FORCES 6. Pressure created in various solid situations, pressure in fluids and hydraulic systems and
P = F/A calculations
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physics, IGCSE physics, O level physics, ~US grades 8, 9 and 10
school science courses or equivalent for ~1416 year old students of
physics
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?
Subindex 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
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(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 interparticle 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 interparticle 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 interparticle 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)
TOP OF PAGE and
subindex
(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 (m^{2})
A force of 1 N
acting on 1 m^{2} 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
PVT pressurevolumetemperature 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 m^{2},
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 m^{2},
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 cm^{2} 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 m^{2}
What is the piston surface area in cm^{2}?
1 m^{2} = 100 cm x 100 cm = 10 000
cm^{2}, so the area of the piston = 10 000 x 0.00667 =
66.7 cm^{2}
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 m^{2}
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 m^{2}
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 m^{2}
Weight force = m x g = 60 x 9.8 = 588
N
Stiletto pressure = P / A = 588 /
0.0005 = ~1 176 000 =
~1.2
x 10^{6}
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 10^{5} Pa)
Q1.6
A person is standing on both feet, with flat trainers, each of which has
an ground contact area of 0.025 m^{2}.
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/m^{2}
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 m^{2}. 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/m^{2}
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/m^{2}, for a single ski surface area of 0.15 m^{2},
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/m^{2}, 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 m^{2}
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/cm^{2}
Conversion: 1 m^{2} = 10^{4} cm^{2}
so, 0.3 x 10^{4} =
3000 Pa (pascals)
Q1.11
A stiletto heal has a base area of 3 cm^{2}.
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/cm^{2}
Conversion: 1 m^{2} = 10^{4} cm^{2}
so, 250 x 10^{4} =
2.5 x 10^{6} N/m^{2}
(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/m^{2}
(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/cm^{2}
Conversion: 1 m^{2} = 10^{4} cm^{2}
so, 1.2 x 10^{4} =
1.2
x 10^{4} N/m^{2}
(Pa)
For specific gas pressure calculations see
PVT pressurevolumetemperature gas
laws and calculations
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subindex
(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/m^{3} and water 1000 kg/m^{3} 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/m^{3}),
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 / m^{2}
= m x kg/m^{3} x 9.8 N/kg
unit analysis: on
the right the kg cancel out, m/m^{3} = 1/m^{2}, you are
left with N/m^{2} !!
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/m^{3} and gravity 9.8 N/kg.
P = hρg
P = 30 x 1000 x 9.8 =
294 000 Pa (2.94 x 10^{5} 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/m^{3}, 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 10^{7} Pa, 8.54 x
10^{4} 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/m^{3}
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/m^{3}
Q2.5 ?
TOP OF PAGE and
subindex
(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 (m^{2})
A force of 1 N
acting on 1 m^{2} 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 rightangles 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 forcepressure.
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 crosssection area of 0.000050 m^{2},
and piston 2 has a crosssection 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 m^{2}. The crosssection area of the master
cylinder piston is 0.00125 m^{2}.
(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 m^{2}, A2 = 4 x 0.01 = 0.04 m^{2}
(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 crosssection area of piston A1 is
0.000400 m^{2}.
The crosssection area of piston A2 is
0.00800 m^{2}.
(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 crosssection 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
m^{2}
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subindex
(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 incompressible liquid as its
fluid, rather than a compressible gas.
The widespread 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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subindex
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