Going swimming

When was the last time you went for a swim? It is a common experience that when you duck your head under the surface of the water you feel that characteristic effect on your ears. We say (quite correctly) that this is due to the pressure of the water.

If you hold your head straight up under the water you feel the effects of the pressure; if you turn it sideways, it's just the same effect. Furthermore, if you go deeper in the water the effect is stronger.

The first-hand experience gained in the sea or a swimming pool is absolutely in line with what physics has to say: The pressure in a fluid, such as water, results in the same force acting on a surface arranged in any direction.

No matter how you hold your head, at a given depth in the pool you feel the same effect on your ears.

In this respect, pressure is a different kind of quantity to force:

• A force is exerted by an environment in a specific direction, which we can show with an arrow, and is called a vector quantity (it has magnitude and direction).
• The pressure in a fluid just is and is called a scalar quantity (it has magnitude only).

For example, it makes no sense to say that the pressure at the bottom of the swimming pool is very large, straight downwards. The pressure is just very large.

Here's a summary:

Force is a vector, so can be represented by an arrow. It's useful to say that a force acts on or is exerted by an object.

Pressure is a scalar, so cannot be represented by an arrow. It's useful to talk about the pressure in a fluid.

Calculating the force

So, if you hold your head under water you can detect the pressure of the surrounding water with your ears. (Your eardrum acts as a difference in pressure detector. The bombarding water molecules exert a force on the exposed area of your ear drums. The air particles in your ear bombard the inside and also exert a force. The resultant of these two forces is what you feel.) The size of this force is given by a simple relationship: force = pressure × area.

You can always equalise the pressure by holding your nose, keeping your mouth shut and exhaling, resulting in more particles bombarding the inside. You can probably adjust the pressure so the force acting on the inside and the force acting on the outside are identical, at least within the swimming pool.

Here's a precise way of writing it out, so that every term is just a number:

forcenewton = pressurenewton / metre² × areametre²

You can also write (making notes to yourself about the units) as force in newton = pressure in newton / metre² × area in metre²

The first form is more concise and precise, as each term reduces to just a number: number  ×  unit divide by unit leaves just number.

(Remember that a physical quantity is a number  ×  a unit.)

Rearrangements

You can rearrange this to make pressure the subject of the formula, but the relationship is the same. You can write, rather fully, (making notes to yourself about the units):

pressure in newton / metre² = force in newtonarea in metre²

Or even express it rather concisely as:

pressure = forcearea

Most concisely, you could write:

P = FA

You'll need to match the representation you choose to the classes you have, so as to allow them to understand the relationship.

In other words, a big pressure in the water will produce a big force acting over the exposed area of your ear drum. If the pressure is too big you may suffer a perforated ear drum due to the force acting.

Pressure and depth

As stated earlier, the pressure in a liquid, such as water, increases with depth. You're also probably aware that this is true of the atmosphere, where altitude sickness sets in at a few thousand metres above sea level, as the density of air is much reduced. So in both liquids and gases, that is fluids, the pressure increases with increasing depth of fluid.

This is an additional complication that is best avoided in an introductory study of pressure. In particular, the relationship between area, force and pressure is simpler to deal with if there is only one pressure throughout the whole fluid. Throughout this episode we shall not deal with these changes of pressure with depth, always choosing our examples carefully so that the depth of fluid is not too great.

Differences in air pressure give resultant forces

Have you seen the demonstration where the air is removed from an aluminium can? The open can sits on the bench totally at rest, but as soon as the air is removed from inside it (with some kind of air pump) the sides of the can cave in spectacularly.

At first the open can is surrounded by air inside and out. The pressure in the air on both the inside and the outside of the can is the same. The air pressure on the outside exerts a force on the outside surface of the can, acting inwards (just as the water of the pool does on your ears). The air pressure on the inside exerts a force on the inside surface of the can, acting outwards. As the two pressures are the same, the two forces on each small section of can are equal also. Each section of the can is in equilibrium as these two forces are equal and opposite.

Although the forces acting on the sides of the can are large the resultant force is zero, so nothing happens. However, when air is removed from inside the can, the can crumples because the large force due to the outside air pressure is still there, acting inwards on the outside of the can. This is a spectacular demonstration of the fact that gases, including the air which is all around us, have a pressure just like liquids.

Pressure results in forces

Look at the forces acting on these surfaces. They are all linked by the pressure of the fluid inside the container and depend on the chosen area of the surface. This is a useful and interesting connection.

Notice that the direction of the force is set by the orientation of the area. The force acts on the area because of the differential bombardment by the particles inside and out (so the pressure difference).

Also note the relationship between pressure in a fluid, the area that you choose to expose to the fluid and the force acting on that surface – always at right angles to that surface.

Hydraulics

Every day on building sites machines move huge amounts of earth that would take several people many days to shift on their own. These machines rely on hydraulics – look out for the large pistons in large cylinders at work.

Just how are such large forces generated to push earth around in this way?

Simple hydraulic systems

The simplest hydraulic system consists of a pair of cylinders with moveable pistons in each and the whole system filled with a fluid. Push one piston in and the other moves out.

If the area of the input piston and the output piston are the same then the forces will be the same. The pressure of the hydraulic fluid is the same throughout the machine and equal to force divided by area. If the pistons have the same surface area there will be the same force acting on them.

• An output piston with a larger area has more force acting on it by the fluid.
• An output piston with a smaller area has less force acting on it by the fluid.

In other words a hydraulic machine can produce a large force by simply having a larger area output cylinder and piston.

Seen in this way, there are clear parallels between hydraulic machines and simple levers.

Designing hydraulic machines

Designing hydraulic machines involves choosing the relative areas of the pistons in just the same way that the design of levers involves choosing the relative lengths of the line of action of the forces from the pivot point.

Being able to tap into the pressure in the fluid at any point makes hydraulic machines quite simple to engineer. You can easily design a machine that exerts an output force of some fixed size compared to the input force, at any angle you choose.

You can also easily have the same effect in two different places, simply by creating two pistons of equal cross-sectional area. This can be used, for example, to exert equal braking forces on all four wheels of a car.

Producing a big force by providing a large area output piston seems to be too good to be true. There must be a catch! The downside is that the output piston which exerts the large force cannot move very far (compared to the input piston). It's just like a lever.

For example:

If the output piston has four times the area of the input piston and the output force is therefore four times the input force then the output piston will move only one quarter the distance of the input piston.

Teacher Tip: Once again, the hydraulic machine acts exactly like a simple lever (where you must move the effort force through four times the distance to increase the load force four times).

A look at how levers and hydraulic systems are comparable

Notice the similarities between the lever systems to the hydraulic systems. The areas of the pistons and the lengths to the pivot have similar effects.

Avoiding compressibility

Liquids are used as the fluid in hydraulic systems. In liquids the particles are close together and so liquids are virtually incompressible. We can therefore assume that the volume of liquid inside the hydraulic system is constant, even when the pistons start moving up and down the cylinders.

In some systems gases are used as the fluid and these are called pneumatic systems. Gases are of low density and cannot therefore be used directly because of their compressibility (the gas in the cylinder would simply compress under the action of the piston). This problem is avoided by using compressed air in pneumatic systems.

The air particles are squeezed closer together to reduce the volume whilst increasing the pressure (there is the same number of particles in a smaller volume, so there are likely to be more collisions with the walls of the container).

Moving fluid: shows how hydraulic systems work in a fundamental way

Now for a thought experiment. Suppose you try moving liquid from the left hand side of the hydraulic system to the right hand side and then adjust the pistons so that they sit on top of the fluid.

The smaller the cross sectional area of the input cylinder, the less liquid needs to be removed before the piston can drop. But this liquid has to go somewhere, and the only place to go, in a sealed system, is to the other output cylinder. Given that the cross-sectional area of this cylinder is larger, the piston will rise less for each cubic metre of liquid fed into it.

These two cylinders constitute a hydraulic system, where a large movement on one side (the input side) creates a smaller movement on the other (the output side).

This confirms the point made earlier that the relative area of the cylinders determines the relative up and down movement of the two pistons. If the second cylinder is bigger the travel of the piston in that cylinder is smaller. As we have just shown, this is due to the fact that no liquid is lost as it moves from one cylinder to the other.

The argument repeated, with algebra

We can follow the argument about shifting fluid with the help of a few simple equations:

volume of liquid pushed out of the left hand side = cross-sectional area of left hand side × height dropped

volume of liquid pushed into the right hand side = cross-sectional area of right hand side × height risen

But the volumes must be equal, therefore:

cross-sectional area of left hand side × height dropped = cross-sectional area of right hand side × height risen

Rewrite this as:

area_left × height_left = area_right × height_right

You can rearrange this to give:

area_leftarea_right = height_rightheight_left

So the changes in the heights are just proportional to the cross-sectional areas.

How fundamental principles of energy conservation apply to hydraulic systems

We can use the idea of conservation of energy to show how the forces produced at the pistons are related to the distances moved by the pistons. There is no need to bother your pupils with this section of argument; we include it here to demonstrate how the working of a hydraulic machine is underpinned by the fundamental principle of conservation of energy.

For the hydraulic machine, three relationships are true:

energy input is equal to energy output and energy input = input force × distance moved by input force and energy output = output force × distance moved by output force.

Therefore:

input force × distance input force moves = output force × distance output force moves.

Rearranging the equation reveals that:

input forceoutput force = distance moved by output forcedistance moved by input force .

So we can design systems that either:

Multiply forces: The relatively small input force is applied through a big distance to produce a big output force but acting over a much smaller distance.

Or:

Multiply distances: The relatively large input force is applied through a small distance to produce a small output force but acting over a much bigger distance.

There is always this trade-off between force and distance.

What is pressure?

Here the macroscopic phenomenon of pressure is related to the actions of the particles.

Fluids – liquids and gases – consist of many small particles in constant random motion. These particles exert forces on any surfaces that they happen to collide with. The force arising from the pressure is therefore due to particle bombardment on surfaces. The pressure is a measure of the state of the fluid: the pressure in the fluid. One thing this measure can predict is the force that will be acting on an area bounding the fluid.

Notice that the direction of the force is set by the orientation of the area. The force acts on the area because of the differential bombardment by the particles inside and out (so the pressure difference). We think its helpful to talk about the pressure in the fluid and the force acting on the area.

It is the bombardment of water molecules on your ears that creates the force on your eardrums. If the water pressure is big (because you dive down to the bottom of the ocean) the collisions of the water molecules on your eardrums are more frequent and a greater force is generated.

We need to picture the action of pressure (always in fluids) as being due to particles colliding with surfaces. Decreasing the volume leads to more collisions, therefore a greater pressure.

Pressure and energy

The conservation of energy helps us to understand pressure. We can also think about pressure as a measure of the compressibility of the fluid. To see what this means read on.

Assume you exert a force on an input piston. For a given volume of fluid being shifted from the input cylinder, the distance moved is calculated by: distance moved by input piston = volume removedarea of piston.

Remembering that energy = input force × distance and Putting these two relationships together, you get:

energy = force  ×  volume removedarea of piston.

Dividing both sides by the volume gives energy inputvolume removed = input forcearea of piston .

The left hand side tells us something about the energy stored for each bit of volume removed, or more formally the energy per cubic metre (measured in joule metre^-3). This measure of the incompressibility of the fluid is called the pressure, and since joule metre^-3 is such a mouthful it is usually measured in a special unit, the pascal, abbreviated to Pa, where 1 pascal = 1 joule metre^-3.

Robert Boyle described pressure as the spring of the air. The following discussion applies to fluids which are quite easy to compress. These are easier to think about, but harder to build reliable engineering structures with. So a way of thinking about hydraulic machines is this: to shift a certain amount of fluid from one cylinder needs an amount of energy for each cubic metre.

As a result of this activity – pushing with a force for a particular distance over a given area – a certain number of joules are shifted. How many joules per metre, the force, depends on both the pressure and the cross-sectional area of the cylinder. To place the fluid in another cylinder needs energy to be transferred. Again the pressure in the fluid and the cross-sectional area of the cylinder fix the force. We are encouraging you to think of this as a two step process – first remove the pressurised fluid from one cylinder – then store it in an interim place, then place it in another cylinder.

This, of course, is not how real hydraulic machines work, but it is an illuminating thought experiment, because it allows you to ask the question: So what is special about the volume of fluid after it has been removed from one cylinder and before it has been placed in the other?

Back to Boyle for the answer: It is just like a compressed spring. There is something special about the state of the fluid. It is squeezed or compressed, just like the compressed spring. As a consequence of this compression it can function as an energy store, and so exert a force when placed in the second cylinder.

Carrying round a lump of compressed fluid is just like carrying round a compressed spring. And compressed fluids do store energy. Think of the energy made available when an air horn sounds, a car air-bag releases or even an inflated, but unsealed, balloon is let go.

Just how are we comparing a spring to the fluid? For a spring it is how far away the spring is from its natural length, the extension, that we use to characterise the compression (or extension). For the fluid it is the excess pressure, the change in pressure from how the fluid was before, that is important in characterising how much energy each cubic metre has available.

The argument above is for compressible fluids. Hydraulic machines mostly use incompressible liquids. But right at the beginning we said that even incompressible liquids compress a bit. Just as very stiff springs are associated with a large store of energy when stretched only a little, so highly incompressible fluids store lots of energy per cubic metre for very little change in volume.

What about solids?

The whole of this section so far has focused on pressure in liquids and gases. What about solids? The simple expression for pressure as force divided by area leads to a bit of a confusion.

Often teaching about pressure in schools starts with one solid resting on another. It might involve camels with big feet walking across the desert or stiletto heels being banned from the school hall. It could be that pupils measure their weight on bathroom scales, find the total area of the soles of their shoes and then calculate the pressure under each foot. These activities (and lots of others) are interesting and offer excellent opportunities for discussion with pupils. Unfortunately:

None of these activities involve pressure!

For camels with big feet, stilettos with sharp heels or snow shoes with big areas the forcearea being measured is not a pressure: It is properly called a stress. The key here is that stress is a vector and that it acts in a particular direction.

Stress is measured as forcearea (exactly the same units as pressure).

So:

• Stiletto heels make a mark on the school hall floor because the gravity force acting on the woman acts on a tiny area giving rise to a big stress.
• The camels' feet do not sink into the sand because the gravity force acting on the camel acts on a big surface area giving rise to a small stress.

In general, if a solid object is placed on a surface and the area of contact between them is small a large stress is produced and since there is a relatively small amount of material supporting the object it makes a deep impression on the surface. If a supporting material is known to be weak then it should be subjected to small stresses, which involves spreading the force over a larger area.

We strongly recommend that you teach these two ideas of pressure and stress separately and furthermore teach them in the order presented here with pressure and hydraulics first (thinking about liquids and gases) and stress very much later (thinking about solids). In fact stress often appears first in post-16 studies in physics.

Why does a drawing pin push into the notice board at the sharp end and not into your finger? This is easily explained in terms of stress: The same force (the same push from your finger) acts on both ends, so where the area is big (at the cap of the pin) the stress is reduced and where the area is tiny (at the point) the stress is much increased.

In similar ways we can also explain the action of stiletto heels and camels' feet.

These stress explanations become much more difficult where one of the surfaces is not a solid. So using large area tyres to move over mud or snowshoes to walk on snow are a little more complex than they seem to explain. A first explanation will be along the same lines as for solid-solid surfaces. But persistent questioning by a class can lead quickly into difficult territory caused by the materials in question (mud, snow, sand) being rather complex in their behaviour. For example, why is it that you can move cautiously over soft sand and mud, but will sink if you stay still for too long or if you try to run? These instances cannot easily be explained simply by talking of force and area.

Here's a more detailed microscopic explanation for what happens to an object (such as a heavy box or a human wearing a snowshoe) when it is placed on the surface of some snow. When the object is first placed on the snow the forces on the object are simple. There is a downwards gravitational force and a negligible upwards force due to the snow. As the object sinks, so it distorts the snow resulting in a net resistive force caused by the cohesive forces in the snow (these are the forces that hold snowballs together). The greater the distortion, the greater is the force. Eventually the resistive force will be equal to the gravitational force at which stage the object will stop accelerating down into the snow and typically stop moving as well. All this means that the more you distort the snow the less you will sink before stopping – so wear snow shoes!

Pressure links input and output

As with the mechanical lever, choices you make about the physical dimensions determine how the machine will work. Energy is again conserved.

The input and output forces are linked by the areas of the pistons. Once these are chosen, conservation of volume and conservation of energy determine the relative distance moved by the pistons.