9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

Definition

Friction

Friction is a force that opposes the relative motion between two surfaces in contact.

Definition

Dissipation

Dissipation is the spreading of transferred energy into the thermal energy stores of the surroundings, which makes the energy less useful for further transfers.

No surface is perfectly smooth, because even polished metal carries microscopic bumps and hollows.
When two surfaces slide over each other these bumps catch, and the surfaces have to be forced past one another, which is where friction comes from.
The moving object therefore has to do work against friction, and doing work is a transfer of energy.
That energy is not destroyed, because it is dissipated into the thermal energy stores of the two surfaces and then spreads out into the surroundings.
The transfer counts as unwanted because the energy is spread so thinly that it cannot be recovered to do a useful job, so less of the input energy reaches the intended output.
Friction also wears the surfaces away and raises their temperature, which shortens the working life of bearings, gears, chains and pistons.
In a bicycle chain, a car engine or a stiff door hinge, every joule dissipated by friction is a joule the machine never delivers.

How lubrication reduces the transfer

Definition

Lubrication

Lubrication is the use of a substance, such as oil or grease, between moving surfaces to reduce friction and the unwanted transfer of energy by heating.

A lubricant is a fluid, usually an oil or a grease, placed between two surfaces that move over each other.
The lubricant is drawn into the gap and forms a thin layer that holds the two solid surfaces apart.
With the bumps no longer catching directly on each other, the frictional force between the surfaces becomes much smaller.
Because the frictional force is smaller, less work is done against friction for the same amount of movement.
Less work done against friction means less energy is dissipated into the thermal energy stores of the parts and the surroundings.
A larger share of the input energy is then transferred usefully, so the machine becomes more efficient, runs cooler and wears more slowly.
The chain of reasoning to write out in an answer runs as follows.
1. The lubricant separates the moving surfaces.
2. The frictional force between them is reduced.
3. Less work is done against friction.
4. Less energy is dissipated by heating.
Lubrication never removes friction completely, because layers of the fluid still resist sliding over each other, although that resistance is far smaller than dry contact between two solids.
A lubricant also carries heat away from the contact and flushes out particles worn from the surfaces, which is why engine oil is pumped around a circuit and then changed.

Comparing rates of energy transfer

The energy wasted by a machine is often quoted as a rate, which is the energy transferred divided by the time taken.
The equation is $taken\text{rate of energy transfer} = \dfrac{\text{energy transferred}}{\text{time taken}}$ .
A rate measured in joules per second, $J/s\text{J/s}$ , is the same quantity as a power measured in watts, $W\text{W}$ .
Rates given per minute or per hour have to be converted before they can be compared, by dividing by $60$ to change minutes into seconds and by $3600$ to change hours into seconds.
Comparing the wasted rate before and after lubrication as a ratio shows how much the change has achieved, and the ratio is simplified by dividing both numbers by their highest common factor.
A ratio of $4 : 1$ means the wasted rate has fallen to a quarter of its original value, so lubrication has cut the waste by a factor of $4$ .
Proportional reasoning works forwards as well, so halving the frictional force over the same distance halves the work done against friction and therefore halves the rate at which energy is dissipated.

Practical

Investigating levers and gears

Aim: to investigate how the distance of an applied force from a pivot changes the force needed to lift a fixed load, and how a pair of meshing gears transmits rotation.
Apparatus: metre rule or wooden lever bar, triangular knife-edge pivot, clamp stand and boss, G-clamp, newtonmeters reading to $N10\ \text{N}$ and to $N50\ \text{N}$ , slotted masses and hanger, string loops, a mounting board carrying gear wheels with different numbers of teeth, marker pen, ruler and a soft mat.
Variables: the distance of the applied force from the pivot is the independent variable, the force needed to just lift the load is the dependent variable, and the size of the load, the position of the load, the position of the pivot, the lever itself and the direction of the pull are all controlled.
Method, setting up the lever:
- Clamp the pivot to the bench and balance the metre rule on it so that it rests level with nothing hanging from it.
- Hang the load from a fixed point on one side, for example a $N5.0\ \text{N}$ weight at $m0.10\ \text{m}$ from the pivot, and record that distance in metres.
- Loop the string of the newtonmeter over the rule on the other side, at a measured distance from the pivot.
Method, taking the lever readings:
- Pull the newtonmeter steadily downwards, keeping it at right angles to the rule so that the distance measured is the normal distance.
- Read the force at the instant the load just lifts clear of the bench, and record it in newtons.
- Repeat the reading three times at that distance and calculate a mean, investigating any anomalous reading rather than deleting it.
- Move the newtonmeter $m0.10\ \text{m}$ further from the pivot and repeat, covering at least six distances out to the end of the rule.
Method, the gears:
- Mount two gears of different sizes so that their teeth mesh without binding, and count the teeth on each one.
- Mark one tooth on each gear with the pen so that complete turns can be counted.
- Turn the smaller gear slowly through a whole number of turns and count the turns made by the larger gear, noting the direction each one turns.
- Repeat each count three times, then place a third gear between the pair and record how the direction of the final gear changes.
Results: the force needed to lift the same load falls as the applied force is moved further from the pivot, and the product of force and distance stays roughly constant because the moment needed to lift the load is fixed.
Results for the gears: two directly meshing gears turn in opposite directions, and the number of turns each gear makes is in inverse proportion to its number of teeth.
Maths:
- calculate the moment of the applied force at each distance using $F\,d$ and compare it with the moment of the load.
- plot applied force against distance from the pivot to give a falling curve, then plot applied force against $1d\dfrac{1}{d}$ , which should give a straight line through the origin.
- Maths for the gears: check that $gear\dfrac{\text{turns of driving gear}}{\text{turns of driven gear}}$ equals $gear\dfrac{\text{teeth on driven gear}}{\text{teeth on driving gear}}$ .
Watch out:
- friction at the pivot and the weight of the rule itself make the measured force slightly larger than the calculated value.
- a newtonmeter that is not perpendicular to the rule means the measured distance is not the normal distance, so the moment is overestimated.
- judging the exact instant the load lifts is a real source of uncertainty, which is why repeats and a mean matter here.
- worn or badly meshed gear teeth can slip, so count whole turns rather than fractions of a turn.
Safety: clamp the stand and the pivot firmly so that the loaded rule cannot topple off the bench.
Safety: keep feet clear of the hanging masses and place a soft mat beneath them.
Safety: turn the gears by the rim rather than by the teeth, so that fingers stay clear of the meshing point.

Example

Comparing the rate of wasted energy

A gearbox transfers $J4800\ \text{J}$ to its surroundings in $min2.0\ \text{min}$ before it is lubricated, and $J1200\ \text{J}$ in the same time afterwards.
Convert the time first, so $s2.0\ \text{min} = 2.0 \times 60 = 120\ \text{s}$ .
The rate before lubrication is $J/s\dfrac{4800}{120} = 40\ \text{J/s}$ .
The rate after lubrication is $J/s\dfrac{1200}{120} = 10\ \text{J/s}$ .
The two rates are in the ratio $40 : 10$ , which simplifies to $4 : 1$ .
Lubrication has cut the rate of unwanted energy transfer to a quarter of its original value.

Example

Converting a wasted energy rate into watts

A stiff hinge dissipates $kJ54\ \text{kJ}$ to the surroundings over $30$ minutes of use.
Convert the energy, so $J54\ \text{kJ} = 54\,000\ \text{J}$ .
Convert the time, so $s30\ \text{min} = 30 \times 60 = 1800\ \text{s}$ .
The rate is $J/s\dfrac{54\,000}{1800} = 30\ \text{J/s}$ .
A rate of $J/s30\ \text{J/s}$ is the same as a power of $W30\ \text{W}$ dissipated by the hinge.

Common Mistake

Do not write that friction uses up or destroys energy, because energy is conserved and friction transfers it into thermal energy stores.
Do not say that oil makes the surfaces smoother, because a lubricant works by holding the surfaces apart rather than by polishing them.
Do not claim that lubrication removes friction, since the layers of lubricant still resist sliding over one another.
Do not compare two rates before converting them to the same unit of time, because a rate per minute and a rate per second are not comparable numbers.

Exam technique

Writing the lubrication explanation

Give the full chain from lubricant to energy, because each link in it tends to carry its own mark.
Use the word friction explicitly, since answers that mention only resistance or smoothness are usually not credited.
Finish on the energy statement, saying that less energy is dissipated into the thermal energy stores of the surroundings.
In a rate comparison, show the time conversion on its own line before dividing, so the method mark is safe even if the arithmetic slips.
Simplify a ratio to its lowest terms and then say in words what it means, such as a quarter of the original rate.

Self review

Explain why two dry sliding surfaces become hotter as they move over each other.
State the four steps that link adding a lubricant to a fall in wasted energy.
Explain why lubrication reduces friction without removing it completely.
Convert an energy transfer of $J3600\ \text{J}$ in $3.0$ minutes into a rate in $J/s\text{J/s}$ .
Describe how the force needed to lift a load on a lever changes as the force is applied further from the pivot.

9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

Definition

Friction

Friction is a force that opposes the relative motion between two surfaces in contact.

Definition

Dissipation

Dissipation is the spreading of transferred energy into the thermal energy stores of the surroundings, which makes the energy less useful for further transfers.

No surface is perfectly smooth, because even polished metal carries microscopic bumps and hollows.
When two surfaces slide over each other these bumps catch, and the surfaces have to be forced past one another, which is where friction comes from.
The moving object therefore has to do work against friction, and doing work is a transfer of energy.
That energy is not destroyed, because it is dissipated into the thermal energy stores of the two surfaces and then spreads out into the surroundings.
The transfer counts as unwanted because the energy is spread so thinly that it cannot be recovered to do a useful job, so less of the input energy reaches the intended output.
Friction also wears the surfaces away and raises their temperature, which shortens the working life of bearings, gears, chains and pistons.
In a bicycle chain, a car engine or a stiff door hinge, every joule dissipated by friction is a joule the machine never delivers.

How lubrication reduces the transfer

Definition

Lubrication

Lubrication is the use of a substance, such as oil or grease, between moving surfaces to reduce friction and the unwanted transfer of energy by heating.

A lubricant is a fluid, usually an oil or a grease, placed between two surfaces that move over each other.
The lubricant is drawn into the gap and forms a thin layer that holds the two solid surfaces apart.
With the bumps no longer catching directly on each other, the frictional force between the surfaces becomes much smaller.
Because the frictional force is smaller, less work is done against friction for the same amount of movement.
Less work done against friction means less energy is dissipated into the thermal energy stores of the parts and the surroundings.
A larger share of the input energy is then transferred usefully, so the machine becomes more efficient, runs cooler and wears more slowly.
The chain of reasoning to write out in an answer runs as follows.
1. The lubricant separates the moving surfaces.
2. The frictional force between them is reduced.
3. Less work is done against friction.
4. Less energy is dissipated by heating.
Lubrication never removes friction completely, because layers of the fluid still resist sliding over each other, although that resistance is far smaller than dry contact between two solids.
A lubricant also carries heat away from the contact and flushes out particles worn from the surfaces, which is why engine oil is pumped around a circuit and then changed.

Comparing rates of energy transfer

The energy wasted by a machine is often quoted as a rate, which is the energy transferred divided by the time taken.
The equation is $taken\text{rate of energy transfer} = \dfrac{\text{energy transferred}}{\text{time taken}}$ .
A rate measured in joules per second, $J/s\text{J/s}$ , is the same quantity as a power measured in watts, $W\text{W}$ .
Rates given per minute or per hour have to be converted before they can be compared, by dividing by $60$ to change minutes into seconds and by $3600$ to change hours into seconds.
Comparing the wasted rate before and after lubrication as a ratio shows how much the change has achieved, and the ratio is simplified by dividing both numbers by their highest common factor.
A ratio of $4 : 1$ means the wasted rate has fallen to a quarter of its original value, so lubrication has cut the waste by a factor of $4$ .
Proportional reasoning works forwards as well, so halving the frictional force over the same distance halves the work done against friction and therefore halves the rate at which energy is dissipated.

Practical

Investigating levers and gears

Aim: to investigate how the distance of an applied force from a pivot changes the force needed to lift a fixed load, and how a pair of meshing gears transmits rotation.
Apparatus: metre rule or wooden lever bar, triangular knife-edge pivot, clamp stand and boss, G-clamp, newtonmeters reading to $N10\ \text{N}$ and to $N50\ \text{N}$ , slotted masses and hanger, string loops, a mounting board carrying gear wheels with different numbers of teeth, marker pen, ruler and a soft mat.
Variables: the distance of the applied force from the pivot is the independent variable, the force needed to just lift the load is the dependent variable, and the size of the load, the position of the load, the position of the pivot, the lever itself and the direction of the pull are all controlled.
Method, setting up the lever:
- Clamp the pivot to the bench and balance the metre rule on it so that it rests level with nothing hanging from it.
- Hang the load from a fixed point on one side, for example a $N5.0\ \text{N}$ weight at $m0.10\ \text{m}$ from the pivot, and record that distance in metres.
- Loop the string of the newtonmeter over the rule on the other side, at a measured distance from the pivot.
Method, taking the lever readings:
- Pull the newtonmeter steadily downwards, keeping it at right angles to the rule so that the distance measured is the normal distance.
- Read the force at the instant the load just lifts clear of the bench, and record it in newtons.
- Repeat the reading three times at that distance and calculate a mean, investigating any anomalous reading rather than deleting it.
- Move the newtonmeter $m0.10\ \text{m}$ further from the pivot and repeat, covering at least six distances out to the end of the rule.
Method, the gears:
- Mount two gears of different sizes so that their teeth mesh without binding, and count the teeth on each one.
- Mark one tooth on each gear with the pen so that complete turns can be counted.
- Turn the smaller gear slowly through a whole number of turns and count the turns made by the larger gear, noting the direction each one turns.
- Repeat each count three times, then place a third gear between the pair and record how the direction of the final gear changes.
Results: the force needed to lift the same load falls as the applied force is moved further from the pivot, and the product of force and distance stays roughly constant because the moment needed to lift the load is fixed.
Results for the gears: two directly meshing gears turn in opposite directions, and the number of turns each gear makes is in inverse proportion to its number of teeth.
Maths:
- calculate the moment of the applied force at each distance using $F\,d$ and compare it with the moment of the load.
- plot applied force against distance from the pivot to give a falling curve, then plot applied force against $1d\dfrac{1}{d}$ , which should give a straight line through the origin.
- Maths for the gears: check that $gear\dfrac{\text{turns of driving gear}}{\text{turns of driven gear}}$ equals $gear\dfrac{\text{teeth on driven gear}}{\text{teeth on driving gear}}$ .
Watch out:
- friction at the pivot and the weight of the rule itself make the measured force slightly larger than the calculated value.
- a newtonmeter that is not perpendicular to the rule means the measured distance is not the normal distance, so the moment is overestimated.
- judging the exact instant the load lifts is a real source of uncertainty, which is why repeats and a mean matter here.
- worn or badly meshed gear teeth can slip, so count whole turns rather than fractions of a turn.
Safety: clamp the stand and the pivot firmly so that the loaded rule cannot topple off the bench.
Safety: keep feet clear of the hanging masses and place a soft mat beneath them.
Safety: turn the gears by the rim rather than by the teeth, so that fingers stay clear of the meshing point.

Example

Comparing the rate of wasted energy

A gearbox transfers $J4800\ \text{J}$ to its surroundings in $min2.0\ \text{min}$ before it is lubricated, and $J1200\ \text{J}$ in the same time afterwards.
Convert the time first, so $s2.0\ \text{min} = 2.0 \times 60 = 120\ \text{s}$ .
The rate before lubrication is $J/s\dfrac{4800}{120} = 40\ \text{J/s}$ .
The rate after lubrication is $J/s\dfrac{1200}{120} = 10\ \text{J/s}$ .
The two rates are in the ratio $40 : 10$ , which simplifies to $4 : 1$ .
Lubrication has cut the rate of unwanted energy transfer to a quarter of its original value.

Example

Converting a wasted energy rate into watts

A stiff hinge dissipates $kJ54\ \text{kJ}$ to the surroundings over $30$ minutes of use.
Convert the energy, so $J54\ \text{kJ} = 54\,000\ \text{J}$ .
Convert the time, so $s30\ \text{min} = 30 \times 60 = 1800\ \text{s}$ .
The rate is $J/s\dfrac{54\,000}{1800} = 30\ \text{J/s}$ .
A rate of $J/s30\ \text{J/s}$ is the same as a power of $W30\ \text{W}$ dissipated by the hinge.

Common Mistake

Do not write that friction uses up or destroys energy, because energy is conserved and friction transfers it into thermal energy stores.
Do not say that oil makes the surfaces smoother, because a lubricant works by holding the surfaces apart rather than by polishing them.
Do not claim that lubrication removes friction, since the layers of lubricant still resist sliding over one another.
Do not compare two rates before converting them to the same unit of time, because a rate per minute and a rate per second are not comparable numbers.

Exam technique

Writing the lubrication explanation

Give the full chain from lubricant to energy, because each link in it tends to carry its own mark.
Use the word friction explicitly, since answers that mention only resistance or smoothness are usually not credited.
Finish on the energy statement, saying that less energy is dissipated into the thermal energy stores of the surroundings.
In a rate comparison, show the time conversion on its own line before dividing, so the method mark is safe even if the arithmetic slips.
Simplify a ratio to its lowest terms and then say in words what it means, such as a quarter of the original rate.

Self review

Explain why two dry sliding surfaces become hotter as they move over each other.
State the four steps that link adding a lubricant to a fall in wasted energy.
Explain why lubrication reduces friction without removing it completely.
Convert an energy transfer of $J3600\ \text{J}$ in $3.0$ minutes into a rate in $J/s\text{J/s}$ .
Describe how the force needed to lift a load on a lever changes as the force is applied further from the pivot.

9.3 Reducing energy transfer

9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

How lubrication reduces the transfer

Comparing rates of energy transfer

Investigating levers and gears

Comparing the rate of wasted energy

Converting a wasted energy rate into watts

Writing the lubrication explanation

Review 9.3 Reducing energy transfer by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

9.3 Reducing energy transfer

9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

How lubrication reduces the transfer

Comparing rates of energy transfer

Investigating levers and gears

Comparing the rate of wasted energy

Converting a wasted energy rate into watts

Writing the lubrication explanation

Review 9.3 Reducing energy transfer by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

9.3 Reducing energy transfer

9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

How lubrication reduces the transfer

Comparing rates of energy transfer

Investigating levers and gears

Comparing the rate of wasted energy

Converting a wasted energy rate into watts

Writing the lubrication explanation

Review 9.3 Reducing energy transfer by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

9.3 Reducing energy transfer Revision Guide

9.3 Reducing energy transfer

9.3.1 Reducing unwanted energy transfer through lubrication

Friction and the energy it wastes

How lubrication reduces the transfer

Comparing rates of energy transfer

Investigating levers and gears

Comparing the rate of wasted energy

Converting a wasted energy rate into watts

Writing the lubrication explanation

Review 9.3 Reducing energy transfer by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

9.3 Reducing energy transfer Revision Guide