\). So \(P = \dfrac{250\ \text{N}}{0.10\ \text{m}^2} = 2500\ \text{N m}^{-2}\).

  3. Thrust \(= 250\ \text{N}\) and pressure \(= 2500\ \text{Pa}\). The check: a tenth of a square metre carries the whole \(250\ \text{N}\), so each full square metre would carry ten times as much — \(2500\ \text{Pa}\) is exactly what we should expect.

4.1.6 Ways Of Increasing And Decreasing Pressure

Since pressure depends on area, we can control the pressure a given thrust produces simply by choosing the area it acts on. Engineers and toolmakers do this all the time, in both directions.

Increasing The Pressure

For a given thrust, the pressure on a surface is increased by reducing the area of the surface on which it acts.

  • The ends of nails and pins are made pointed. The tiny tip has a very small area, so a modest thrust produces an enormous pressure there, and the nail drives in with less effort.
  • Cutting tools are given sharp or pointed edges. Even a small thrust then causes a great pressure along the thin edge, so cutting takes less effort.
Decreasing The Pressure

For a given thrust, the pressure on a surface is reduced by increasing the area of the surface.

  • Wide wooden sleepers are laid below railway tracks, so that the thrust from the rails is spread over a large area of ground and the pressure on the ground becomes less.
  • The foundation walls of a building are made wider than the walls above them, so the building’s thrust is spread over more ground and the pressure on the soil is reduced.

Both ideas are drawn together in Fig. 4.1-2.

Diagram

Fig. 4.1-2 — Choosing The Area To Control The Pressure Two panels. Left panel headed ‘small area, large pressure’: a nail being driven into a wooden plank, with a downward arrow at the head labelled ‘thrust’, a magnified circle showing the pointed tip in contact with the wood, and the label ‘very small contact area’. Right panel headed ‘large area, small pressure’: a cross-section of a railway track showing two rails resting on a broad wooden sleeper lying on the ground, with downward arrows from the rails labelled ‘thrust’ and a row of shorter downward arrows spread along the underside of the sleeper labelled ‘thrust spread over a large area’.

Why A Camel Does Not Sink In Sand A camel is a heavy animal, yet it walks over soft desert sand where a much lighter animal with narrow hooves would sink. Its foot ends in a broad, spreading pad that flattens out further as the animal puts its weight on it. The thrust on the sand is the camel’s weight, which is large, but that thrust is shared over a wide contact area — so the pressure on the sand stays small enough for the surface to hold. Snowshoes and the wide, flat tracks fitted to bulldozers work on exactly the same reasoning.

The Same Thumb, Two Very Different Pressures

1. Given: a drawing pin is pushed into a board with a thrust \(F = 20\ \text{N}\). The flat head under the thumb has area \(A_1 = 1\ \text{cm}^2\); the sharp tip touching the board has area \(A_2 = 0.05\ \text{mm}^2\). To find: the pressure on the thumb and the pressure on the board.

  1. The same thrust acts on both ends of the pin, so \(P = F/A\) applies twice, once with each area. The areas must first be converted to m², since the thrust is in N and we want the pressure in Pa.

  2. Converting: \(A_1 = 1\ \text{cm}^2 = 1 \times 10^{-4}\ \text{m}^2\), and \(A_2 = 0.05\ \text{mm}^2 = 0.05 \times 10^{-6}\ \text{m}^2 = 5 \times 10^{-8}\ \text{m}^2\).

On the thumb: \(P_1 = \dfrac{20\ \text{N}}{1 \times 10^{-4}\ \text{m}^2} = 2 \times 10^{5}\ \text{N m}^{-2}\).

On the board: \(P_2 = \dfrac{20\ \text{N}}{5 \times 10^{-8}\ \text{m}^2} = 4 \times 10^{8}\ \text{N m}^{-2}\).

  1. The pressure on the thumb is \(2 \times 10^{5}\ \text{Pa}\) and the pressure on the board is \(4 \times 10^{8}\ \text{Pa}\) — two thousand times greater. The check: the tip’s area is two thousand times smaller than the head’s, and the same thrust on a two-thousandth of the area must give two thousand times the pressure. That is why the pin pierces the board but not the thumb.

Hold a sharpened pencil between your finger and thumb, with the point against your thumb and the blunt end against your finger. Press gently, and increase the press slowly. Which side complains first? The push on the two ends is the same — only the contact area differs. Stop as soon as you feel anything sharp.

4.2 Pressure In Fluids

4.2.1 What A Fluid Is

Tip a bucket of water and it runs across the floor, taking the shape of whatever it meets. Open a bottle of scent in one corner of a room and the vapour spreads to every corner. Both substances share one property: they can flow.

A substance which can flow is called a fluid. All liquids and all gases are therefore fluids, and everything said about fluid pressure in this chapter applies to both.

NoteDefinition

Fluid A fluid is a substance which can flow. All liquids and gases are fluids.

Liquid and fluid A liquid is one kind of fluid, not the whole of it. Anything that flows is a fluid — so water, oil and mercury are fluids, and so are air, steam and cooking gas. Every liquid is a fluid; not every fluid is a liquid.

4.2.2 A Fluid Exerts Pressure In All Directions

A solid block resting on a table has weight, and that weight presses down on the table. The pressure it exerts acts on one surface only — the surface it stands on — and in one direction, downwards.

A fluid also has weight, and it also presses because of that weight. But a fluid cannot be stacked up and left alone; it flows sideways as readily as it settles downwards. Pour water into a vessel and it presses not only on the base but also outwards on the walls, wherever it touches them.

This is the key difference between the two, set out fully in Table 4.2.

Table 4.2 — Pressure Exerted By A Solid Compared With That Exerted By A Fluid | Property | Solid | Fluid | |—|—|—| | Cause of the pressure | Its own weight | Its own weight | | Surfaces on which it presses | Only the surface on which it rests, that is, its base | The base of the vessel and its walls as well | | Directions in which it presses | Downwards only | In all directions | | Reason | A solid keeps its shape and does not flow | A fluid flows, so it pushes against whatever confines it |

So we can state the behaviour of a fluid in one line.

NotePrinciple / Law

Pressure In A Fluid A fluid contained in a vessel exerts pressure at all points and in all directions.

Why An Inflated Tube Keeps Its Shape The air inside a cycle tube, a football bladder or an air mattress is a fluid, and it presses outwards on every part of the rubber that holds it in — the top, the bottom and the sides alike. That is why an inflated tube stays firm and round instead of sagging into a puddle at the bottom of its casing. Press a finger into it anywhere and you feel the air pushing straight back at you, whichever way round you turn it. If the pressure acted downwards only, an inflated ball would be flat on top.

4.2.3 Experimental Demonstration Of Liquid Pressure

The statement above can be checked with a very simple set-up, shown in Fig. 4.2-1.

Take a vessel filled with a liquid, say water, and place it on a horizontal surface. Make several small holes in the wall of the vessel, at different heights, anywhere below the free surface of the liquid. The free surface is simply the open upper surface of the liquid, the one in contact with the air. Three things are observed.

Diagram

Fig. 4.2-1 — Liquid Pressure Increases With Depth Below The Free Surface Sectional view of a tall vessel standing on a horizontal table, filled with water and shown with horizontal shading lines; mark the open top surface as ‘free surface’. Three small holes in the side wall, one near the top, one near the middle and one near the bottom. From each hole draw a curved jet of water falling to the table, the top jet landing nearest the vessel and the bottom jet landing farthest away; draw the three landing points on the table and mark the horizontal distance of each from the base of the vessel. Label the depth of each hole below the free surface as h, and add the note ‘greater depth, greater throw’.

  • Liquid spurts out through every one of the holes. This shows that the liquid exerts pressure at each point on the wall of the vessel, not merely on its base.
  • If you place a finger over any hole, the finger feels a thrust from the liquid. So the liquid contained in the vessel exerts thrust at all points below its free surface. The thrust on unit area at a point is the pressure due to the liquid at that point.
  • Note where each jet strikes the table, measuring from the foot of the vessel. The deeper the hole lies below the free surface, the farther out its jet lands.

That third observation carries the most information, so it is worth reading carefully. A jet that leaves its hole faster is thrown farther before it falls, and what pushes it out is the pressure of the liquid at that hole. A greater throw therefore means a greater pressure. Since the deepest hole throws its jet farthest, the pressure at a point in a liquid increases as the depth of that point below the free surface increases.

Q3. If you punch several holes down the side of a tall can of water, all the jets shoot out — but the lowest one lands farthest away. What is different about the water down there?

A. The water at the lowest hole is at the greatest depth below the free surface, and liquid pressure increases with depth. That larger pressure drives the water out of the lowest hole faster than out of the holes above it, so its jet travels farther before landing.

Take an empty plastic bottle and, with an adult’s help, make three pinholes in a vertical line down one side. Cover them with a strip of tape, fill the bottle with water at a sink, stand it on the draining board and peel the tape off. Watch where the three jets land. Which travels farthest, and does the pattern change as the bottle empties?

A tall cylindrical drum of water has two identical small holes in its wall, one at a depth of 10 cm below the free surface and the other at a depth of 40 cm. State which jet leaves its hole faster, and give the reason. Then state what happens to both jets as the water level in the drum slowly falls.

4.3 Pressure Exerted By A Liquid Column

4.3.1 The Relation P = hρg

The last section showed that a liquid presses harder the deeper you go. Now we ask a sharper question: exactly how much pressure does a liquid exert at a depth \(h\) below its free surface?

Think about what the answer should depend on. The pressure at a point comes from the weight of the liquid standing above that point. Go deeper and there is a taller column above you, so the pressure should rise. Use a heavier liquid and the same column weighs more, so the pressure should rise again. And weight itself depends on gravity. Three quantities, then — depth, density and \(g\) — and the pressure is simply their product.

\[P = h\rho g\]

\[\text{pressure} = \text{depth} \times \text{density of liquid} \times \text{acceleration due to gravity}\]

Here \(h\) is the depth of the point below the free surface of the liquid in m, \(\rho\) is the density of the liquid in kg m⁻³, \(g\) is the acceleration due to gravity in m s⁻², and \(P\) is the pressure due to the liquid column in Pa. Density is the mass of unit volume of a substance, so a volume \(V\) of a liquid of density \(\rho\) has mass \(\rho V\). The relation holds for a stationary liquid of uniform density, and it gives only the pressure due to the liquid itself.

P = hρg The pressure at a point in a liquid is the weight of the liquid column standing above one square metre of that point. Go twice as deep and the column is twice as tall, so the pressure doubles. Use a liquid twice as dense and the same column weighs twice as much, so again the pressure doubles.

4.3.2 Deriving The Relation

Setting Up The Column

Consider a vessel containing a liquid of density \(\rho\), with the liquid at rest. To find the pressure at a depth \(h\), imagine a small horizontal circular surface \(PQ\) of area \(A\) lying at that depth, as in Fig. 4.3-1.

Diagram

Fig. 4.3-1 — Pressure At A Depth In A Liquid Sectional view of a vessel filled with liquid, shown with horizontal shading lines. Mark the free surface as the line XY across the top. Inside the liquid, draw a vertical cylinder: its top face RS drawn as an ellipse lying on the free surface XY, and its bottom face PQ drawn as an ellipse at depth h. Join the two ellipses with vertical lines to show the cylinder PQRS. Label the free surface XY, the top face RS, the bottom face PQ, the area of PQ as A, and mark the vertical distance between RS and PQ with a double-headed arrow labelled h. Label the vessel contents ‘liquid of density ρ’.

Directly above \(PQ\) stands a cylinder of liquid, \(PQRS\). Its base is \(PQ\), its height is \(h\), and its top face \(RS\) lies on the free surface \(XY\) of the liquid. Everything resting on \(PQ\) is contained in this cylinder, so it is the weight of this cylinder that produces the thrust on \(PQ\).

Finding The Thrust, Then The Pressure

We build the thrust up in three easy steps — first the volume of the cylinder, then its mass through the density, then its weight.

\[\text{thrust on } PQ = \text{weight of the liquid column } PQRS\] \[= \text{volume of } PQRS \times \text{density} \times g\] \[= (\text{area of base } PQ \times \text{height}) \times \rho \times g\] \[= (A \times h) \times \rho \times g = Ah\rho g\]

This thrust acts on the surface \(PQ\), whose area is \(A\). Dividing the thrust by the area gives the pressure:

\[P = \frac{\text{thrust on the surface}}{\text{area of the surface}} = \frac{Ah\rho g}{A} = h\rho g\]

Notice what happened in that last step. The area \(A\) cancels out completely. We never had to say how big the imaginary surface was, which is why the answer holds at every point at that depth, whatever the surface we choose there.

Pressure At The Bottom Of An Oil Drum

1. Given: depth of oil \(h = 2.5\ \text{m}\); density of the oil \(\rho = 800\ \text{kg m}^{-3}\); \(g = 9.8\ \text{m s}^{-2}\). To find: the pressure at the bottom of the drum due to the oil.

  1. The oil is at rest and the pressure asked for is the pressure due to the liquid column alone, so \(P = h\rho g\) applies directly.

  2. \(P = 2.5\ \text{m} \times 800\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 19\,600\ \text{N m}^{-2}\).

  3. The pressure is \(1.96 \times 10^{4}\ \text{Pa}\). The check: this is about a fifth of atmospheric pressure, which sounds right for a couple of metres of a liquid lighter than water.

4.3.3 Total Pressure At A Depth In A Liquid

The relation \(P = h\rho g\) counts only the liquid. But the free surface of the liquid is not pressed on by nothing — the atmosphere sits on it and pushes down as well, and that push is passed on to every point below.

So to find the total pressure at a depth \(h\), the atmospheric pressure must be added in. If the atmospheric pressure acting on the free surface is \(P_0\), then

\[\text{total pressure at a depth } h = P_0 + h\rho g\]

where \(P_0\) is the atmospheric pressure in Pa, and \(h\rho g\) is the pressure due to the liquid column in Pa. At sea level \(P_0 = 1.013 \times 10^{5}\ \text{Pa}\) unless a different value is given.

Whether you include \(P_0\) depends on what the question asks for. If it asks for the pressure due to the liquid, use \(h\rho g\) alone; if it asks for the total pressure, add \(P_0\).

Pressure due to the liquid and total pressure \(h\rho g\) is only the share contributed by the liquid column. The total pressure at that point is \(P_0 + h\rho g\), because the atmosphere presses on the free surface too. Read the question: ‘pressure due to water’ means leave \(P_0\) out, ‘total pressure’ means put it in.

Total Pressure And Thrust On The Base Of A Tank

1. Given: depth of water in the tank \(h = 1.5\ \text{m}\); density of water \(\rho = 10^{3}\ \text{kg m}^{-3}\); area of the base \(A = 0.40\ \text{m}^{2}\); atmospheric pressure \(P_0 = 1.013 \times 10^{5}\ \text{Pa}\); \(g = 9.8\ \text{m s}^{-2}\). To find: (a) the pressure at the base due to the water, (b) the thrust on the base due to the water, and (c) the total pressure at the base.

  1. Part (a) needs \(P = h\rho g\), the pressure of the liquid column. Part (b) reverses the definition of pressure: since \(P = F/A\), the thrust is \(F = P \times A\). Part (c) adds the atmospheric pressure acting on the free surface.

    1. \(P = 1.5\ \text{m} \times 10^{3}\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 14\,700\ \text{N m}^{-2}\).

  1. \(F = P \times A = 14\,700\ \text{N m}^{-2} \times 0.40\ \text{m}^{2} = 5880\ \text{N}\).

  2. Total pressure \(= P_0 + h\rho g = (1.013 \times 10^{5}) + (0.147 \times 10^{5}) = 1.160 \times 10^{5}\ \text{N m}^{-2}\).

  1. Pressure due to the water \(= 1.47 \times 10^{4}\ \text{Pa}\); thrust on the base \(= 5880\ \text{N}\); total pressure \(= 1.160 \times 10^{5}\ \text{Pa}\). The check: \(5880\ \text{N}\) is the weight of \(600\ \text{kg}\) of water, and a base of \(0.40\ \text{m}^2\) carrying water \(1.5\ \text{m}\) deep does hold \(0.60\ \text{m}^3\) of it — so the thrust on the base is just the weight of the water above it, as it should be.

4.3.4 Factors Affecting The Pressure In A Stationary Liquid

Read \(P = h\rho g\) again as a list of ingredients. The pressure at a point inside a stationary liquid depends directly on three factors, and on nothing else:

  1. the depth of the point below the free surface, \(h\);
  2. the density of the liquid, \(\rho\);
  3. the acceleration due to gravity, \(g\).

At a particular place on the earth’s surface, \(g\) has a fixed value. Only two of the three can then vary, so at a given place the pressure at a point in a liquid is directly proportional to the depth \(h\) of the point below the free surface, and directly proportional to the density \(\rho\) of the liquid. That proportionality is shown in Fig. 4.3-2.

Diagram

Fig. 4.3-2 — Pressure Against Depth For Two Liquids Graph on plain axes. Horizontal axis labelled ‘depth h below the free surface (m)’, marked 0 to 4 in steps of 1. Vertical axis labelled ‘pressure due to the liquid P (Pa)’, marked 0 to 4 × 10⁴ in steps of 1 × 10⁴. Draw two straight lines, both starting at the origin: line A, labelled ‘water, ρ = 1000 kg m⁻³’, passing through the point (4 m, 3.92 × 10⁴ Pa); line B, steeper, labelled ‘glycerine, ρ = 1260 kg m⁻³’, passing through the point (4 m, 4.94 × 10⁴ Pa). Show dotted guide lines from h = 2 m up to line A and across to the pressure axis.

Fig. 4.3-2 — Pressure Against Depth For Two Liquids Start with the axes. Depth below the free surface, in metre, runs along the horizontal axis; the pressure due to the liquid, in pascal, runs up the vertical axis. Every point on a line is one depth paired with the pressure found there. Both lines start at the origin. At zero depth you are at the free surface, where the liquid column above you is zero, so the pressure due to the liquid is zero as well. Both lines are straight. A straight line through the origin means the two quantities are in direct proportion — double the depth and you double the pressure. This is the graph’s way of saying \(P \propto h\). Now the gradient. Reading line A, at a depth of \(2\ \text{m}\) the pressure is \(1.96 \times 10^{4}\ \text{Pa}\), so the gradient is \(\dfrac{1.96 \times 10^{4}\ \text{Pa}}{2\ \text{m}} = 9800\ \text{Pa m}^{-1}\). Compare that with \(\rho g = 1000\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 9800\). The gradient of the line is \(\rho g\) — it tells you the density of the liquid, once you know \(g\). That is why line B is steeper. Glycerine is denser than water, so each extra metre of depth adds more pressure, and the line climbs faster. The area under either line has no physical meaning here; on this graph it is the gradient that carries the physics.

Two things are missing from that list of three, and their absence surprises most people. The pressure inside a liquid does not depend on the shape or the size of the vessel holding it, and it does not depend on the area of the surface on which it acts. Fig. 4.3-3 makes the first of these visible.

Diagram

Fig. 4.3-3 — Same Depth, Same Pressure, Whatever The Shape Three vessels of very different shapes standing side by side on the same horizontal bench, all connected at their bases by a common horizontal tube and all containing the same liquid: (i) a narrow straight cylinder, (ii) a wide vessel that flares outwards towards the top, (iii) a vessel with a bulge in the middle and a narrow neck. Show the liquid standing at exactly the same level in all three, and draw a single horizontal dashed line across all three vessels at that level, labelled ‘free surface — same level in every vessel’. Draw a second horizontal dashed line at a depth h below the first, cutting all three vessels, and mark a point on it in each vessel labelled ‘pressure the same at all three points’. Mark the vertical distance between the two dashed lines as h.

Look at the wide vessel and the narrow one. There is far more liquid in the wide one, and so a far greater total weight — yet at the same depth the pressure is identical. The reason is in the derivation: what pressed on the surface \(PQ\) was only the column standing directly above it, of height \(h\). Liquid off to one side rests on the vessel, not on that point.

Finding A Depth From A Pressure

1. Given: density of the liquid \(\rho = 1.25 \times 10^{3}\ \text{kg m}^{-3}\); pressure due to the liquid at a certain point \(P = 4.9 \times 10^{4}\ \text{Pa}\); \(g = 9.8\ \text{m s}^{-2}\). To find: the depth \(h\) of that point below the free surface.

  1. The pressure given is the pressure due to the liquid column, so \(P = h\rho g\) applies. Here the unknown is \(h\), so the relation must first be rearranged to make \(h\) the subject: dividing both sides by \(\rho g\) gives \(h = \dfrac{P}{\rho g}\).

  2. \(h = \dfrac{4.9 \times 10^{4}\ \text{N m}^{-2}}{1.25 \times 10^{3}\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2}} = \dfrac{4.9 \times 10^{4}}{1.225 \times 10^{4}}\ \text{m} = 4.0\ \text{m}\).

  3. The point lies \(4.0\ \text{m}\) below the free surface. The check: this liquid is denser than water, and \(4\ \text{m}\) of water alone would give about \(3.9 \times 10^{4}\ \text{Pa}\) — so a slightly larger pressure at the same depth is exactly what a denser liquid should give.

A vertical pipe stands full of water. A pressure gauge fitted at a point in the pipe reads \(2.94 \times 10^{4}\ \text{Pa}\) for the pressure due to the water alone. Taking the density of water as \(10^{3}\ \text{kg m}^{-3}\) and \(g = 9.8\ \text{m s}^{-2}\), find how far that point lies below the water level at the top of the pipe.

4.4 Laws Of Liquid Pressure

Everything established so far about a liquid at rest can be gathered into five short statements. Together they are called the laws of liquid pressure.

NotePrinciple / Law

The Laws Of Liquid Pressure For a liquid at rest: 1. Inside the liquid, pressure increases with the increase in depth from its free surface. 2. In a stationary liquid, pressure is the same at all points on a horizontal plane. 3. Pressure is the same in all directions about a point inside the liquid. 4. Pressure at the same depth is different in different liquids. It increases with the increase in density of the liquid. 5. A liquid seeks its own level.

The first and the fourth you have already met — both are contained in \(P = h\rho g\), since the pressure rises with \(h\) and with \(\rho\). The second follows too: every point on one horizontal plane sits at the same depth \(h\), so each has the same weight of liquid standing above it.

The third law is about direction rather than depth, and Fig. 4.4-1 shows what it means. Choose any single point inside the liquid. The liquid there pushes upwards, downwards and sideways with the same pressure — which is why a jet spurts sideways out of a hole in a wall just as readily as it would from a hole in the base.

Diagram

Fig. 4.4-1 — Pressure At A Point And On A Horizontal Plane Sectional view of a vessel of liquid shown with horizontal shading lines and a labelled free surface. Mark a point X well inside the liquid, and draw six equal-length arrows radiating from X — up, down, left, right and two diagonal — all the same length, with the label ‘equal pressure in every direction about a point’. Lower down, draw a horizontal dashed line across the vessel at a depth h below the free surface, and mark three points on it, L, M and N, with a short equal arrow at each; label the line ‘same horizontal plane — same depth h, same pressure at L, M and N’. Mark the depth h with a double-headed vertical arrow from the free surface to the dashed line.

The fifth law is the one that needs unpacking, because it is stated so briefly. To say that a liquid seeks its own level means this: if a liquid is free to flow between two connected regions, it settles until its free surface stands at the same height everywhere. Should one side stand higher, the pressure at the bottom of that side would be greater, and liquid would be pushed across until the two heights matched. Fig. 4.4-2 shows the everyday version of this.

Diagram

Fig. 4.4-2 — A Liquid Seeks Its Own Level Two panels. Left panel: a teapot in side view, cut away so that the tea inside and inside the spout are both visible; show the tea standing at exactly the same height in the body of the pot and in the spout, with a horizontal dashed line joining the two surfaces labelled ‘same level’. Right panel: the same teapot tilted forwards to pour, with the tea again level in body and spout but now overflowing at the lip of the spout; keep the dashed horizontal level line. Label both panels ‘the free surface stays horizontal and at the same height in every connected part’.

The Mason’s Water Level Long before laser levels were sold, builders marked a straight horizontal line around a room using nothing but a length of clear plastic tubing filled with water. One end is held against a mark on the first wall; wherever the other end is carried, the water surface in it settles at exactly the same height as in the first end, because the water in the tube is connected and seeks its own level. Marking that height on each wall gives a true horizontal line right round the room — around corners, past doorways, and over distances no straight edge could span. The tube must be free of trapped air bubbles, since a bubble breaks the connection along the water column.

Half fill a clear bottle with water and stand it upright on a table. Note the level of the surface against the side of the bottle. Now tilt the bottle slowly, watching the surface. Does the water surface stay parallel to the table, or does it tilt with the bottle? Try the same with the bottle resting on a slope.

A vessel of water has two points, \(L\) and \(M\), marked inside it. \(L\) lies \(30\ \text{cm}\) below the free surface near the left wall; \(M\) lies \(30\ \text{cm}\) below the free surface near the right wall, close to the base. State, with a reason, which of the two has the greater pressure. Then state which law of liquid pressure you have used.

Level, not amount ‘A liquid seeks its own level’ is a statement about height, not about quantity. Joined vessels of very different sizes hold very different amounts of liquid, yet the free surface settles at the same height in all of them, because it is the height of the column that fixes the pressure at the bottom.

4.5 Some Consequences Of Liquid Pressure

The single relation \(P = h\rho g\) explains a surprising number of things that are built, designed or simply observed around us. Five of them follow.

4.5.1 Pressure In Sea Water And In River Water

Go down to the same depth in the sea and in a river, and the sea presses harder. The reason lies in the density term of \(P = h\rho g\). Sea water carries a large quantity of dissolved salts, which makes it denser than river water. With \(h\) and \(g\) the same at both places, the greater density gives the greater pressure.

Sea Water Against River Water At The Same Depth

1. Given: depth \(h = 20\ \text{m}\) in each case; density of sea water \(\rho_1 = 1030\ \text{kg m}^{-3}\); density of river water \(\rho_2 = 1000\ \text{kg m}^{-3}\); \(g = 9.8\ \text{m s}^{-2}\). To find: the pressure due to the water at that depth in each, and the difference between them.

  1. Both liquids are at rest, and only the pressure due to the liquid column is asked for, so \(P = h\rho g\) is used twice with the same \(h\) and \(g\) but different \(\rho\).

  2. In sea water: \(P_1 = 20\ \text{m} \times 1030\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 2.019 \times 10^{5}\ \text{N m}^{-2}\).

In river water: \(P_2 = 20\ \text{m} \times 1000\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 1.960 \times 10^{5}\ \text{N m}^{-2}\).

Difference: \(P_1 - P_2 = 0.059 \times 10^{5}\ \text{N m}^{-2} = 5880\ \text{N m}^{-2}\).

  1. The pressure is \(2.019 \times 10^{5}\ \text{Pa}\) in the sea and \(1.960 \times 10^{5}\ \text{Pa}\) in the river, the sea being greater by \(5880\ \text{Pa}\). The check: sea water is denser by \(3\ \text{per cent}\), and the pressures differ by about \(3\ \text{per cent}\) — as direct proportionality to \(\rho\) requires.

4.5.2 The Wall Of A Dam Is Thicker At Its Base

Fig. 4.5-1 shows the side view of a dam. Its wall is comparatively thin at the top and grows steadily thicker towards the base.

Diagram

Fig. 4.5-1 — Wall Of A Dam With Its Thickness Increasing Towards The Bottom Side sectional view. On the left, a body of water shown with horizontal shading lines, labelled ‘water’, with a labelled free surface at the top. On the right, the wall of the dam drawn in cross-section as a wedge — narrow at the top, broadening steadily towards its base where it rests on the ground; hatch the wall and label it ‘wall of dam’, and label the ground to the right ‘land’. From the water side, draw a series of horizontal arrows striking the wall at several depths, the arrows growing longer from top to bottom; label this set ‘pressure increasing downwards’. Along the right edge of the wall, mark a vertical double-headed arrow labelled ‘width of wall increasing downwards’.

The reason is the depth term of \(P = h\rho g\). Near the top of the reservoir the water is shallow and presses gently on the wall. Deeper down, the water column above is taller, so the water pushes on the wall harder and harder. In Fig. 4.5-1 the lengthening arrows show that increase.

A wall must be strong enough to withstand the pressure it meets, and greater pressure calls for a thicker wall. Since the pressure is greatest at the bottom, the dam is built with its thickness increasing towards the base — putting the material where the water pushes hardest.

Q2. Look at any large dam. Its wall is thin at the top and thick near the base. Why is the extra material needed only at the bottom?

A. Because the water’s pressure on the wall is not the same all the way down. Pressure in a liquid rises with depth, \(P = h\rho g\), so the deepest part of the wall is pushed on far harder than the part near the surface. A thicker wall is needed to withstand that greater pressure, and there is no point in spending material higher up where the push is small.

4.5.3 A Water Supply Tank Is Placed High

To supply water to a town or a colony, the storage tank is built at a height, as in Fig. 4.5-2.

Diagram

Fig. 4.5-2 — Why The Supply Tank Is Placed High Simple street-scene section. On the left, a large water storage tank raised on a tall framework of columns, with its water level marked and labelled ‘level of water in the supply tank’. A pipe leads down from the tank and runs to a house on the right, ending in a tap on an upper floor and a second tap on the ground floor. Draw a horizontal dashed line from the tank’s water level across the picture, and mark the vertical distance from that line down to each tap with a double-headed arrow labelled h; make the arrow to the ground-floor tap clearly longer. Add the labels ‘greater h, greater pressure at the tap’.

The pressure of water at a tap is the pressure due to the column of water standing between the tap and the water level in the tank. That column has height \(h\), so the pressure at the tap is \(h\rho g\). The higher the tank stands above the tap, the greater \(h\) is, and the stronger the flow from the tap.

This is why the height of the supply tank must always be a few metres above the level at which the water is to be delivered. A tank at or below the level of a tap gives no column of water above the tap, and so no pressure to drive the flow.

Why A Hospital Drip Is Hung High In a hospital ward the bag of fluid on a drip stand is always hung well above the patient’s arm, and never set down beside the bed. Fluid will only enter a vein if the pressure at the lower end of the tube is greater than the blood pressure inside that vein. That pressure comes from the height of the fluid column, \(h\rho g\) — so raising the bag raises \(h\) and raises the delivery pressure. Nursing staff adjust the flow partly by changing the height of the bag on the stand, which is the same physics that sends water from a rooftop tank down to a tap.

Pressure At A Ground-Floor Tap

1. Given: the water level in an overhead tank stands \(12\ \text{m}\) above a ground-floor tap, so \(h = 12\ \text{m}\); density of water \(\rho = 10^{3}\ \text{kg m}^{-3}\); \(g = 9.8\ \text{m s}^{-2}\). To find: the pressure of the water at the tap due to the water column.

  1. The water in the connected pipe is at rest when the tap is shut, and the tap lies at a depth \(h\) below the free surface of the water in the tank. So \(P = h\rho g\) applies, with \(h\) measured vertically, not along the pipe.

  2. \(P = 12\ \text{m} \times 10^{3}\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 1.176 \times 10^{5}\ \text{N m}^{-2}\).

  3. The pressure at the tap is \(1.176 \times 10^{5}\ \text{Pa}\). The check: this is slightly more than one atmosphere, which is a sensible supply pressure for a domestic tap — and a tap on an upper floor, with a smaller \(h\), would receive correspondingly less.

4.5.4 The Diver’s Suit

A diver working deep in the sea is surrounded by water at very high pressure, since \(h\) there is large. The total pressure on the diver’s body — atmospheric pressure plus the pressure of the water column above — becomes much greater than the blood pressure inside the body.

To withstand it, a deep-sea diver wears a special protective suit, made from glass reinforced plastic or cast aluminium. The pressure inside the suit is maintained at one atmosphere, so the diver’s body works under the conditions it is built for while the rigid suit takes the load from outside.

A diver is working at a depth of \(30\ \text{m}\) in sea water of density \(1030\ \text{kg m}^{-3}\). Taking \(g = 9.8\ \text{m s}^{-2}\) and atmospheric pressure as \(1.013 \times 10^{5}\ \text{Pa}\), calculate the total pressure on the diver’s suit at that depth, and state how many times the atmospheric pressure this is.

4.5.5 Growth Of A Gas Bubble Rising In Water

A bubble of gas released at the bottom of a lake grows steadily larger as it rises, as shown in Fig. 4.5-3. Nothing is added to the bubble on the way up, so its growth has to be explained by the water around it.

Diagram

Fig. 4.5-3 — A Gas Bubble Grows As It Rises Sectional view of a lake, with the bed at the bottom and the free surface at the top marked ‘free surface’. Show one bubble at four stages along a rising path: very small at the lake bed, larger part way up, larger still near the top, and largest just as it reaches the surface. Beside the path, mark the depth h of each stage below the free surface with double-headed arrows, and label the column of arrows ‘depth decreasing’. Add a second label alongside: ‘total pressure on bubble = atmospheric pressure + hρg, decreasing as the bubble rises’. Show downward arrows above the free surface labelled ‘atmospheric pressure’.

At the bed of the lake, the total pressure on the bubble is the sum of the atmospheric pressure and the pressure due to the water column above it, and that water column is tall. As the bubble rises, its depth \(h\) falls, so the term \(h\rho g\) falls, and the total pressure squeezing the bubble decreases.

Now use what you know about gases. By Boyle’s law, the volume of a given mass of gas is inversely proportional to the pressure on it. Less pressure therefore means more volume, so the bubble expands as it climbs. When it reaches the surface, the water column above it has vanished and the total pressure on it is at its minimum, just equal to the atmospheric pressure — so the bubble’s size there is at its maximum.

Q. Does the bubble grow because gas is added to it as it rises?

A. No. The same amount of gas is inside it throughout. What changes is the pressure squeezing it, and a smaller squeeze lets the same gas occupy a larger volume.

Q4. A bubble of gas leaves the muddy bed of a lake as a tiny speck, yet it is noticeably bigger by the time it reaches the surface. Nothing has been added to it. So what grew?

A. Only its volume grew, and it grew because the pressure on it fell. At the lake bed the bubble is squeezed by the atmospheric pressure plus a tall column of water. As it rises, \(h\) shrinks and so does \(h\rho g\), so the total pressure on the trapped gas falls — and by Boyle’s law, less pressure on a fixed mass of gas means a greater volume.

Boyle’s law, in one line For a fixed mass of gas kept at a constant temperature, pressure and volume are inversely proportional: halve the pressure and the volume doubles. It is the gas law you will use whenever something trapped and gaseous changes size as the pressure around it changes.

4.6 Transmission Of Pressure In Liquids; Pascal’s Law

4.6.1 Why Pressure Is Transmitted Through A Liquid

So far the pressure inside a liquid has come from the liquid’s own weight. Now suppose we press on the liquid from outside — with a piston, say — and ask what happens deep inside it.

Take two points \(x\) and \(y\) inside a stationary liquid of density \(\rho\), as in Fig. 4.6-1, with \(y\) lying a vertical distance \(\Delta h\) below \(x\). Since the pressure at any point is fixed by its depth, the difference in pressure between the two points is \(\rho g \Delta h\) — it depends on the vertical height between them and on nothing else.

Diagram

Fig. 4.6-1 — Two Points In A Confined Liquid Sectional view of a closed vessel completely filled with liquid, shown with horizontal shading lines, with a piston fitted air-tight into an opening at the top and a downward arrow on the piston labelled ‘extra pressure applied here’. Inside the liquid mark two points, x higher up and y lower down, joined by a vertical double-headed arrow labelled Δh. Beside x write ‘pressure at x’, beside y write ‘pressure at y’, and alongside the pair write ‘difference in pressure = ρgΔh, fixed by Δh alone’.

Now push the piston down, so that the pressure at \(x\) is increased by some amount. The vertical distance \(\Delta h\) has not changed, and neither has the density of the liquid — so the difference in pressure between \(x\) and \(y\) must still be exactly \(\rho g \Delta h\).

There is only one way that can happen. The pressure at \(y\) must increase by precisely the same amount as the pressure at \(x\). In other words, the pressure applied at one point in an enclosed liquid is passed on, undiminished, to every other point in it. This is Pascal’s law.

NotePrinciple / Law

Pascal’s Law Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid. A confined liquid means one completely enclosed in a vessel, with no free surface open to escape through.

Pressure due to depth and pressure transmitted The two act together, they do not replace one another. The \(h\rho g\) part still differs from point to point, because depths differ. The pressure applied from outside is added on top of it, and that added part is the same everywhere in the liquid.

Q. If pressure is transmitted equally everywhere, is the total pressure the same at every point in a confined liquid?

A. No. The added pressure is the same everywhere, but each point keeps its own \(h\rho g\) as well. It is the increase that is equal, not the total.

4.6.2 Experimental Demonstration Of Pascal’s Law

Pascal’s law can be shown with a single piece of apparatus, drawn in Fig. 4.6-2.

Diagram

Fig. 4.6-2 — Demonstration Of Pascal’s Law Sectional view of a round glass flask with several narrow open tubes leading out of it — two from opposite sides, one slanting from the shoulder, and one from the bottom, each turned so that its open end points upwards. The flask is filled with water, shown with horizontal shading lines and labelled ‘water’. An air-tight piston is fitted into the neck at the mouth of the flask, with a downward arrow above it labelled ‘push’. Draw a dotted black horizontal line across all the tubes at the level of the base of the piston, labelled ‘initial level of water in each tube’. From each tube, draw a jet of water rising and curving over, and draw a second dotted line above the first, touching the top of every jet, labelled ‘jets of water reaching the same height’. Label the piston, the flask and the tubes.

Take a glass flask with several narrow tubes leading out of its sides and its bottom, and fit an air-tight piston into its mouth. Fill the flask with water. The water in each tube settles at the same level, which is also the level of the base of the piston — the liquid seeking its own level, as it must.

Now push the piston gently down into the flask. Jets of water rise out of every tube, and they all reach the same height, marked by the upper dotted line in Fig. 4.6-2.

The equal heights are the result that matters. A jet rises higher when the pressure driving it is greater, so jets of equal height mean equal pressures at the mouths of all the tubes — at the sides and at the bottom alike. The pressure applied to the enclosed liquid by the piston has been transmitted equally, and in all directions, everywhere inside the liquid.

Pressure At A Point After The Piston Is Pushed

1. Given: a closed vessel is completely filled with water of density \(\rho = 10^{3}\ \text{kg m}^{-3}\). A point \(Y\) lies \(2.0\ \text{m}\) vertically below the piston at the top. Pushing the piston applies an extra pressure of \(2.0 \times 10^{4}\ \text{Pa}\) to the water at the piston face. Take \(g = 9.8\ \text{m s}^{-2}\) and ignore the atmospheric pressure. To find: the pressure at \(Y\) before and after the piston is pushed.

  1. Before the push, the pressure at \(Y\) is due to the water column alone, so \(P = h\rho g\). After the push, Pascal’s law says the applied pressure is transmitted undiminished to every point, so the same \(2.0 \times 10^{4}\ \text{Pa}\) is added at \(Y\).

  2. Before: \(P = 2.0\ \text{m} \times 10^{3}\ \text{kg m}^{-3} \times 9.8\ \text{m s}^{-2} = 1.96 \times 10^{4}\ \text{N m}^{-2}\).

After: \(P' = (1.96 \times 10^{4}) + (2.0 \times 10^{4}) = 3.96 \times 10^{4}\ \text{N m}^{-2}\).

  1. The pressure at \(Y\) rises from \(1.96 \times 10^{4}\ \text{Pa}\) to \(3.96 \times 10^{4}\ \text{Pa}\). The check: the increase is \(2.0 \times 10^{4}\ \text{Pa}\), exactly what was applied at the piston — which is what ‘transmitted undiminished’ means. A point at any other depth would gain the same \(2.0 \times 10^{4}\ \text{Pa}\), though its total would differ.

Fill a small polythene bag with water, hold its mouth tightly closed, and make three or four pinholes in different parts of it — one low down, one on the side, one near the top. Hold it over a sink and squeeze the bag anywhere you like. Water comes out of every hole at once. Does it matter which part of the bag you squeeze?

4.7 Application Of Pascal’s Law

Hydraulic machines such as the hydraulic press, the hydraulic jack and hydraulic brakes are all built on Pascal’s law of transmission of pressure in liquids. Before looking at the machines themselves, we work out the idea they share.

4.7.1 Principle Of A Hydraulic Machine

The principle of every hydraulic machine is this: a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.

Fig. 4.7-1 shows the arrangement. Two cylindrical vessels \(P\) and \(Q\) are connected by a horizontal tube \(R\). They contain a liquid, and each is fitted with a water-tight piston — piston \(A\) in the narrower vessel \(P\), piston \(B\) in the wider vessel \(Q\).

Diagram

Fig. 4.7-1 — Principle Of A Hydraulic Machine Sectional view. On the left, a narrow cylindrical vessel labelled P fitted with a water-tight piston labelled ‘piston A’ of area A₁; a weight rests on this piston, with a downward arrow through it labelled F₁, and inside the liquid just below it mark the pressure P₁ with short arrows pointing in all directions. On the right, a much wider cylindrical vessel labelled Q fitted with a water-tight piston labelled ‘piston B’ of area A₂, with an upward arrow above it labelled F₂ and short arrows below it marking the pressure P₂. Both vessels contain the same liquid, shown with horizontal shading lines, and their bases are joined by a horizontal connecting tube labelled R, also full of liquid. Mark the widths of the two pistons to show A₂ much larger than A₁.

Let the area of cross section of the piston in vessel \(P\) be \(A_1\) and that of the piston in vessel \(Q\) be \(A_2\), with \(A_2\) the larger of the two. Place a weight on piston \(A\), so that it exerts a force \(F_1\) on the liquid below it. The pressure applied on the piston \(A\) is then

\[P_1 = \frac{F_1}{A_1}\]

where \(F_1\) is the force applied on the smaller piston in N, \(A_1\) is its area of cross section in m², and \(P_1\) is the pressure produced in the liquid in Pa.

By Pascal’s law, this pressure is transmitted through the liquid, undiminished, to the piston \(B\). It acts upwards on that piston as a pressure \(P_2\) equal to \(P_1\):

\[P_2 = P_1\]

If the upward force exerted on piston \(B\) is \(F_2\), then \(F_2 = P_2 A_2\), that is

\[P_2 = \frac{F_2}{A_2}\]

Putting the three relations together, \(\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}\), which rearranges to

\[\frac{F_2}{F_1} = \frac{A_2}{A_1}\]

This holds for a liquid that is confined and at rest, with no leakage and no friction at the pistons. The chain of reasoning is set out end to end in FC 4.1.

Flowchart

FC 4.1 — From A Small Force To A Large Force Small force F₁ applied on the smaller piston A → pressure P₁ = F₁/A₁ produced in the liquid → pressure transmitted equally and undiminished through the liquid (Pascal’s law) → same pressure P₂ = P₁ acts on the larger piston B → large upward force F₂ = P₂ × A₂ on piston B

F₂/F₁ = A₂/A₁ Both pistons feel the same pressure, so each one gets the force that its own area is worth. Make the second piston ten times the area of the first, and it collects ten times the force. It applies to a confined liquid with no leakage and no friction.

Since \(A_2 > A_1\), it follows that \(F_2 > F_1\). A small force \(F_1\) applied on the smaller piston \(A\) produces a large force \(F_2\) on the bigger piston \(B\) — which is the principle of a hydraulic machine.

Force Obtained On The Larger Piston

1. Given: force applied on the smaller piston \(F_1 = 25\ \text{N}\); area of the smaller piston \(A_1 = 5\ \text{cm}^2\); area of the larger piston \(A_2 = 250\ \text{cm}^2\). To find: the force \(F_2\) obtained on the larger piston.

  1. The liquid is confined, so the pressure applied at the small piston reaches the large one undiminished, giving \(\dfrac{F_2}{F_1} = \dfrac{A_2}{A_1}\), that is \(F_2 = F_1 \times \dfrac{A_2}{A_1}\). The two areas appear only as a ratio, so they may both be left in cm² provided they are in the same unit.

  2. \(F_2 = 25\ \text{N} \times \dfrac{250\ \text{cm}^2}{5\ \text{cm}^2} = 25\ \text{N} \times 50 = 1250\ \text{N}\).

  3. The force on the larger piston is \(1250\ \text{N}\). The check: the larger piston has fifty times the area, and it collects fifty times the force — the gain in force is exactly the ratio of the areas.

When The Pistons Are Given By Their Diameters

1. Given: the two pistons of a hydraulic machine have diameters \(d_1 = 4\ \text{cm}\) and \(d_2 = 20\ \text{cm}\); a load requiring an upward force of \(F_2 = 2000\ \text{N}\) rests on the larger piston. Friction is neglected. To find: the force \(F_1\) that must be applied on the smaller piston.

  1. The relation is \(\dfrac{F_2}{F_1} = \dfrac{A_2}{A_1}\), but the pistons are given by diameter, not area. A piston face is a circle of area \(\dfrac{\pi d^2}{4}\), so in the ratio the factor \(\dfrac{\pi}{4}\) cancels and \(\dfrac{A_2}{A_1} = \left(\dfrac{d_2}{d_1}\right)^2\).

  2. \(\dfrac{A_2}{A_1} = \left(\dfrac{20\ \text{cm}}{4\ \text{cm}}\right)^2 = 5^2 = 25\). Therefore \(F_1 = \dfrac{F_2}{25} = \dfrac{2000\ \text{N}}{25} = 80\ \text{N}\).

  3. A force of \(80\ \text{N}\) on the smaller piston is enough. The check: the diameter is five times larger, so the area is twenty-five times larger — and the force needed is twenty-five times smaller, as a force multiplier should give.

4.7.2 A Hydraulic Machine As A Force Multiplier

A machine that turns a small force into a large one sounds too good to be true, so it is worth asking what it costs. Nothing comes free: in every hydraulic machine, the effort is less than the load, but the distance moved by the effort is more than the distance moved by the load.

In the ideal situation, where there is no energy lost to friction or leakage, the two balance exactly:

\[\text{work done by the effort} = \text{work done by the load}\]

The gain in force is measured by the mechanical advantage, and the loss in distance by the velocity ratio:

\[\text{M.A.} = \frac{\text{load}}{\text{effort}} > 1 \qquad \text{and} \qquad \text{V.R.} = \frac{\text{distance moved by the effort}}{\text{distance moved by the load}} > 1\]

Both are ratios of like quantities, so neither carries a unit. Since the mechanical advantage of a hydraulic machine is greater than one, it acts as a force multiplier.

Force multiplier and energy multiplier A hydraulic machine multiplies force, never energy. The big piston rises through only a small distance while the small piston travels a long way down. In the ideal case the work put in equals the work got out — you have traded distance for force, not created something from nothing.

In a hydraulic machine the two pistons have areas of cross section \(8\ \text{cm}^2\) and \(400\ \text{cm}^2\). A force of \(60\ \text{N}\) is applied on the smaller piston.

(a) Calculate the pressure produced in the liquid, in Pa.

(b) Calculate the force obtained on the larger piston.

(c) State the mechanical advantage of the machine.

4.8 Examples Of Hydraulic Machines

Three machines built on the principle of the last section are described here. In each of them, a modest force applied at one piston appears as a much larger force at another, because the pressure between them travels through a confined liquid.

4.8.1 Hydraulic Press

A hydraulic press, also called a Bramah press, works on the principle of Pascal’s law. It is used to squeeze and compress things very hard.

Construction Of The Press

Fig. 4.8-1 shows a hydraulic press. It consists of two hollow cylinders \(P\) and \(Q\), fitted with valves \(V_1\) and \(V_2\) respectively at their bottom. Cylinder \(P\) has a tank, or reservoir, at the bottom connected to it through the valve \(V_1\). The area of cross section of cylinder \(Q\) is larger than that of \(P\).

Diagram

Fig. 4.8-1 — Hydraulic Press Sectional working diagram. On the right, a narrow cylinder labelled ‘cylinder P’ containing a water-tight piston labelled ‘piston A (pump plunger)’; above it a lever pivoted at a fulcrum, with a handle H at its free end, the pivot labelled ‘pivot (fulcrum)’ and the bar labelled ‘lever’. At the bottom of cylinder P a valve labelled ‘valve V₁’ connects down to an open tank at the lower right labelled ‘water supply tank (reservoir)’, with downward arrows on the water surface in the tank labelled ‘atmospheric pressure’. A horizontal pipe labelled ‘connecting pipe R’ runs from the base of cylinder P leftwards to the base of a much wider cylinder labelled ‘cylinder Q’, through a valve labelled ‘valve V₂’. Inside cylinder Q is a large piston labelled ‘piston B (press plunger or ram)’, shown part way up, with a bale of cotton resting on it and a rigid beam above labelled ‘fixed roof’. At the base of cylinder Q, mark a tap-like valve labelled ‘release valve’ connected back towards the reservoir. Show liquid in both cylinders and the pipe with horizontal shading lines.

Water-tight pistons \(A\) and \(B\) are fitted in these cylinders, of areas of cross section \(A_1\) and \(A_2\) respectively, and the two cylinders are connected by a pipe \(R\). The piston \(A\) of the smaller cylinder is called the pump plunger, and the piston \(B\) of the larger cylinder is called the ram, or press plunger. A lever arrangement with a handle \(H\) is used to press down or raise up the pump plunger \(A\). At its bottom, cylinder \(Q\) is provided with a release valve which joins it to the reservoir.

How The Press Works

When a force \(F_1\) is applied on the piston \(A\), it exerts a pressure on the liquid in cylinder \(P\). By Pascal’s law this pressure is transmitted through the liquid in the tube \(R\) to the piston \(B\) of cylinder \(Q\), and piston \(B\) tends to move upwards. Since the area of cross section of cylinder \(P\) is less than that of cylinder \(Q\), a small force applied on piston \(A\) can lift a large weight kept on piston \(B\).

When no weight is placed on piston \(B\), it rises up against a fixed roof with a force \(F_2\) greater than \(F_1\). If a bale of cotton is kept on the press plunger \(B\), it is crushed between the plunger and the roof.

The pumping action itself works in two strokes. When the pump plunger \(A\) is raised by raising the handle \(H\), the pressure in cylinder \(P\) decreases and the valve \(V_1\) opens upwards. Water from the reservoir tank is then pushed up into cylinder \(P\) by the atmospheric pressure acting on the free surface of the water in the supply tank.

When the pump plunger \(A\) is pushed down by lowering the handle \(H\), the pressure in cylinder \(P\) increases, and this closes the valve \(V_1\). The pressure from cylinder \(P\) is now transmitted to the connecting pipe \(R\). As soon as the pressure in the pipe \(R\) becomes greater than that in cylinder \(Q\), the valve \(V_2\) opens, so water from cylinder \(P\) is forced into cylinder \(Q\). The press plunger \(B\) is raised against the fixed roof, and the bale of cotton on it is compressed.

Repeated strokes of the handle push more and more liquid into cylinder \(Q\), raising the ram further each time. When the work of the machine is over, the release valve is opened, so the ram \(B\) is lowered and the water of cylinder \(Q\) runs out into the reservoir.

Uses Of The Hydraulic Press

A hydraulic press is used mainly for the following purposes.

  1. For pressing cotton bales and goods like quilts and books.
  2. For extracting the juice from sugarcane and sugar beet.
  3. For squeezing oil out of linseed and cotton seeds.
  4. For engraving monograms on goods.

4.8.2 Hydraulic Jack

A hydraulic jack, or hydraulic lift, is used for lifting heavy vehicles such as cars and trucks in service stations, so that they can be worked on from below. It too works on Pascal’s principle, that is, on the principle of a hydraulic machine.

Construction Of The Jack

A simple form of a hydraulic jack is shown in Fig. 4.8-2. It consists of two cylindrical vessels \(P\) and \(Q\) connected to each other by a tube \(R\) having a valve \(V\). The piston \(A\) in the narrow cylinder \(P\) is attached to a lever, and the piston \(B\) of the wider cylinder \(Q\) carries a platform for lifting the vehicle. The vessels are filled with a liquid.

Diagram

Fig. 4.8-2 — Hydraulic Jack Sectional diagram. On the right, a narrow cylinder labelled P containing a piston labelled ‘piston A’, connected at its top to a lever pivoted at a point labelled ‘fulcrum F’, with a handle labelled ‘handle H’ at the free end and a downward arrow at the handle labelled ‘effort’. On the left, a much wider cylinder labelled Q containing a large piston labelled ‘piston B’, with an upward arrow on it; above piston B a flat platform labelled ‘platform’ carrying a car. The bases of the two cylinders are joined by a horizontal tube labelled ‘tube R’ fitted with a valve labelled ‘valve V’. Show the liquid filling both cylinders and the tube with horizontal shading lines and label it ‘liquid’.

How The Jack Works

When the handle \(H\) of the lever is pressed down by applying an effort, the pressure in cylinder \(P\) increases and the valve \(V\) opens. The liquid then runs out from cylinder \(P\) into cylinder \(Q\). As a result, the piston \(B\) rises and lifts the car standing on the platform.

When the car has reached the required height, the handle \(H\) is no longer pressed. The pressure on either side of the valve \(V\) then becomes the same, so the valve closes, and the liquid cannot run back from cylinder \(Q\) into cylinder \(P\). This is what holds the car steady at that height.

Hydraulic press and hydraulic jack Both are force multipliers, and both use Pascal’s law. The difference is what the large force is used for. In a press, the ram pushes up against a fixed roof and crushes whatever lies between. In a jack, the large piston is free to rise, and it lifts a load instead.

4.8.3 Hydraulic Brakes

The brakes used in cars and other vehicles are also based on Pascal’s principle.

Construction Of The Brake System

Fig. 4.8-3 shows the hydraulic brake arrangement of a vehicle. It consists of a pipe line \(R\) containing a liquid — an oil — one end of which is connected to the master cylinder \(P\), fitted with a piston \(A\) attached to the foot pedal. The other end of the pipe \(R\) is connected to the brake arrangement at the wheels of the vehicle; Fig. 4.8-3 shows only one wheel connected to the pipe line.

Diagram

Fig. 4.8-3 — Hydraulic Brake Sectional diagram. At the top right, a foot pedal labelled ‘foot pedal’ with an arrow showing it being pressed, connected to a piston labelled ‘piston A’ inside a cylinder labelled ‘master cylinder P’. From the master cylinder, a horizontal pipe labelled ‘pipe line R’ runs leftwards, filled with liquid and labelled ‘liquid’; a branch continues off to the left labelled ‘to other wheel’. Below, draw the wheel as a large circle labelled ‘rim of wheel’, with a hinge at the bottom labelled ‘hinge’. Inside the rim, two curved brake shoes labelled ‘brake shoe’ face outwards towards the rim. Between the upper ends of the two shoes, a small cylinder labelled ‘wheel cylinder Q’ containing two pistons labelled ‘piston B₁’ and ‘piston B₂’ pushed outwards in opposite directions by arrows. Between the lower ends of the shoes, draw a coiled spring labelled ‘spring’. Mark the wheel cylinder as clearly wider in cross section than the master cylinder.

For each wheel there is a wheel cylinder \(Q\) having two pistons \(B_1\) and \(B_2\), attached to the brake shoes. The area of cross section of the wheel cylinder \(Q\) is greater than that of the master cylinder \(P\). The brake shoes press against the rim of the wheel.

How The Brakes Work

To apply the brakes, the foot pedal is pressed. This exerts a pressure on the liquid in the master cylinder \(P\), so liquid runs out from the master cylinder \(P\) into the wheel cylinder \(Q\), and the pressure is transmitted equally and undiminished through the liquid to the pistons \(B_1\) and \(B_2\).

The two pistons are therefore pushed outwards, and the brake shoes are pressed against the rim of the wheel, so the motion of the vehicle is retarded. Since the area of cross section of piston \(A\) in the master cylinder is less than that of the pistons in the wheel cylinder, a small force applied at the foot pedal produces a large force on the pistons \(B_1\) and \(B_2\). It is this large force that stops the wheel.

Because the pressure is transmitted through a single connected pipe line \(R\), equal pressure is exerted at every wheel of the vehicle connected to it. All the wheels are therefore braked together and by the same amount, which is what keeps the vehicle straight as it slows.

On releasing the pressure on the pedal, the spring pulls the brake shoes back to their original position and forces the pistons \(B_1\) and \(B_2\) back into the wheel cylinder \(Q\). The liquid runs back from the wheel cylinder \(Q\) to the master cylinder \(P\), and the brakes are released.

This whole arrangement depends on the liquid filling the pipe line completely. If air leaks into the line, the pedal pushes on a pocket of gas, and a gas is easily squeezed — so much of the movement of the pedal is spent compressing the air instead of pressing the shoes against the rim. A soft or spongy pedal is a warning that the braking force has been lost, and the vehicle must not be driven until it has been repaired.

Q5. A driver presses one small pedal with one foot, and a car weighing over a tonne comes to a stop. How can such a small push be turned into such a large force?

A. The pedal presses a small piston in the master cylinder, and the pressure this produces travels undiminished through the brake oil to a much wider piston at each wheel. Since \(F_2 = P \times A_2\), the wider piston collects a far larger force than was applied at the pedal, and that force presses the brake shoes hard against the rims of all the wheels at once.

Force At A Brake Shoe

1. Given: area of cross section of the piston in the master cylinder \(A_1 = 2\ \text{cm}^2\); area of cross section of each piston in the wheel cylinder \(A_2 = 30\ \text{cm}^2\); force applied at the pedal on the master cylinder piston \(F_1 = 40\ \text{N}\). Friction is neglected. To find: the force on each piston of the wheel cylinder.

  1. The brake oil is confined, so the pressure produced in the master cylinder reaches the wheel cylinder undiminished, and \(\dfrac{F_2}{F_1} = \dfrac{A_2}{A_1}\). As before, the areas may stay in cm² because only their ratio is used.

  2. \(F_2 = F_1 \times \dfrac{A_2}{A_1} = 40\ \text{N} \times \dfrac{30\ \text{cm}^2}{2\ \text{cm}^2} = 40\ \text{N} \times 15 = 600\ \text{N}\).

  3. Each piston of the wheel cylinder pushes its brake shoe outwards with a force of \(600\ \text{N}\). The check: the wheel piston has fifteen times the area, so it collects fifteen times the force — and this happens at every wheel on the same pipe line at the same time.

Hydraulics On A Building Site The arm and bucket of an excavator are moved by hydraulic rams — closed cylinders in which oil, pushed by a pump, drives a piston out along a polished steel rod. A machine of a few tonnes can lift and swing a bucket loaded with rubble because the oil pressure acts over the large area of the piston face, giving a force far greater than the pump alone applies. The same arrangement raises the tipping body of a lorry and closes the doors of a bus. Oil is used rather than water throughout, since it will not rust the steel it works against and it lubricates the sliding surfaces as it goes.

Why brake fluid is not water The oil in a brake line has a very high boiling point and a very low freezing point. Braking makes the parts near the wheels hot, and water there would boil into steam — a gas, easily compressed — leaving the pedal soft and the brakes weak. Water would also freeze in a cold winter and rust the steel parts.

In the hydraulic brake system of a car, the piston of the master cylinder has an area of cross section \(3\ \text{cm}^2\) and each piston in a wheel cylinder has an area of cross section \(24\ \text{cm}^2\). A force of \(90\ \text{N}\) is applied on the master cylinder piston through the foot pedal.

(a) Calculate the force exerted on each piston of the wheel cylinder.

(b) State whether the force at the other wheels of the car is larger, smaller or the same, and give the reason.

4.9 Summary

  1. Thrust is the force acting normally on a surface. It is a vector quantity. Its SI unit is the newton (N) and its CGS unit is the dyne, where \(1\ \text{N} = 10^{5}\ \text{dyne}\).

  2. The thrust exerted by a body placed on a surface is equal to the weight of the body, and it is the same whichever way round the body is placed.

  3. The gravitational units of thrust are the kgf and the gf, where \(1\ \text{kgf} = 9.8\ \text{N}\) and \(1\ \text{gf} = 980\ \text{dyne}\).

  4. Pressure is the thrust per unit area of a surface. It is a scalar quantity. \[P = \frac{F}{A}\] where \(F\) is the thrust in N, \(A\) is the area on which it acts in m², and \(P\) is the pressure in Pa.

  5. The SI unit of pressure is the pascal (Pa), where \(1\ \text{Pa} = 1\ \text{N m}^{-2}\). One pascal is the pressure exerted on a surface of area \(1\ \text{m}^2\) by a force of \(1\ \text{N}\) acting normally on it. The CGS unit is dyne cm⁻², where \(1\ \text{dyne cm}^{-2} = 0.1\ \text{N m}^{-2}\).

  6. Other units of pressure are the bar and the millibar, where \(1\ \text{bar} = 10^{5}\ \text{N m}^{-2}\) and \(1\ \text{millibar} = 10^{-3}\ \text{bar} = 10^{2}\ \text{N m}^{-2}\).

  7. Atmospheric pressure at sea level supports a mercury column of \(0.76\ \text{m}\), and this is taken as one atmosphere: \[1\ \text{atm} = 0.76\ \text{m of Hg} = 1.013 \times 10^{5}\ \text{Pa}, \qquad 1\ \text{torr} = 1\ \text{mm of Hg}, \qquad 1\ \text{atm} = 760\ \text{torr}\]

  8. The pressure on a surface depends on two factors only — the magnitude of the thrust, and the area on which the thrust acts. For a given thrust, a larger area gives a smaller pressure; for a given area, a larger thrust gives a greater pressure.

  9. For a given thrust, pressure is increased by reducing the area of the surface, as in the pointed tips of nails and the sharp edges of cutting tools. It is decreased by increasing the area, as in the wide wooden sleepers below railway tracks and the wide foundation walls of buildings.

  10. A fluid is a substance which can flow. All liquids and gases are fluids. A solid exerts pressure only on the surface on which it rests, but a fluid contained in a vessel exerts pressure at all points and in all directions, on the walls as well as on the base.

  11. Liquid spurting from holes in the wall of a vessel shows that a liquid exerts pressure at every point on the wall, and that the jet from a deeper hole is thrown farther — so liquid pressure increases with depth below the free surface.

  12. The pressure exerted by a liquid column at a depth \(h\) is \[P = h\rho g\] where \(h\) is the depth below the free surface in m, \(\rho\) is the density of the liquid in kg m⁻³, \(g\) is the acceleration due to gravity in m s⁻², and \(P\) is the pressure in Pa. It holds for a stationary liquid of uniform density and gives the pressure due to the liquid alone.

  13. Taking the atmospheric pressure \(P_0\) acting on the free surface into account, \[\text{total pressure at a depth } h = P_0 + h\rho g\] where \(P_0\) and \(h\rho g\) are both in Pa.

  14. The pressure at a point in a stationary liquid depends on three factors: the depth \(h\), the density \(\rho\), and \(g\). At a given place \(g\) is constant, so the pressure is directly proportional to \(h\) and directly proportional to \(\rho\). It does not depend on the shape or size of the vessel, nor on the area of the surface on which it acts.

  15. The five laws of liquid pressure: pressure increases with depth below the free surface; in a stationary liquid it is the same at all points on a horizontal plane; it is the same in all directions about a point; at the same depth it is different in different liquids, increasing with density; and a liquid seeks its own level.

  16. Consequences of \(P = h\rho g\): the pressure at a given depth is greater in sea water than in river water, because sea water is denser; the wall of a dam is made thicker towards its base, where the water presses hardest; a water supply tank is placed high, so that the tall water column gives a good pressure at the taps; a deep-sea diver wears a protective suit of glass reinforced plastic or cast aluminium, kept at one atmosphere inside, because the total pressure outside far exceeds the blood pressure; and a gas bubble grows as it rises in a lake, because the total pressure on it falls and, by Boyle’s law, its volume increases.

  17. Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid. It is demonstrated by the equal heights of the jets rising from tubes at the sides and bottom of a flask when its piston is pushed in.

  18. In a hydraulic machine, a small force applied on a smaller piston produces a large force on a bigger piston. If a force \(F_1\) acts on a piston of area \(A_1\) and a force \(F_2\) is obtained on a piston of area \(A_2\), then \[\frac{F_1}{A_1} = \frac{F_2}{A_2} \qquad \text{or} \qquad \frac{F_2}{F_1} = \frac{A_2}{A_1}\] where \(F_1\) and \(F_2\) are in N and \(A_1\) and \(A_2\) are in m². Since \(A_2 > A_1\), it follows that \(F_2 > F_1\). This holds for a confined liquid with no leakage and no friction.

  19. In a hydraulic machine the effort is less than the load, while the distance moved by the effort is more than the distance moved by the load, so that in the ideal case the work done by the effort equals the work done by the load. Since \[\text{M.A.} = \frac{\text{load}}{\text{effort}} > 1\] a hydraulic machine acts as a force multiplier. The velocity ratio, the distance moved by the effort divided by the distance moved by the load, is also greater than one. Both ratios have no unit.

  20. Machines working on Pascal’s law include the hydraulic press, used for pressing cotton bales, extracting juice from sugarcane, squeezing oil from seeds and engraving monograms; the hydraulic jack, used for lifting vehicles in service stations; and hydraulic brakes, in which a small force at the foot pedal produces a large force at the brake shoes of every wheel connected to the pipe line.

Flowchart

FC 4.2 — The Spine Of The Chapter Force acting normally on a surface (thrust) → thrust shared over the area it acts on, P = F/A (pressure) → a fluid flows, so it presses at all points and in all directions → the weight of the liquid column above a point gives P = hρg, increasing with depth and with density → pressure applied to a confined liquid is transmitted equally and undiminished (Pascal’s law) → the same pressure acting on a larger piston gives a larger force, F₂/F₁ = A₂/A₁ → hydraulic press, hydraulic jack and hydraulic brakes

4.10 Key Terms

  1. Atmospheric pressure — the pressure exerted by the air of the atmosphere on everything below it; usually stated as the height of the mercury column it supports, where one atmosphere is \(0.76\ \text{m}\) of mercury, that is \(1.013 \times 10^{5}\ \text{Pa}\). SI unit: pascal (Pa).

  2. Density — the mass of unit volume of a substance; it decides how heavy a given column of a liquid is. SI unit: kilogramme per cubic metre (kg m⁻³).

  3. Fluid — a substance which can flow, and which therefore presses on the walls of its container as well as on its base. All liquids and gases are fluids.

  4. Pascal — the SI unit of pressure; one pascal is the pressure exerted on a surface of area \(1\ \text{m}^2\) by a force of \(1\ \text{N}\) acting normally on it, so \(1\ \text{Pa} = 1\ \text{N m}^{-2}\).

  5. Pascal’s law — the statement that pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid; it is the basis of every hydraulic machine.

  6. Pressure — the thrust acting on unit area of a surface, \(P = F/A\); a scalar quantity. SI unit: pascal (Pa), or N m⁻².

  7. Thrust — the force acting normally, that is at right angles, on a surface; a vector quantity, equal to the weight of a body resting on a horizontal surface. SI unit: newton (N).

4.11 Points To Ponder

  1. The derivation of \(P = h\rho g\) never asked how wide the vessel was, yet the answer applies to a narrow test tube and to a reservoir behind a dam alike. What does that tell you about which liquid is actually ‘holding up’ a point deep inside a lake?

  2. A dam holds back a lake that stretches for kilometres behind it. Would the wall have to be built any thicker if the lake were twice as long, but no deeper? Argue your answer from the relation for liquid pressure.

  3. Pressure is defined as thrust divided by area, and both thrust and area have directions associated with them — yet pressure turns out to be a scalar. What is it about the way a fluid pushes that makes a single number enough to describe the pressure at a point?

  4. A hydraulic machine can turn a force of \(50\ \text{N}\) into a force of \(5000\ \text{N}\). If forces can be multiplied so freely, why can no machine be built that multiplies energy in the same way? What would have to be true of the world for such a machine to work?

  5. Pascal’s law is stated for a confined liquid. Air is also a fluid, and it can be confined in a cylinder too. Would a hydraulic press work just as well if the oil in it were replaced by air? Think about what a gas does when you press on it.

  6. Suppose the same experiment with a vessel of water and holes in its wall were carried out far out in space, in a spacecraft where objects float freely. Which of the observations would still be seen, and which would not? Use \(P = h\rho g\) to say why.

4.12 Objective Questions

  1. The SI unit of thrust is: (1 mark)
    1. pascal
    2. newton
    3. newton per square metre
    4. dyne per square centimetre
  2. One kilogramme-force is equal to: (1 mark)
    1. \(0.98\ \text{N}\)
    2. \(9.8\ \text{N}\)
    3. \(980\ \text{N}\)
    4. \(10^{5}\ \text{N}\)
  3. The pressure at a point inside a stationary liquid does not depend on: (1 mark)
    1. the depth of the point below the free surface
    2. the density of the liquid
    3. the shape of the vessel holding the liquid
    4. the acceleration due to gravity
  4. One bar is equal to: (1 mark)
    1. \(10^{2}\ \text{N m}^{-2}\)
    2. \(10^{3}\ \text{N m}^{-2}\)
    3. \(10^{5}\ \text{N m}^{-2}\)
    4. \(10^{6}\ \text{N m}^{-2}\)
  5. Two identical small holes are made in the wall of a tall vessel of water, one at a depth of \(10\ \text{cm}\) and the other at a depth of \(40\ \text{cm}\) below the free surface. The jet from the deeper hole is thrown: (1 mark)
    1. a shorter distance
    2. a greater distance
    3. the same distance
    4. vertically upwards
  6. In a hydraulic machine, the area of the larger piston is twenty times that of the smaller piston. A force of \(30\ \text{N}\) applied on the smaller piston gives a force on the larger piston of: (1 mark)
    1. \(1.5\ \text{N}\)
    2. \(20\ \text{N}\)
    3. \(30\ \text{N}\)
    4. \(600\ \text{N}\)
  7. Thrust and pressure are respectively: (1 mark)
    1. scalar and scalar quantities
    2. vector and vector quantities
    3. vector and scalar quantities
    4. scalar and vector quantities

For questions 8 to 10, choose the answer from the codes given below. a. Both A and R are true, and R is the correct explanation of A b. Both A and R are true, but R is not the correct explanation of A c. A is false but R is true d. A is true but R is false

  1. Assertion (A): A sharp knife cuts more easily than a blunt one. Reason (R): For a given thrust, reducing the area of the surface increases the pressure exerted on it. (1 mark)

  2. Assertion (A): An overhead water tank is built several metres above the taps it supplies. Reason (R): The pressure due to a liquid column is directly proportional to the height of the column. (1 mark)

  3. Assertion (A): A gas bubble shrinks as it rises from the bed of a lake to the surface. Reason (R): The total pressure on the bubble decreases as its depth below the free surface decreases. (1 mark)

  4. Match each item in Column A with the correct item in Column B. (4 marks)

Column A Column B
a. Pascal i. unit of atmospheric pressure equal to 1 mm of Hg
b. Torr ii. SI unit of pressure
c. Ram iii. the wider piston of a hydraulic press
d. Dyne iv. CGS unit of thrust
  1. Fill in the blanks.
    1. The force acting normally on a surface is called __________. (1 mark)
    2. For a given thrust, the pressure on a surface is decreased by __________ the area of the surface. (1 mark)
    3. The pressure at a point inside a liquid is __________ proportional to the density of the liquid. (1 mark)
    4. Pressure applied to a confined liquid is transmitted __________ and __________ in all directions throughout the liquid. (1 mark)
  2. State whether the following statements are true or false.
    1. The thrust exerted by a brick on a table changes when the brick is turned to rest on a smaller face. (1 mark)
    2. A fluid contained in a vessel exerts pressure on the walls of the vessel as well as on its base. (1 mark)
    3. A hydraulic machine multiplies the energy supplied to it. (1 mark)
  3. Name the following.
    1. Name the quantity whose SI unit is the pascal. (1 mark)
    2. Name the law on which the working of hydraulic brakes is based. (1 mark)
    3. Name the two factors on which the pressure exerted on a surface depends. (1 mark)
  4. Give one word or phrase for each of the following.
    1. A substance which can flow. (1 mark)
    2. The piston of the larger cylinder of a hydraulic press, on which the goods are compressed. (1 mark)
  5. State the SI unit of the following.
    1. Thrust. (1 mark)
    2. Pressure. (1 mark)
    3. Density of a liquid. (1 mark)
  6. Pick the odd one out and give a reason for your choice.
    1. Pascal, bar, newton, torr. Give a reason for your choice. (1 mark)
    2. Hydraulic press, hydraulic jack, hydraulic brake, railway sleeper. Give a reason for your choice. (1 mark)

4.13 Check Your Concepts

  1. Define thrust and state its SI unit. (2 marks)

  2. What is a fluid? Give one example of a fluid which is a liquid and one which is a gas. (2 marks)

  3. State the two factors on which the pressure exerted by a thrust on a surface depends. (2 marks)

  4. Write the relation for the pressure at a depth inside a liquid, and explain the meaning of each symbol used, along with its unit. (2 marks)

  5. Differentiate between thrust and pressure, stating the basis of difference in each case. (3 marks)

  6. State the three factors on which the pressure at a point inside a stationary liquid depends. Name two things on which it does not depend. (3 marks)

  7. Give a reason for each of the following. (3 marks)

    1. The tip of a drawing pin is made sharp.
    2. Wide wooden sleepers are laid below railway tracks.
    3. The foundation walls of a building are made wider than the walls above.

  8. State Pascal’s law, including the condition under which it applies. (3 marks)

  9. Why does a sea diver working at a great depth need a special protective suit? At what pressure is the inside of such a suit maintained? (3 marks)

  10. State the five laws of liquid pressure. (4 marks)

  11. How does the pressure exerted by a solid differ from that exerted by a fluid? Explain the reason for the difference. (4 marks)

  12. Deduce the expression for the pressure at a depth \(h\) inside a liquid of density \(\rho\). State clearly what happens to the area of the chosen surface in the course of the derivation, and why that matters. (5 marks)

  13. Describe a simple experiment to show that a liquid enclosed in a vessel transmits pressure equally in all directions. State the observation and the conclusion drawn from it. (5 marks)

  14. Explain the principle of a hydraulic machine and obtain the relation between the forces on the two pistons and their areas of cross section. Why is such a machine called a force multiplier, and what is given up in return for the gain in force? (5 marks)

4.14 Diagrams, Graphs & Constructions

  1. Fig. 4.1-1 shows the same brick resting on the ground in two different positions. State in which position it exerts the greater pressure on the ground, give the reason, and state whether the thrust differs between the two positions. (3 marks)

  2. Draw a labelled diagram of a tall vessel of water with three small holes made in its wall at different depths, showing the jets of water leaving them. Mark the depth of each hole below the free surface, and state what your diagram shows about liquid pressure. (4 marks)

  3. Fig. 4.3-2 shows how the pressure due to a liquid varies with depth below the free surface for two liquids. (4 marks)

    1. Read from line A the pressure in water at a depth of \(3\ \text{m}\).
    2. Calculate the gradient of line A, and state the physical quantity it represents.
    3. State, with a reason, which of the two liquids is denser.

  4. Sketch the graph you would expect for the pressure due to a liquid, plotted against the depth below its free surface, for a liquid of uniform density. Label both axes with their units, and state why the line passes through the origin. (3 marks)

  5. Fig. 4.6-2 shows the apparatus used to demonstrate Pascal’s law. Draw the apparatus, label the piston, the flask and the tubes, and show what is observed when the piston is pushed gently down. (4 marks)

  6. Draw a simple labelled diagram of a hydraulic jack, marking the two cylinders, the two pistons, the connecting tube, the valve and the platform. State briefly how the load is held in position once it has been raised. (5 marks)

  7. Fig. 4.8-3 shows the hydraulic brake arrangement of a vehicle. Name the parts labelled as the master cylinder, the wheel cylinder and the pistons \(B_1\) and \(B_2\), and state the function of each. (4 marks)

4.15 Practice With Numericals

  1. A force of \(30\ \text{N}\) acts normally on a surface of area \(0.02\ \text{m}^2\). Calculate the pressure exerted on the surface. (2 marks)

  2. A wooden box weighing \(400\ \text{N}\) rests on a floor on a face measuring \(0.80\ \text{m} \times 0.50\ \text{m}\). Calculate (a) the thrust on the floor and (b) the pressure on the floor. (3 marks)

  3. Calculate the pressure due to water at a depth of \(8\ \text{m}\) below the surface of a lake. Take the density of water as \(10^{3}\ \text{kg m}^{-3}\) and \(g = 9.8\ \text{m s}^{-2}\). (2 marks)

  4. A tank of base area \(0.25\ \text{m}^2\) contains water to a depth of \(2.0\ \text{m}\). Taking the density of water as \(10^{3}\ \text{kg m}^{-3}\) and \(g = 9.8\ \text{m s}^{-2}\), calculate (a) the pressure at the base due to the water and (b) the thrust on the base. (3 marks)

  5. At what depth below the free surface of kerosene of density \(800\ \text{kg m}^{-3}\) is the pressure due to the kerosene equal to \(1.568 \times 10^{4}\ \text{Pa}\)? Take \(g = 9.8\ \text{m s}^{-2}\). (3 marks)

  6. Calculate the total pressure at a depth of \(5.0\ \text{m}\) below the surface of a lake. Take the atmospheric pressure as \(1.013 \times 10^{5}\ \text{Pa}\), the density of water as \(10^{3}\ \text{kg m}^{-3}\) and \(g = 9.8\ \text{m s}^{-2}\). (3 marks)

  7. A vessel of base area \(0.60\ \text{m} \times 0.40\ \text{m}\) is filled to a depth of \(50\ \text{cm}\) with a liquid of density \(1.2 \times 10^{3}\ \text{kg m}^{-3}\). Taking \(g = 9.8\ \text{m s}^{-2}\) and ignoring the atmospheric pressure, calculate (a) the pressure at the base and (b) the thrust on the base. (4 marks)

  8. In a hydraulic machine, the pistons have areas of cross section \(4\ \text{cm}^2\) and \(100\ \text{cm}^2\). A force of \(50\ \text{N}\) is applied on the smaller piston. Calculate the force obtained on the larger piston, and state the mechanical advantage of the machine. (3 marks)

  9. A nail is driven into a plank with a thrust of \(60\ \text{N}\). The area of its pointed tip is \(0.1\ \text{mm}^2\) and the area of its flat head is \(1\ \text{cm}^2\). Calculate the pressure at the tip and the pressure at the head, and state how many times greater one is than the other. (4 marks)

  10. The two pistons of a hydraulic machine have diameters \(5\ \text{cm}\) and \(30\ \text{cm}\). A load requiring an upward force of \(3600\ \text{N}\) rests on the larger piston. Neglecting friction, calculate the force that must be applied on the smaller piston, and state the assumption you have made. (4 marks)