FORCES 5. Turning forces and moment calculations

 from spanners to wheelbarrows & equilibrium situations

 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?

Sub-index 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)

(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.

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(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

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(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 (left-hand side of pivot), clockwise moment = w2 x d2 (right-hand 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.

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(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

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(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 pull-down 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!

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(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.

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(g) Some physics of the human skeleton and muscle system

1. The knee joint

Cartilage of joints in your skeletal-muscle 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 bone-cartilage 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.

TOP OF PAGE and sub-index

WHAT NEXT? Forces revision notes index

FORCES 1. What are contact forces & non-contact 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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