FORCES 6. Pressure created in various solid situations, pressure in fluids and hydraulic systems and P = F/A calculations

This page will answer help you answer questions like e.g.  

What is a fluid? What is the formula for pressure?  What causes pressure in liquids?   How do you calculate pressure in a liquid?   How do hydraulic systems work?  What do we use hydraulic systems for?

Sub-index for this page

(a) Particle theory - revision of states of matter and density

(b) Pressure created by standing objects and in fluids, calculations using P = F/A formula

(c) Pressure in a liquid - density, depth factors and calculations

See also 7. Pressure & upthrust in liquids, why objects float/sink?

(d) Hydraulic systems - mechanical devices, transmission of forces in liquids, calculations

(e) Some applications of hydraulic systems

(a) Particle theory - revision of states of matter and density

Gases and liquids are fluids because the particles are free to move around from place to place

In solids, the particles can just vibrate from a fixed position and cannot move to another position in the lattice of atoms.

very low density FLUID	high density FLUID	highest density, but NOT a FLUID

Solids have the highest density, the particles are the closest together.

In fluids the inter-particle attractive forces are sufficiently weak to prevent a solid forming, allowing free random movement of the molecules of a liquid or gas.

Because of the weaker inter-particle force, the particles of gas will spread out to fill any space available giving gases by far the lowest density of the three states of matter.

The particles in a liquid are held much closer together because of greater inter-particle forces giving liquids a much greater density than gases.

Apart from water, liquid densities of materials are usually a few % less than that of their solid form.

The density of liquids is so high, with so little space between the particles, they are almost impossible to compress to a smaller volume.

Gases have so much space between the particles that they are readily compressed to a smaller volume.

The closer together the particles are, the more compact the substance is, the greater density of it.

See also Particle theory models of gases, liquids and solids and the particle diagram above.

Liquids have a uniform density (same throughout its bulk) which only increases very slightly under extremely high pressures because there is so little free space to squeeze the molecules into.

However, there is considerable space between gaseous particles and its is relatively easy to compress the particles closer together e.g a bicycle pump filled with air or water.

The air is easily compressed and the water isn't. The water can ruin the pump but you can't compress the water in it. High pressure water pistols rely on compressed air NOT compressed water.

Liquid fluids have a much greater density than gaseous fluids

So, for similar depths (or heights) of gas and liquid, liquids will create a much greater pressure because of the greater weight (due to gravity) of substance acting on the same surface area.

The greater the density of a material, the greater the number of collisions can take place, creating a greater pressure.

This concept is most applicable to gases, which are so easily compressed under pressure, so considerably increasing their density - this happens with the Earth's atmosphere.

Also, the greater the column of fluid e.g. water, the greater the pressure created - the greater weight acting on a given area.

See later section on pressure in liquids.

See also

Particle models of gases–liquids–solids, explaining properties, state changes (GCSE chemistry notes)

The density of materials and the particle model of matter (GCSE physics notes)

Gas law calculations - P, V and T relationships (GCSE chemistry notes)

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(b) Pressure created by standing objects and in fluids, calculations, P = F/A formula including where pressure is deliberately applied to an object

Pressure is defined as force per unit area and is calculated from the simple formula

pressure = force normal to the surface ÷ area of that surface,

P = F / A,  F = P x A, A = F / P

P, pressure in pascals (Pa);   F, contact force in newtons (N);   A, area on which force acts in square metres (m²)

A force of 1 N acting on 1 m² creates a pressure of 1 Pa

A variety of situations to increase or decrease pressure in various situations

F = weight on applying the P = F/A equation

Camels have feet of large surface area, this reduces the F/A ratio (reducing pressure), so their feet do not sink to deeply into sand.

Similarly, skis or snow shoes have a relatively large area to reduce the F/A ratio, hence decreasing the impact pressure on snow, so the wearer does not sink to deeply into the snow.

Scissors have a sharp edge to create high F/A ratio to create a high pressure to cut cleanly through materials.

Drawing pins and hypodermic needles have very sharp end points to create a very high F/A ratio, so the high 'impact' pressure enables them to easily pass into materials when pushed.

Similar arguments apply to ...

Vehicles with large wide tyres or wide caterpillar tracks to minimise sinking in soft ground.

Crampons are sharp to penetrate ice or snow to get a good grip, but the rest of the broad surface area of the boot prevents the wearer from sinking too deeply into the snow.
 

Pressures caused by standing objects - some simple calculations, if occasionally painful !

Any solid object standing on a solid surface will, due to the force gravity, create a pressure on the surface due to its weight.

This is a normal contact force, balanced by the compressed atoms of the solid surface pushing back up.

(This complies with Newton's 1st Law of Motion)

Ignoring the different weights of people (the weight 'force'), you should realise from the formula why it is better to be trodden on by a broad shoe sole than a stiletto heel !!!! (CLUE !!! P = F / A, no exam pressure here !!!)

In moving around the particles of a fluid collide with each other and with any surface they are in contact with.

Although the mass of an individual particle is minute and each collision involves the transfer of an equally minute amount of kinetic energy, collectively the trillions of collisions cause a pressure to be exerted in both gases and liquids.

The combined effects of these particle collisions produces a net resultant force at right angles to the surface of contact with an object or side of a container.

e.g. the pressure of gases in a container or the pressure from the atmosphere of air pressure around you.

It is the same anywhere in a liquid, pressure is exerted against the side of a container wherever the liquid is in contact and it is the same for an object immersed in a liquid.

Remember that the force of the pressure acts in all directions in a fluid.

The maximum pressure exerted in a fluid is considered to be due to the collective force of the particle collisions acting at right angles (normal, 90^o, perpendicular) to the surface on which the collisions take place i.e. any surface in contact with the fluid.

 

For specific gas calculations see P-V-T pressure-volume-temperature gas laws and calculations

Examples of pressure calculations based on P = F/A

Q1.1 If a weight of fluid of 200 N acts on a surface of 5 m², calculate the pressure created.

pressure = force ÷ area,    P = F / A = 200 / 5 = 40 Pa

Q1.2 What force must be applied to a surface area of 0.0025 m², to create a pressure of 200,000 Pa?

P = F / A, rearranging gives: F = P x A = 200000 x 0.0025 = 500 N

Q1.3 In a hydraulic lift system, what must the surface area of a piston be in cm² if a pressure of 300 kPa is used to give a desired upward force of 2000 N?

P = F / A, rearranging gives: A = F / P = 2000 / 300000 = 0.00667 m²

What is the piston surface area in cm²?

1 m² = 100 cm x 100 cm = 10 000 cm², so the area of the piston = 10 000 x 0.00667 = 66.7 cm²

Q1.4 The average standard rectangular building brick has a mass of 3.10 kg and dimensions of 225 mm x 112 mm x 75 mm.

The gravitational field constant g = 9.8 N/kg.

(a) Calculate the pressure the brick creates when standing on its surface of smallest area.

smallest surface area = 112 mm x 75 mm

area = (112/1000) x (75/1000) =  0.0084 m²

downward force (weight) = mass x g = 3.1 x 9.8 =  30.38 N

pressure = force / area = 30.38 / 0.0084 = 3620 Pa (3 sf)

(b) Calculate the pressure the brick creates when standing on its surface of greatest area.

largest surface area = 225 mm x 112

area = (225/1000) x (112/1000) =  0.0252 m²

downward force (weight) = mass x g = 3.1 x 9.8 =  30.38 N

pressure = force / area = 30.38 / 0.0252 = 1210 Pa (3 sf)

Q1.5 A stiletto heel has a surface area of approximately 2.5 cm x 2.0 cm.

If a lady puts all of her 60 kg 'weight' on it, what pressure is created on the floor surface.

The gravitational field constant g = 9.80 N/kg.

Stiletto area = (2.5/100 x 2.0/100) =  0.0005 m²

Weight force = m x g = 60 x 9.8 = 588 N

Stiletto pressure = P / A = 588 / 0.0005 =  ~1 176 000 = ~1.2 x 10⁶ Pa

Ad hoc notes!

This is over a million pascals! Painful if you are trodden on!

If trodden on, on the Moon, its not so bad, with its 1/6th the gravity of Earth.

Atmospheric pressure is ~101 000 Pa, so its over 10 x greater than the pressure of the air around you!

Its also the same pressure you experience at depth of 100 m in water!

(See section (c) where in fluids P = hρg = 100 x 1000 x 9.8 = 9.8 x 10⁵ Pa)

Q1.6 A person is standing on both feet, with flat trainers, each of which has an ground contact area of 0.025 m².

If the person weighs 800 N, what pressure is created on the ground by each foot when the person is standing still?

Answer: P = F/A = pressure = force or weight / area = 800 / (2 x 0.025) = 800 / 0.05 = 32000 N/m² or Pa (pascals), divide by 2 because of two feet.

Q1.7 An elephant is standing on all of its feet, each of which has an ground contact area of 0.08 m². If the elephant weighs 50000 N, what pressure is created on the ground?

Answer: P = F/A = pressure = force or weight / area = 50000 / (4 x 0.08) = 50000 / 0.32 = 156250 N/m² or Pa (pascals)

Q1.8 A ski design team has to take into account the pressure created by the skier on the surface of deep snow.

They need to the calculate the effects of the variables which are (i) area of one ski, (ii) weight of skier, (iii) pressure created by the skier on the snow and (iv) the pressure the snow can take without the skier sinking in too much!

If the maximum acceptable snow pressure is 5000 N/m², for a single ski surface area of 0.15 m², what is the maximum weight the skier can be?

P = F/A, so max. weight = F = P x A = pressure x area = 5000 x 015 = 750N

Q1.9 A ski design team has to take into account the pressure created by the skier on the surface of deep snow.

They need to the calculate the effects of the variables which are (i) area of one ski, (ii) weight of skier, (iii) pressure created by the skier on the snow and (iv) the pressure the snow can take without the skier sinking in too much!

If the maximum acceptable snow pressure is 4000 N/m², what is the minimum single ski surface area acceptable, for a skier of weight 800 N?

P = F/A, so min. area A = F/P = weight/pressure = 800/4000 = 0.2 m²

Q1.10 A bag of sugar has a base of 6 cm x 10 cm. If it weighs 18 N, what pressure does it create standing on a shelf?

P = F/A = pressure = force or weight / total area = 18 / (6 x 10) = 18/60 = 0.3 N/cm²

Conversion: 1 m² = 10⁴ cm²

so, 0.3 x 10⁴ = 3000 Pa (pascals)

Q1.11 A stiletto heal has a base area of 3 cm².

If the woman weighs 750 N, what pressure does she create on the floor when standing on one heel?

P = F/A = pressure = force or weight / total area = 750 / 3 = 250 N/cm²

Conversion: 1 m² = 10⁴ cm²

so, 250 x 10⁴ = 2.5 x 10⁶ N/m² (Pa)

 

Q1.12 A waste skip a base of 2m x 4m. If it weighs 10000 N when full, what pressure does it create when standing on the road?

P = F/A = pressure = force or weight / total area = 10000/(2 x 4) = 10000/8 = 1250 N/m² (Pa)

Q1.13 A brick has a base of 10 cm x 25 cm and weighs 30 N. what pressure does a stack of ten bricks create simulating the pressure created by a low wall?

P = F/A = pressure = force or weight / total area = (30 x 10) / (10 x 25) = 300 / 250 = 1.2 N/cm²

Conversion: 1 m² = 10⁴ cm²

so, 1.2 x 10⁴ = 1.2 x 10⁴ N/m² (Pa)

 

For specific gas pressure calculations see P-V-T pressure-volume-temperature gas laws and calculations

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(c) Pressure in a liquid - density and depth factors - calculations

Density is a measure of how close the particles are together. The more compact they are, the greater the density.

As already mentioned, in liquids the density is uniform throughout and because there is so little space between the particles the density only slightly decreases with increase in temperature with the increased kinetic energy of the particles.

However, the volume shows almost no change with increased pressure (so here you can consider liquids to be virtually incompressible).

All liquids expand on heating - observe a mercury or alcohol thermometer.

The pressure in a fluid varies AND increases with depth - it doesn't matter whether you are dealing with gases like the atmosphere or liquids like the water of a lake or ocean.

The greater the height/depth of fluid, the greater the weight of particles that gravity is pulling down, hence the increase in force per unit area at a particular level, hence the increase in pressure.

The pressure in a fluid acts in all directions (← → ↑  ↓) because the particles are moving and colliding with each other, and the sides of the container, at random in all directions.

Liquid pressure significantly increases with depth as the weight of the column of liquid increases.

A simple experiment can demonstrate this rule using a tall container with holes in the side. When you fill it with water, the water gushes out of the holes, but the lower the hole, the greater the water pressure, the faster the water comes out and travels a greater distance.

A note on dam construction (e.g. reservoir for water supply or hydroelectric power plant)

Since water pressure increases with depth, to resist this increase in pressure, the width of a dam must also be increased to combat the increased force on the dam wall.

The width of the dam is much greater at its base compared the top of the dam wall.

 

The pressure also increases with increase in density of the fluid - so air and water differ significantly for a pressure created at a specific depth of fluid (densities: air density 1.2 kg/m³ and water 1000 kg/m³ at room temperature).

From your own experience you may have observed:

Sometimes in a car descending or ascending a few hundred metres on a road can make your ears 'pop' and hurt slightly because of the change in pressure with height.

BUT, you only have to dive into a few metres of water to experience the same effect on your ears. As you suddenly into the water, the pressure is suddenly increased on your ear drums - the pain comes from the greater external water pressure than the internal body pressure on the other side of your ear drum. However, for most people, when under the water at shallow depths, the pressures become equal quite quickly

You can calculate the pressure at a given depth created by the weight of liquid in the earth's gravitation field using the following formula:

pressure in a liquid = depth of liquid x density of liquid x gravitational field strength

P = hρg

P, pressure in pascals (Pa);   h = depth in metres (m);   ρ = density (kg/m³),

and the gravitational field strength =  g = 9.8 N/kg (on the Earth's surface)

Unit connections

Taking the formula P = h x ρ x g 'apart' in terms of units.

pressure = force per unit area = height of column of material x density of material x gravitational constant

N / m²  =  m  x  kg/m³  x  9.8 N/kg

unit analysis: on the right the kg cancel out, m/m³ = 1/m², you are left with N/m² !!

 

Note: Upthrust force in fluids and flotation etc. are covered in

FORCES 7. Pressure & upthrust in liquids, why do objects float or sink in a fluid?, variation of atmospheric pressure with height

Example calculations involving liquid pressure  

 (the gravitational field effect is taken as 9.8 kg/N in these questions).

Q2.1 Divers have to be careful when working at depth in water and need to carefully control the dissolving of gases in their blood stream.

(a) Calculate the pressure created by a 30 m depth of water given the density of water is 1000 kg/m³ and gravity 9.8 N/kg.

P = hρg

P = 30 x 1000 x 9.8 = 294 000 Pa (2.94 x 10⁵ Pa, 294 kPa)

(b) Comment on the dangers when diving at great depths and how to avoid dangerous problems.

Atmospheric pressure is about 101 kPa, so a diver at these depths will experience a much greater pressure than on the surface of the water

 Increase in pressure causes more gases to dissolve in the blood stream (this is a general rule for gases in contact with a liquid that can act as a solvent).

This can have serious consequences if time isn't allowed for the body pressure to adjust to the new external pressure, particularly when returning back to the surface.

The bends, also known as decompression sickness disease, occurs in divers when dissolved gases (mainly nitrogen) come out of solution in bubbles and can affect any body area including joints, lung, heart, skin and brain.

The effects can be fatal unless time is allowed for the body to adjust in a decompression chamber.

 

Q2.2 The density of sea water is ~1025 kg/m³, the maximum depth of the Atlantic ocean is ~8500 m (8.5 km).

(a) Calculate the water pressure at this depth.

P = hρg

P = 8500 x 1025 x 9.8 = 85 400 000 Pa (to 3 sf, 85.4 MPa, 85400 kPa, 8.54 x 10⁷ Pa, 8.54 x 10⁴ kPa)

(b) By what factor is the pressure greater at these depths compared to the ocean surface?

Atmospheric pressure is ~101 kPa

Pressure at bottom of ocean ÷ pressure at surface = 85400 ÷ 101 = 846 (3 sf).

Note: This extraordinary increase in pressure mean to explore this 'alien' world you need a very strong submersible craft. However, evolution has allowed all sorts of creatures to live down at these depths, all fully pressure adjusted over time! If you (theoretically) brought any such creatures rapidly to the surface and exposed them to normal pressure, it would kill them!

 

Q2.3 At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?

5 x 101 = 505 kPa, 505000 Pa, density of water 1000 kg/m³

P = hρg, rearranging gives h = P/ρg = 505000/(1000 x 9.8) = 51.5 m

Note: The pressure increase in water increases by about the value of atmospheric pressure for every 10 m.

Q2.4 At a depth of 12.5 m of a chemical solvent the pressure at the bottom of the storage tank due to the solvent was 306 kPa

Calculate density of the solvent.

P = hρg, rearranging gives ρ = P/hg = 306000/(12.5 x 9.8) = 2498 kg/m³

 

Q2.5 ?

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(d) Hydraulic systems - mechanical devices for the transmission of forces in liquids and calculations

Hydraulics is how we can use fluids (e.g. air or oil) to transfer force to achieve some useful work e.g. operating machinery like car jacks, car brakes or JCB digging machines.

As we have said above, pressure in fluids is caused by particle collisions with themselves and the surface of a container.

These collisions cause a net force at right angles to all surfaces the fluid is in contact with. Hence the equation

pressure = force ÷ area,    P = F / A

P, pascals (Pa);   F, force in newtons (N);   A, area on which force acts in square metres (m²)

A force of 1 N acting on 1 m² creates a pressure of 1 Pa

(I've repeated these details because there are more calculations coming up!)

However, there is an important new idea to understand the applications of hydraulics.

The pressure in a fluid is transmitted equally in all directions.

AND, for the purposes of calculations, you consider the force involved to act at right-angles to the surface.

Since liquids are effectively incompressible, if a force is applied to any point in an enclosed fluid system, the net force is transmitted to any other point in the fluid (gas or liquid). This is the basis of all hydraulic systems.

It is no good using a gas in a hydraulic system, because a gas is readily compressed, affecting the action of the pistons in a hydraulic system - see diagram below.

A lot of the force and energy of the compression action is used to compress the gas and so reducing the transmitted force-pressure.

The principles and calculations involving hydraulic systems.

Diagram of the basic principles of a hydraulic system

The diagram above illustrates the idea of using a hydraulic system to apply a small force to produce a large force - an example of a force multiplier system to effect a mechanical advantage (remember moments and lever systems!)

The formula for pressure is P = F/A (already dealt with in detail on this page).

Therefore: P1 = F1_input/A1 and output P2 = F2_output/A2   (where = pressure, A = area, F = force)

BUT, P1 = P2 because the pressure at any given moment in time is the same throughout the hydraulic system.

Therefore: F1_input/A1 = F2_output/A2,  rearranging gives the output force F2_output = F1_input x A2/A1

So the ratio of the two piston areas (A2/A1) gives you the force multiplying effect (as long as A2 > A1).

Note:  In hydraulic systems, piston 1 is in the master cylinder and piston 2 is in a slave cylinder (can be several of the latter e.g. in a car braking system of a system for raising a car in a garage).

Some examples of how to do hydraulic system calculations

Q3.1 In a simple hydraulic system piston 1 has a cross-section area of 0.000050 m², and piston 2 has a cross-section area of 0.00025.

If a force of 35 N is applied to piston 1,

(a) What is the force multiplying ratio?

force ratio = A2/A1 = 0.00025/0.000050 = 5.0

(b) What output force is generated by piston 2?

F2 = F1 x A2/A1 = 35 x 0.00025/0.000050 = 35 x 5 = 175 N

Q3.2 A medium sized car typically weighs 1000 kg.  (gravity force 9.8 N/kg)

A hydraulic system to jack the car up above an inspection pit has four slave cylinders whose pistons have a cross section area of 0.01 m². The cross-section area of the master cylinder piston is 0.00125 m².

(a) What is the weight of the car?

w = mg = 1000 x 9.8 = 9800 N

(b) What is the minimum force that must be applied to the master cylinder to raise the car upwards?

F1/A1 = F2/A2, rearranging gives F1_input = F2_output x A1/A2

The output force must at least match the weight of the car, so F2 = 9800 N

A1 = 0.00125 m², A2 = 4 x 0.01 = 0.04 m² (remember there are 4 connected cylinders).

Therefore F1 = 9800 x 0.00125/0.04 = 306 N (3 sf)

So relatively small force can be used to raise a much larger weight by means of a hydraulic system.

 

Q3.3 Solve the following problem with a little help from the diagram on the right of a simple hydraulic system of two pistons and cylinders connected together.

The cross-section area of piston A1 is 0.000400 m².

The cross-section area of piston A2 is 0.00800 m².

(a) If a force of 40 N (F1) is applied to piston 1, calculate the pressure created in the fluid.

P = F / a = 40 / 0.0004 = 100 000 Pa

(b) Calculate the force F2 created by the force of 40 N on piston A1.

Since the pressure is the same throughout the hydraulic system, we can say the pressure acting on piston 2 is also 100 000 Pa.

P = F / A, F = P x A = 100 000 x 0.008 = 800 N

(c) What is the multiplying or mechanical advantage factor in this hydraulic system?

force multiplying factor = final force / initial force = 800 / 40 = 20

(d) In order to create a force of 2000 N from cylinder 2, what should the cross-section area of piston 2 be if the force applied to piston 1 is still 40 N?

The pressure P2 remains the same at 100 000 N

F2 is now 2000 N, need to solve for A2

P = F / A,  A = F / P = 2000 / 100 000 = 0.02 m²

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(e) Some applications of hydraulic systems

Reminders and uses:

Hydraulic machines use liquid fluid power to perform light or heavy work in many situations.

Heavy construction/demolition/digging vehicles are a common examples (illustrated below).

In these machines, hydraulic fluid is pumped to various hydraulic motors and hydraulic cylinders throughout the machine and becomes pressurized sufficiently to overcome resistance and effect movement.

The fluid is controlled directly or automatically by control valves and distributed through tubes/pipes.

Hydraulic systems, like pneumatic systems, are based on the fact that any pressure applied to a fluid inside a closed system will transmit that pressure equally everywhere and in all directions.

To be effective in transmitting force, hydraulic systems must use an in-compressible liquid as its fluid, rather than a compressible gas.

The wide-spread use of hydraulic machinery is because large amounts of power that can be transferred mechanically, and relatively simply, through small tubes and flexible hoses to achieve movement e.g. lifting or digging.

 (1) Hydraulic jack system for lifting vehicles in garages

Four hydraulic piston and cylinder systems are used to jack up the red van. The fluid can be oil or compressed air.

The Land Rover (left) and the red  van (right) both have hydraulic brake systems

Key for the 5 photographic diagrams: S = suspension spring;  H = the pipe conveying the hydraulic brake fluid force.

D = the brake drum and disc on which the brake pads in casing P are forced into contact with the smooth disc by the hydraulic transmitted force when you press the brake pedal.

A conveys speed of rotation information for the correct functioning of the advanced ABS braking system.

How does the brake system of a motor vehicle work?

When you press on the brake pedal of motor vehicle, the force is transmitted through the brake fluid liquid by a pipe system (in modern cars it is aided by an electric pump system).

The braking mechanism involves several pistons and cylinders, known as the master cylinder (activated by the pedal) and slave cylinders on the brake drums (4 of the latter in the case of 4 wheeled vehicles).

The transmitted force pushes the brake pads onto the brake disc on the brake drum and the resulting friction reduces the speed of the vehicle.

When you take you foot off the brake pedal, the force is no longer transmitted and springs move the brake pads away from the brake discs to avoid unnecessary friction.

 

'Hydraulic' photographs by courtesy of Mark Raw of M T R Autotech Ltd garage, Castleton, North Yorkshire, England

 

(2) Plant machinery - examples of using hydraulic machinery

A huge range of excavators ('diggers'), elevated work platforms are used on demolition sites, construction sites use liquid power in hydraulic systems.

In the above photographs you can see the shiny steel systems of the hydraulic systems.

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