Machines and Mechanical Advantage

A machine is a device that makes a task easier by changing the magnitude, direction, or distance over which a force acts.

Machines do not create energy. Instead, they allow us to transfer energy in a more useful way.

For example, a machine may allow you to:

• lift a heavy object using a smaller force
• change the direction of a force
• increase the distance through which a force acts
• increase the speed or distance of an output movement
• apply a force more conveniently

Common simple machines include:

• levers
• pulleys
• inclined planes
• wheel and axle systems
• gears
• screws
• wedges

 

 

 

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Input Force and Output Force

When using a machine, we can identify two important forces.

Input Force

The input force is the force applied to the machine.

It is sometimes called the effort force.

For example:

• pulling on a rope
• pushing on a lever
• turning a handle
• pedaling a bicycle

Output Force

The output force is the force the machine applies to the load.

It is sometimes called the load force.

For example:

• the force lifting a box
• the force moving a rock
• the force raising a vehicle
• the force pulling a load

A machine can allow a relatively small input force to produce a larger output force.

Mechanical Advantage

Mechanical advantage (MA) compares the output force produced by a machine with the input force applied to the machine.

MA = output force / input force

or:

MA = Fout / Fin

Mechanical advantage has no unit because it is a ratio of two forces measured in the same units.

Interpreting Mechanical Advantage

Suppose:

MA = 4

This means:

output force = 4 × input force

If you apply:

100 N

the machine can ideally produce:

400 N

of output force.

The machine has multiplied the force by a factor of four.

Worked Example: Mechanical Advantage

A person applies an input force of:

150 N

to a machine.

The machine produces an output force of:

600 N

Calculate the mechanical advantage.

MA = Fout / Fin

MA = 600/150

MA = 4

The machine multiplies the input force by a factor of 4.

Finding Output Force

The equation can be rearranged:

Fout = MA × Fin

Suppose a machine has:

MA = 5

and the input force is:

80 N

Then:

Fout = 5 × 80

Fout = 400 N

Finding Input Force

We can also rearrange:

Fin = Fout / MA

Suppose a machine must produce:

1200 N

and has:

MA = 6

Then:

Fin = 1200/6

Fin = 200 N

Only 200 N of input force is required.

Machines Do Not Reduce the Work in an Ideal System

A machine can reduce the force needed, but there is a trade-off.

Recall:

W = Fd

where:

• W = work done
• F = force
• d = displacement in the direction of the force

In an ideal machine:

input work = output work

Therefore:

Fin × din = Fout × dout

If a machine increases the force, the input force must usually act through a greater distance.

This is one of the most important ideas in simple machines:

smaller force → greater distance

Force-Distance Trade-Off

Suppose we need to perform:

1000 J

of work.

Without a Machine

Force:

500 N

Distance:

2 m

Work:

W = Fd

W = 500 × 2

W = 1000 J

With a Machine

Suppose the machine allows us to use only:

100 N

To perform the same 1000 J of work:

1000 = 100 × d

Therefore:

d = 10 m

The force has been reduced from 500 N to 100 N, but it must act through a much greater distance.

Machines Trade Force for Distance

This gives us a central rule for ideal machines:

What you gain in force, you give up in distance.

A machine does not provide free energy.

If it multiplies force by five, the input typically moves approximately five times farther than the output.

For example:

Input force = 100 N

Input distance = 5 m

Output force = 500 N

Output distance = 1 m

Input work:

Win = 100 × 5 = 500 J

Output work:

Wout = 500 × 1 = 500 J

Energy is conserved.

Ideal Mechanical Advantage

For an ideal machine:

Fin × din = Fout × dout

Rearrange:

Fout / Fin = din / dout

Since:

MA = Fout / Fin

we obtain:

MA = din / dout

Therefore, ideal mechanical advantage can also be understood as a distance ratio.

A machine that multiplies force does so because the input moves farther than the output.

Worked Example: Force and Distance

A machine has an ideal mechanical advantage of:

MA = 8

The output moves:

0.5 m

The input distance is:

din = MA × dout

din = 8 × 0.5

din = 4 m

The input must move 4 m while the load moves only 0.5 m.

Levers

A lever is a rigid bar that rotates around a fixed point called a fulcrum.

Important parts include:

• effort force
• load force
• fulcrum
• effort arm
• load arm

A lever can:

• multiply force
• change the direction of force
• increase movement or speed

 

 

 

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Mechanical Advantage of a Lever

For an ideal lever:

MA = effort arm / load arm

The effort arm is the distance from the fulcrum to the input force.

The load arm is the distance from the fulcrum to the load.

A longer effort arm can provide a greater mechanical advantage.

Worked Example: Lever

A lever has:

Effort arm = 1.5 m

Load arm = 0.30 m

Mechanical advantage:

MA = 1.5/0.30

MA = 5

If the input force is:

120 N

then:

Fout = 5 × 120

Fout = 600 N

Why a Long Lever Helps

Imagine trying to loosen a tight bolt.

Using a longer wrench makes the task easier because the applied force acts farther from the pivot.

Similarly, a long crowbar can allow a relatively small input force to produce a much larger force on a load.

This is why long handles are common on:

• wrenches
• crowbars
• bolt cutters
• pliers
• bottle openers

Classes of Levers

Levers can be classified according to the positions of the:

• fulcrum
• effort
• load

First-Class Lever

The fulcrum lies between the effort and load.

Examples:

• seesaw
• crowbar
• scissors

Second-Class Lever

The load lies between the fulcrum and effort.

Examples:

• wheelbarrow
• bottle opener
• nutcracker

Third-Class Lever

The effort lies between the fulcrum and load.

Examples:

• tweezers
• fishing rod
• human forearm

Third-class levers often have mechanical advantage less than 1, but they can increase the speed and distance of movement.

Mechanical Advantage Less Than 1

A machine does not always multiply force.

Suppose:

Fin = 200 N

and:

Fout = 50 N

Then:

MA = 50/200

MA = 0.25

The machine reduces the force.

Why would this be useful?

Because it may increase:

• movement distance
• speed
• precision

Human limbs often work this way.

Pulleys

A pulley consists of a wheel with a rope or cable passing around it.

Pulleys can:

• change the direction of a force
• provide mechanical advantage
• make lifting loads easier

 

 

 

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Fixed Pulley

A single fixed pulley mainly changes the direction of the force.

For an ideal fixed pulley:

MA = 1

For example, instead of lifting a load upward directly, you can pull downward on the rope.

The force is not multiplied, but the task may be more convenient.

Movable Pulleys

Movable pulleys can provide mechanical advantage.

In an ideal pulley system, mechanical advantage can often be determined by counting the number of rope sections supporting the moving load.

For example:

2 supporting sections:

MA = 2

4 supporting sections:

MA = 4

6 supporting sections:

MA = 6

The exact arrangement of the pulley system must always be considered.

Worked Example: Pulley

A pulley system has an ideal mechanical advantage of:

4

A load weighs:

800 N

Required ideal input force:

Fin = Fout / MA

Fin = 800/4

Fin = 200 N

However, the person must pull a greater length of rope than the distance the load rises.

Pulley Distance Trade-Off

Suppose the pulley has:

MA = 4

and the load rises:

1 m

Ideally, the input rope must move:

4 m

Check the work:

Input:

Win = 200 × 4

Win = 800 J

Output:

Wout = 800 × 1

Wout = 800 J

Again:

input work = output work

for an ideal machine.

Inclined Planes

An inclined plane is a sloped surface used to raise or lower objects.

Examples include:

• ramps
• loading ramps
• wheelchair ramps
• sloping roads

An inclined plane reduces the force needed to raise an object by increasing the distance over which the force acts.

 

 

 

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Mechanical Advantage of an Inclined Plane

For an ideal inclined plane:

MA = length of slope / vertical height

For example:

Ramp length = 6 m

Height = 1.5 m

Then:

MA = 6/1.5

MA = 4

The ramp ideally reduces the required force by a factor of four.

Worked Example: Ramp

A 1000 N load must be raised vertically by:

1 m

using a:

5 m

ramp.

Ideal mechanical advantage:

MA = 5/1

MA = 5

Ideal input force:

Fin = 1000/5

Fin = 200 N

Check the work:

Input:

Win = 200 × 5

Win = 1000 J

Output:

Wout = 1000 × 1

Wout = 1000 J

The ramp reduces the required force but increases the distance.

Wheel and Axle

A wheel and axle consists of a large wheel connected to a smaller axle.

Examples include:

• steering wheels
• door handles
• screwdrivers
• winches
• water wheels

When the input force acts on the larger wheel, the system can multiply force at the axle.

 

 

 

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Gears

Gears are rotating wheels with teeth that transfer force and motion.

Gears can change:

• force
• rotational speed
• direction of rotation

A small gear driving a larger gear can increase output force but reduce rotational speed.

A large gear driving a smaller gear can increase rotational speed but reduce output force.

Again, there is a trade-off between:

force and movement

Screws

A screw is essentially an inclined plane wrapped around a cylinder.

Turning the screw requires the input force to move through a relatively large distance.

The screw itself advances only a small distance.

This can produce a large mechanical advantage.

Examples include:

• bolts
• screws
• screw jacks
• vises

Wedges

A wedge can be thought of as a moving inclined plane.

Examples include:

• knives
• axes
• chisels
• nails

The input force moves the wedge forward, producing forces that push material apart.

A wedge changes both the direction and magnitude of forces.

Real Machines

So far, many examples have assumed an ideal machine.

An ideal machine has:

input work = output work

Real machines experience:

• friction
• deformation
• vibration
• sound
• heating
• air resistance

Therefore:

useful output energy < input energy

Efficiency of Machines

Machine efficiency can be calculated using:

efficiency = useful output energy / input energy

or:

efficiency (%) = useful output energy / input energy × 100

Since work is an energy transfer:

efficiency (%) = output work / input work × 100

Worked Example: Real Machine

A machine receives:

1000 J

of input work.

It produces:

800 J

of useful output work.

Efficiency:

efficiency = 800/1000 × 100

efficiency = 80%

The remaining:

200 J

has been dissipated into other forms.

Actual and Ideal Mechanical Advantage

In an ideal machine, there is no friction.

In a real machine, friction means that the actual output force may be lower than predicted by the ideal geometry.

For example:

Ideal MA = 5

Input force = 100 N

Ideal output:

500 N

But suppose the real output is:

420 N

Actual mechanical advantage:

MA = 420/100

MA = 4.2

The actual mechanical advantage is lower because some energy is dissipated.

Worked Example: Real Ramp

A ramp is:

5 m long

and raises a load:

1 m

Ideal mechanical advantage:

IMA = 5/1

IMA = 5

The load weighs:

1000 N

Ideally, the required force would be:

1000/5 = 200 N

But suppose the actual force required is:

250 N

Actual mechanical advantage:

AMA = 1000/250

AMA = 4

The actual mechanical advantage is lower than the ideal value because of friction.

Efficiency from Mechanical Advantage

For many simple machines:

efficiency = AMA / IMA

As a percentage:

efficiency (%) = AMA / IMA × 100

Using the previous ramp:

AMA = 4

IMA = 5

Therefore:

efficiency = 4/5 × 100

efficiency = 80%

Work Analysis of the Same Ramp

We can verify this using work.

Input force:

250 N

Input distance:

5 m

Input work:

Win = 250 × 5

Win = 1250 J

Output force:

1000 N

Output distance:

1 m

Output work:

Wout = 1000 × 1

Wout = 1000 J

Efficiency:

efficiency = 1000/1250 × 100

efficiency = 80%

Both methods give the same result.

Why Machines Make Work Easier

It may seem strange to say machines make work easier if they do not reduce the total work in an ideal system.

Machines make tasks easier because humans and motors often have force limitations.

For example, you might not be able to lift:

1000 N

directly.

But you may be able to push:

200 N

over a longer distance.

The total work may remain approximately the same, but the task becomes physically possible.

Direction Can Also Make Work Easier

Machines can also make tasks easier without multiplying force.

A fixed pulley is a good example.

Instead of pulling upward on a load, you can pull downward.

This allows you to:

• use your body weight
• work in a more comfortable position
• apply the force in a convenient direction

So machines can make work easier by changing either the magnitude or direction of force.

Force Multiplication Does Not Mean Energy Multiplication

Suppose a machine has:

MA = 10

Input force:

100 N

Output force:

1000 N

It may appear that the machine has created extra energy.

It has not.

If the output moves:

0.20 m

then ideally the input must move:

2.0 m

Input work:

100 × 2.0 = 200 J

Output work:

1000 × 0.20 = 200 J

The machine multiplies force, not energy.

Machines and Power

Machines can also affect the relationship between force and speed.

Recall:

P = Fv

For an ideal machine:

input power = output power

Therefore:

Fin vin = Fout vout

If the machine produces a larger output force, the output generally moves more slowly.

For example:

Input:

100 N at 4 m/s

Input power:

P = 100 × 4

P = 400 W

Suppose output force is:

400 N

Then:

400 = 400 × vout

vout = 1 m/s

The machine multiplied the force by four but reduced the speed by four.

Mechanical Advantage and Speed

This leads to another important trade-off.

A machine can produce:

high force + low speed

or:

low force + high speed

This is why gears are useful.

For example:

• low gears on a bicycle provide greater force at the wheel but lower speed
• high gears allow greater speed but provide less force multiplication
• vehicle transmissions adjust force and speed according to driving conditions

 

 

 

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Comparing Simple Machines

All obey the same basic physical principles:

energy is conserved

and:

force advantage requires a trade-off

Choosing a Machine

Suppose a worker needs to raise a heavy object.

Possible solutions might include:

• a long lever
• a ramp
• a pulley system
• a hydraulic or mechanical jack

The best machine depends on factors such as:

• available input force
• available distance
• available space
• desired lifting height
• speed
• efficiency
• safety
• friction

There is no single best machine for every situation.

Worked Multi-Step Problem

A worker uses a 6 m ramp to raise a 1200 N load by 1.5 m.

The worker applies:

400 N

along the ramp.

Ideal Mechanical Advantage

IMA = ramp length / height

IMA = 6/1.5

IMA = 4

Actual Mechanical Advantage

AMA = output force / input force

AMA = 1200/400

AMA = 3

Input Work

Win = Fin × din

Win = 400 × 6

Win = 2400 J

Useful Output Work

Wout = Fout × dout

Wout = 1200 × 1.5

Wout = 1800 J

Efficiency

efficiency = 1800/2400 × 100

efficiency = 75%

Or:

efficiency = AMA/IMA × 100

efficiency = 3/4 × 100

efficiency = 75%

Energy dissipated:

2400 − 1800 = 600 J

Energy Flow in a Real Machine

Consider a machine receiving:

2000 J

of mechanical input energy.

Suppose:

1600 J → useful mechanical output

300 J → thermal energy

100 J → sound and vibration

Energy conservation:

2000 = 1600 + 300 + 100

Efficiency:

efficiency = 1600/2000 × 100

efficiency = 80%

The machine does not destroy the other 400 J.

That energy has been transferred into less useful forms.

Improving Machine Performance

Engineers can improve machine efficiency by reducing unwanted energy transfers.

Methods include:

• lubrication
• ball bearings
• smoother surfaces
• better alignment
• reducing unnecessary vibration
• reducing air resistance
• using appropriate materials
• maintaining moving components

Reducing friction is particularly important in machines with many moving parts.

Machines in the Human Body

The human body contains biological lever systems.

Bones act as:

levers

Joints act as:

fulcrums

Muscles provide:

input forces

Loads may include:

• body parts
• objects being carried
• external forces

Many human-body levers have mechanical advantages less than 1.

This may seem inefficient in terms of force, but it provides advantages in:

• speed
• range of movement
• precision

 

 

 

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Common Mistakes

Thinking Machines Create Energy

Machines cannot create energy.

They redistribute force, distance, speed, and direction while obeying conservation of energy.

Thinking Mechanical Advantage Has Units

Mechanical advantage is:

force / force

Therefore, the units cancel.

Mechanical advantage has no unit.

Reversing the Mechanical Advantage Equation

Correct:

MA = output force / input force

Not:

input force / output force

Thinking MA Must Be Greater Than 1

Mechanical advantage can be:

greater than 1

equal to 1

or:

less than 1

A machine with MA < 1 may provide advantages in speed, distance, or precision.

Thinking Smaller Force Means Less Work

In an ideal machine:

input work = output work

A smaller input force usually acts through a greater distance.

Ignoring Friction

Real machines require additional input energy because some energy is dissipated through friction, heating, sound, and vibration.

Confusing Force Advantage with Energy Advantage

A machine can multiply force.

It cannot multiply energy.

Confusing Ideal and Actual Mechanical Advantage

Ideal mechanical advantage is based on machine geometry.

Actual mechanical advantage uses measured input and output forces.

Friction usually causes:

AMA < IMA

Assuming a Machine Must Multiply Force

Some machines mainly:

• change force direction
• increase speed
• increase movement distance

Force multiplication is only one possible function.

Check Your Understanding

1. Define a machine.
2. Explain two ways a machine can make a task easier.
3. Define input force.
4. Define output force.
5. Define mechanical advantage.
6. A machine receives an input force of 100 N and produces an output force of 500 N. Calculate its mechanical advantage.
7. A machine has MA = 6 and receives 80 N. Calculate the output force.
8. A machine must produce 1200 N and has MA = 4. Calculate the required input force.
9. Explain why mechanical advantage has no unit.
10. What does MA = 5 mean physically?
11. Can mechanical advantage be less than 1? Explain why this might be useful.
12. A machine uses 200 N acting through 5 m. Calculate the input work.
13. If the machine is ideal and produces 1000 N, how far does the output move?
14. Explain the relationship between force and distance in an ideal machine.
15. A lever has an effort arm of 2 m and a load arm of 0.5 m. Calculate its ideal mechanical advantage.
16. If 150 N is applied to the lever in Question 15, calculate the ideal output force.
17. A pulley system has four supporting rope sections. What is its ideal mechanical advantage?
18. A 1200 N load is lifted by an ideal pulley system with MA = 4. Calculate the required input force.
19. If the load rises 2 m, how much rope must ideally be pulled?
20. A ramp is 8 m long and 2 m high. Calculate its ideal mechanical advantage.
21. A 1600 N load is pushed up this ideal ramp. Calculate the ideal input force.
22. Explain why a longer, less steep ramp requires less force.
23. A machine receives 1500 J and provides 1200 J of useful output. Calculate its efficiency.
24. How much energy is dissipated in Question 23?
25. A ramp has IMA = 5 and AMA = 4. Calculate its efficiency.
26. Explain why actual mechanical advantage is usually less than ideal mechanical advantage.
27. A machine has an input force of 300 N moving at 4 m/s. Calculate its input power.
28. If its output force is 1200 N and the machine is ideal, calculate the output speed.
29. Explain why increasing output force generally decreases output speed in an ideal machine.
30. Explain why machines make tasks easier even though an ideal machine does not reduce the total work required.

Key Terms

Machine — Device that changes the magnitude, direction, distance, or speed associated with a force.

Simple machine — Basic mechanical device used to change how a force is applied.

Input force — Force applied to a machine.

Effort force — Another name for input force.

Output force — Force produced by a machine on a load.

Load — Object or resistance upon which a machine performs work.

Mechanical advantage (MA) — Ratio of output force to input force.

Ideal mechanical advantage (IMA) — Mechanical advantage predicted from the geometry of an ideal machine.

Actual mechanical advantage (AMA) — Mechanical advantage calculated from actual measured forces.

Input work — Energy transferred to a machine by the input force.

Output work — Useful mechanical energy transferred by a machine to its load.

Fulcrum — Fixed pivot point around which a lever rotates.

Effort arm — Distance from the fulcrum to the input force on a lever.

Load arm — Distance from the fulcrum to the load force.

Lever — Rigid bar that rotates around a fulcrum.

Pulley — Wheel and rope system used to change force direction or provide mechanical advantage.

Inclined plane — Sloping surface that reduces required force by increasing the distance over which the force acts.

Wheel and axle — Simple machine consisting of connected rotating components of different radii.

Gear — Toothed wheel used to transfer rotational motion and change force, speed, or direction.

Screw — Inclined plane wrapped around a cylinder.

Wedge — Moving inclined plane used to apply forces to materials.

Efficiency — Fraction of input energy transferred into useful output energy.

Energy dissipation — Transfer of energy into less useful forms such as thermal energy and sound.

Key Takeaways

• Machines make tasks easier by changing the magnitude, direction, distance, or speed associated with forces.
• The input force is the force applied to the machine.
• The output force is the force the machine applies to the load.
• Mechanical advantage compares output force with input force.
• MA = Fout / Fin.
• Mechanical advantage has no unit.
• An MA greater than 1 means the machine multiplies force.
• An MA of 1 means the output and input forces are equal.
• An MA less than 1 can provide advantages in speed, distance, or precision.
• Machines cannot create energy.
• In an ideal machine, input work = output work.
• Since W = Fd, reducing force requires increasing the distance over which the force acts.
• A machine therefore trades force for distance.
• Ideal mechanical advantage can often be determined from the geometry of the machine.
• Levers can multiply force depending on the relative lengths of their effort and load arms.
• Pulley systems can multiply force by distributing the load across supporting rope sections.
• Inclined planes reduce the force required to raise an object by increasing the distance traveled.
• Gears and wheel-and-axle systems trade force for rotational speed.
• Real machines are less than 100% efficient because of friction, heating, vibration, sound, and deformation.
• Actual mechanical advantage is generally lower than ideal mechanical advantage.
• Machine efficiency can be determined by comparing useful output work with input work.
• For many simple machines, efficiency = AMA/IMA × 100%.
• Force multiplication does not mean energy multiplication.
• Increasing output force generally means decreasing output speed or increasing input distance.
• Energy conservation provides the underlying explanation for how all simple machines operate.

The central trade-off is:

smaller input force → greater input distance

or:

greater output force → smaller output distance

A machine does not allow us to get more energy than we put in. It allows us to use the available energy and force in a way that makes the task more practical, controllable, or convenient.

Useful Friction

Friction is not always an unwanted energy loss.

Without friction, many everyday activities would be impossible.

Useful friction allows:

• shoes to grip the ground
• tyres to grip roads
• brakes to stop vehicles
• pencils to write
• screws and nails to remain in place
• objects to be held
• belts to transfer motion between machine components

The goal in engineering is therefore usually not to eliminate all friction.

Instead, engineers try to:

increase friction where it is useful

and:

reduce friction where it causes unwanted energy transfers.

Walking

Walking depends on friction between your shoes and the ground.

Your foot pushes backward against the ground.

The ground provides a forward frictional force on your foot.

Without sufficient friction, your foot slips.

This is why:

• shoes have textured soles
• sports shoes use specialized materials
• hiking boots have deep tread patterns

Friction is essential for controlled movement.

Tyres and Roads

Vehicles depend on friction between tyres and the road.

This friction allows vehicles to:

• accelerate
• turn
• brake
• climb hills

Tyre tread helps maintain effective contact, especially when water is present.

Too little friction can cause a vehicle to slide.

However, some resistive effects associated with tyre deformation and road contact also contribute to energy losses.

 

 

 

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Braking

Brakes deliberately use friction.

When brake pads press against a disc:

kinetic energy → thermal energy

The vehicle slows because its kinetic energy decreases.

The brakes become hot because energy has been transferred into thermal energy.

In this situation, friction is useful because stopping the vehicle is the intended result.

Worked Example: Braking

A car initially has:

300,000 J

of kinetic energy.

It brakes until it stops.

Ignoring other energy transfers, approximately:

300,000 J

must be transferred into other forms.

Much of this becomes thermal energy in:

• brakes
• tyres
• road
• surrounding air

Energy has not disappeared.

It has changed form and spread into the surroundings.

Regenerative Braking

Some electric and hybrid vehicles can recover part of the vehicle's kinetic energy.

During regenerative braking, the electric motor operates as a generator.

Energy transformation:

kinetic energy → electrical energy → stored chemical energy

Some energy is still dissipated, so the process is not 100% efficient.

However, regenerative braking can recover energy that would otherwise largely become thermal energy in conventional friction brakes.

Friction in Machines

Machines often contain many moving components.

Examples include:

• shafts
• gears
• bearings
• pistons
• chains
• belts
• wheels

Contact between moving parts can produce friction.

This causes:

• unwanted heating
• energy dissipation
• component wear
• reduced efficiency
• increased energy consumption

 

 

 

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Friction and Wear

Friction can gradually remove material from surfaces.

This is called wear.

Wear can:

• change the shape of components
• increase gaps between parts
• increase vibration
• reduce efficiency
• eventually cause component failure

Machines therefore require maintenance.

This may include:

• lubrication
• cleaning
• adjustment
• replacing worn components

Lubrication

One of the most common ways to reduce unwanted friction is lubrication.

Lubricants include:

• oils
• greases
• specialized synthetic fluids

A lubricant forms a layer between surfaces, reducing direct contact.

This can:

• reduce friction
• reduce heating
• reduce wear
• increase efficiency
• extend machine lifetime

However, some friction still remains.

Bearings

Bearings are designed to reduce friction between moving components.

Instead of surfaces sliding directly against each other, bearings may allow rolling motion.

Examples include:

• ball bearings
• roller bearings

Rolling friction is often much smaller than sliding friction.

Bearings are widely used in:

• bicycles
• cars
• electric motors
• turbines
• industrial machinery

Rolling Instead of Sliding

Consider moving a heavy object across the ground.

Dragging it causes substantial sliding friction.

Placing rollers underneath can make movement much easier.

This is one reason wheels are such an important technological development.

Wheels allow much of the interaction to involve rolling rather than sliding.

Air Resistance

Objects moving through air experience air resistance, also called drag.

Air resistance acts opposite the direction of motion.

Examples include:

• cars
• cyclists
• trains
• aircraft
• falling objects
• projectiles

Energy is transferred from the moving object to the surrounding air.

Some of this energy eventually becomes thermal energy and turbulent motion.

 

 

 

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Air Resistance and Speed

Air resistance generally increases as an object's speed increases.

At high speeds, air resistance can become a major energy loss.

This is especially important for:

• highway vehicles
• high-speed trains
• racing bicycles
• aircraft

A faster vehicle must often use significantly more power to overcome drag.

This is one reason aerodynamic design is so important.

Streamlining

Streamlining means designing an object so air or water can flow around it with less resistance.

Examples include:

• aerodynamic cars
• aircraft
• high-speed trains
• racing bicycles
• streamlined helmets

Reducing drag means less energy is needed to maintain a particular speed.

This can improve efficiency.

Worked Example: Work Against Drag

A vehicle experiences an average resistive force of:

600 N

while traveling:

1000 m

Energy transferred against the resistance:

E = Fd

E = 600 × 1000

E = 600,000 J

or:

600 kJ

The engine must supply at least this amount of energy simply to overcome the resistive force over that distance.

Power Needed to Overcome Resistance

Recall:

P = Fv

If a vehicle moves at constant velocity, the engine's driving force balances the resistive forces.

Suppose:

Resistive force = 500 N

Speed = 20 m/s

Power needed just to overcome resistance:

P = Fv

P = 500 × 20

P = 10,000 W

or:

10 kW

This power is continuously being transferred into the surroundings.

Worked Example: Higher Speed

Suppose a vehicle experiences:

800 N

of total resistance while moving at:

30 m/s

Power required:

P = Fv

P = 800 × 30

P = 24,000 W

P = 24 kW

The engine must provide 24 kW merely to balance the resistance and maintain constant speed under these simplified conditions.

Transportation Efficiency

Transportation systems experience several sources of energy dissipation.

These may include:

• air resistance
• rolling resistance
• friction in bearings
• friction in gears
• tyre deformation
• engine losses
• electrical resistance
• braking

Reducing these losses can reduce the energy required to move people and goods.

Reducing Energy Losses in Cars

Engineers can reduce unwanted energy transfers by using:

• aerodynamic body shapes
• properly inflated tyres
• efficient bearings
• low-friction lubricants
• efficient transmissions
• lightweight materials
• regenerative braking
• efficient electric motors

Each improvement may reduce the amount of input energy required for a journey.

Tyre Pressure

Tyres deform as they roll.

This deformation causes energy to be transferred into thermal energy.

If tyre pressure is too low, deformation can increase.

This can increase rolling resistance.

Proper tyre pressure can therefore:

• reduce energy losses
• improve fuel economy or driving range
• reduce unnecessary tyre heating
• improve tyre performance

Bicycles

Bicycles provide a useful example of energy losses.

A cyclist must provide energy to overcome:

• air resistance
• rolling resistance
• bearing friction
• chain and gear friction

At higher speeds, aerodynamic drag can become particularly important.

Engineers and athletes therefore use:

• streamlined riding positions
• aerodynamic helmets
• narrow profiles
• efficient bearings
• lubricated chains
• appropriate tyre pressures

to reduce unwanted losses.

Aircraft

Aircraft must overcome drag while moving through the atmosphere.

The engines must continuously provide energy to balance drag during steady flight.

Aircraft designers reduce drag through:

• streamlined fuselages
• carefully shaped wings
• smooth surfaces
• optimized aircraft geometry

Reducing drag can reduce fuel consumption and increase range.

Thermal Energy in Machines

If a machine becomes warm while operating, this often indicates that some input energy is being transferred into thermal energy.

Examples include:

• warm electric motors
• hot engines
• heated bearings
• warm bicycle brakes
• hot tyres

Not all heating is caused by mechanical friction. Electrical resistance and other processes can also produce heating.

However, unwanted heating often indicates energy that is not reaching the desired output.

Sound and Vibration

Energy can also be dissipated through:

• sound
• vibration

A noisy, vibrating machine may be transferring more energy into unwanted forms.

Engineers may reduce these transfers using:

• better alignment
• balanced rotating components
• damping materials
• improved bearings
• stronger structural support

Reducing vibration can also reduce wear and improve machine lifespan.

Useful and Wasted Transfers

Whether an energy transfer is useful depends on the purpose of the system.

Consider friction.

Car Brakes

Thermal energy produced by friction is part of the intended process of slowing the vehicle.

Friction is useful.

Car Engine Bearings

Thermal energy caused by friction is unwanted.

Friction reduces efficiency.

Shoes

Friction provides grip.

Useful.

Skiing

Excessive friction between skis and snow can slow the skier.

Often unwanted.

The same physical process can therefore be useful in one situation and unwanted in another.

Comparing Two Machines

Machine A receives:

1000 J

and provides:

700 J

of useful output.

Energy dissipated:

300 J

Efficiency:

70%

Machine B receives:

1000 J

and provides:

900 J

of useful output.

Energy dissipated:

100 J

Efficiency:

90%

Machine B dissipates less energy and is more efficient.

Worked Example: Machine with Friction

A machine uses an input force of:

300 N

over:

4 m

Input work:

Win = Fd

Win = 300 × 4

Win = 1200 J

The machine provides:

900 J

of useful output work.

Energy dissipated:

1200 − 900 = 300 J

Efficiency:

efficiency = 900/1200 × 100

efficiency = 75%

Reducing Friction Does Not Always Mean Eliminating It

Suppose engineers completely eliminated friction between car tyres and the road.

The result would not be an extremely efficient car.

The car would be unable to:

• accelerate properly
• steer
• brake effectively

This illustrates an important engineering principle:

friction should be controlled, not simply eliminated.

Useful friction should be maintained while unwanted friction is reduced.

Methods for Reducing Unwanted Energy Losses

Lubrication

Reduces direct contact between moving surfaces.

Bearings

Replace much sliding friction with lower rolling friction.

Streamlining

Reduces air or fluid resistance.

Smooth Surfaces

Can reduce friction and turbulence in appropriate systems.

Proper Alignment

Prevents unnecessary rubbing and vibration.

Proper Tyre Pressure

Reduces unwanted tyre deformation and rolling resistance.

Lightweight Design

Can reduce some energy requirements in transportation.

Regenerative Braking

Recovers part of a vehicle's kinetic energy instead of dissipating all of it thermally.

Regular Maintenance

Keeps machines operating close to their intended performance.

Engineering Trade-Offs

Reducing friction can sometimes create other problems.

For example:

• extremely smooth tyres may provide poor grip
• very thin lubricants may not protect surfaces adequately
• lightweight components may be less durable
• aerodynamic designs may reduce cargo or passenger space

Engineers therefore balance:

• efficiency
• safety
• cost
• durability
• performance
• reliability

The goal is not simply to minimize friction at all costs.

Friction, Efficiency, and Sustainability

Reducing unnecessary energy losses can have environmental benefits.

If a transportation system requires less energy:

• less fuel may be consumed
• battery range may increase
• energy costs may decrease
• emissions associated with energy production may decrease
• components may last longer

Small improvements in efficiency can become significant when applied to millions of vehicles or machines.

Multi-Step Example

A vehicle experiences an average total resistive force of:

700 N

while traveling:

5 km

Convert distance:

5 km = 5000 m

Energy dissipated:

E = Fd

E = 700 × 5000

E = 3,500,000 J

E = 3.5 MJ

Suppose the vehicle travels at:

25 m/s

Power needed to overcome this resistance:

P = Fv

P = 700 × 25

P = 17,500 W

P = 17.5 kW

This means the vehicle must continuously provide 17.5 kW just to overcome these resistive forces while maintaining that speed under the stated conditions.

Common Mistakes

Saying Friction Destroys Energy

Friction does not destroy energy.

It transfers energy, usually into thermal energy.

Saying Energy Is "Lost"

Energy is conserved.

When we talk about an energy loss, we usually mean energy has been transferred into a form that is less useful for the intended purpose.

Assuming Friction Is Always Bad

Friction is essential for:

• walking
• braking
• gripping
• steering
• many machines

Only unwanted friction should generally be reduced.

Assuming an Object Must Slow Down if Friction Acts

If another force balances friction, an object can move at constant velocity.

Confusing Friction with Air Resistance

Both are resistive forces, but ordinary surface friction involves interacting surfaces in contact.

Air resistance results from movement through air.

Forgetting Distance in Energy Calculations

The energy dissipated by a constant frictional force depends on both force and distance:

E = Fd

Confusing Energy and Power

Energy dissipated is measured in:

joules (J)

Power dissipated is measured in:

watts (W)

Power tells us how quickly energy is being transferred.

Assuming Reducing Friction Always Improves a System

Some friction is necessary for safe and effective operation.

For example, reducing tyre-road friction too much would make a vehicle dangerous.

Check Your Understanding

1. Define friction.
2. In what direction does friction act relative to relative motion?
3. Explain how friction can cause an object to slow down.
4. Explain what happens when the driving force on a car equals the total resistive force.
5. Describe what happens to mechanical energy when friction acts.
6. Why is it incorrect to say friction destroys energy?
7. A sliding object loses 300 J of kinetic energy. Where has this energy gone?
8. A frictional force of 50 N acts over 6 m. Calculate the energy dissipated.
9. A frictional force of 120 N acts over 20 m. Calculate the energy dissipated.
10. Explain the difference between mechanical energy loss and total energy loss.
11. Give three examples in which friction is useful.
12. Give three examples in which friction causes unwanted energy transfers.
13. Explain how friction reduces machine efficiency.
14. A machine receives 1500 J and provides 1200 J of useful output. Calculate its efficiency.
15. Calculate the energy dissipated in Question 14.
16. Explain how lubrication can improve machine efficiency.
17. Why can bearings reduce energy losses?
18. Explain why rolling an object is often easier than sliding it.
19. Define air resistance.
20. Explain why aerodynamic design can improve transportation efficiency.
21. A vehicle experiences a resistive force of 500 N over 2 km. Calculate the energy transferred against resistance.
22. A car experiences 600 N of resistance while moving at 20 m/s. Calculate the power required to balance the resistance.
23. Explain why high-speed transportation can require large amounts of power to overcome air resistance.
24. Describe how regenerative braking differs from conventional friction braking.
25. Explain why correct tyre pressure can improve vehicle efficiency.
26. Why might reducing tyre-road friction too much be dangerous?
27. Identify two ways sound and vibration can reduce machine efficiency.
28. Suggest four methods engineers can use to reduce unwanted energy losses.
29. Explain why a machine becoming hot can be evidence of energy dissipation.
30. Explain the statement: "The engineering goal is usually to control friction, not eliminate it."

Key Terms

Friction — Force that opposes relative motion or the tendency for relative motion between surfaces.

Resistive force — Force acting against motion.

Air resistance — Resistive force acting on an object moving through air.

Drag — Resistive force caused by motion through a fluid such as air or water.

Energy dissipation — Transfer and spreading of energy into less useful forms.

Thermal energy — Energy associated with the internal motion and interactions of particles.

Mechanical energy — Energy associated with motion and position, including kinetic and potential energy.

Kinetic energy — Energy associated with motion.

Useful energy transfer — Energy transfer that contributes to the intended purpose of a system.

Wasted energy transfer — Energy transfer that does not contribute to the intended purpose.

Efficiency — Fraction of input energy or power transferred into useful output.

Lubrication — Use of a material between surfaces to reduce unwanted friction and wear.

Lubricant — Substance used to reduce friction between moving surfaces.

Bearing — Component designed to reduce friction and support relative motion.

Rolling resistance — Resistance associated with rolling objects, often involving deformation of wheels, tyres, or surfaces.

Streamlining — Designing an object to reduce resistance as it moves through a fluid.

Aerodynamic drag — Air resistance acting on an object moving through the atmosphere.

Regenerative braking — Braking system that recovers some kinetic energy and transfers it into stored energy.

Wear — Gradual removal or deformation of material caused by repeated interactions between surfaces.

Key Takeaways

• Friction is a force that opposes relative motion between interacting surfaces.
• Friction affects both the motion of objects and the way energy is transferred.
• Friction often transfers mechanical energy into thermal energy.
• Energy dissipated by friction is not destroyed.
• Total energy is always conserved.
• A decrease in mechanical energy does not mean total energy has disappeared.
• For constant friction, the magnitude of energy dissipated can be calculated using E = Fd.
• Friction can reduce the efficiency of machines by transferring energy into unwanted forms.
• Friction can also be useful for walking, gripping, braking, steering, and transmitting forces.
• Whether friction is useful or unwanted depends on the purpose of the system.
• Air resistance is another important source of energy dissipation.
• At high speeds, aerodynamic drag can require substantial power to overcome.
• Streamlining can reduce drag and improve transportation efficiency.
• Lubrication reduces unwanted friction between moving surfaces.
• Bearings can reduce friction by replacing sliding interactions with rolling motion.
• Correct maintenance can reduce friction, vibration, wear, and unnecessary energy losses.
• Regenerative braking can recover some kinetic energy that would otherwise be dissipated.
• Sound, vibration, deformation, electrical resistance, and heating can also reduce system efficiency.
• Reducing unwanted energy losses can lower energy consumption and operating costs.
• Engineers must balance efficiency with safety, reliability, durability, and performance.

The central energy idea is:

friction does not destroy energy — it transfers mechanical energy into other forms, usually thermal energy.

The central engineering idea is:

keep friction where it is useful and reduce it where it wastes energy.

Energy in Transportation and Engineering

Transportation systems move people and materials by transferring energy from an energy source into useful mechanical energy.

Whether the system is a car, train, aircraft, bicycle, ship, or electric vehicle, the same physics applies:

energy is transferred → forces produce motion → some energy becomes useful output → some energy is dissipated

Engineers aim to maximize the useful energy transfer while reducing unnecessary losses.

Important concepts include:

• energy stores
• energy transfer pathways
• work
• kinetic energy
• power
• efficiency
• friction
• air resistance
• energy density
• mechanical advantage
• conservation of energy

 

 

 

7

Energy Flow in Transportation

Every transportation system requires an energy source.

Examples include:

The energy must ultimately be transferred into the kinetic energy of the vehicle or used to overcome forces opposing its motion.

Energy Flow in a Gasoline Vehicle

A simplified energy pathway is:

chemical store in fuel

↓

heating during combustion

↓

mechanical transfer in engine

↓

mechanical transfer through transmission

↓

kinetic store of vehicle

However, not all the chemical energy becomes useful kinetic energy.

Energy is also transferred into:

• engine thermal stores
• exhaust gases
• cooling system
• tyres
• road
• surrounding air
• sound

Therefore, the engine and drivetrain are not 100% efficient.

 

 

 

5

Energy Flow in an Electric Vehicle

A battery-electric vehicle has a different energy pathway:

chemical store in battery

↓

electrical transfer

↓

electric motor

↓

mechanical transfer

↓

kinetic store of vehicle

Energy losses occur through:

• electrical resistance
• battery heating
• motor heating
• friction
• tyre deformation
• air resistance

Electric drivetrains generally have fewer major energy-conversion stages than combustion drivetrains.

Conservation of Energy

Energy cannot be created or destroyed.

For any vehicle:

total input energy = useful energy + dissipated energy

Suppose a vehicle receives:

1000 kJ

of energy.

If:

700 kJ → useful output

then:

300 kJ

must have been transferred elsewhere.

For example:

1000 kJ input = 700 kJ useful + 300 kJ dissipated

The energy has not disappeared.

Useful Energy

The useful energy output depends on the intended purpose.

For a moving vehicle, useful transfers can include:

• increasing kinetic energy
• increasing gravitational potential energy while climbing
• moving passengers or cargo
• operating necessary systems

Some energy may also be required simply to overcome resistive forces.

Energy Losses

Common sources of unwanted energy transfer in transportation include:

• friction
• rolling resistance
• air resistance
• engine heating
• electrical resistance
• vibration
• sound
• braking
• deformation of tyres

These processes reduce the fraction of input energy available for the desired output.

Efficiency

Efficiency measures how much input energy becomes useful output.

efficiency = useful energy output / total energy input

As a percentage:

efficiency (%) = useful output / total input × 100

Efficiency can also be calculated using power:

efficiency (%) = useful power output / total power input × 100

Worked Example: Vehicle Efficiency

A vehicle receives:

5000 kJ

from its energy source.

Useful mechanical output:

3500 kJ

Efficiency:

efficiency = 3500/5000 × 100

efficiency = 70%

Energy dissipated:

5000 − 3500 = 1500 kJ

Comparing Two Transportation Systems

Vehicle A:

Input = 1000 kJ

Useful output = 300 kJ

Efficiency:

30%

Vehicle B:

Input = 1000 kJ

Useful output = 850 kJ

Efficiency:

85%

Vehicle B produces much more useful output from the same input energy.

However, efficiency alone does not determine which vehicle is best.

Other considerations include:

• energy-storage mass
• energy density
• range
• cost
• charging or refueling time
• infrastructure
• environmental impact
• reliability

Power in Transportation

Power is the rate at which energy is transferred.

P = E/t

where:

• P = power in watts
• E = energy in joules
• t = time in seconds

Recall:

1 W = 1 J/s

A vehicle with a high-power motor can transfer large amounts of energy quickly.

This can allow:

• rapid acceleration
• climbing steep hills
• moving heavy loads
• maintaining high speeds against resistance

Worked Example: Motor Power

A motor transfers:

600,000 J

in:

20 s

Power:

P = E/t

P = 600,000/20

P = 30,000 W

P = 30 kW

Mechanical Power

When a constant force acts on an object moving at constant velocity:

P = Fv

where:

• P = power
• F = force
• v = velocity

This equation is particularly useful for transportation problems.

Worked Example: Car at Constant Speed

A car experiences:

800 N

of total resistance while traveling at:

25 m/s

At constant speed:

driving force = resistance

Therefore:

F = 800 N

Power required:

P = Fv

P = 800 × 25

P = 20,000 W

P = 20 kW

The car must provide at least 20 kW of mechanical power simply to overcome resistance under these conditions.

Why High-Speed Travel Requires Power

Vehicles experience resistive forces including:

• rolling resistance
• mechanical friction
• air resistance

Air resistance becomes particularly important at high speeds.

As speed increases, aerodynamic drag generally increases.

Therefore, a vehicle traveling faster often requires much more power to maintain its speed.

This is a major consideration in:

• cars
• aircraft
• high-speed trains
• bicycles

 

 

 

8

Aerodynamic Drag

Drag is a resistive force caused by movement through a fluid.

Air is a fluid.

Drag depends on factors including:

• speed
• shape
• frontal area
• air density
• surface characteristics

Streamlined designs allow air to flow around a vehicle more effectively.

Reducing drag can reduce the power needed to maintain speed.

Streamlining

Engineers reduce aerodynamic losses using:

• smooth body shapes
• smaller frontal areas
• carefully shaped vehicle surfaces
• aerodynamic fairings
• optimized underbody designs
• streamlined mirrors and accessories

Examples include:

• aerodynamic cars
• high-speed trains
• aircraft
• racing bicycles

Reducing drag can increase:

• range
• fuel economy
• maximum speed
• energy efficiency

Rolling Resistance

Tyres deform as they roll.

The road surface may also deform slightly.

These deformations transfer energy into thermal stores.

This creates rolling resistance.

Rolling resistance can be reduced through:

• suitable tyre design
• correct tyre pressure
• appropriate materials
• reduced unnecessary vehicle mass

However, tyres must still provide enough grip for safe acceleration, steering, and braking.

Friction in the Drivetrain

Mechanical systems contain moving components such as:

• gears
• shafts
• bearings
• chains
• axles

Friction between these components transfers mechanical energy into thermal energy.

Engineers reduce these losses using:

• lubrication
• efficient bearings
• precise alignment
• appropriate materials
• improved gear design

 

 

 

5

Vehicle Mass

Mass affects transportation energy requirements.

Recall:

Eₖ = ½mv²

A more massive vehicle requires more energy to reach the same speed.

For example:

Vehicle A:

m = 1000 kg

Vehicle B:

m = 2000 kg

Both travel at:

20 m/s

Vehicle A:

Eₖ = ½(1000)(20²)

Eₖ = 200,000 J

Vehicle B:

Eₖ = ½(2000)(20²)

Eₖ = 400,000 J

Doubling the mass doubles the kinetic energy at the same speed.

Speed Has an Even Greater Effect

Because:

Eₖ = ½mv²

kinetic energy depends on the square of speed.

Consider a 1000 kg vehicle.

At 10 m/s:

Eₖ = ½(1000)(10²)

Eₖ = 50,000 J

At 20 m/s:

Eₖ = ½(1000)(20²)

Eₖ = 200,000 J

Doubling speed increases kinetic energy by a factor of:

4

This has major implications for:

• acceleration
• braking
• collision energy
• energy consumption

Climbing Hills

Vehicles climbing hills gain gravitational potential energy.

Recall:

ΔEₚ = mgΔh

Suppose a:

1500 kg

car climbs:

100 m

Using:

g = 9.8 N/kg

Increase in gravitational potential energy:

ΔEₚ = 1500 × 9.8 × 100

ΔEₚ = 1,470,000 J

ΔEₚ = 1.47 MJ

Additional energy must also be supplied to overcome resistance.

Descending Hills

When a vehicle travels downhill:

gravitational potential → kinetic

Without braking, the vehicle may accelerate.

Conventional brakes transfer much of this energy into thermal energy.

Electric and hybrid vehicles may recover some of it using regenerative braking.

Regenerative Braking

During conventional braking:

kinetic energy → thermal energy

During regenerative braking:

kinetic energy → mechanical transfer → electrical transfer → chemical store in battery

Some energy is still dissipated, so regenerative braking is not 100% efficient.

However, it allows part of the vehicle's kinetic energy to be recovered.

 

 

 

5

Worked Example: Regenerative Braking

A vehicle has:

600 kJ

of kinetic energy before braking.

Its regenerative system recovers:

420 kJ

Recovery efficiency:

efficiency = 420/600 × 100

efficiency = 70%

Energy not recovered:

600 − 420 = 180 kJ

This energy is transferred into other stores.

Internal Combustion Vehicles

Internal combustion engine vehicles store energy in fuels such as:

• gasoline
• diesel
• biofuels

Advantages can include:

• high fuel energy density
• long driving range
• rapid refueling
• established infrastructure

Challenges include:

• substantial thermal losses
• exhaust emissions
• mechanical complexity
• friction
• noise
• dependence on combustible fuels

Electric Vehicles

Battery-electric vehicles store energy chemically in rechargeable batteries.

Potential advantages include:

• efficient electric motors
• regenerative braking
• fewer major moving drivetrain components
• no direct exhaust emissions
• quiet operation

Challenges can include:

• battery mass
• charging time
• charging infrastructure
• raw material requirements
• battery manufacturing
• battery degradation
• lower stored energy per kilogram compared with liquid hydrocarbon fuels

A complete comparison therefore requires more than simply comparing motor efficiency.

Energy Density and Transportation

Energy density describes the energy stored per unit mass or volume.

For transportation, high energy density can be particularly valuable because the vehicle must carry its energy source.

A high-energy-density fuel can provide:

• longer range
• lower fuel mass
• smaller storage volume

This is particularly important in aviation.

Batteries and Energy Density

Modern batteries have much lower stored energy per kilogram than gasoline or aviation fuels when comparing raw energy content.

However, electric motors can convert a high proportion of battery energy into useful mechanical output.

Therefore, transportation comparisons should consider:

energy density + drivetrain efficiency

rather than either quantity alone.

Hydrogen Transportation

Hydrogen can be used in:

• fuel-cell vehicles
• some experimental combustion systems

Hydrogen has high energy content per kilogram.

However, hydrogen gas has low density, so storage may require:

• high-pressure tanks
• cryogenic liquid storage
• specialized infrastructure

The complete system must include the mass and volume of the storage equipment.

 

 

 

5

Trains

Rail transportation has an important mechanical advantage:

low rolling resistance

Steel wheels rolling on steel rails can produce relatively low rolling resistance compared with rubber tyres on roads.

Trains can also carry large numbers of passengers or large quantities of cargo.

Electric trains can receive energy continuously from:

• overhead wires
• electrified rails

This means they may not need to carry a large onboard energy store.

High-Speed Trains

At lower speeds, rolling resistance may be important.

At very high speeds, aerodynamic drag becomes increasingly significant.

High-speed trains therefore use:

• streamlined noses
• smooth surfaces
• covered equipment
• carefully designed connections between cars

Reducing drag lowers the power required to maintain high speed.

Aircraft

Aircraft present one of the most demanding energy-engineering problems.

Aircraft must carry:

• passengers
• cargo
• structure
• engines
• fuel or energy storage

Every additional kilogram must be supported in flight.

This makes energy density by mass particularly important.

 

 

 

8

Improving Aircraft Efficiency

Engineers can improve aircraft efficiency through:

• improved aerodynamics
• efficient engines
• lightweight materials
• optimized flight paths
• winglets
• reduced unnecessary mass
• improved surface design

Small improvements can produce large energy savings across thousands of flights.

Ships

Ships can carry enormous loads.

Their large size means mass energy density may be less restrictive than for aircraft.

However, ships must overcome:

• water resistance
• air resistance
• mechanical friction

Engineers improve ship efficiency through:

• hull design
• efficient propellers
• engine improvements
• route optimization
• reduced friction
• improved loading

Bicycles

Bicycles are highly efficient transportation machines.

Energy flow:

chemical store in food

↓

muscle contraction

↓

mechanical transfer through pedals

↓

chain and gears

↓

wheel rotation

↓

kinetic energy of bicycle and rider

Energy losses include:

• air resistance
• rolling resistance
• chain friction
• bearing friction

At higher speeds, aerodynamic drag becomes increasingly important.

Gears and Transportation

Gears allow engineers to trade:

force for speed

or:

speed for force

Low gears provide:

• greater force at the wheel
• lower vehicle speed for a given motor or engine rotation

High gears provide:

• greater wheel speed
• less force multiplication

This allows an engine, motor, or cyclist to operate effectively under different conditions.

Machines and Mechanical Advantage

Mechanical advantage is:

MA = output force / input force

Machines can reduce the input force required by increasing the distance over which that force acts.

For an ideal machine:

input work = output work

Therefore:

Fin × din = Fout × dout

Machines multiply force, not energy.

This principle appears throughout transportation engineering.

Power and Hill Climbing

Suppose a vehicle climbs vertically at a steady rate.

The power required to increase gravitational potential energy is:

P = E/t

Since:

ΔEₚ = mgΔh

we can write:

P = mgΔh/t

If vertical speed is:

vvertical = Δh/t

then:

P = mgvvertical

This gives the minimum useful power required to gain gravitational potential energy.

Worked Example: Truck Climbing a Hill

A:

10,000 kg

truck gains:

20 m

of height in:

40 s

Using:

g = 9.8 N/kg

Increase in gravitational potential energy:

ΔEₚ = 10,000 × 9.8 × 20

ΔEₚ = 1,960,000 J

Useful power:

P = 1,960,000/40

P = 49,000 W

P = 49 kW

This ignores air resistance, rolling resistance, and drivetrain losses.

The actual engine or motor power required would be higher.

Including Efficiency

Suppose the drivetrain is:

70% efficient

and the useful mechanical power required is:

49 kW

Input power:

efficiency = useful power / input power

Therefore:

input power = useful power / efficiency

input power = 49/0.70

input power = 70 kW

The energy source must therefore supply approximately 70 kW.

Combined Resistance and Hill Climbing

A vehicle climbing a hill may need power for both:

• increasing gravitational potential energy
• overcoming resistive forces

Suppose:

Power for climbing = 40 kW

Power against resistance = 20 kW

Total useful mechanical power:

Ptotal = 40 + 20

Ptotal = 60 kW

If the drivetrain is:

80% efficient

Input power:

Pin = 60/0.80

Pin = 75 kW

Engineering for Lower Mass

Reducing vehicle mass can reduce the energy required for:

• acceleration
• climbing hills

Engineers may use:

• aluminum
• high-strength steels
• composites
• optimized structures

However, reducing mass must not compromise:

• safety
• strength
• durability
• cost
• crash protection

Engineering decisions involve trade-offs.

Recovering Energy

Engineers increasingly attempt to recover energy that would otherwise be dissipated.

Examples include:

• regenerative braking
• waste-heat recovery
• energy recovery from exhaust
• regenerative elevators
• hybrid power systems

Energy recovery can improve overall system efficiency.

Waste Heat

Engines, motors, batteries, brakes, and electronic systems can become warm during operation.

This indicates energy transfer into thermal stores.

In many systems, this thermal energy is difficult to recover.

Engineers therefore try to reduce unnecessary heating using:

• lubrication
• low-resistance conductors
• efficient motors
• improved cooling systems
• better materials

Cooling itself may require energy, so preventing unnecessary heating can be especially valuable.

Engineering Design and Efficiency

An energy-efficient design tries to maximize:

useful output / total input

Possible improvements include:

• reducing friction
• reducing drag
• reducing unnecessary mass
• improving motor efficiency
• improving engine efficiency
• optimizing gears
• recovering braking energy
• improving bearings
• using appropriate tyres
• reducing electrical resistance
• improving thermal management

No single modification solves every problem.

Whole-System Efficiency

Engineers must consider the complete system.

Suppose energy passes through three stages:

Battery discharge = 95% efficient

Motor = 90% efficient

Transmission = 95% efficient

Overall efficiency:

0.95 × 0.90 × 0.95

= 0.812

Approximately:

81%

This is why efficiencies of different stages must be multiplied rather than added.

Worked Example: Multi-Stage Vehicle

A battery initially supplies:

20 MJ

of energy.

Battery output efficiency:

95%

Energy leaving battery:

20 × 0.95 = 19 MJ

Motor efficiency:

90%

Mechanical output:

19 × 0.90 = 17.1 MJ

Drivetrain efficiency:

95%

Useful output:

17.1 × 0.95 ≈ 16.25 MJ

Overall useful fraction:

16.25/20 ≈ 0.8125

Overall efficiency:

≈ 81%

Approximately:

3.75 MJ

is transferred into less useful forms.

Comparing Transportation Technologies

No transportation technology is automatically "best."

Engineers must compare many factors.

Operational Versus Life-Cycle Energy

Transportation systems should not be evaluated only during operation.

A complete analysis may consider:

raw materials → manufacturing → energy production → operation → maintenance → recycling/disposal

For an electric vehicle, this can include:

• mineral extraction
• battery manufacturing
• electricity generation
• charging
• driving
• battery recycling

For a combustion vehicle:

• oil extraction
• fuel refining
• fuel transportation
• vehicle manufacturing
• fuel combustion
• maintenance

This is called life-cycle analysis.

Designing for the Intended Purpose

Different vehicles have different requirements.

City Car

Priorities may include:

• efficiency
• regenerative braking
• compact size
• low-speed performance

Long-Distance Truck

Priorities may include:

• range
• payload
• energy density
• reliability

Aircraft

Priorities include:

• very low mass
• high energy density
• aerodynamic efficiency

Train

Priorities may include:

• low rolling resistance
• large passenger or cargo capacity
• continuous electrical supply

Therefore, the most appropriate energy technology depends strongly on the application.

Practical Problem: Electric Car

An electric car has a mass of:

1600 kg

and accelerates from rest to:

25 m/s

Increase in kinetic energy:

Eₖ = ½mv²

Eₖ = ½(1600)(25²)

Eₖ = 500,000 J

Eₖ = 500 kJ

Suppose the drivetrain is:

80% efficient

Required battery energy:

input energy = useful energy / efficiency

input energy = 500/0.80

input energy = 625 kJ

Energy dissipated:

625 − 500 = 125 kJ

Practical Problem: Delivery Truck

A 5000 kg truck climbs 50 m while traveling a route.

Increase in gravitational potential energy:

ΔEₚ = mgΔh

Using:

g = 9.8 N/kg

ΔEₚ = 5000 × 9.8 × 50

ΔEₚ = 2,450,000 J

ΔEₚ = 2.45 MJ

If the drivetrain is 70% efficient:

input energy = 2.45/0.70

input energy = 3.5 MJ

This ignores resistance, acceleration, and other energy requirements.

Practical Problem: Train

A train experiences:

20,000 N

of total resistance while traveling at:

30 m/s

Required mechanical power:

P = Fv

P = 20,000 × 30

P = 600,000 W

P = 600 kW

If the motor system is:

90% efficient

Electrical input power:

Pin = 600/0.90

Pin ≈ 667 kW

Practical Problem: Energy Used During a Journey

Suppose a vehicle motor operates at an average useful power of:

40 kW

for:

30 minutes

Convert time:

30 min = 1800 s

Energy transferred:

E = Pt

E = 40,000 × 1800

E = 72,000,000 J

E = 72 MJ

If the system is 80% efficient:

input energy = 72/0.80

input energy = 90 MJ

Therefore:

18 MJ

is transferred into less useful forms.

Safety and Energy

Transportation engineering must also consider the large quantities of kinetic energy carried by moving vehicles.

Recall:

Eₖ = ½mv²

Because speed is squared, increasing speed greatly increases kinetic energy.

This affects:

• braking distance
• collision energy
• structural design
• restraint systems
• crash protection

Energy management is therefore important not only for efficiency but also for safety.

Crash Engineering

During a collision, kinetic energy must be transferred.

Vehicle structures may be designed to deform in controlled ways.

Energy can be transferred into:

• deformation
• thermal energy
• sound
• movement of other objects

Crumple zones increase the distance and time over which a vehicle slows down.

This can reduce the average force experienced by occupants.

The energy still has to be transferred, but engineers control how the transfer occurs.

Common Mistakes

Saying Vehicles "Use Up" Energy

Energy is not destroyed.

It is transferred between stores and eventually becomes increasingly dispersed.

Confusing Energy and Power

Energy tells us how much.

Power tells us how quickly.

P = E/t

Assuming a More Powerful Vehicle Is More Efficient

High power does not necessarily mean high efficiency.

A machine can transfer energy quickly while wasting a large fraction of it.

Ignoring Air Resistance

At high speeds, aerodynamic drag can become one of the most important energy losses.

Ignoring Vehicle Mass

More massive vehicles require more energy to accelerate and climb.

Thinking Speed and Kinetic Energy Are Directly Proportional

Kinetic energy depends on:

v²

Doubling speed produces four times the kinetic energy.

Assuming Electric Vehicles Have No Energy Losses

Electric vehicles still experience:

• motor losses
• electrical resistance
• battery heating
• rolling resistance
• air resistance
• friction

Assuming Regenerative Braking Recovers All Energy

Regenerative braking is not 100% efficient.

Some energy is still dissipated.

Comparing Technologies Using Only Efficiency

Efficiency is important, but so are:

• energy density
• range
• mass
• infrastructure
• cost
• environmental impacts

Adding Stage Efficiencies

Overall efficiency is found by multiplying decimal efficiencies.

For example:

90% × 80%

means:

0.90 × 0.80 = 0.72

Overall efficiency:

72%

Check Your Understanding

1. Describe the main energy transfers in a gasoline-powered vehicle.
2. Describe the main energy transfers in a battery-electric vehicle.
3. State the principle of conservation of energy.
4. Give four examples of unwanted energy transfers in transportation.
5. Define efficiency.
6. A vehicle receives 4000 kJ and produces 3000 kJ of useful output. Calculate its efficiency.
7. Calculate the dissipated energy in Question 6.
8. Define power.
9. A motor transfers 900,000 J in 30 s. Calculate its power.
10. A car experiences 700 N of resistance while traveling at 20 m/s. Calculate the power required to overcome resistance.
11. Explain why aerodynamic design is particularly important at high speeds.
12. Explain how tyre deformation causes energy losses.
13. Describe two ways engineers can reduce drivetrain friction.
14. A 1200 kg car travels at 20 m/s. Calculate its kinetic energy.
15. Calculate the kinetic energy if its speed increases to 40 m/s.
16. By what factor did the kinetic energy increase?
17. Explain why reducing vehicle mass can reduce energy consumption.
18. A 2000 kg vehicle climbs 50 m. Calculate its increase in gravitational potential energy using g = 9.8 N/kg.
19. Explain regenerative braking.
20. A vehicle has 800 kJ of kinetic energy and regenerative braking recovers 600 kJ. Calculate the recovery efficiency.
21. Explain why energy density is important in transportation.
22. Why is energy density especially important for aircraft?
23. Give two advantages and two challenges associated with battery-electric vehicles.
24. Explain why trains can be energy-efficient transportation systems.
25. A system has stages that are 90%, 80%, and 95% efficient. Calculate the overall efficiency.
26. Explain why a more powerful motor is not necessarily a more efficient motor.
27. Describe three engineering changes that could improve vehicle energy efficiency.
28. Explain why transportation technologies should be evaluated using life-cycle analysis.
29. Explain how energy concepts are important in crash safety.
30. Evaluate the statement: "The most efficient transportation technology is automatically the best transportation technology."

Key Terms

Transportation system — System designed to move people or materials from one location to another.

Energy transfer — Movement of energy between stores or systems.

Drivetrain — Components that transfer mechanical power from an engine or motor to a vehicle's wheels or other propulsion system.

Efficiency — Fraction of input energy or power converted into useful output.

Power — Rate at which energy is transferred.

Mechanical power — Rate at which mechanical work is performed.

Drag — Resistive force acting on an object moving through a fluid.

Air resistance — Drag caused by movement through air.

Rolling resistance — Resistance associated with deformation and interaction between rolling objects and surfaces.

Streamlining — Shaping an object to reduce fluid resistance.

Regenerative braking — System that recovers part of a vehicle's kinetic energy during braking.

Energy density — Amount of energy stored per unit mass or volume.

Drivetrain loss — Energy transferred into less useful forms as energy passes through drivetrain components.

Life-cycle analysis — Evaluation of environmental and energy impacts throughout production, operation, and disposal.

Crumple zone — Vehicle structure designed to deform during a collision and control energy transfer.

Mechanical advantage — Ratio of output force to input force.

Energy dissipation — Transfer and spreading of energy into less useful stores.

Key Takeaways

• Transportation systems transfer stored energy into useful mechanical energy.
• Energy is conserved throughout every transportation system.
• Some input energy is inevitably transferred into less useful forms.
• Efficiency measures the fraction of input energy reaching the desired useful output.
• Power measures how quickly energy is transferred.
• P = E/t.
• For constant-velocity motion against a force, P = Fv.
• Vehicle mass affects the energy required for acceleration and climbing.
• Kinetic energy is calculated using Eₖ = ½mv².
• Because kinetic energy depends on speed squared, speed has a major effect on energy requirements.
• Gravitational potential energy changes according to ΔEₚ = mgΔh.
• Friction, rolling resistance, air resistance, heating, sound, and vibration can reduce transportation efficiency.
• Streamlining reduces aerodynamic drag.
• Lubrication and efficient bearings reduce mechanical losses.
• Regenerative braking can recover some energy that would otherwise be dissipated.
• Electric and combustion vehicles have different energy-transfer pathways and different engineering trade-offs.
• High energy density is particularly valuable when a vehicle must carry its energy source.
• Aircraft are especially sensitive to energy-storage mass.
• Rail systems can benefit from low rolling resistance and external electrical supplies.
• Engineers often reduce mass to decrease energy requirements, but safety and strength must still be maintained.
• Overall efficiency in a multi-stage system is calculated by multiplying the efficiencies of the individual stages.
• A high-power machine is not necessarily an efficient machine.
• The best transportation technology depends on the intended application.
• Energy efficiency must be considered alongside range, energy density, infrastructure, safety, cost, reliability, and environmental impact.
• Life-cycle analysis provides a more complete comparison of transportation technologies.
• Energy concepts are important not only for transportation efficiency but also for braking and collision safety.

The central engineering goal is not simply to provide more energy or more power. It is to transfer the available energy to the desired output as effectively as possible while controlling losses, mass, cost, safety, and environmental impacts.