FORCES
5. Turning forces and moment calculations
from spanners to wheelbarrows & equilibrium situations
IGCSE AQA GCSE Physics Edexcel GCSE Physics OCR GCSE
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Doc Brown's school physics revision notes: GCSE
physics, IGCSE physics, O level physics, ~US grades 8, 9 and 10
school science courses or equivalent for ~1416 year old students of
physics
This page will help you answer questions such
as:
What is a moment? What is a mechanical
advantage? How do you calculate the turning effect
of a force? Why are the turning effects of a force
so important? Where do we apply the advantages of the
turning effect of a force?
Subindex for this page
(a)
An
introduction
to moments and mechanical forces of rotation
(b)
Some simple moment calculations
(c)
Moment calculations and a balancing situation (equilibrium)
(d)
More complex moment and equilibrium calculations
(e)
Some simple applications of turning effects of forces
including levers
(f)
Gears and cog
wheels  a means of transmitting rotational effects
(g)
Some physics of the
human skeleton and muscle system
See also
Mass  effect of gravity on it 
weight, (mention of work done, GPE and circular motion)
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(a) An
introduction
to moments and mechanical forces of rotation
Forces can cause an object to rotate and the
turning effect of the force is called a
moment.
If a resultant force acts on an object
about a fixed turning point (the pivot) it will cause the object to
rotate e.g. turning a nut with a spanner, applying a screwdriver, opening a door
fixed on hinges. The pivot might also be called the
fulcrum.
The rotational or turning effect, the moment,
has a magnitude easily
calculated from the formula:
M = F x d, where M
= the moment of a force (Nm), F = force applied (N)
and d (m) is the
perpendicular distance from the pivot point to the line of action of the
force.
You
get the maximum moment by pushing/pulling the spanner at a right angle
(at 90^{o})
to the line (d) between the pivot point and the line of action where the
force is applied.
Screwing or unscrewing a nut on a bolt
This is illustrated by the simple diagram
of a spanner above. The turning
force is F x d.
The pivot point is the central axis of
the bolt on which the nut is being turned by the spanner.
To tighten or loosen the nut you apply a
force, to the best mechanical advantage, at 90^{o} to the spanner
arm itself.
Applying the force at any other angle
less than 90^{o} reduces d and so reduces the effective moment of
the force.
You determine the force F by how hard you
push/pull the end of the spanner, but d is a fixed distance for a given
spanner.
This is one of many situations where you
are applying a force to increase the effect of your hand muscles.
The size of the moment increases with
increase in distance d or applied force F.
The longer the spanner, the greater is d,
hence the greater the
turning force generated  the greater the mechanical advantage of the
lever system.
More on the physics of unscrewing a tight
nut!
Spanner
'situations' A to D
Relative comments on the three 'moment' situations A to C
Situation A
With the longer spanner, and applying
the force perpendicular (at 90^{o}) to the line from the point
of applying the force (end of 'handle') to the pivot point (centre of the nut or bolt),
you generate the maximum moment (F x d).
The perpendicular distance d, is the shortest distance between the pivot
and the line of action of the force
Situation
B
If you apply the force at any other
angle than at 90^{o} to the perpendicular distance line (d), d
will always be shorter and hence a smaller moment is generated by
applying the same force as in situation A.
Note that when the line of action of
the force is down the 'spine' of the spanner, d is zero and the moment
is zero.
This is shown by the blue force
arrow at situation D.
Situation
C
With the shorter spanner, despite
applying the force at 90^{o} (perpendicular), d is smaller and
you cannot generate as large a moment as in situation A.
So, for the same applied force,
the moment is smaller for the smaller length spanner.
Situations A and B and riding
a bicycle
You get situation B (from above) pedalling a
bicycle when your foot and pedal are near the top or bottom of the
crank's cycle  the pivot point is the crank axle. If you press
downwards, you are creating the minimum moment  the minimum driving force forward.
At the top of the 'pedal cycle', if the direction of force is
downwards through the crank, the turning force, the moment, is very small.
BUT, at the top of the 'pedal
cycle', you soon learn to push
forwards to utilise the maximum moment and generate the maximum
force so that you mimic situation A where you are generating a force
at 90^{o} to the pivoting axle and the line of action of the
force 
maximum moment = F x d on the above diagram.
You get the maximum force
transferred when the crank and pedal are parallel to the ground and
you exerting your maximum downward force at 90^{o} to the
pivot point (axle) and the line of action of the force.
The centre of mass
and the stability of a free standing object
The centre of mass is a single point
in the object through which the whole weight of an object is considered to
act.
Its quite easy to envisage where it
is for a regular shape e.g. a rectangular block  shown in profile in
the diagram below. It is coincident with what is termed the 'centre
of gravity' of an object.
A standing object becomes unstable
when the vertical line through its centre of mass falls outside its
base, which effectively acts as a base  this happens if it is
tilted over on one edge, thereby creating a moment  a turning force.
Under these conditions, the weight of the object causes a
turning effect about the pivotal base.
The idea is illustrated by the
diagram below of a regular shaped block, shown in profile, and tilted at various angles (but it could be a
bus going round a corner!).
1. The vertical line from the centre
of mass passes right through the centre of the block's base.
The object
is completely stable  no moment (turning force) is generated.
2. The vertical line from the centre
of mass still passes through the base, but not its centre, and the block is unstable, so it
will wobble a bit from side to side, and eventually settle down in an
stable upright position as in 1.
The edge of the block touching the
surface acts as a pivot point.
The weight of the block creates an
anticlockwise moment (turning force) that makes the block fall back in
an anticlockwise direction, but not sufficient to topple the block over
on its longer side.
3. The vertical line from the centre
of mass passes outside of the block's base. The block won't even
wobble, it is highly unstable and will just topple over on its longer side (to
become stable!).
Again, the edge of the block
touching the surface acts as a pivot point.
Again, the weight of the block
creates a clockwise moment (turning force) that makes the block fall
over in a clockwise direction, and sufficient to topple the block
over on to its longer side.
Tests on stability in terms of the
centre of mass are important e.g. road vehicles like buses are safety
tested to see the maximum angle allowed when tilted over without
toppling over in an accident.
TOP OF PAGE and
subindex
(b)
Some simple moment calculations
Q1
Calculate the moment if a force of 5.0 N
is applied to a spanner 15 cm long.
F = 5, d = 15/100 = 0.15 m
moment M = F x d = 5 x 0.15
=
0.75 Nm
Q2
A force of 20 N is applied to a door causing a moment of 5 Nm.
Calculate the distance in cm from the
hinge axle to the point on the door where the force was applied.
moment M = F x d, so d = M/F =
5/20 = 0.25 m, therefore d = 0.25 x 100 =
25 cm
Q3
What force must applied to a 30 cm long spanner to generate a moment of 6.0 Nm?
moment M = F x d, so F = M/d, d =
30 cm = 30/100 = 0.30 m
F = M/d = 6.0/0.30 =
20 N
TOP OF PAGE and
subindex
(c)
Moment calculations and a balancing situation (equilibrium)
The left diagram illustrates a balanced situation (equilibrium) where a ruler is
pivoted in the middle and two weights w1 and w2 are placed at distances d1 and
d2 from the pivot point.
Remember weight = force in newtons.
The weights hang vertically so the force due to gravity is
acting perpendicularly (at 90^{o}) to the ruler
For the ruler to be balanced in a perfect horizontal position
the two turning forces must be equal.
Here we use the terms clockwise moment and
anticlockwise moment for the two turning effects of the forces involved.
anticlockwise moment = w1 x d1 (lefthand side of pivot),
clockwise moment = w2 x d2 (righthand side of pivot)
so when
w1d1 = w2d2
the ruler is balanced horizontally, at
equilibrium when the turning effects of the forces are equal.
This situation conforms with the principle
of moments which
states that when the total sum of the anti clockwise
moments is
equal to the total sum of the clockwise
moments the system is in equilibrium
and the object (system) will NOT turn.
When a system is stable (no movement) or balanced it is said to be in
equilibrium as all the forces acting on the system cancel each other out.
You
see this when you do a simple experiment balancing a rule on a pencil and
putting small weights on either side until balanced.
Similarly, when a the nut of a bolt is tightened, there comes a point when
the moment you are applying is balanced by the opposing moment of the bolt
and nut and you cannot tighten the nut anymore.
The
middle of a seesaw is the pivot point. If two people of equal weight
sit on either end, the seesaw is balanced horizontally  the
clockwise and anticlockwise moments are equal. If the two people differ in
weight, the seesaw will drop down on the side of the heaviest person because
the clockwise and anticlockwise moments are unequal.
The direction of rotation i.e. clockwise or anticlockwise, will be
decided on the relative weights (forces) at each end of the seesaw. One
end will fall in the direction of the largest moment
An example of using the principle of
moments  old fashioned kitchen scales
The beam of the scales should be horizontal
when the bowl and weights plate are empty (d1 = d2, w1 = w2).
In other words d1w1 = d2w2
When the object to be weighed is placed in
the dish, the scales tip anticlockwise down on the left.
You then add weights until the beam is
horizontally balanced again, thus giving the weight of the material e.g. flour
in the bowl.
Examples of simple calculations using the above situations
Predict what happens in the following situations Q1(a) to (c)
1 kg = 1000 g and 100 cm = 1 m
and for
simplicity assume g = 10 N/kg (weight = mass x force of gravity)
Q1(a) Suppose d1 = 20 cm, w1 = mass of 25 g, d2 = 10 cm,
w2 = mass of 50 g
but is it balanced?
anticlockwise moment = d1w1 = (20/100) x
(10 x 25/1000) = 0.05 Nm
clockwise moment = d2w2 = (10/100) x (10
x 50/1000) = 0.05 Nm
In this case the anticlockwise moment
= clockwise moment, so the ruler is balanced horizontally.
identical twins on a seesaw will be balanced !!
BUT, by applying an alternating extra muscle turning force, you can have great
fun !!!
Q1(b) Suppose d1 is 14 cm, w1 = mass of 52 g, d2 = 12
cm, w2 = mass of 60 g
but is it balanced?
anticlockwise moment = d1w1 = (14/100) x
(10 x 52/1000) = 0.073 Nm
clockwise moment = d2w2 = (12/100) x (10
x 60/1000) = 0.072 Nm
In this case the anticlockwise moment
> clockwise moment, so the ruler will rotate anticlockwise.
Q1(c) Suppose d1 is 2.5 m, w1 = mass of 55 kg, d2 = 3.0
m, w2 = mass of 50 kg
but is it balanced?
anticlockwise moment = d1w1 = 2.5 x (10 x
55) = 1375 Nm
clockwise moment = d2w2 = 3.0 x (10 x 50)
= 1500 Nm
In this case the clockwise moment >
clockwise moment, so the ruler will turn clockwise.
TOP OF PAGE and
subindex
(d)
More complex moment and equilibrium calculations
Q2
If w1 is 12.5 N and 3.5 m from the pivot
point, what weight w2 is required if placed at 2.5 m from the pivot to balance
the beam?
anticlockwise moment = 12.5 x 3.5 = 43.75
Nm
clockwise moment = w2 x 2.5
To balance the moments must be equal so:
w2 x 2.5 = 43.75, therefore w2 =
43.75/2.5 = 17.5
N
Q3
A beam is placed evenly on a pivot point
(fulcrum).
On one side a 10 N weight is placed 2 m from
the pivot point and a 40 N weight a further 4 m from the pivot point.
How far from the pivot point must the centre
of gravity of an 80 N weight be placed to perfectly balance the beam
horizontally?
The principle of moments states that the sum
of the clockwise moments must equal the sum of the anticlockwise moments to
attain equilibrium. A moment (Nm) = F (N) x d (m)
The sum of the clockwise moments = (10 x 2) +
(40 x {2 + 4}) = 20 + 240 = 260 Nm
To balance this the anticlockwise moment must
= 240 Nm, 240 = 80 x d, d = 260/80 = 3.25
Therefore the 80 N weight must be placed
on the left 3.25 m from the pivot point.
Q4
This calculation is the sort of thing civil engineers and architects have to
consider in the construction of 'modern' buildings.
A 5.0 m aluminium beam is suspended by a
steel cable from a concrete beam and 3.0 m along rests on a steel pole.
Assuming gravitational field force is 9.8
N/kg, calculate T, the tension in newtons on the supporting steel cable.
The weight of the aluminium beam = 9.8 x 200
= 1960 N.
moment = force x perpendicular distance from
turning point.
You consider the weight of the aluminium beam
to act through its centre of mass 2.5 m from either end, but it is 0.5 m from
the steel rod which is effectively the pivot point on which you base your moment
calculations.
Its actually the same situation as a
wheelbarrow described further down the page!
anticlockwise moment = beam weight x
distance from steel pole to centre of mass of aluminium beam = 1960 x 0.5 = 980 Nm
clockwise moment = tension in the steel cable x distance
from steel cable to steel pole = T x 3.0
at equilibrium, i.e. balanced, the clockwise
moment = anticlockwise moment
therefore: T x 3.0 = 980, so T = 980/3 =
327 N
(3 sf)
Q5
The diagram on the right shows downward forces (4000 N and 6000 N) acting on a
concrete beam that is held in position by a wall at P  effectively
acting as a pivot point.
Another 2nd supporting wall is to be
erected at X.
Calculate the force F that this
2nd wall needs to be able to withstand to produce a structurally stable
balanced equilibrium situation.
Clockwise moment = 6000 x (40 + 20) = 360
000 N
Anticlockwise moments = (4000 x 32) + (40
x F)
360 000 = 128 000 + (40 x F)
232 000 = 40F
F = 232 000 / 40 =
5800 N
TOP OF PAGE and
subindex
(e)
Some simple applications of turning effects of forces including levers
Many of the examples described below involve
a lever, which is a means of increasing the rotational effect of a force.
You push down on one end of a lever, and the
rotation about the pivot point can cause the other end to rise with a greater
force.
The diagram below illustrates the principle
of lever to gain a mechanical advantage  its all about the d1/d2 ratio.
F is the force involved (N) and d (m)
is the (shortest) perpendicular distance from the pivot point to the point where
the force is applied OR generated.
F1 = the applied input force, d1 = the
distance from the pivot to where the force F1 is perpendicularly applied
F2 = the output force, d2 = the distance
from the pivot to where the force F2 is perpendicularly generated
When balanced, i.e. just before something is
'levered' and caused to move
input moment = F1 x d1 = output moment =
F2 x d2
From F1 x d1 = F2 x d2, rearranging gives
F2 = F1 x d1/d2
Therefore by making d1 much bigger than d2
you can produce a much bigger output force compared to the original input
force.
Generally speaking you make the distance d1
much bigger than distance d2  you can see this using scissors, levering a lid
off a can and its a very similar situation when using a fork to lift deep
tough roots out of the soil or moving a heavy stone with a pole.
So, levers are very useful because they
make lifting and moving things much easier by reducing the input force
necessary to perform the task.
Example
of a lever question
Suppose a heavy manhole cover requires a
force of 160N to open it. Imagine you have a steel bar 1 metre long pivoted
at a point 0.1 m (10 cm) from the end to lever open the heavy manhole cover.
If you press down with a force of 20 N,
will the upward force you generate be enough to lift the manhole cover?
referring
to the diagram again
input moment = F1 x d1 = 20 x 0.9 = 18 Nm
output moment = F2 x d2 = F2 x 0.1 Nm
Since F1 x d1 = F2 x d2
F2 = (F1 x d1) / d2 = 18 / 0.1 =
180 N
The
output force 180 N exceeds the 160 N force required, so the manhole cover
can be lifted.
A hole punch of some description
The diagram shows a simple machine to punch holes in a
sheet of material.
The lever is pulled down to produce the hole.
The pivot point (turning point) is on the left.
We can analyse this situation in terms of
turning forces.
Applying the principle of moments: F1 x d1 =
F2 x d2
Rearranging the equations gives: F1 = F2 x d2
/ d1
Therefore by making d2 'long' and d1 'short'
you considerably multiply the force F1 compared to F2.
So you are able to easily punch holes in a
strong material e.g. sheet of metal.
For example, suppose d2 is 0.5 m (50 cm)
and d1 0.05 m (5 cm)
F1 = F2 x 0.5/0.05, so F1 = 10 x F2
So the force you manually apply is multiplied 10
ten times, not bad for a little effort!
In other words you need less force to get
the same moment.
Calculations based on the punch machine illustrated
above
You can treat the moments as Ncm, but you may be asked
to give the answer in Nm.
(a) If d1 is 10 cm, d2 120 cm and the pull down force F2
required is 2 N, what punch force F1 is produced?
moment = force x perpendicular distance to the
turning point:
moment 1 = moment 2 = F1 x d1 = F2 x d2
so F1 x 10 = 2 x 120 = 240 Ncm (2.4 Nm)
therefore F1 = 240/10 =
F1 = 24 N
(b) If d2 is 80 cm and the pull force F2 used is 4 N, what is the maximum length d1 can be to produce a minimum required punch force F1 of 40 N?
moment = force x perpendicular distance to the
turning point:
moment 1 = moment 2 = F1 x d1 = F2 x d2,
so 40 x d1 = 4 x 80 = 320 Ncm, so d1 = 320/40 =
d1 = 8 cm
(c) If d1 is 9 cm, d2 108 cm and the punch force F1
required is 60 N, what is the minimum pull down force F2 required?
moment = force x perpendicular distance to the
turning point:
moment 1 = moment 2 = F1 x d1 = F2 x d2
9 x 60 = F2 x 108 = 540 Ncm
so F2 = 530/108 =
F2 = 5 N
(d) When d1 is 12 cm, the punch force required F1 is 36
N if a pulldown force F2 of 3N is applied. What is the minimum length of the lever d2 required to punch the hole?
moment = force x perpendicular distance to the
turning point:
moment 1 = moment 2 = F1 x d1 = F2 x d2,
so 36 x 12 = 3 x d2 = 432 Ncm, so d2 = 36 x 12 / 3 =
d2 = 144 cm
Scissors
When you press the scissor hands together you create a powerful turning force
effect close to the pivot point.
F1 x d1 = F2 x d2, rearranging gives F2 = F1
x d1/d2
So, by making d1 >> d2, you create a much
bigger out force F2, sufficient to crisply cut through paper or card.
That's why you apply the blades to whatever
you are cutting as near as possible to the pivot point.
You don't cut using the ends of the scissor
blades where you gain little mechanical advantage i.e no significant multiplication of the force
you apply. Its the same principle as described in the whole punch machine
described in (a) above.
Levering the lid of a can
You can use a broad bladed screwdriver to get
the lid off a can of paint. The pivot point is the rim of the can.
The length of the screwdriver to the pivot
point (d2) is much greater than tip of the screwdriver beyond the rim (d1).
F1 x d1 = F2 x d2, F1 = F2 x d2/d1, so if d2
is much bigger than d1, you get a great magnification of the force you apply
(F2) to give a much greater up force (F1) to force the lid off.
Another example of needing less force to get
the same moment to do the job of opening the can.
The relatively long handle of a spanner
Spanners have long handles to give a strong
turning force effect.
Generally speaking, the larger the nut to be
tightened, the longer the spanner.
Spanners were discussed in detail at the
start of the page.
A cork screw
The radius of the handle is much greater than
the boring rod.
The great difference in radius gives you a much greater torque
(turning force effect) to bore into the cork stopper of a wine bottle.
The screw driver
The argument for a screwdriver is the same as for the corkscrew
above.
The greater the diameter of the screwdriver 'handle' compared to the
diameter of the screw head, the greater the force
(torque) you
can apply to drive a screw into wood.
A wheelbarrow
The handles of the wheelbarrow are much further away from the wheel axis than the centre of gravity of the full wheelbarrow
(shown by the
yellow blob!). The wheel axis is the pivot point about which you calculate
the two moments involved.
F1 is the weight of the loaded wheelbarrow acting from
its centre of mass (centre of gravity).
F2 is the force you exert to lift the loaded
wheelbarrow.
The two moments are as follows:
The 'weight' moment F1 x d1, is a small moment to
manage the weight of the wheelbarrow.
(F1 acts down from the centre of mass/gravity)
However the 'lift' moment is F2 x d2 and so a
smaller force F2 is needed operating at the longer perpendicular distance
d2 to lift up the wheelbarrow and its load.
F1 x d1 = F2 x d2, so F2 = F1 x d1/d2
So F2 is << F1
The magnitude of the lifting force F2 is much
less than the weight of the load, so you can lift the barrow and move it
along. Another example of needing less force to get the same moment to do
the job of lifting the wheel barrow to move it along.
Note that since
Work = force x
distance, you can think of this lift as doing work against the
Earth's gravitational field.
In the case of moments involving lifting
objects vertically, each moment = gain in GPE = mgh,
GPE = gravitational potential energy (J), m = mass
(kg), g = 9.8 N/kg, h = height lifted in m, and for more on GPE
see
Mass and
the effect of gravity force on it  weight, (mention of work done and
GPE )
The crucial experimental result is six year old granddaughter Niamh can barely lift
the wheelbarrow off the ground (just a few cm) but Granny Molly has no
trouble lifting the barrow to move it along!
At an
earlier age Niamh wasn't quite as interested in science!
TOP OF PAGE and
subindex
(f) Gears and cog
wheels  a means of transmitting a rotational force effect
Some simple calculations are included at the
end of section (f)
Cog wheels
are circular discs with teeth and components of many machines in transport and
industry.
They are a means of transmitting the rotational effect of
a force from one part to another part of a mechanical device e.g. industrial
machine, motor vehicle or a bicycle.
When several of them are
combined together (linked in contact via the teeth) a rotational force can be transmitted when fitted in contact
with each other. The cog/gear wheels when in direct contact will rotate in
opposite directions.
Through the interlocking, one cog wheel can turn another in the
opposite direction i.e. if one gear wheel goes clockwise, the gear
wheel in contact with it will go anticlockwise, no matter how many gear
wheels are connected together.
By using different size cog wheels differing in the number
of teeth, you can increase or decrease the force generated by the turning effect
of the gear wheels.
From the gear ratio you can work the
revolutions of one gear wheel with respect to the neighbour wheel e.g.
in the diagram gear wheel 1 has 12
teeth and gear wheel 2 has 18 teeth.
If the smaller wheel 1
turns once, wheel 2 turns 2/3 third of a revolution, teeth ratio
12/18 = 2/3 (0.66).
It the larger wheel 2 turns once,
wheel 1 turns 1.5 revolutions, teeth ratio 18/12 = 3/2 = 1.5.
The teeth ratio gives you the
gear ratio.
A force applied to a smaller gear wheel
creates a small moment
 shorter distance from the teeth to the axle pivot
point.
A force applied to a larger gear wheel
creates a larger moment
 longer distance from the teeth to the axle pivot
point.
As moment = force x distance,
the
ratio of the two moments of the ears is equal to the ratio of the gear radii
which equals the ratio of the teeth.
If you transfer the force from a larger cog wheel (gear wheel
of more teeth) to a smaller cog wheel (gear wheel of less teeth) you decrease
the moment of the 2nd as you have decreased the distance from the
applied force to axle pivot point.
No mechanical advantage is gained
 you haven't increased the output force of the smaller cogwheel.
The smaller cog wheel will be made to
turn faster than the larger cog wheel.
This is a way to increase rotational
speeds in machines.
If the first cog wheel has 20 teeth and
the second cog wheel 5 teeth, one rotation of the first wheel causes the
smaller wheel to rotate 20/5 times = 4 x more  a gear ratio of 1 : 4.
If it was the other way round and you
rotated the 2nd smaller cog wheel first, one turn of it would only
rotate the larger cog wheel by 1/4 (5/20) of a turn  a gear ratio of 4
: 1, decreasing the speed of rotation.
If you transfer the force from a smaller cog wheel (gear
wheel of less teeth) to a larger cog wheel (gear wheel of more teeth) you
increase the moment as you have increased the distance from the applied
force to the axle pivot point.
A mechanical advantage is gained

you have increased the output force of the larger cogwheel.
By using a set of interlocking gears that
get bigger and bigger you can multiply the moment of the first small gear.
The larger cog wheel will turn more
slowly than the smaller one.
This is the way a relatively low powered
machine can be made to lift heavy weights.
If the first cog wheel has 8 teeth and
the second wheel 56 teeth, you need to turn the first wheel 56/8 = 7 times
to completely rotate the second wheel.
Examples of gear wheel (cogwheel)
applications
An old fashioned manual drill
The large cog wheel turns a smaller cog wheel at much greater
speed with good old fashioned muscle power!
The force is transmitted from one cog wheel to another.
Since a larger cog wheel of more teeth drives a smaller cog
wheel of less teeth, the output is a high turning speed of the drill.
Gearing in mill wheel systems
Complex machines such as you find in older flour mills and
textile mills, use
gears to utilise the power of the e.g. water wheel, to transmit a force to drive the
machinery with the required speed and power.
A slow rotating water wheel driving a cog wheel system ( a
gear system)
can produce high speeds of rotation to drive a spinning machine  an important
mechanical feature in the Industrial Revolution from the 18th to the 19th
century.
Clocks
Clocks use gear wheels to transmit
the potential energy of the spring and move the arms around at the
correct speed to indicate the correct time.
Different sized gear wheels
are need to operate the minute and hour hands.
The minute hand must go
round 60 x faster than the hour hand, so the gear ratios will take this
into account.
Gears on bicycles
Gears (cogwheels) are used in
bicycles to transfer the force from pedalling the front gears to the
gears on the rear wheels.
The gears are not in contact with
each other, but the cogs are connected by a continuous chain mechanism.
The force of your foot applied to the
pedal rotates the first gear (front cog) and, via the chain, the rear
gear (rear cog) is rotated in the same clockwise direction.
If the cogs are the same size (same
number of teeth), they both rotate at the same speed.
Bicycles often have complex sets of gears
for efficiently transferring the force generated by pedal action to driving
the rear wheel.
The following
two 'simplified' diagrams and notes explain the 'physics' principles behind
gear changes on a bicycle.
'Speeding up'  especially
downhill !!!
Apart from greater physical
exertion, in order to speed up when cycling, you change to a higher
gear.
You do this by switching to a
smaller rear gear wheel on the back wheel, the smaller gear wheel
rotates faster, but with a smaller force e.g.
The front gear has 12 teeth (or
cogs) and the rear cogwheel 8 in the diagram above.
The ratio is 12/8 = 1.5, so every
time you pedal the front cog round once, the rear cog and wheel are
turned 1.5 times  this assumes the gear teeth of both
cogwheels are the same size so the radius ratio of the cogwheels is
3 : 2.
The gear ratio change can be as high as a
53 to 11 ratio to go fast, so one pedal cycle rotates the rear wheel nearly
five times, but this gear ratio can be hard work!
With a triple chainring on the
front gear and 10 gear set on the rear wheel, you have a choice of
30 gear ratios for fastest speeds on the flat or downhill or
climbing the steepest hills.
'Tackling a steep hill'
Apart from greater physical
exertion, in order to climb up a steep hill when cycling, you change
to a lower gear.
You do this by switching to a
larger rear gear wheel on the back wheel, the larger gear wheel
rotates more slowly, but with a larger force e.g.
The front gear has 8 teeth (or
cogs) and the rear cogwheel 12 (ratio 2/3) in the diagram above, so
one turn of the front gear by the pedals only produces 2/3rds of
turn of the rear cog and rear wheel.
If the ratio the front cog : rear
cog was 1 : 3, the mechanical advantage is 3  this assumes the gear
teeth of both cogwheels are the same size so the radius ratio of the
cogwheels is 1 : 3.
This means the input force
produces an output force 3 x greater, but you do a lot of
pedalling to generate a continuous output of a force sufficient
to get up a steep hill.
Gear wheel questions
Q1 A 25 toothed gear wheel is in
contact with a 2nd gear wheel with 5 teeth.
When the first wheel is turned
twice clockwise, in what direction does the 2nd wheel turn and by
how many times?
The 2nd gear wheel must turn
anticlockwise  it must turn in the opposite
direction.
The 2nd gear wheel will turn 25/5
x 2 = 10
revolutions
Argument: The 25 teeth of the
first gear wheel will move 25 teeth of the smaller gear wheel. Since
the 2nd smaller wheel only has 5 teeth, the 25 teeth of the larger
gear wheel will move it 5 times per revolution and this is doubled
for two revolutions of the 1st larger wheel. The gear ratio is 5 :
1.
(g)
Some physics of the human skeleton and muscle system
1. The knee joint
Cartilage of joints
in your skeletalmuscle system, has many functions, including
the ability to resist compressive forces, enhance bone
resilience, and provide support on bony areas where there is a
need for flexibility e.g.
The cartilage is a softer material between
the bones of a joint and acts as a shock absorber to
minimise the chance of bone fracture damage.
Cartilage being a softer material between
the bones of a joint, reduces friction that would cause pain
and wear of the bones.
Cartilage connective tissue also acts as a
binding agent to hold the bones in place, but not too
rigidly, the bonecartilage system must be flexible.
2. The arm joint
The comments on cartilage in 1. apply here too.
The arm joint is a lever system operated by the
expansion and contraction of muscles e.g.
to bend and raise your forearm the biceps
contract and the triceps relax, and,
to lower or straighten your forearm the
biceps relax and the triceps contract
3. The pelvic hip joint
The comments on cartilage in 1. apply here too.
WHAT NEXT?
Forces revision notes index
FORCES 1. What are contact forces &
noncontact forces?, scalar & vector quantities, free body force diagrams
FORCES
2. Mass and the effect of gravity force on it  weight, (mention of work done and
GPE)
FORCES 3. Calculating resultant forces using vector
diagrams and work done
FORCES
4.
Elasticity and energy stored in a spring
FORCES 5. Turning forces and moments  from spanners
to wheelbarrows and equilibrium situations
FORCES 6. Pressure in liquid fluids and hydraulic
systems
FORCES 7. Pressure & upthrust in liquids, why objects float/sink in a fluid?, variation of atmospheric pressure
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