TECHNOLOGICAL STUDIES
Intermediate 2
Mechanical Systems
Section 2
Outcome 2 – Mechanisms
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OUTCOME 2
Outcome 2 – Describe the operation and performance of mechanisms.
The purpose of this unit of work is to introduce students to mechanisms and
mechanical systems. Students will be exposed to a wide range of mechanisms and
will be expected to be able to describe how to modify or adjust them to affect their
input/output conditions.
When the students have completed this unit of work they should be able to:
• Describe the operation and performance of a range of mechanisms.
• Understand methods used to adjust mechanisms.
• Describe how to adjust mechanisms.
• Perform calculations related to mechanisms e.g. velocity ratio, moments etc.
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MECHANICAL SYSTEMS
Our every day lives are made much easier by a variety of mechanical systems that
help us do jobs of work and leisure. Most of the mechanisms you use are so familiar
you take them for granted.
In the home for example, from simple mechanical mechanisms such as scissors and
those in vacuum cleaners, to more complex mechanisms such as sowing machines, all
help us do work in the house.
Mechanisms at home
fig1.1
Mechanisms play a vital role in transporting people to school, college or work and
mechanical systems included in bicycles, cars, buses, aeroplanes and ships make all
this possible.
Mechanisms in transportation
fig1.2
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In the factories, the work involved in making all these consumer products so
necessary for our modern society is made easier by the large and complex machines
using mechanical systems in their manufacture.
Mechanisms in the workplace
fig1.3
All machines have one thing in common, to do their work they require energy.
Mechanical Systems within Machines in fact can be considered as systems which
process energy to produce useful work.
ENERGY
MECHANICAL
SYSTEM
WORK
fig1.4
Common forms of energy you may already know:
Electrical, Heat, Chemical, Potential and Kinetic Energies.
In this unit we intend to look at that energy most closely associated with machines,
called work.
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The main purpose of most mechanical systems is to make work easier to do.
In fact, if a mechanical system doesn’t make work easier to do then there is little point
in having it.
Just think what hard work it must have been to wash clothes before washing machines
were invented.
fig 2.1
Modern machines like washing machines are made up from many different parts or
subsystems such as electronic devices to control their operation. However, the parts
of the machine which do the hard work of washing clothes, such as turning the drum
and pumping out the water, are mechanical systems, the simplest of which are called
mechanisms.
DRUM
E LEC TR O N IC
C O N TR O LLER
H E AT IN G
E LEM E N T
PUMP
fig 2.2
In your project work in this subject you will require to know about a range of
mechanical subsystems and be able to use them to solve problems.
Through completing this learning outcome you will be introduced to the most
common of mechanical subsystems.
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Although different mechanical systems or mechanisms are all designed to perform
different jobs, they all have two similar functions.
Mechanisms are systems, which transmit force and transmit and/or convert motion.
FORCE IN
FORCE OUT
MECHANISMS
MOTION IN
MOTION OUT
fig 3.1
Common place items such as bicycles, door handles and kitchen taps all involve some
form of motion and force. This force and motion is transmitted to provide a desired
output force and motion.
Motion
The four types of motion most basic to machines are:
Rotary; Linear; Oscillating and Reciprocating
An example of a device, which uses each of these types of motion, is shown.
Rotary motion
Rotary motion is motion through a circular path, such as is produced when a
sharpener handle rotates.
fig 3.2
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Linear Motion
Linear motion is motion in straight line, such as is produced when a person bowls a
bowling ball.
fig 3.3
Oscillating Motion
Oscillating motion is motion back and forward through the arc of a circle, as can be
produced by the gymnast.
fig 3.4
Reciprocating Motion
Reciprocating motion is motion back and forward through a straight line, such as is
produced by jigsaw.
fig 3.5
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Conversion of Motion
Most mechanisms are used to convert one of these types of motion into another.
An example of this is in an electric sewing machine.
CR AN K / SLIDE R
NE EDLE
RE CIP RO CATING
M O TIO N
fig 4.1
The input motion from the electric motor, which drives the machine, is rotary. This
rotary motion is converted by a mechanism called a crankshaft to produce the
reciprocating motion required by the sewing machine needle.
ROTARY
MOTION
SEWING
MACHINE
RECIPROCATING
MOTION
fig 4.2
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Levers
The main factor that has led to mankind dominating the world in which we live has
been the ability to harness energy to do work through developing machines and
mechanical systems.
An early example of this was demonstrated in the ability to lift and move heavy
objects using a simple mechanism called a Lever.
SMALL
FORCE
SMALL
DISTANCE
LARGE
DISTANCE
LARGE
FORCE
fig 5.1
The idea behind a lever is simple enough. It is a system which will produce a large
output force from a smaller input force.
SMALL INPUT
FORCE
LARGE OUTPUT
FORCE
LEVER
fig 5.2
For this to happen a rigid beam is required to act as the lever which must pivot about
a point closer to the output force than the input force.
SMALL INPUT
FORCE
LARGE OUTPUT
FORCE
PIVOT
fig 5.3
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Force Ratio (Mechanical Advantage)
There is an obvious advantage in using a lever for this purpose, in that it increases the
effect of the small input force to produce a larger output force to lift the load.
The advantage gained from using a lever mechanism for this purpose is called Force
Ratio (Mechanical Advantage).
Force Ratio (Mechanical Advantage) is the ratio of the output force to the input force.
MA =
Force out (Fo)
Force in (Fi)
Example
The lever shown in the diagram requires a 10 N force to raise a load weighing 100 N.
Calculate the Force Ratio (Mechanical Advantage) of the lever.
10N (F i)
100N (F o)
P IV O T
fig 7.1
MA =
Fo
Fi
MA =
100
10
= 10 or 10:1
(i.e. the input force is ten times smaller than the output force.)
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Velocity Ratio (VR)
Most mechanical systems are designed to produce a Force Ratio (Mechanical
Advantage) (i.e. a small input force producing a large output force). There is,
however, a price to pay for this advantage because the smaller input force has to be
moved through a much larger distance than the large output force.
Just as there is a name given to the ratio between the input and output forces, i.e.
Force Ratio (Mechanical Advantage), there is a name given to the ratio between the
distances moved by the input and output force.
This ratio is called the Velocity Ratio.
Velocity Ratio =
Distance moved by the input (di)
Distance moved by the output (do)
VR =
di
do
VR =
di
do
For our lever then:
1000
10
= 10 or 101
i.e. the input force has to be moved ten times further than the output force.
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Efficiency
The efficiency of a mechanism is a measure of how effectively it does its work.
Efficiency =
Work out × 100%
Work in
For our lever then:
Efficiency =
=
Wo
Wi
10 × 100
10
= 100%
The efficiency of a mechanical system can also be calculated from the ratio of the MA
to the VR.
Mechanical Advantage (MA)
Efficiency =
Velocity Ration (VR)
Again for the lever:
Efficiency =
=
MA
× 100
VR
10 × 100
10
= 100%
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Friction
From our calculations it would appear that our lever is an ideal mechanism. It
converts all the energy input to do the work at the output. In reality however this
would not be the case.
Probably the lever would bend slightly as the input force was applied. Therefore the
distance which this force would have to be moved would be greater thus increasing
the VR and reducing the efficiency.
Even if the lever was perfectly rigid it still could not be 100% efficient because of
friction.
Friction occurs when any two surfaces come into contact. In the case of the lever,
friction between the pivot and the underside of the lever will generate heat energy.
This means that some of the input energy will be ‘lost’ at the pivot and the amount of
work done at the output will be less.
H EAT EN ER G Y O UT
Fi
W O RK
IN
W O RK
OUT
Fo
di
do
P IV O T
SY STE M B OU N DA RY
fig 8.1
In more complex systems friction can be a major problem and engineers have
developed various methods of reducing friction.
The simplest way to reduce friction is to use a lubricant such as oil or grease.
fig 8.2
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In moving parts such as at the hub of a wheel, ball bearings get over the problem of
surfaces rubbing together.
O U T ER R AC E
INN ER RA CE
BAL L B EAR IN G
fig 8.3
The bearing consists of an outer and an inner ‘race’ which have grooves machined in
them. Hardened steel spheres, or ‘balls’, are fitted between the outer and inner race
which can then rotate freely. To work effectively the ball bearing must be well
lubricated.
Friction is not all bad however and without friction many mechanical systems would
not work.
It is friction between the soles of our shoes and the pavement, which allows us to
walk. If it were not for friction it would feel a bit like walking on ice and we would
find it impossible to get about. Tyres on cars are designed to increase friction with the
road surface to ensure that the force from the engine can be transmitted effectively.
A CANTILEVER-ACTION BICYCLE BRAKE
CABLE O PERATED
FROM HANDLEBAR
FIXED PIVO T
BRAKE BLO CK
MADE O F RUBBER
fig 8.4
Without friction it would be impossible to stop our cars or bicycles because the
brakes also rely on friction.
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Moments
With the lever we saw that applying it further away from the pivot point can increase
the effect of a force.
Fi
Fo
di
do
P IV O T
fig 8.5
This effect is used when tightening a bolt with a spanner. To get the bolt really tight
the force must be applied at the end of the spanner.
d
F
fig 8.6
The effect of producing a large output force from a small input force is called a
moment.
The moment of a force is dependent on the size of the force and the distance at which
it is applied from the pivot point.
MOMENT = FORCE × DISTANCE
M=F×d
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Example
The spanner shown in the diagram is 250 mm long. If a force of 40 N is exerted at its
end, find the moment of the force.
F = 40 N
fig 8.7
M
=
F×d
M
=
40 × 250
=
10,000 Nmm
=
10 Nm
Equilibrium
Moments are very important in the design of some mechanisms. Large and complex
mechanical systems such as cranes use moments to balance either side of the arm,
otherwise the crane would topple over.
6m
3m
LO AD
5kN
C O U N TE R BA LA N C E
10 kN
fig 8.8
i.e. for Equilibrium
COUNTER BALANCE MOMENT = LOAD MOMENT
10 × 3 = 6 × 5
30 = 30 Nm
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Linkages
Linkages are mechanical systems which transmit force and movement. Like levers
they can be used to increase the effect of an input force.
INPUT FORCE
MOVEMENT
OUTPUT
MOVEMENT
LINKAGE
fig 9.1
Some common types of linkage are shown in the diagrams.
Bell Crank Linkage
Reverse Motion Linkage
INP UT M O TIO N
F IX ED PIVO T
O U T PU T M O T IO N
fig 9.2
fig 9.3
Bell cranks and reverse motion linkages can be used to alter the direction of
transmission of the input force.
If the pivot point on the reverse motion linkage is changed a larger output force is
transmitted.
INP UT M O TIO N
SM ALL IN PU T
F O R CE
O U T PU T M O T IO N
LAR G E O UT PU T
F O R CE
fig 9.4
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The mechanism that operates calliper brakes on bicycles uses two bell cranks to
transmit braking force. These mechanisms transmit the force from the brake cable at
right angles onto the wheel rim.
Fi
Fi
Fo
Fo
fig 9.5
Parallel linkages are used to make two or more parts of a mechanical system move
together in parallel. These systems are based on a parallelogram and are used in
folding devices such as extension mirrors and tool boxes.
Extension Mirror
Tool Box
fig 9.6a
fig 9.6b
A common application of a parallel linkage can be seen in some car windscreen
wipers. The diagram shows how this is achieved using a rotating linkage to drag a
parallel linkage back and forth.
IN PUT M O TIO N
P US H-PULL
M O TIO N
O UTPUT MO TION
fig 9.7
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Windscreen Wipe
Build and test a model which shows the action of a windscreen wiper linkage.
fig 9.8
1.
(a) Draw a systems diagram which shows the input and output motion to the
mechanism.
(b) Draw a line diagram of the device and explain how it operates.
2.
(a) Develop an electronic control system which would switch on the wipers
automatically when it rains and can also be operated manually.
(b) Draw a systems diagram for the control system.
(c) Build and test the control system and draw a block diagram explaining how
it operates.
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Assignment
A hand operated crusher for aluminium cans is shown below. A schematic diagram of
the lever system is also shown.
80
200
D
A
PIN JO INT C
PIVO T B
fig 9.9
(a) Write a few sentences to explain why a lever is useful for this application.
(b) In the situation above, if the operator applies a force of 100N on the handle at A,
calculate the size of the total force applied to the can at D.
(c) Identify one part of the mechanism which is in tension.
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Assignment
Waste material is transported to a site by a cab with a trailer supported on a ball joint
tow bar at P as shown on the diagram. The centre of gravity of the cab acts through
point A and the centre of gravity of the loaded trailer acts through point B.
A
R2
P
B
R3
R1
4.5kN
6.25kN
fig 9.10
A line diagram of the system is shown below.
6.25kN
3.0
2.0
4.5kN
2.0
1.5
2.0
R3
N OTE: A LL D IS TAN C E S IN M E TRE S
R1
R2
fig 9.11
(a) By considering the trailer, calculate the force acting at the ball joint at P.
(b) Calculate the reaction forces between the wheels of the cab and the ground at R1
and R2.
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Assignments: Moments and Levers
1. Explain in your own words what the term Moment means?
?
1 .5m
1m
2. A garage door is 2 metres high and is hinged at the top. The door begins to open
when a force of 65 newtons is applied to the door handle at right angles, which is
1.5 meters down from the hinge.
Find the least amount of force which, when applied to the middle of the door, will
open it?
65N
MS. Int2 02 Fig 9.16
3. In a theatre stage a horizontal aluminum beam 5 metres long carries lighting
weighing 70 newtons at one end and a counter balance of 80N at the other end.
Calculate the position of a single pivot support to allow the system to balance?
5m
MS. Int2 02 Fig 9.17
4. In the wheelbarrow shown a load of 500 newtons is to be removed. Calculate the
amount of force that is needed to lift the handle to transport the load.
F
500N
MS. Int2 02 Fig 9.18
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5. During a cross-country rally the jacking system to remove a punctured tyre could
not be placed on a level surface. A new arrangement using a lever to change the
tyre was attempted. If the weight of the car was 4,800 newtons, calculate the
position of the pivot from the car if the effort applied is 600 newtons.
6. A clawhammer was used to extract a nail from a piece of wood. The necessary
force was 45 newtons applied at 300mm from the back of the hammer.
Find the resistance force offered by the nail, which was 40mm away from the
pivot point of the hammer.
300
45N
45
F
MS. Int2 02 Fig 9.20
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7. A pneumatic clamp is used to hold a workpiece in place during a machining
process. The double acting cylinder has an upward thrust of 100 newtons.
a) Calculate the force exerted on the workpiece by the clamp when the cylinder is
positive?
200 m m
100 m m
W ORKPIECE
MS. Int2 02 Fig 9.21
8. A truck weighs 60kN with its weight acting as shown. If the pivot point is the rear
wheels,
a) Calculate the weight of the skip (W)
b) What effect would a loaded skip have on the lifting process?
1.2m
60kN
4m
7.4m
W
MS. Int2 02 Fig 9.22
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9. The diagram below shows part of a tractor handbrake system. The line diagram
indicates dimensions and forces within the system; a spring and a force of 120N
will return the brake handle.
For the position shown, calculate the force F required at the brake handle.
FO R C E F
F
36
0
60
P IV O T P
75
5N
B RA K E R O D
116 N
MS. Int2 02 Fig 9.23
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Lifting Systems
Screw Thread
If you take a piece of card cut into a wedge like an inclined plane and wrap it around a
cylinder you produce a screw thread as shown in fig10.1.
fig 10.1
The screw thread is another useful lifting device with a high Force Ratio (Mechanical
Advantage).
A common application of this device is a car jacking system in which a small input
force is required to produce a large output force to overcome the weight of the car
when a wheel has to be removed.
fig 10.2
Pulley Systems
Although screw threads and inclined planes are useful aids in lifting heavy objects a
far more flexible system is to be found in the pulley.
A pulley is a wheel which rotates freely or pivots about a central point.
LO AD
AR M
EF F O RT
AR M
F
fig 10.3
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In fact a pulley can be thought of as a simple lever. When an equal input and output
force is applied to the pulley, it will be in balance or equilibrium.
FU LC R U M
Fo = LO AD
E FFO RT = Fi
fig 10.4
To raise a load, the output force due to its weight, will have to be overcome by the
input force.
T
T
Fi = E FFO RT
LO A D = F o
Fi
Fo
fig 10.5
There is no Force Ratio (Mechanical Advantage) to using a pulley system in this way
because the input force = the output force.
MA =
Fout
F in
=1
Single pulley systems arranged in this way are used for convenience reasons, because
it is easier, to pull a rope downwards than upwards.
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To obtain a Force Ratio (Mechanical Advantage), pulley systems have to be arranged
so that a pulley or pulleys are connected to the load as shown below.
EFFO R
Fi
LO A D
Fo
Fi
Fo
fig 10.6
In the pulley arrangement shown in the diagram a Force Ratio (Mechanical
Advantage) of 2:1 is achieved. This is because the force due to the weight of the load
F out is supported evenly between the rope to the left and right of the pulley.
To raise the load the input force Fin only needs to be 1/2 F out.
However, as with the lever, to raise the load through a distance din, the input force
will have to be pulled down a distance of 2 (times) d out.
Greater Force Ratio (Mechanical Advantage) s can be achieved by arranging the
pulleys in such a way that the rope is divided into more parts to support the load.
In the diagram the load is being supported by two pulleys which divide the weight of
the pulley between four ropes.
MA =
Fout
F in
= 4 or 4:1
VR =
d in
d out
= 4 or 4:1
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R O PE A
R O PE B
EF F O RT = F i
PU LLE Y B
PU LLE Y A
fig 10.7
1m
Example
The pulley system shown in the diagram is used to lift a load of 50 N a distance of
0.5 m. The input force required to lift the load is 30 N, which has to be pulled down
through 1 m. Calculate the efficiency of the pulley system.
0.5m
Fi = 15N
fig 10.8
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Efficiency =
Mechanical Advantage (MA)
× 100
Velocity Ratio (VR)
Fout
F in
MA =
=
50
30
= 1.67
VR =
Di
Do
1
0.5
=
=2
Efficiency =
=
MA
VR
1.67
× 100
2
= 83%
In this example we can see that the pulley is only 83% efficient. Energy losses due to
friction at the pulleys, and the fact that we have not taken into account the weight of
the pulleys and the rope, account for the system not being 100% efficient.
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The Windlass
150
400
Fo
Fi = ?
Fo = 60N
Fi
fig 11.1
fig 11.2
If we look at the device side on, we can see that in fact it too is a form of lever.
The Force Ratio (Mechanical Advantage) of a windlass will depend on how much
longer the crank handle is than the radius of the drum.
Example
Calculate the MA of the windlass shown in the diagram.
400
150
fig 11.3
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If we consider the windlass as a lever:
400
Fi = ?
150
F o = 60N
fig 11.4
By applying equilibrium to the windlass, we can find the magnitude of F in.
Fin × Din = Fout × Dout
Fin × 400 = 60 × 150
Fin =
60 × 150
400
MA =
=
F out
F in
60
20
= 3:1
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Ratchet and Pawl
You will notice that if you release the windlass handle whilst raising the load, the lift
will fall back down again due to its own weight.
A mechanical system which behaves like this (i.e. when the input force is removed the
output force moves the system) is said to overhaul. A useful device, which can be
used to prevent this from happening, is a ratchet and pawl.
PAW L
R AT C H E T
A LLO W S R O TATIO N TH IS W AY O N LY
fig 12.1
A ratchet and pawl system allows rotation of the drum in one direction but not the
other.
C ABLE
W IN C H D R U M
R AT CH ET
C RA NK
HA ND LE
PAW L BA R
fig 12.2
This function of the device is a useful safety feature when used in lifting systems.
Ratchet and pawl systems are also used in devices such as clockwork toys, fishing
reels and socket sets.
R AT CH ET
PAW L
fig 12.3
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Rotary Systems
Most modern mechanical systems are powered by some form of motor. For this
reason, mechanical systems which process rotary motion and transmit force are very
important.
R OTARY
M O TIO N
R OTARY
M O TIO N
R OTARY
SY STEM
FO R CE
FO R CE
fig 13.1
Pulley Systems
The simplest form of system, which achieves this, is the Pulley and Belt.
During the industrial revolution pulley and belt systems were used extensively
throughout industry to transmit power to machinery. Below, you can see a type of
pulley system used in DAF cars to transmit high and low speeds.
S P R IN G P U LLE Y AT
M A X IM U M D IA M E TE R
D R IV IN G P U LLE Y
D R IV E N P U LLE Y
LO W SPEED
C E N TR IFU G A L
W E IG H TS AT RE S T
S P R IN G P U LLE Y AT
M IN IM U M D IA M E TE R
H IG H SPEED
C E N TR IFU G A L
W E IG H TS FU L LY
E X TE N D E D
VARIOMATIC BELT-DRIVE SYSTEM
fig 13. 2
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In a pulley system, force and motion are transmitted from the input pulley to the
output pulley by means of a belt.
G R APH IC S YM BO L FO R
PU LLEY & B ELT S YS TEM
fig 13.3
When the input pulley is turned in a clockwise direction the output pulley turns in the
same direction.
G RA PH IC SY M BO L FO R
PU LLEY & C RO S SED
BE LT SY STEM
fig 13.4
However, if the belt is crossed over, then the direction of the output pulley is reversed.
The speed of rotation of the output pulley depends on the ratio of its diameter to that
of the input pulley.
0$&+,1(
6+$)7
#
%
287387
38//(<
fig 13.5
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PULLEY &
BELT SYSTEM
ROTARY SPEED
OF MOTOR
MACHINE SHAFT
ROTATES SLOWER
fig 13.6
If the output pulley is bigger than the input pulley then the output will run more
slowly than the input.
287387
38//(<
&
75,
&
5
(/( 272
0
0$&+,1(
6+$)7
,1387
38//(<
fig 13.7
R O TARY SPE ED
O F M O TO R
PU LLEY &
BE LT SY STEM
M A CH IN E S HA FT
R O TATES FAST ER
fig 13. 8
If the output pulley is smaller than the input pulley then the output will run faster than
the input. This relationship between the speed of rotation of the input and output
pulleys is called the Velocity Ratio
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Velocity Ratio (VR)
The velocity ratio for a rotary transmission system is the ratio of the number of
revolutions made by the input (Ni) to the number of revolutions made by the output
(No).
VR =
Ni
No
Example
In the pulley system shown the input has a diameter of 20mm and the output diameter
is 60mm.
12 0 R EV/M IN
O U TP U T
PU LL EY
IN PU T
PU LL EY
Ø 20m m
Ø 60m m
fig 13.9
Calculate the VR of the pulley system.
If the input pulley turns through one revolution then the output pulley must rotate
through 3 revolutions.
VR =
=
Ni
No
3
1
=3
i.e. 3:1
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If the input pulley is being rotated at a constant speed of 120 rev/min calculate the
speed of the output pulley.
We can use the VR already calculated to find the speed of rotation of the output
pulley.
VR =
speed of rotation of the input (Ni)
speed of rotation of the output (No)
VR =
Ni
No
No =
Ni
VR
=
120i
3
= 40 rev/min
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Assignment
Build and test the pulley systems shown in the diagram.
fig 13.10
Using the correct graphic symbols draw a line diagram for each.
On the diagram indicate the input and output pulley and its size.
Calculate the VR for each of the systems. Assuming that in each case the input pulley
rotates at 100 rev/min, calculate the rotational speed of the output pulley.
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Torque
As well as transmitting and processing rotary motion, pulley systems also transmit
force. In a rotary system the force is transmitted to produce a turning effect. This
turning force, and the effect it has, is called Torque.
0$&+,1(
6+$)7
287387
72548(
&
75,
&
5
(/( 272
0
,1387
72548(
fig 14.1
PULLEY DRIVE
SYSTEM
INPUT TORQUE
Ti
OUTPUT TORQUE
To
fig 14. 2
To understand torque we can consider a motor such as that shown in the diagram.
#
%
38//(<
/2$'
fig 14.3
The motor’s ability to transmit a force through a pulley to lift a load will depend upon
the torque which the motor can produce and the radius of the pulley.
TORQUE = FORCE X RADIUS
T=F×r
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As the motor torque remains constant this means that the motor can be used to lift a
small load quickly using a large diameter pulley or a large load, slowly, using a small
diameter pulley.
R=
SMALL
R=
LARGE
Fo = LARGE
Fi = SMALL
fig 14.4
Example
The motor shown can produce a torque of 100 Nm and the radius of the pulley is 0.1
m, calculate the load which can be raised by the motor.
% # % #"&
2
% #
38//(< 5$',86
P
/2$'
4
fig 14.5
T=F×r
F =
T
r
F =
100
0.1
= 1000 N
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If we increase the size of the pulley then a smaller load will be lifted because the same
motor produces a fixed torque.
% # % #"&
2
#
%
38//(< 5$',86
P
/2$'
4
fig 14.6
T=F×r
F =
T
r
F =
100
0.5
= 200 N
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Transmission of Torque
In a rotary system the torque at the input pulley will not necessarily be the same as the
torque transmitted to the driven pulley.
Any changes in speed between the input and output pulley will result in a change in
the torque transmitted.
If the output is smaller than the input pulley then the output will rotate faster than the
input and the torque at the output will be smaller.
0$&+,1(
6+$)7
287387
38//(<
&
75,
&
5
(/( 272
0
,1387
38//(<
fig 15.1
IN PU T S PE ED
SLO W
H IG H
TO R Q UE
R OTARY
SY STEM
O UT PU T SP EED
FAS T
LO W
TO R Q UE
fig 15.2
If the input pulley is smaller than the output pulley then the torque at the output will
be greater than at the input.
In effect, the ratio of the input and output torque for a rotary system is dependent on
the velocity ratio for the system.
Velocity Ratio =
VR =
Torque at output
Torque at input
To
Ti
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This effect can be seen in the use of stepped pulleys on pillar drilling machines.
fig 15.3
When the belt is on the top set of pulleys the output pulley rotates the drill very fast.
The torque transmitted to the drill will be small. Therefore with the belt in this
position only smaller diameter drills can be used as there will not be enough torque to
turn large drills.
38//(<
38//(<
02725
'5,//
fig 15.4
SLO W S PEE D
H IG H TO R Q UE
PU LLEY DR IV E
SY STEM
FAS T SP EED
LO W TO R QU E
fig 15.5
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Example
The motor, which drives the pulley system on a drilling machine, produces a torque of
100 Nm. The diameter of the motor pulley is 200 mm and the diameter of the drill
head pulley is 100 mm.
Calculate the torque transmitted to the drill head.
'LD
'LD
PP
PP
72548(
1P
02725
'5,//
fig 15.6
VR =
Torque out (To)
Torque in (Ti)
Velocity ratio can be calculated from the size of the pulleys.
VR =
diameter output Ni
=
diameter input
No
VR =
100
200
= 1:2
= 0.5
VR =
To
Ti
To = VR × Ti
To = 0.5 × 100
= 50 Nm
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The belt is now moved onto the bottom pair of pulleys. The motor pulley is now
diameter 100 mm and the drill head pulley is 200 mm. Calculate the torque
transmitted to the drill head.
'LD PP
'LD PP
72548(
1P
02725
'5,//
fig 15.7
VR
=
diameter output
diameter input
VR
=
200
100
=
2:1
VR
=
To
Ti
To
=
VR × Ti
=
2 × 100
=
200 Nm
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Belt Slip
Wherever a belt is used in a pulley system, the belt must be properly tensioned. If the
belt is too tight it will probably break. If it is too loose it will slip and/or come off the
pulley.
Tensioning a belt correctly can be achieved by adjusting the position of a pulley as is
shown in the diagram of the car fan-belt system.
7(16,2 1(5
fig 15.8
Another way a belt can be tensioned is by the addition of an extra, spring loaded,
pulley called a jockey pulley. The jockey pulley keeps the maximum amount of belt
in contact with the in pulley at all times.
OUTPU T
PULLEY
INPUT
PULLEY
JOCKEY
PULLEY
fig 15.9
Belt slip in pulley systems can be an advantage. If a machine jams or seizes and the
belt slips then the drive motor is unlikely to be damaged or overloaded.
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Assignment: Belt and Pulley Systems
1. A pulley drive system is shown below.
a) Explain in your own words why pulley C is needed in this system?
Pulley A rotates at 120rpm
b) Calculate the speed of rotation of pulley B.
c) Calculate the speed of rotation of pulley C
P ULLE Y B
P ULLE Y A
Ø 300m m
Ø 200m m
P ULLE Y C
Ø 50m m
MS. Int2 02 Fig15.10
2. A belt driven machine is shown below.
O VERH EAD PULLEY
Ø 450m m
LATH E PU LLEY
Ø 150m m
MS. Int2 02 Fig15.11
The driving pulley is 450mm diameter running at 150 revs per min and the
machine pulley is 200mm.
Determine the speed of the machine
a) If there is no slip?
b) If there is a 3% slip?
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3. The polishing machine shown is used to buff and polish plastic materials in a
workshop. The process requires the machine to operate at three different speeds
and the arrangement for this is shown using a belt and pulley system.
With the belt in the position shown, the speed of rotation is 1200 rpm.
Ø 100
BEA RIN G
Ø 50
MOP
Ø 25
AXL E
BELT
M O TO R
Ø 25
Ø 50
MS. Int2 02 Fig15.12
Determine the speed of the motor
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4. The arrangement shown below is the drive system for a tumble dryer drum with a
fan blowing hot air into the dryer drum.
The drum is required to rotate at 150rpm.
a) Calculate the rotational speed of the motor.
The fan is driven by the same motor as the drum through a pulley system.
b) State two advantages in using a pulley arrangement in this application?
c) Calculate the rotational speed of the fan.
d) A name two type of belt drives?
DRUM
DRUM PULLE Y
(Ø 500m m )
HEATER
CIR CUIT
FAN
FAN P ULLEY
(Ø 50mm )
M O TO R PU LLEYS
(Ø 150m m & Ø 50m m
CO M PO UN D PU LLEYS
(Ø 200m m Ø 50mm
MS. Int2 02 Fig15.13
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5. Find the transmission factor for A to B for each of the pulleys shown below. All
diameters are in millimeters.
100
20
80
A
B
A
30
180
B
90
60
A
B
A
40
B
20
120
80
B
A
20
100
A
B
50
30
100
120
80
B
20
30
160
40
A
MS. Int2 02 Fig15.14
6. On most car engines a pulley on the crankshaft is connected by a fan belt to the
pulley that drives the alternator, which in turn keeps the battery charged. The
typical arrangement is shown below. When the engine is running at 2800 rpm at
what speeds will the fan and the alternator be turning?
FA N P ULLE Y
Ø 75m m
A LTE R NATO R
P ULLE Y
Ø 50m m
C RA N K SH A FT
P ULLE Y
MS. Int2 02 Fig15.15
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7. A selection of pulleys is available in a construction kit as transmission aids. The
sizes available are 12mm, 24mm, 36mm, 48mm, 72mm, and 144mm. Show haw
they may be used to produce the following ratio multipliers.
a)
b)
c)
d)
e)
+4
–6
+1/3
–2/3
+72
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Chain and Sprocket Systems
The main problem with belt drive systems is that even when properly tensioned they
will slip.
Where it is undesirable for this to happen, the pulley can be replaced by a toothed
wheel called a sprocket, and the belt by a chain.
6352&.(7
&+$,1
fig 16.1
Where a quiet positive drive is required toothed belts can be used rather than chains.
7227+(' %(/7
fig 16.2
The most common application of a sprocket and chain is in the drive mechanism of a
bicycle.
Just as with a pulley system, a chain and sprocket needs to be properly tensioned. On
a bicycle this is done by moving the position of the back wheel of the bike or if the
bike has ‘Deraileur’ gears by the spring loaded jockey wheels.
-2&.(<6352&.(7
fig 16.3
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In much the same way as the stepped pulleys used in drilling machines, sprockets and
chain systems are used to control the speed and torque of the output.
IG H TO RQ UE
SL0W SPEED
fig 16.4
FIX ED TOR Q U E
FR O M C RA NK
C HA IN & SPR O C KET
SY STEM
LO W TO R Q UE
AT BAC K W HE EL
fig 16.5
If you want to travel up a hill you need high torque at the back wheel and therefore
the chain is moved onto the largest sprocket on the back wheel. This means that a
larger force can be transmitted through the back wheel but that wheel rotates slowly.
O W TO RQU E
FAST SPEED
fig 16.6
FIX ED TOR Q U E
FR O M C RA NK
C HA IN & SPR O C KET
SY STEM
H IG H TO R Q UE
AT BAC K W HE EL
fig 16.7
If you are travelling on the flat, less force is required at the rear wheel and you will
want to travel fast. To do this, the chain is moved onto the smallest sprocket on the
rear wheel.
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Example
The diagram below represents the chain drive system of a pedal cycle. The crank
sprocket has a diameter of 200mm and the rear (wheel) sprocket a diameter of 50mm.
',$
PP
',$
PP
fig 16.8
Calculate the VR of the sprocket system.
Driver speed x driver Diameter = Driven Speed x Driven Diameter
For one turn of the crank:
=
1 × 200
50
=
4 turns
VR
=
Ni
No
VR
=
1
4
=
0.25
Driven Speed
If the cyclist turns the crank at 50rev / min, calculate the speed of rotation of the rear
wheel.
VR
=
Ni
No
VR
=
50
0.25
Rear wheel
Speed
=
200rev/min
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If the rear wheel of the bike has a circumference of 2m, how far would the bike travel
in one minute?
Every time the wheel rotates the bike moves 2 m.
If the bike makes 200 turns in one minute then the distance(s) moved would be:
S = Rate of rotation (no) × Wheel circumference (circ)
S = no × circ
S = 200 × 2
= 400 m
The cyclist produces a torque of 1000 Nm, calculate the torque transmitted to the rear
wheel from the crank.
VR =
To
Ti
To = VR × Ti
To = 0.25 × 1000
= 250Nm
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Assignment
The mechanism shown is used to hoist loads on a site. The motor drives pulley A,
which is connected to pulley B by a V-shaped belt. Pulley B is on the same shaft as
gear C, which meshes with gear D. Fixed to gear D is a winding drum around which
is fixed a rope which passes over a pulley at the end of a jib. A load-carrying bucket
is attached to the other end of the rope.
G EA R C
M O TO R
W IN DING DR UM
PU LLEY A
G EA R D
PU LLEY B
W IN DING
D RU M
G EA R D
PU LLEY B
M O TO R
JIB
PU LLEY A
G EA R C
fig 16.9
State one advantage of using a belt drive instead of a chain in a power driven
mechanism.
Belt stretch can occur in belt drives. Sketch a method of counteracting belt stretch in
the diagram below. Name the device.
If the hoist motor fails, the load might fall. Show by means of a sketch a safety
device which could be used to prevent this happening. Name the device.
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Gear Systems
For very accurate transmission of rotary motion and force, gear wheels are used.
R O TARY
M O TIO N
FO R CE
G EA R
SY STEM
R O TARY
M O TIO N
FO R CE
fig 17.1
Gear wheels transmit force and motion by means of teeth which are very accurately
machined around their circumference.
fig 17.2
Two gear wheels in mesh
Gears are used in groups of two or more and when linked together to drive each other,
they are said to be in mesh.
fig 17.3
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We have already seen that the VR of a system is calculated from:
VR =
distance moved by input (di)
distance moved output (do)
VR =
do
di
When using gears we can also calculate VR by using the number of teeth on the gears.
VR =
Number of teeth on ouput gear (to)
Number of teeth on imput (ti)
VR =
to
ti
Example
Calculate the VR of the gear system shown in the diagram.
40 TEE TH
0 TEE TH
A
B
fig 17.4
VR =
to
ti
VR =
20
40
= 0.5 or 1.2
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Simple Gear Trains
Assignment
Build and test the gear system shown in the diagram.
fig 17.5
(a) Using the correct symbols, draw a diagram of the system showing the input and
output shafts, the number of teeth on each gear wheel and the direction of rotation of
both gears.
(b) Calculate the VR of the gear system.
(a) If the input gear were to rotate at 100 rev/min at what rate would the output gear
rotate?
(b) If the input gears shaft produces a torque of 50Nm what will be the torque
transmitted to the output shaft?
Copy and complete the table shown, either by building and testing gear trains with
the required sizes of wheels OR by showing calculations of your answers.
N UM B ER O F TE ETH
O N DRIVEN G EA R
N UM B ER O F TE ETH
O N DRIVING G EA R
30
20
?
?
10
4
20
?
2/3
40
20
?
40
10
?
?
20
1/2
VE LO C ITY R ATIO
fig 17.6
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Idler Gears
If an intermediate gear is used in a simple gear train then the direction of rotation of
the output can be changed. An additional gear used in this way is called an idler gear.
fig 17.7
The idler gear does not affect the VR of the system. Its main function is to change the
direction of rotation of the output gear.
The MA of the system will be reduced, due to an increase in friction caused by having
more moving parts in the system.
Any number of idler gears can be added to a system in this way. If there is an odd
number of idlers then the direction of rotation of the input and output gears will be the
same. If there is an even number of idlers then the direction of rotation will be
opposite.
The greater the number of idler gears the greater the friction produced, therefore
reducing the Force Ratio (Mechanical Advantage) and overall efficiency of the
system of Idler Gears.
If an intermediate gear is used in a simple gear train then the direction of rotation of
the output can be changed. An additional gear used in this way is called an idler gear.
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Gear Train with Idler
Assignment
Modify your last gear train to include an idler gear.
'5,9(5 *($5
7((7+
G R APH IC S YM B OL FO R G EAR TR AIN W ITH ID LER
fig 17.8
Using the correct graphic symbols, draw a diagram of the system showing the input
and output shafts, the number of teeth on each gear wheel and the direction of
rotation of both gears.
Calculate the VR of the gear system.
Compound Gear Trains
Whereas simple gear trains are useful in producing a relatively small velocity ratio in
a gear system, if we want to achieve large velocity ratios, simple gear trains become
impractical.
For example, if we require a VR of 1000 in a system then even using a small input
gear wheel with 10 teeth would require the output gear wheel to have 10,000 teeth. If
the input gear has a diameter of 20 mm then the output gear would have to have a
diameter of 20 m.
10 0 00 TO O TH E D
GEAR W HEEL
10 TO O TH E D
GEAR W HEEL
The diagram shows a 2m diameter gear relative to a mini
fig 17.9
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Obviously gear systems which require gears of this size are totally impractical.
The problems of producing high velocity ratios whilst keeping gear sizes down can be
overcome by using Compound Gear Trains.
Compound gear trains consist of a pair of meshed gears where the shafts are in
parallel. The input and output gears are connected by means on intermediate gears
fixed together on common shafts. The intermediate gears are not idlers, they do affect
the output of the gear train.
fig 17.10
The compound gear train shown in the diagram has two pairs of meshed gears on
parallel shafts. Although standard graphic symbols can be used to represent
compound gear trains it is better to represent this type of system with a symbol which
the top of the gear system.
IN TER M ED IATE
SH AFT
O UT PUT
SH AFT
B = 40t
A = 20t
C = 20t
D = 40t
fig 17.11
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Calculating Compound Gear Ratios
Build and test the compound gear train shown in fig 17.10.
Copy the graphic symbol for the system shown in fig 17.11.
(a) Use a model OR simulation to help you understand the calculation that follow.
(b) Write down your calculations as you complete the assignment.
Example
To calculate the total gear ratio of the gear system between gear A and gear D we
must find the velocity ratios of the two sets of gears in mesh, i.e. AB and CD.
VR AB =
number of teeth B
number of teeth A
VR AB =
40
20
=2 or 2:1
Similarly
VR CD =
number of teeth D
number of teeth C
VR CD =
30
20
= 3:2 or 1:5
To find the total VR of the system AD, the velocity ratios of the two gear systems
must be multiplied together.
Total VR AD = VR AB × VR CD
= 2 × 1:5
= 3 or 3:1
If the crank handle is rotated at 180 rev/min calculate the speed at which the output
shaft D rotates.
VR =
ni
no
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VR =
180i
3
= 60 rev/min
If the crank handle produces a torque of 15 Nm, calculate the torque transmitted to the
output shaft.
VR =
To
Ti
To = VR × Ti
= 3 × 15
= 45 Nm
With this system the output is greater than that at the input. In effect this
system will be able to deliver more force at the output.
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Gear Boxes
The gearing systems looked at so far have all produced a fixed output speed and
torque.
For exactly the same reasons as ‘Deraileur’ gears are used on bicycles it is useful to
be able to vary the output torque and speed of gear systems. This is the case in
vehicles such as cars which on a given journey a high torque may be required to drive
the vehicle. When travelling fast on the motorway a low torque system will be
required. A variable output speed and torque system is provided by means of a gear
box.
VA RIABLE O UTP UT SPE ED
FIX ED INPU T SPEE D
G EA R BO X
VA RIABLE O UTP UT TO R QU E
FIX ED INPU T TO RQ U E
fig 17.12
INPUT SHA FT
O UTP UT S HAFT
INPUT SHA FT
BOTTOM GEAR
NEUTRAL
INPUT SHA FT
O UTP UT S HAFT
O U T PU T S PE ED LO W
O U T PU T T O R Q U E H IG H
O UTP UT S HAFT
INPUT SHA FT
SECOND GEAR
O UTP UT S HAFT
TOP GEAR
O U T PU T S PE ED H IG H
O U T PU T T O R Q U E LO W
The diagrams show the operation of a car’s gearbox
fig 17.13
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Reversing Gear Box
The other advantage of this type of gear box is that it can be arranged to provide a
reverse gear.
When we investigate simple gear systems we found that it was possible to reverse the
direction of rotation of the output by adding an idler gear to the system.
A similar result can be achieved by including an idler gear on an intermediate shaft
into the gear box as shown on fig 17.14
INPU T SH AFT
OUTPU T SHAFT
REVERSE
IDLER GEAR REVERSES
DIRECTION OF ROTATION
OF OUTPUT SHAFT
fig 17.14
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Compound Gear Design
Assignment
(a) Design and build OR simulate a compound gear train with a VR of 12:1.
(b) Using the correct graphic symbols draw a diagram of your system indicating the
input and output and number of teeth on each gear used.
(a) Connect the input gear shaft to a 6 volt motor and gear box.
(b) Connect the output shaft
(c) An electrical circuit to control the lift is shown. Explain how it works.
6V
M O TO R
G EA R
BO X
R O PE TO
LIFT
W IN DING DR UM
fig 17.15
(a) Test the system to make sure that it can raise the lift.
(b) Draw a block diagram of the whole system.
The work done in raising the lift will be equal to the force due to the lift times the
distance it is raised.
WD = F × d
Calculate the work done in raising the lift 200 mm. You should already know the
weight of the lift from your previous work.
(a) The power required to raise the lift 200 mm will equal the work done in raising
the lift, times the time it takes to raise it.
Pm =
W
t
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Calculate the power required to raise the lift.
(b) The power to raise the lift is being supplied by the 6 volt motor.
Using a digital multimeter measure the current supplied to the motor.
Electrical power Pe can be calculated from the product of voltage and current.
Pe = I × V
Calculate the electrical power supplied to the motor.
The efficiency of a system can be calculated by comparing the power input to the
system to the
Power output.
Efficiency =
Pi
× 100
Po
Calculate the overall efficiency of the lift system.
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Assignments: Gear Systems
1. In the compound gear train, gears A and C are drivers and gears B and D are
driven gears. If gear A rotates at 1 revolution how many revolutions will gear D
rotate?
150 T EET H
INP UT
(D R IVE R)
A
200 T EET H
B
C
50 T EET H
D
100 T EET H
O U T PU T
(D R IVE N)
MS. Int2 02 Fig17.16
2. A gear wheel with 90 teeth revolves at 5 revs per second while driving a gear with
15 teeth. Determine the force multiplier and the speed of the driven gear?
5 R EV/S
15 TEETH
90 TEETH
MS. Int2 02 Fig17.17
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3. A driving gear in a simple gear train has 40 teeth. If the velocity ratio has to be 5
how many teeth will the driven gear have? If the speed of the driven gear is 25rev
per min, what is the speed of the driver?
4. The gearing of a hand-operated winch is shown below. A crank of radius 0.5-m
drives gear, A which meshes with gear B. Gears B and C are on the same shaft
where Gear C drives. The gear D, which is fixed to a winding drum, is NOT fixed
to the crank handle. The drum diameter is 0.3m with the rope diameter 20mm.
Determine the overall velocity ratio of the system?
D
Ø 0.3m
A
B
C
D
= 20T
= 100T
= 30T
= 90T
A
0.5m
1
2
C
B
MS. Int2 02 Fig17.19
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5. A motor, which runs at 2000 revs per minute, drives a pumping system at 400 revs
per minute. The compound gear train that is used is shown in the diagram below.
Find the number of teeth, which must be on gear wheel C to run the system.
D
IN P U T
2000 R E V /m in
A
G EA R
A
B
C
D
E
F
E
F
B
N o O F TEETH
30
40
?
50
25
45
C
O UTP U T
400 R E V /m in
MS. Int2 02 Fig17.20
6. A pupils computer simulation kit contains gears with 8, 16, 24 and 40 teeth
respectively. Show how these may be used to give gear trains with the following
multiplier ratios.
•
•
•
15
6
2.4 : 1
MS. Int2 02
Fig17.21
Calculate the output speeds if the input speeds are:
•
•
•
40rpm
10rpm
15rpm
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
73
7. The gearing of a roller arrangement for squeezing excess dye from soaked
material in an industrial process is shown. A flywheel is rotated and drives the
lower roller through a gear train consisting of 4 gears: A and C with 12 teeth Band
D with 32 teeth. The top roller is meshed to turn at the same speed as the lower
roller, but in opposite directions, by gears E and F which both have 16 teeth.
a) What is the transmission factor from the flywheel to the lower roller?
b) To the top roller?
F
TO P R O L LE R
FLY W H E E L
D
A
B O TTO M R O LLE R
E
C
B
MS. Int2 02 Fig17.22
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
74
8. A simple gear train is used in a machine when it is required to increase the speed
of rotation of an output shaft. If the speed of the driving gear is 1 rev/sec and the
number of teeth on each gear is 160 on the driver and 40 on the driven, calculate
the speed of rotation on the driven gear by comparing:
a) Machine speeds.
b) Teeth on the gear wheels.
9. The gear train shown was part of a remote control car’s gearing system. This
gearing greatly reduced the speed of rotation of the driver and turned it through
90°. If the driver is being turned by an electric motor at 3000 rpm, what will be
the speed of rotation of the follower?
50
50
25
19
57
FO LLO W E R
D RIV ER
MS. Int2 02 Fig17.24
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
75
Assignment
Recycled material is being compressed into bales and loaded onto a conveyor system
by the power driven winch shown below.
D RIVIN G
W HE EL
80T
R O PE DR UM
R O PE
D RU M
M O TO R
15T
20T
PU LLOE Y
75T
LO AD
LO AD
fig 17.16
If the motor runs at 1250 rev/min and the rope drum is 250mm diameter, calculate the
distance the load will be lifted in 1 second.
What is the tension in the rope if the load is 3kN?
Sketch a mechanism which would prevent the load from falling back if the motor
failed.
Name the parts of this mechanism.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
76
Worm Gears
W O RM
W O RM W H EE L
G RA PH IC S YM BO L FO R
W O RM & W O R M W HE EL
fig 18.1
Another system which can produce a large VR and MA is a worm gear and wheel.
The worm, which looks like a screw thread, can be thought of as a gear with one tooth
and is fixed to the input shaft. It meshes with the wormwheel which is fixed to the
output shaft. The output shaft runs at right-angles to the input shaft.
R OTARY
M O TIO N
W O RM &
W O RM W H EE L
FO R CE
R OTARY M OT IO N
O
AT 90
FO R CE
fig 18.2
One of the main advantages of worm systems is that they will not overhaul. This
means that they are very useful when used in lifting systems, when the input force is
removed the load will then stay in position.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
77
Worm systems are also used in tensioning the strings on instruments such as guitars
and some members of the violin family.
fig 18.3
Assignment
Build and test the worm and wormwheel system shown in the diagram below.
fig 18.4
Using the correct graphic symbols, produce a diagram for the system. On the diagram
indicate the direction of rotation of the gears and the number of teeth.
Calculate the gear ratio for the system.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
78
Assignment
A diagram of a drive gear for a time switch is shown below. The motor shaft is fitted
with a single start worm, this is connected to a compound gear train as shown.
Calculate the motor speed required for the clock dial to revolve once in every 24
hours.
O N / O FF CO N TA CTS
M O TO R
10 TE E TH
360
TE E TH
12 TE E TH
W O RM
D RIV E
100
TE ETH
240
TE ETH
C LO C K D IA L
240
TE E TH
fig 18.5
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
79
Right Angled Drive Systems
Two other gear systems which produce a right angled drive are shown.
G R A P H IC S YM B O L
B EV E L G E A R S
fig 19.1
R O TARY
M O TIO N
R IG HT AN G LE
D RIVE SYS TEM
FO R CE
R O TARY M O TIO N
O
AT 90
FO R CE
fig 19.2
Bevel Gears
Bevel gears have their teeth machined at an angle of 45 degrees. When the meshed
bevel gears are the same size and have the same number of teeth, they are called mitre
gears.
Face Gears
Gear teeth can be cut on the top surface of a gear wheel to produce a right angled
drive. A gear wheel machined in this way is called a face gear.
G R A PH IC S Y M B O L
FA C E G E A R S
fig 19.3
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
80
Rack and Pinion System
A rack is like a gear wheel that has been rolled out flat. When a rack is meshed with a
pinion or small gear wheel, a system is produced which can convert rotary motion to
linear motion or linear motion to rotary motion.
R O TARY M O TIO N
(LINE AR M O TIO N)
R AC K & PINIO N
SY STEM
FO R CE
LIN EA R M O TIO N
(RO TA RY M O TIO N)
FO R CE
fig 20.1
fig 20.2
Rack and pinion systems are commonly used on drill and lathe feed systems and car
steering mechanisms.
:,3(5 $50 *($56
029(' %< 5$&.
5(&,352&$7,1* 027,21
0(&+$1,60 3529,',1*
527$5< 72
/,1($5 027,21
02725
fig 20.3
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
81
The rack and pinion can be represented by a graphic symbol as shown.
G RA PH IC SY M BO L
R AC K and PINIO N
fig 20.4
When a pinion is turned the linear movement of the rack is determined by the number
of teeth per metre on the rack and the number of teeth on the pinion.
Example
A rack with 200 teeth per metre is meshed with a pinion that has 10 teeth. If the
pinion is rotated through 15 revolutions how far does the rack move?
92
9
fig 20.5
The distance moved by the rack will be found from the number of teeth per metre of
rack, the number of revolutions made and the number of teeth on the pinion.
Distance moved (Dr) =
Dr =
Number of teeth on pinion × Number of revs made
Number of teeth per metre of rack
10 × 15
200
= 750mm
= 0.75m
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
82
Assignment
A conveyor system designed for processing coal is shown. In order to control the
flow - rate of coal on the belt, a variable height gate is fitted into the system.
P O W ER
S UP P LY
R AC K
A ND
P IN IO N
C O N TRO L
U NIT
VA R IA B LE H E IG HT
G ATE
P
C O N V E YO R
FR AM E P IV O TED
H ER E
C O N S TA N T S PE E D
D RIV E M O TO R
LO AD SE N S O R
fig 20.6
What device could be used at A to drive the rack, and how would it work?
Explain the need for the frame to be pivoted at P.
Complete a Block diagram for the system.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
83
The Cam and Follower
A Cam and follower is a mechanism that is designed to convert rotary motion into
reciprocating motion.
R OTARY
M O TION
R EC IPR O CATIN G
M O TION
C AM &
C AM FOLLO W ER
FO R CE
FO R CE
fig 21.1
A cam is a specially shaped piece of metal or plastic. The edge or profile of the cam
guides the motion of the follower.
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027,21
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)$//
5,6(
)2//2:(5
527$5<
027,21
':(//
&$0 )2//2:(5
fig 21.2
The diagram shows the motion of a cam over one revolution.
The ‘Dwell’, where no motion takes place, is followed by the ‘Rise’ and the ‘Fall’.
The length of travel of the follower is fixed by the ‘Stroke’ of the cam.
Pear shaped cams are used in valve control mechanisms.
&$0
6+$)7
fig 21.3
Cam used to operate a pneumatic 3/2 roller trip valve
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
84
The valve is actuated when the highest point of the valve is in contact with the roller.
In a similar way a cam can be used to operate a braking system in the wheel drum of a
car.
CAM
FO LLO W E R
B R A K E A D JU S TIN G M E C H A N ISM
Cam used to operate a car’s breaking system
fig 21.4
In a car engine a cam and follower arrangement is used to control the operation of the
valves which allow the petrol air mixture into and exhaust gases out off the engine
cylinders.
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7$33(7
)2//2:(5
&$0
7$33(7
)2//2:(5
fig 21.5
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
85
Assignment
The system shown in the diagram is to be built so that microswitches can be operated
by the rotation of the 4 cams on the camshaft. The switches should be connected to
individual bulbs in order that switch ‘1’ controls bulb ‘1’ etc.
fig 21.6
Build and test a cam system that will operate the following sequence, for one
revolution of the camshaft
BULB 1
BULB 2
BULB 3
BULB 4
ON
O FF
O FF
ON
O FF
ON
O FF
ON
O FF
O FF
ON
O FF
O FF
ON
ON
O FF
fig 21.7
Use the correct graphic symbols to produce a diagram that shows the operation of the
system.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
86
Crank Slider
A crank slider is a device by which torque can be transmitted to a shaft.
The torque transmitted to the shaft = force applied to the handle x
radius length of the handle.
T = F x r
TO R Q U E
fig 22.1
A crank handle can be incorporated into a shaft and a shaft with several cranks is
called a crankshaft.
fig 22.2
A child’s pedal car uses a crankshaft to transmit the torque produced by the child’s
legs into rotary motion to drive the wheels.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
87
In a similar way, a crankshaft is used in car engines to convert the reciprocating
motion of the cylinders into rotary motion to drive the wheels.
fig 22.3
Most cars have a four-stroke engine. This means that each piston goes through a
cycle of 4 movements or strokes, thousand of times a minute.
The four strokes are:
FIRST STROKE
SECOND STROKE
THIRD STROKE
FORTH STROKE
TH E P IS TO N M O V ES
D O W N THE C Y LIN D ER
S UC K IN G IN FUE L AN D
A IR M IX TU RE
TH E P IS TO N M O V ES UP
C O M PR E S SIN G THE
M IX TUR E
A S PA R K S E TS LIG H T TO
TH E M IX TU R E . TH E
E X PLO S IO N P US H E S THE
P IS TO N D O W N
TH E P IS TO N M O V ES UP
P US H IN G TH E B UR N T
G AS E S O UT O F THE
C YLIN DE R
fig 22.4
A crank slider consists of a rotating crank that is connected to a slider by a connecting
rod.
ROTARY
M OTION
RECIPROCATING
M OTION
G RA P HIC AL S Y M B O L FO R
C RA N K S LID E R M E C H AN ISM
fig 22.5
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
88
The mechanism is used to transmit force and to convert rotary motion to reciprocating
motion or reciprocating motion to rotary motion.
R OTA RY
M O TIO N
FO RC E
C RAN K / SLID ER
M EC H AN ISM
R ECIPR O CATIN G
M O TIO N
FO RC E
fig 22.6
During the industrial revolution most machines where powered by steam engines and
the crank slider was used extensively to convert the reciprocating motion produced by
the steam engine into rotary motion.
fig 22.7
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
89
Assignment
A waste disposal plant sorts, recycles, and burns unusable waste. The waste is first
transported to a size sorting plant where it is passed over a series of different sizes of
mesh. These meshes are vibrated by means of the mechanism shown in the diagram
below.
MESH BOX
R 50
M E C H AN ISM
N O T TO S CA LE
fig 22.8
Name the mechanism.
Identify the type of motion with which the mesh box will move.
The system is driven by a motor running at 1200 rev/min. The motor shaft has a
worm drive, engaging with a 40 tooth wheel. The wheel is fixed to the shaft of the
mechanism.
Calculate how many seconds it takes for the mechanism to complete one full cycle of
its operation.
Calculate the total distance traveled by the mesh box during one full cycle of
operation of the mechanism.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
90
Mechanical Systems used in keep fit equipment.
It was stated earlier that our lives are made easier by the using mechanical systems in
our jobs of work and leisure. In this outcome some mechanical systems within leisure
equipment will be looked at.
Exercise is a fundamental source of physical and mental wellbeing and mechanical
systems play a part in how fitness equipment operate. Exercise machines such as
bicycles, treadmills, steppers, rowing machines etc are used in schools, sports centres
and leisure facilities to allow individuals to improve their fitness.
This unit of work will look at some of the knowledge you have gained about
mechanical systems and apply it to exercise equipment with a view to how the
equipment benefits fitness and the individual.
For example the work done by an individual on exercise equipment relates to the
amount of energy used in completing a fitness exercise.
MS. Int2. 04 fig 1
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
91
Work
The best way to understand work is to look at a simple example.
A person keeps fit by pulling weights inwards. This involves doing work. How hard
the work done is will depend on two factors:
1.The mass of the weights .
2 The distance she moves the pads together.
Work Done = Force (Weights) x Distance (for the pads to meet)
As the units of Force is the Newton and the unit of Distance is the Metre. Work is
measured in Newton Metres (NM). The NM is a measurement of energy which is
called the Joule (J).
Therefore it follows that:
Work is a form of energy and energy measured in Joules is equivalent to 0.2388
Calories.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
92
Assignment
Calculate the work done during a fitness exercise with the above equipment for a
person lifting 3 weights. Each weight weighs 1OON and the distance the weights are
lifted is 1.2 m.
a. Work done in raising 1 weight:
W=FxD
b. Work done in raising 3 weights:
W=3FxD
c. How many Calories did the individual use?
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
93
Power
Is the rate of working, the amount of energy used per second.
Power = Work Done / Time J/s
P =W/t
Power is measured in Joules/second or watts.
Lets look at the example of keeping fit.
Assignment
a. If it takes the keep fit person 10 seconds to lift 3 weights then the power used to
do this work will be:
Power = Work/time
If the exerciser is more powerful then she / he may be able to lift double the weight at
same time.
Of course if she / he does this it will only take him 20 seconds to do the same work.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
94
Assignment
a. What power is used this time to do the work?
P = W/t
From this simple example we should be able to see that a powerful person is capable
of doing work quickly.
The same is true of machines, powerful machines are capable of doing an enormous
amount of work very quickly and as a consequence of this, require a great deal of
energy.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
95
Exercise Cycle
The following example looks at an exercise bike with pedals applying the effort of the
exerciser. The pedals rotate a chain wheel with a chain drive to rotating the flywheel.
The flywheel creates a resistance (Torque) which can be changed depending on the
level of exercise required and effort applied by the exerciser.
The example below involves a speed of 50 revs/min, with an effort of 200N and a
pedal radius of 0.175 metres.
MS. Int2. 04 fig 3
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
96
Assignment.
a. Calculate the distance moved by the pedals in one minute?
b. Calculate the work done in one minute?
c. What power has been applied?
d. Indicate the torque on the pedals by the exerciser?
e. If the efficiency of the exerciser is 20% for cycling, what energy is used per
minute?
f. How many calories does the exerciser use?
Exercise Stepper
The stepper is a commonly used item within fitness and leisure facilities. The
mechanism to control the reciprocating pedal movement varies from machine to
machine. Foot pedal resistance can be controlled by a variety of ways, e.g. link cables,
belts, hydraulics, electronically controlled magnetic brakes etc.
MS. Int2. 04 fig 4
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
97
Assignment
Show how you would use a pneumatic system to operate the foot pedals resistance.
a. Using your pneumatic symbols complete a circuit diagram.
b. Show calculations how varied resistance levels can be applied.
c. Name all the components.
d. Complete a systems diagram for the exercise stepper.
e. Describe how the system operates.
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
98
Exercise Treadmill
The surface of the treadmill has good traction and the shock absorbing rubber mat can
absorb 86% of the impact that stresses muscles, bones and joints. The running surface
can be horizontal or inclined and various speeds can be selected.
Usually the speed is electronically controlled using a rheostat, which makes manual
adjustment easy.
MS. Int2. 04 fig 5
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
99
Assignment
a. Complete a systems diagram for the exercise treadmill.
b. Add additional systems to develop a closed looped system.
c. Show how a mechanical system can decrease the output speed of the motor.
d. Show calculations how your mechanical system can decrease the output speed of a
b. 1200 rpm motor.
e. What effect is created if the running surface is inclined to 15 degrees?
DET Technological Studies Support Materials:
Applied Electronics (Int 2) Outcome 2
100