Revised Standard Grade Technological Studies
Mechanical Systems
Contents
Preface
Structure
Resources
Assessment
Introduction
Motion
Rotary
Linear
Reciprocating
Oscillating
Forces
Static forces
Dynamic forces
Bending forces
Shear forces
Torsion forces
Compression forces
Tension forces
Vectors
Equilibrium
Levers
Force multiplier ratio
Movement multiplier ratio
Efficiency
Classes of lever
Moments
The bell-crank lever
Linkages
Free-body diagrams
Beams
Beam reactions
Types of beam support
Gears
Simple gear train
Movement multiplier ratio in gears
Idler gears
Ratchet and pawl
Compound gears
Worm and wheel
Bevel gears
Torque and Drive Systems
Power transmitted by a belt drive
Belt and chain drives
Multiplier ratio for belt drives
Toothed belts
Chain drives
Multiplier ratio for chain drives
Chain tension
Converting motion
Cams
Cam motion
Crank slider
Rack and pinion
Couplings
Bearings
Clutches
Mechanical Systems – Homework
Unit Assessment
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Standard Grade Technological Studies: Mechanical Systems
Preface
Mechanisms are widely used in industry and are included in a large number of devices
and equipment commonly used in society. It is important, therefore, that students have
a good understanding of how mechanical systems assist and aid everyday tasks.
Throughout the coursework, students should gain knowledge of how mechanical
systems operate and how they fit into common everyday devices. It is not only
necessary to understand how the mechanical system carries out its task, but also
important to know hot the systems can be altered to improve performance and
efficiency. It is therefore recommended that this unit of work be completed towards
the end of the course to allow students to grasp the requirements and rigour of the
mechanisms calculations.
Owing to the nature of course modifications, it is not necessary for students to
construct mechanical or compound systems. If, however, the class teacher sees a need
for construction to assist the learning process, then this can be undertaken. The
construction of ‘rigs’ or working models to display and convey important mechanical
observations is also acceptable.
Structure
The material provides the student with natural progression throughout the course.
There is, however, flexibility within the material so that the teacher can determine
which sections can be taught to suit the classroom needs.
The following areas are covered.








Introduction
Motion
Forces
Vectors
Levers
Moments
Linkages
Free-body diagrams






Beams
Gears
Torque and drive systems
Converting motion
Rack and pinion
Couplings, bearings, clutches,
friction
The content of the unit fulfils all the requirements of the course and the structured
homework exercises can be distributed at the end of a course topic. Some teachers
may, however, add additional well-tried and tested worksheets to benefit student
learning and understanding of topics.
The use of the CD-ROM gives students an interactive way of understanding
mechanical devices in real-life situations. This classroom methodology is a very
important part of students’ development and their knowledge of how mechanical
systems work can assist in everyday situations. The tasks associated with the CDROM allow pupils to search and interact at suitable times within the coursework.
There is a variety of suitable CD-ROMs on mechanisms and these could also be used
at the discretion of the teacher.
The interactive computer simulation associated with the tasks:
Standard Grade Technological Studies: Mechanical Systems
iii
 6 Levers
 4 Gears and Belts
 12 Gears and Belts
 2 Cams and Cranks
 2 Friction
refers to The New Way Things Work CD-ROM.
The ‘Simulation of Gear Systems: Task 1  Simple gear train’ refers to Crocodile
Clips  Mechanical components.
Resources
The majority of resources required to assist in the understanding of this unit should be
available if Technological Studies has already been undertaken. The resources
comprise:
 examples of classes of levers
 a balance lever
 a model of a gear train
 a model of a compound gear system
 a computer simulation of gear systems
 an interactive computer simulation of mechanisms
 a model of a bevel-gear system
 a model of belt/chain-drive systems
 a model of a rack-and-pinion system
 examples of couplings, bearings and clutches
 individual components.
Assessment
Internal assessment
Internal assessment can be measured against the tasks that the students complete
during the coursework allied with their homework at the end of topics.
A twelve-question end-of-unit assessment prepares students to undertake assessable
elements in knowledge and understanding and reasoning and numerical analysis. This
assessable work can be used to supplement students’ grades in the event of appeals.
Marks are awarded for each question and are associated with elements.
External assessment
This unit of work and the exercises within prepare the students for any mechanisms
questions that appear in the 90-minute exam at the end of the course. It will enable all
students to gain the knowledge and understanding required and give them suitable
practice in reasoning and numerical analysis.
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Standard Grade Technological Studies: Mechanical Systems
Introduction
Mechanisms are still a large part of modern society. Most of the mechanisms that we
use every day are so familiar that we never think twice about them, for example door
handles, light switches, scissors, etc.
Mechanisms play a vital role in industry. While many industrial processes now have
electronic control systems, it is still mechanisms that provide the muscle to do the
work. They provided the forces to press steel sheets into the shape of car body panels,
to lift large components from place to place and to force power hacksaws to cut
through thick metal bars – the list of jobs is endless. It is only by using mechanisms
that industry can make products you use every day.
Some machines are easy to understand, but many are hidden away from sight behind
glossy panels and covers. In the past, machines were much easier to see, as with the
old steam engine, for example, but as people became more concerned about safety, it
was necessary to fit guards over moving parts. Today, guards are often replaced by
styled covers that make it much harder to see what is happening, but whether you can
see them or not, mechanisms are still playing a vital part in everyday life.
All mechanisms:
 involve some kind of motion
 involve some kind of force
 make a job easier to do
 need some kind of input to make them work
 produce some kind of output.
Standard Grade Technological Studies: Mechanical Systems
1
Motion
There are four basic kinds of motion.
Rotary
Turning in a circle. This is the most common
type of movement, for example wheels, clock
hands, compact discs, CD-ROMs.
Linear
Movement in a straight line, for example
movement of a paper trimmer cutting a straight
edge on paper or a lift moving between floors.
Reciprocating
Backwards and forwards movement in a straight
line, for example the needle in a sewing machine or
the piston in a car engine.
Oscillating
Swinging backwards and forwards in an arc, for
example the pendulum of a clock, a playground
swing or a rocking horse.
2
Standard Grade Technological Studies: Mechanical Systems
Motion: task 1
What types of motion do the following sports or leisure activities show when they are
being used or carried out? Complete a systems diagram for each.

Swing

100 metres’ sprint

Golfing

Bungee jump

See-saw

Fire button on a computer game
Standard Grade Technological Studies: Mechanical Systems
3
Motion: task 2
The machines and tools that are used in your practical rooms in school use all types of
motion. The four types of motion are listed; now list as many machines/tools as
possible for each type of motion.
Rotary
Linear
Reciprocating
Oscillating
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Standard Grade Technological Studies: Mechanical Systems
Forces
Forces affect structures in a variety of different ways depending on how they are
applied to the structure. Forces can move a structure slightly or cause damage by
changing the shape of the structure.
Sometimes when forces are applied to a structure, it may be almost impossible to see
changes happening. For example, a bridge will sag slightly when a vehicle drives over
it, but this is not visible to the human eye. Nevertheless, the vehicle causes
downwards movement of the bridge structure. Loads such as vehicles on a bridge can
be deemed examples of forces acting on the bridge. Forces can stop an object from
moving or they can make it change direction. When a football is kicked, the forces
applied from the player cause the dimensions of the ball to change on impact. It
happens so quickly that it is not visible.
Forces are measured in newtons and the symbol is the letter ‘N’.
There are a number of different types of forces that can be applied to and which affect
bodies and structures.
Static forces
When static loads or forces are applied to structures, the structures do not normally
move. Normally the total downwards force comprises the weight of the structure plus
the load it is carrying. The runner below is in his starting position; his weight is a
static or stationary downwards force.
Dynamic forces
When dynamic loads or forces are applied to a structure, the structure does move and
the forces applied can be varied. Dynamic forces are visually more noticeable and are
produced by a variety of means and effects: machines, wind directions, people, etc.
The picture below shows the sprinter after the starting gun has been fired; he is
creating a dynamic impact to gain momentum.
Standard Grade Technological Studies: Mechanical Systems
5
Bending forces
Structures that carry loads across their length are subject to bending forces. The
weightlifter lifting a weighted bar feels the effect of the downward forces of the
weights and these cause the bar to bend.
A car driving across a bridge will cause bending forces on the structure but often they
are not visible.
Shear forces
Shear forces can be described as tearing or cutting forces affecting structures. Simple
examples are a pair of scissors used to cut a ribbon at an opening ceremony and a
mower cutting the grass.
Torsion forces
Torsion or torque forces have the effect of trying to turn or twist a structure or a piece
of material. A screwdriver being twisted to apply a force to a screw and a spanner
turning a bolt to lock it into place are examples of torque being applied.
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Standard Grade Technological Studies: Mechanical Systems
Compression forces
The figure below shows a column with a weight pressing down on it, but the column
does not disappear into the ground because the ground exerts an upwards reaction
force on the column’s base. The downward pressure of the weight and the upward
reaction are external forces trying to squash or shorten the column. Forces that act like
this are called compressive forces and the column is said to be in compression.
For example, when you sit on a stool in the classroom, your weight acts as a
downward force on the chair. However, there must be an
upward force on the legs of the chair; therefore the legs are
said to be in compression.
The same can be said about the weightlifter’s arms and legs.
WEIGHT FORCE W
(EXTERNAL FORCE
ON COLUMN)
W
COLUMN
R
GROUND REACTION R
(EXTERNAL FORCE ON COLUMN)
Figure 1
Tension force
We have noted that compression occurs when things are being pushed together. The
opposite of compression is ‘tension’ – when a structure is being pulled apart. In a tug
of war, the two sides are pulling the rope in opposite directions. The forces applied by
the teams are called tensile forces and cause the rope to be in tension. It could also be
said that the arms of team members are in tension.
Figure 2
The wire rope holding the net in volleyball is also in
tension.
Figure 3
Standard Grade Technological Studies: Mechanical Systems
7
Force: task 1
Against each of the six forces mentioned make a list of ‘real life’ situations where
these types of forces may be found. Ask the teacher if you are unsure which category
the situations fit into.
(a) Static
(b) Dynamic
(c) Bending
(d) Shear
(e) Torsion
(f) Compression
(g) Tension
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Standard Grade Technological Studies: Mechanical Systems
Vectors
Force is a vector quantity and has both magnitude and direction. This means it is often
convenient to represent a force by a line, that is, a vector quantity, which is sometimes
easier to understand visually. The direction of the force may be indicated by an arrowheaded line, with the length of the line drawn to scale to represent the size of the
force. This line is called a vector.
Example
The cyclist pedalling with a force of 800 N is being assisted by a tail-wind of 400 N,
but the friction from the road surface measures 200 N.
800 N
400 N
200 N
Figure 1
The overall effect will be 800 N + 400 N – 200 N = 1000 N (or 1 kN).
A suitable scale would be selected – possibly 10 mm to represent 20 N – and using
this scale each force is drawn in turn, one following on from the other.
800 N + 400 N – 200 N = 1000 N or 1 kN
Figure 2: a vector diagram
When the three forces are added together, they can be replaced by a single force that
has the same effect, called the ‘resultant’.
RESULTANT = 1 kN
Figure 3
Standard Grade Technological Studies: Mechanical Systems
9
Vectors are also used to find the resultant of two forces that are inclined at an angle to
each other.
25 N
60 º
35 N
Figure 4
In the example above the resultant of the two forces can be found by drawing two
vectors. First choose a suitable scale and draw the two vectors CA and CB.
Scale: 10 mm = 10 N
25 mm = 25 N = CA
35 mm = 35 N = CB
The bigger the scale the more accurate the vectors.
From A draw a line parallel to CB, and from B draw a line parallel to CA. Call the
point where the two lines intersect point D. Now draw a line from C to D. A line
drawn from C to D is the resultant of the two forces CA and CB.
Figure 5
The resultant has a magnitude of 46 N by measurement.
Equilibrium
Certain conditions must apply within structures in order to create stability. The
resultant is made up of the combined forces that are trying to move an object or
structure in a set direction. If such a force were applied without an opposing force
then major problems could occur. Structures have to remain in a stable or balanced
state called ‘equilibrium’, which simply means ‘balanced’. There are three types of
balancing that must exist if structures, bodies, objects, etc. are to remain in
equilibrium: horizontal, vertical and rotational forces must all balance.
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Standard Grade Technological Studies: Mechanical Systems
The general conditions of equilibrium are as follows.



upward forces
leftward forces
clockwise moments
=
=
=
downward forces
rightward forces
anticlockwise moments
Example 1
Consider again the same two forces in figure 5. Are they in equilibrium? It is easily
seen that a force must be added acting downwards to the left, but we cannot tell from
this exactly how large this force must be or its exact direction (figure 6).
Figure 6
The resultant has been drawn and it can be seen that to balance it, the equilibrant CE
is required. If the forces F1 and F2 are drawn as in figure 7 then it is much easier to
obtain the equilibrant by completing the triangle, as shown in figure 7.
Figure 7
Standard Grade Technological Studies: Mechanical Systems
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Example 2
A crane is fixed against a wall, as shown in figure 8. Find the forces in the
compression and tension members.
Figure 8
To find the forces created in the tension and compression members by the 1000 N
load, draw the triangle from the area circled. Select a suitable scale and then draw the
known force first, the 1000 N load (figure 8). A line is drawn through one end of the
load line parallel to one of the unknown forces. Another line is drawn through the
other end, parallel to the second unknown force. By measuring each line, the size of
each force can be found. (Note: the arrowheads must follow round the triangle.)
Figure 9
Scale: 10 mm = 200 N
The compression member = 2000 N
The tension member = 1733 N
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Standard Grade Technological Studies: Mechanical Systems
Equilibrium: task 1
1. Study the following statements and cross out the incorrect answer.

A body that is accelerating is in a state of equilibrium.
TRUE/FALSE

For a body to be in a state of equilibrium it is necessary only for the vector
sum of the forces acting on it to be zero.
TRUE/FALSE

A resultant force is a single force that can replace two or more forces.
TRUE/FALSE

If two or more forces are replaced by a resultant force, the effect on the body
is changed.
TRUE/FALSE

An equilibrant force is the force that, if applied to a body, will cause the body
to be in a state of equilibrium.
TRUE/FALSE

The equilibrant force is identical to the resultant force.
TRUE/FALSE
2. Try to explain two conditions necessary for a structure or body to be in
equilibrium.
3. Two forces are acting on a body as shown.
(a) Graphically indicate their size and direction.
(b) Graphically indicate the resultant of the two forces.
4. Two forces are acting on a body as shown.
(a) Graphically indicate their size and direction.
(b) Calculate the resultant and direction of the two forces.
5. What are the resultant and equilibrant of the two forces affecting the system
below?
Standard Grade Technological Studies: Mechanical Systems
13
6. A small crane is used on a fishing trawler to lift cases of fish to the dock. The
weight of the lift is 1200 N. Determine the size and direction of the forces in each
of the crane members. (Use a scale where 10 mm represents 200 N.)
1200 N
7. A weight of 2000 N is suspended by a rope attached to a hook firmly fixed to a
roof joist. A second rope is attached to the vertical rope and pulled horizontally
until the rope makes an angle of 30 to the vertical as shown. Determine the
horizontal pull on the rope and the force on the hook.
2000 N
8. The figure below shows a cranked lever that is part of a gear-change mechanism.
Find the resultant force FR acting on the hinge pivot and the angle .
750 N
FR
600 N
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Standard Grade Technological Studies: Mechanical Systems
Levers
Figure 1(a) shows an early lever. The large boulder is too heavy to move by pushing
it. By using a small boulder as a pivot point and a branch as a lever, it is possible to
amplify the force applied to the large rock. The further from the pivot the effort is
applied, the easier it is to move the large rock or load.
Figure 1(a)
Figure 1(b)
When a weight is attached to one side of a lever to assist the user, it is known as a
counterbalance.
A universal systems diagram of a lever is shown in figure 2. A lever system changes
an input force and an input motion into an output force and an output motion.
INPUT FORCE
INPUT MOTION
OUTPUT FORCE
LEVER SYSTEM
OUTPUT MOTION
Figure 2
The point that a lever pivots about is called a fulcrum. A line diagram of a lever is
shown in figure 3. The input force is called the effort and the input motion is the
distance moved by the effort force. The output force is called the load and the output
motion is the distance moved by the load.
EFFORT
DISTANCE
MOVED
BY EFFORT
LOAD
DISTANCE
MOVED
BY LOAD
Figure 3
The lever is a force multiplier and is normally used to get a large output force from a
Standard Grade Technological Studies: Mechanical Systems
15
small input force. However, it can also be used as a distance multiplier, giving a large
output movement for a small input motion; but it cannot do both at the same time.
Figure 4 below shows a lever system designed to move heavy machine castings from
a lower level to a position of installation.
The castings must be lifted 200 mm.
EFFORT = 260 N
LOAD = 750 N
600 mm
Figure 4: machine-loading lever system
Force multiplier ratio
In the lever system shown in figure 4 above, the load being lifted is about three times
more than the effort being applied. The load divided by the effort gives a ratio. This
ratio is a force multiplier, or how much more load can be lifted compared to the effort.
The lever in figure 4 therefore has a force-multiplier ratio of 2.88 (a ratio has no units
of value).
Example 1
Find the force-multiplier ratio for the lever in figure 4 above.
Force-multiplier ratio =
load
effort
=
750 N
260 N
=
2.88
Movement-multiplier ratio
The force multiplier ratio appears to give the user something for nothing. The user is
only applying about a third of the force to move the load. However, it can be seen
from figure 4 that the effort side of the lever has to move much further than the load
side. The ratio of the distance moved by the effort, divided by the distance moved by
the load, is known as the distance-multiplier ratio.
The lever in figure 4 therefore has a distance-multiplier ratio of three (again a ratio
has no units of value).
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Standard Grade Technological Studies: Mechanical Systems
Example 2
Find the distance-multiplier ratio for the lever in figure 4 above.
Movement-multiplier ratio = distance moved by the effort
distance moved by the load
= 600 mm
200 mm
=3
Efficiency
Owing to the effects of friction and inertia associated with the movement of any
object, some of the input energy to a machine is converted into heat, and losses occur.
Since losses occur, the energy output of a machine is less than the energy input; thus
the mechanical efficiency of any machine cannot reach 100 per cent.
The efficiency of a lever system is found by dividing the force ratio by the movement
ratio, with the efficiency given as a percentage. The result of the above division is
multiplied by 100 to give the percentage efficiency value.
Example 1
Find the efficiency of the lever system shown in figure 4.
Efficiency (η) = Force Ratio
Movement Ratio
 100
η = 2.88  100
3
η = 96
The system shown in figure 4 has an efficiency of nearly 100 per cent. No system can
be 100 per cent efficient; there are always losses. The losses in a lever system consist
of energy lost to friction at the fulcrum of the lever and the energy lost in strain as the
lever bends slightly. In some cases a small amount of energy will also be lost in the
form of sound.
Remember, no machine is 100 per cent efficient. Common energy losses include heat
energy due to friction, strain energy and sound energy.
Standard Grade Technological Studies: Mechanical Systems
17
Levers: task 1
Draw a universal system diagram of a lever system. Label the diagram input, process
and output.
Complete the line diagram of a lever shown below. You should identify the load,
effort and fulcrum.
INPUT
OUTPUT
Levers: task 2
Calculate the force-multiplier ratio of the following levers. Show all working.
EFFORT
1OO N
18
EFFORT
300 N
EFFORT
200 N
LOAD
400 N
EFFORT
50 N
LOAD
100 N
Standard Grade Technological Studies: Mechanical Systems
Levers: task 3
A diagram of a lever system is shown below.
(a)
(b)
(c)
(d)
Find the force-multiplier ratio of the lever system.
Calculate the movement-multiplier ratio of the lever.
Calculate the efficiency of the system.
Identify possible efficiency losses in the system.
Show all calculations.
EFFORT = 150 N
LOAD = 450 N
650 mm
200 mm
(a) Force ratio =
(b) Movement ratio =
(c) Efficiency (η) =
(d) Possible efficiency losses in a lever system =
Standard Grade Technological Studies: Mechanical Systems
19
Classes of levers
Levers can be divided into three distinct types (classes) determined by the position of
the load, effort and fulcrum. Applications of their use are found almost everywhere,
from the home or school to equipment on the space shuttle. The classes of levers are
as follows.
Class 1
In class 1 levers the effort is on one side of the fulcrum and the load is on the opposite
side (figure 5). Class 1 levers are the simplest to understand: the longer the crowbar
the easier it is to prise open the lid.
LOAD
EFFORT
FULCRUM OR PIVOT
Figure 5
Class 2
In class 2 levers the fulcrum is at one end of the lever and the load and the effort are
spaced out on the other end of the bar. The load must be closer to the fulcrum than the
effort (figure 6). A wheelbarrow is a good example of a class 2 lever. The wheel is the
fulcrum, the load is in the container area and the effort is applied to the handles.
Similarly, a door has a hinge (fulcrum), the load can be considered as acting in the
door’s centre of gravity and the effort is applied as far from the hinge as possible.
LOAD
EFFORT
FULCRUM
Figure 6
Class 3
Class 3 levers are similar to class 2 levers except that now the effort is closer to the
fulcrum than the load (figure 7). This means that more effort has to be applied to
move the load. This type of lever is used when mechanisms require a large output
movement for a small input movement.
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Standard Grade Technological Studies: Mechanical Systems
EFFORT
FULCRUM
LOAD
Figure 7
Examples of various types of lever are shown below; in some cases it is difficult to
tell exactly into which class they fit.
E
E
F
E
F
L
L
C
A
B
F
L
F
E
L
E
E
L
F
L
L
Standard Grade Technological Studies: Mechanical Systems
E
21
Levers: task 4
Complete the following list in a table format.



22
Name the equipment.
Draw a line diagram with arrows showing the fulcrum, effort and load.
Name the class of lever.
Standard Grade Technological Studies: Mechanical Systems
Levers: task 5
Calculate the force multiplier ratios for the following levers and state which class of
lever each one belongs to.
1..
F.M.R.
Class of lever _________
Class of lever _________
10N
Standard Grade Technological Studies: Mechanical Systems
23
Levers: task 6
In the printed version of these materials, issued to Scottish schools in August 2001, this page contained
embedded copyright material. For copyright reasons that material has been removed for this website
version. In order to see the completed text for this page Scottish schools are advised to refer to their
copy of the printed version. For other users the complete pack is available from Learning and Teaching
Scotland, priced £24.00.
Using your CD-ROM The New Way Things Work, try to answer the following
questions from Principles of Science (Levers) in an interactive way.
1. When rotating a lever mechanism what other name can be used when effort is
applied?
2. Name two related machines that use levers.
3. Explain in your own words how a lever mechanism is used in a car’s clutch
system.
4. What class of lever is a can opener? Sketch a line diagram to show effort, fulcrum
and load.
5. How does a lever system work in a fire extinguisher?
6. What class of lever is used in an aneroid barometer?
24
Standard Grade Technological Studies: Mechanical Systems
Moments
The sketch in figure 8 shows a weight attached to a metal rod, and the rod is free to
rotate around a hinge (pivot) P. If the rope holding the weight stationary is cut, what
happens to the rod? If the rope is cut, the force on the weight causes the rod to swing
or turn around the pivot. This ‘turning effect’ is called a moment.
The weight in figure 8 shows a moment of 20 Nm (10 N  2 m). A moment is
measured in newton-metres.
As long as the rope is not cut, the weight and rod are held in balance by the force in
the tie rope.
When any system is in a steady state it is said to be in equilibrium.
ROPE
WEIGHT
HINGE P
2m
TURNING EFFECT
Figure 8
The lever system in figure 9 shows a lever that is in a state of equilibrium (balance).
The input force is tending to turn the lever anticlockwise; the load is tending to turn
the lever clockwise. The forces on each end of the lever are exerting a moment: one
clockwise, the other anticlockwise. If the beam (lever) is in equilibrium, both of these
moments must be equal.
FORCE (10 N)
Figure 9
Standard Grade Technological Studies: Mechanical Systems
25
The principle of moments states that the sum of the moments must equal zero or the
sum of the clockwise moments must equal the sum of the anticlockwise moments.
The Greek letter  stands for ‘the sum of’ and can be used as a shorthand way of
writing the principle of moments:
CWM = ACWM
F1¹ d1 = F2  d2
The force times the distance turning the lever clockwise is equal to the force times the
distance turning the lever anticlockwise. As stated, moments are measured in newtonmetres. It can be seen that the moment on one side of the lever is equal to the moment
on the other side. (Work done = force  distance in the direction of motion.)
Example 4
Using the lever system in figure 10, use the principle of moments to show that the
lever is in equilibrium.
Figure 10
Answer
For equilibrium, the CWM = ACWM. A moment is a force multiplied by a
distance
CWM = ACWM
F1¹ d1 = F2  d2
The load is exerting a clockwise moment; that is, it is tending to make the lever turn
clockwise.
Clockwise moment = 200 N  2 m = 400 Nm
The effort is exerting an anticlockwise moment.
Anticlockwise moment = 400 N  1 m = 400 Nm
 CWM = ACWM
Therefore the lever is in a state of equilibrium.
26
Standard Grade Technological Studies: Mechanical Systems
Moments: task 1
Use a balanced lever similar to the one in the line diagram shown below, which is
available from your teacher. Use a set of weights as a load. Use the spring balance to
apply the effort to the system for each of the load positions.
LOAD
PIVOT
EFFORT
1
2
3
Suspend the load from position 1 on the load side of the lever. Measure the effort
required to balance the lever using a spring balance. Record the effort in the table
below. Move the load to positions 2 and 3 and record the effort required for balance
each time.
Position 1
Position 2
Position 3
Load
Effort
Calculate the force multiplier ratio for each position.
Position 1
Position 2
Position 3
Complete this statement. As the load gets further away from the fulcrum, the effort
required to balance it …
Standard Grade Technological Studies: Mechanical Systems
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Example 5
A car footbrake uses a lever action to amplify the force transmitted by the driver to
the braking system when the driver’s foot presses the foot-pedal. If the driver’s foot
can exert a force of 5000 N, what force will be transmitted to the braking system?
5000 N INPUT
500 mm
FORCE TO BRAKING
SYSTEM (LOAD)
100 mm
FULCRUM
Figure 11
This is a class 2 lever. Take moments about the fulcrum to find the force on the
braking system. Notice the distance from the fulcrum to the input is 600 mm.
The input tends to make the lever turn clockwise; the braking system is opposing the
input and so acts to turn the lever anticlockwise.
The principle of moments states that
CWM = ACWM
F1  d1 = F2  d2
5000 N  0.6 m = braking force  0.1 m
braking force = 5000 N  0.6 m
0.1 m
braking force = 30,000 N or 30 kN
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Standard Grade Technological Studies: Mechanical Systems
Moments: task 2
Use the principle of moments to find the missing force or distance in the following
problems. Show all working.
E=?
L = 5 kN
0.9 m
CWM = ACWM
0.3 m
FULCRUM
E = 50 N
L=?
200 mm
CWM = ACWM
40 mm
FULCRUM
E = 480 N
L = 960 N
300 mm
CWM = ACWM
d=?
FULCRUM
E = 400 N
L = 1200 N
d=?
CWM = ACWM
0.2 m
FULCRUM
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Moments: task 3
The hand-cutters shown are used to cut thin metal with the effort and load shown.
(a) Draw a suitable line diagram.
(b) What effort will have to be applied if the force required in the hand-cutters to
shear metal is 1.5 kN?
Moments: task 4
The diagram below shows a tower crane carrying a load of 90,000 N. At the other end
a counterbalance load is applied.
(a) Explain why the crane would be unstable without the counterbalance.
(b) Is it an advantage for the counterbalance to be able to move, either towards the
centre of the crane or away from its centre?
(c) The crane in the diagram is lifting a load of 90,000 N, which is 6.3 m away from
the tower. How far from the tower should a 100,000 N counterbalance be placed
so that the crane remains stable?
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Standard Grade Technological Studies: Mechanical Systems
Moments: task 5
A single-lever monobloc tap is shown below.
(a) If the length of the handle is 250 mm and the effort to turn it is 15 N, what
moment would close the tap valve?
(b) What is the benefit of this type of tap?
(c) Where would this type of tap be very useful?
Moments: task 6
When a fish has been hooked, the pull from the fish is 22 newtons at right angles to
the fishing rod. The pivot is at the end of the rod, which is 2.4 metres long. The angler
applies an effort at 0.4 metres from the end of the rod.
(a) Draw a line diagram with dimensions, loads, pivots, etc.
(b) Calculate the anticlockwise turning moment applied by the fish.
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(c) Calculate the effort the angler must apply to stop the rod from turning
anticlockwise.
(d) The angler has to exert a greater effort than the load applied by the fish to
maintain equilibrium. Is this an advantage or disadvantage to the angler?
The bell-crank lever
The bell-crank lever shown in figure 12 is used to transmit the input force and motion
through a right angle. It gets its name from part of the bell mechanism used to
summon servants in Victorian houses. By varying the lengths of the two arms of the
bell crank, it is possible to use it either to magnify an input force or to magnify an
input motion.
Figure 12
Example 6
Use the principle of moments to determine the length of the output side of the bellcrank lever in figure 12.
Calculate the force-multiplier ratio of the lever.
Answer
This is a class 1 lever with a right-angled bend. To find the distance ‘d’, take moments
about the fulcrum. Assume the lever is in equilibrium so that the output force opposes
the input force.
CWM = ACWM
F1 d1 = F2  d2
600 N  d = 400 N  0.15 m
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Standard Grade Technological Studies: Mechanical Systems
d = 400 N  0.15 m
600 N
d = 0.1 m
The force-multiplier ratio = load
effort
= 600 N
400 N
= 1.5
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Linkages
Levers are often linked together to transmit force or motion. A linkage consists of two
or more levers connected together. Linkages are useful for changing the direction of
an input or for giving greater force or distance amplifications.
Five common linkages found in many machines are shown below.
Figure 1
Reverse motion output;
distance from fulcrum
is the same, therefore,
same force.
Figure 2
Reverse motion output,
but fulcrum is nearer
the output so the
force is amplified.
Figure 3
Input and output motion are the same, but there is a large amplification of force.
Figure 4
Reciprocating motion transformed
to rotary motion.
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Figure 5
Lazy tongs linkage
for extra reach.
Standard Grade Technological Studies: Mechanical Systems
Linkages: task 1
A system diagram of a lever mechanism is shown below. The requirements state that
when the lever is pushed down, the output should rise.
The force-multiplier ratio should be 2:1.
INPUT
DOWNWARD
FORCE
OUTPUT
LEVER MECHANISM
UPWARD
FORCE
SYSTEM DIAGRAM
INPUT
OUTPUT
MECHANISM
Design a suitable linked lever system that will achieve the desired output.
A sketched diagram should show:
 a line diagram
 the load, effort and fulcrum
 your notional load and effort indicated in newtons
 your calculation showing the force ratio.
Answer
Evaluate your solution by stating whether the solution satisfied the requirements
identified from the specification in the question.
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Linkages: task 2
A counterbalance lever is required for a small city-centre car park.
Requirements
The car park is a small, one-person business. The operator wants a system that can be
opened easily by one person. The system must stay open until closed by the operator
and the system must be safe for the operator’s and customers’ cars.
The following criteria have been identified from the requirements.
The car park barrier must:
 lift with a small effort
 be operated initially by hand
 be able to be locked in an upright position
 be improved to operate with a simple electronic circuit.
Design a suitable system to satisfy the design requirements. State how you would test
for the following features.

A suitable lever system

Manual lift and lower, showing the force-multiplier ratio

Lock in raised position

Electronic circuit to automate the system

Safety
Draw a sketch of your solution and state whether your system satisfied the criteria.
Write down how you think you could have improved your solution.
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Standard Grade Technological Studies: Mechanical Systems