Curriculum/DP Design/B3.3 Mechanical Systems Application and Selection

Mechanical Systems Application and Selection | B3.3

Guiding questionHow can mechanical systems be incorporated into product design?

B3.3 is the calculating half of mechanisms, where the ideas from A3.3 turn into design decisions instead of descriptions. Mechanical advantage, velocity ratio and efficiency are the three numbers that let you state what a mechanism is doing for the user. They are the difference between "this gear train makes it easier" and "this gear train trades a factor of four in speed for a factor of four in torque, which is why the user can lift it at all".

Two pieces of advice. Efficiency is the objective students skip most often and the one that matters most in practice, because no real mechanism gives back everything you put into it and the losses are exactly where the design choices live. And do the problems by hand until the relationships feel obvious. These questions are multi-step, the marks sit in the working rather than the final figure, and a habit of showing what you did is worth building now rather than in May.

Students must be able toCalculate mechanical advantage in gear, pulley, belt and lever systems.

Belt drive diagram showing drive pulley, driven pulley, and belt

Mechanical advantage (MA) is the ratio of the output load to the input effort in a machine. For a belt drive or pulley system, MA is determined by the ratio of pulley diameters, or equivalently by the inverse ratio of rotational speeds:

MA = Load / Effort = d₂ / d₁ = N₁ / N₂

where d₂ is the driven (output) pulley diameter, d₁ is the drive (input) pulley diameter, N₁ is the input speed and N₂ is the output speed.

Effect on torque and speed:

  • If d₂ > d₁ (MA > 1): output torque increases, output speed decreases: a force multiplier.
  • If d₂ < d₁ (MA < 1): output speed increases, output torque decreases: a speed multiplier.

Characteristics of belt drives: Belts transmit power by friction between belt and pulley. They are quiet, cost-effective, and can span large distances between shafts. Under heavy loads, belts can slip: this protects components from shock overload but reduces efficiency and speed accuracy. V-belts wedge into grooves to increase friction and reduce slip.

Worked example (belt drive MA)

GivenFormulaResult
Drive pulley d₁ = 100 mm, driven d₂ = 300 mmMA = d₂ / d₁MA = 300 / 100 = 3
MA = 3, input speed N₁ = 300 rpmN₂ = N₁ / MAN₂ = 300 / 3 = 100 rpm
The driven pulley produces 3× the torque but rotates at 1/3 the speed.
Interactive
Gear / Velocity-Ratio Calculator

Enter driver and driven sizes (teeth count or diameter, any consistent unit) for one stage, or add stages for a compound drive. Add an input speed and/or torque to see the output.

Stage 1
rpm
N·m

Students must be able toCalculate velocity ratios for gear-, pulley- and belt-driven systems.

Crossed belt drive and compound belt drive diagrams

The velocity ratio (VR) describes how much the rotational speed changes from input to output. It is always written as driven over driver. In a perfect machine with no friction, VR and MA are equal, so a system that multiplies force by 3 also divides speed by 3. In a real machine some input is lost to friction, so the actual MA is always a little lower than the VR. That gap is what efficiency measures.

For a simple belt drive or pulley:

VR = d₂ / d₁ = N₁ / N₂

Crossed belt drives reverse the rotation direction of the driven pulley: the belt crosses between the two pulleys. The VR formula is unchanged; only the direction differs.

Compound belt drives connect multiple pulley pairs in series. Two pulleys share a common shaft so that when the first driven pulley rotates, it drives the next stage. The total VR is the product of each stage's VR:

VR_total = VR₁ × VR₂ × VR₃ …

For gear trains, VR is calculated using the number of teeth instead of pulley diameters:

VR = T_driven / T_driver = N_driver / N_driven

System typeVR formulaDirection change?
Simple belt / pulleyd₂ / d₁No
Crossed beltd₂ / d₁Yes: reverses
Compound beltVR₁ × VR₂ × …Depends on stages
Gear pairT_driven / T_driverYes: reverses
Compound gear trainVR₁ × VR₂ × …Depends on stages

Students must be able toCalculate efficiency for gear- and belt-driven systems.

Power flow diagram showing input power, losses and output power

Efficiency (η) is the fraction of input power that reaches the output as useful work. No real machine reaches 100%: some energy is always lost to friction and heat.

η = (P_out / P_in) × 100%

Power in gear systems is the product of torque and angular velocity:

P = τ × ω    where ω = N × (2π / 60) rad/s

To convert rpm to rad/s, multiply by 2π and divide by 60.

Power in belt drives is transmitted by the difference in belt tensions between the tight side and the slack side:

P = F_E × v_belt    where F_E = T₊ − T₋

T₊ is the tight-side tension and T₋ is the slack-side tension. Belt speed v_belt = π × d₁ × N₁ / 60 (m/s, with d₁ in metres).

Why real systems fall short of 100%:

  • Gear tooth friction: as teeth slide against each other, energy converts to heat. This is the primary loss in most gear trains.
  • Churning losses: gears and belts must push through lubricating oil or grease, creating resistance.
  • Bearing friction: shaft bearings absorb some input power.
  • Belt slip: the belt slides slightly on the pulley under heavy load, reducing speed accuracy and transmitting less power.

Worked example (gear power and efficiency)

StepWorkingResult
Convert speed to rad/sω = 120 × (2π / 60)12.57 rad/s
Calculate input powerP = τ × ω = 1.2 Nm × 12.5715.1 W
Calculate efficiencyη = (13.5 / 15.1) × 100%89%

Work in the order shown: convert rpm to rad/s first, then find power, then compare output with input. Forgetting the rpm conversion is the most common mistake in this calculation.

Discussion
Chasing the last few percent

A well-lubricated gear train can run at 95–98% efficient; a worm gear can sit under 50%. A petrol engine typically wastes 70–80% of its fuel energy as heat before any of it reaches the wheels; an electric drivetrain converts roughly 85–90% of its input into motion. Closing that gap almost always costs something: precision-ground gears cost more than stamped ones, better bearings cost more than bushings, and most loss-reduction measures add mass, complexity or price.

Find a real machine's published efficiency figure. Is the gap between that number and 100% a design failure, or the correct trade-off given what a more efficient version would cost, weigh or take to manufacture? At what point does chasing the last few percent of efficiency stop being worth it?

Students must be able toCalculate gear ratios and belt-driven system ratios, calculate the speed of rotation of a gear system at several points including initial input and final output speed, and construct systems that use gears to increase or decrease speed and motion.

Compound gear train diagram showing multiple gear pairs on shared shafts

Gear and belt systems are selected by designers to achieve specific combinations of speed, torque and direction. Choosing the right system requires calculating what will happen at each stage of the mechanism.

Direction of rotation in gear trains: In a simple gear train, adjacent gears rotate in opposite directions. An idler gear placed between two gears reverses direction again: the output then turns the same way as the input without changing the speed ratio. Belts and pulleys do not reverse direction (unless crossed).

Compound gear trains mount two gears on a shared shaft. This allows large speed changes in a compact space. The total velocity ratio is the product of each gear pair's ratio:

VR_total = VR₁ × VR₂ × VR₃ …

Calculating speed at multiple stages: Work stage by stage. For each gear pair: N_out = N_in × (T_driver / T_driven). For each belt stage: N_out = N_in × (d_drive / d_driven).

Overdrive gearboxes have VR < 1, meaning output speed is greater than input speed but output torque is reduced. Used in bicycle derailleur top gears and vehicle transmissions during motorway cruising, where higher speed and less force are needed.

Worked example (compound belt drive, multi-stage)

StagePulley diametersVROutput speed
Stage 1d₁ = 400 mm → d₂ = 200 mmVR₁ = 200/400 = 0.5N₂ = 60 × (400/200) = 120 rpm
Stage 2d₃ = 80 mm → d₄ = 50 mmVR₂ = 50/80 = 0.625N₄ = 120 × (80/50) = 192 rpm
OverallVR_total = 0.5 × 0.625 = 0.3125N₄ = 192 rpm (input N₁ = 60 rpm)

For compound belt drives, you cannot divide the final pulley diameter by the first: you must multiply individual stage VRs. A common exam mistake is treating it like a simple two-pulley system.

Product Spotlight
A bicycle rear derailleur shifting between sprockets

The Bicycle Derailleur

A dozen gear ratios, chosen with a lever instead of a gearbox.

Read spotlight →

Students must be able toAnalyse how cam systems translate rotary motion into reciprocating motion, construct mechanical systems that use cams, and interpret diagrams that represent the use of cams in a system.

Cam and follower diagram showing cam profile with rise, fall and dwell phases

A cam is a specially shaped rotating component. A follower is held against its surface and is pushed up and down as the cam rotates, converting continuous rotary motion into reciprocating (back-and-forth) motion. The cam's profile (its outline shape) determines exactly how the follower moves.

Cam motion phases:

  • Rise: the follower moves away from the cam centre as the cam radius increases.
  • Fall: the follower returns toward the centre as the radius decreases.
  • Dwell: the follower remains stationary while the cam radius stays constant.

Follower types: knife-edge (precise but wears quickly), roller (reduces friction, suited to most applications), flat-faced (suited to high-speed cams with gentle profiles).

Historical development: Trip hammers in Han Dynasty China (206 BCE–220 CE) used cams driven by water wheels to produce a hammering action: one of the earliest uses of rotary-to-reciprocating conversion. In the 12th century, Ismail al-Jazari described programmable camshafts in his Book of Knowledge of Ingenious Mechanical Devices. His "Mechanical Servant" automaton used cam lobes on a hidden shaft to produce a sequence of controlled movements: an early form of mechanical programming.

Modern applications:

  • Sewing machine stitch cams: interchangeable cam profiles produce different reciprocating needle patterns, creating different stitches (zigzag, blind hem) without altering any other part of the machine. Different cam shapes encode different stitch programs.
  • Internal combustion engine camshaft: driven at half the crankshaft speed, each cam lobe pushes open an intake or exhaust valve at a precisely timed moment. The cam profile controls valve lift (how far it opens), duration (how long), and timing (when). Once the lobe rotates past, a valve spring closes the valve.

Students must be able toAnalyse the Load (L), Effort (E) and Fulcrum, calculate Load (L) and Effort (E), construct mechanical systems that use levers, and interpret diagrams that represent the use of levers in a system.

Three classes of levers showing positions of fulcrum, effort and load with examples

A lever is a rigid beam that rotates about a fixed point called the fulcrum (F). By varying the positions of the effort (E) and load (L) relative to the fulcrum, a lever can multiply force, speed, or distance.

Equilibrium condition: for a lever in balance, the turning moments on each side of the fulcrum must be equal:

E × effort arm = L × load arm    →    MA = load arm / effort arm

Oblique forces: when a force acts at angle θ to the lever rather than perpendicular to it, only the perpendicular component creates a turning moment. Giovanni Batista Benedetti (16th century) recognised this and showed the effective force can be resolved using trigonometry:

(E × sin θ) × effort arm = L × load arm

Biomechanics (Giovanni Alfonso Borelli, 1608–1679), the father of biomechanics, proved that most joints in the human body act as third-class levers. Muscle insertion points sit close to the joint (very short effort arm), so large muscular forces produce small loads but move the limb end through a large distance. Human body levers are speed and distance magnifiers, not force multipliers: a design essential for throwing, writing, reaching and all fine motor skills. If the body used second-class levers at the elbow, we would be extraordinarily strong but too slow for daily tasks.

ClassPosition of F, E, LMAExample中文示例
First classF between E and LCan be >1 or <1Crowbar, scissors, seesaw撬棍、剪刀、跷跷板
Second classL between F and EAlways >1Wheelbarrow, nutcracker独轮车、胡桃夹
Third classE between F and LAlways <1Bicep curl, tweezers, shovel肱二头肌弯举、镊子、铁锹

Worked example (oblique force, bicep curl)

StepWorkingResult
Load torque50 N × 0.35 m17.5 Nm
Equilibrium with oblique effort (θ = 75°)(E × sin 75°) × 0.04 = 17.5E × 0.966 × 0.04 = 17.5
Solve for EE = 17.5 / (0.966 × 0.04)E ≈ 453 N
Mechanical advantageMA = 50 / 453MA ≈ 0.11

MA = 0.11 confirms this is a third-class lever. The body exerts ~453 N to lift a 50 N load, but the hand moves ~35 cm for every ~3 cm of muscle contraction. Speed and range of motion are gained at the expense of force.

Ten questions covering all six learning objectives. Select one answer per question, then click "Check all answers" to see your score and the explanations.

Q1 · 3.3.1 Mechanical Advantage
A belt drive has a drive pulley of 100 mm diameter and a driven pulley of 300 mm. The mechanical advantage and its effect are:
MA is the ratio of driven to drive diameter, 300 / 100 = 3. Torque rises by that factor and speed falls by the same one, because power cannot be created by the mechanism. An input at 300 rpm would leave the driven pulley turning at 100 rpm.
Q2 · 3.3.2 Velocity Ratios
In a crossed belt drive, the driven pulley:
Crossing the belt between the pulleys reverses the output direction without altering the diameters, so the speed and torque relationship is exactly as it would be with an open belt. An ordinary belt drive keeps both shafts turning the same way, whereas a meshing gear pair always reverses direction.
Q3 · 3.3.2 Velocity Ratios
How is the overall ratio of a compound belt drive with two stages found?
Each stage passes its output on as the input to the next, so the ratios compound. Treating the arrangement as a single pair by comparing only the first and last pulley ignores the intermediate pair entirely and is one of the most common errors in this topic. The same rule applies to compound gear trains.
Q4 · 3.3.3 Efficiency
In a belt drive, the effective force that transmits power is:
Only the net difference in tension does useful work, so power transmitted is that difference multiplied by belt speed. The slack side still carries tension, which is what keeps the belt seated on the pulley, but it contributes nothing to the power delivered.
Q5 · 3.3.3 Efficiency
A shaft transmits a torque of 1.8 N·m at 120 rpm. The power transmitted is approximately:
Convert the speed first: ω = 120 × 2π / 60 = 12.57 rad/s. Then P = τω = 1.8 × 12.57 ≈ 22.6 W. Forgetting to convert rpm into rad/s is the usual source of a wrong answer here.
Q6 · 3.3.4 Gear & Belt Systems
A compound gear train is chosen in preference to a simple gear train when:
Mounting two gears on a shared shaft lets each pair contribute its own ratio, and the ratios multiply. Achieving the same overall change with one pair would need a gear of impractical size, which is why gearboxes, drills and watches all use compound arrangements.
Q7 · 3.3.4 Gear & Belt Systems
A 20-tooth gear drives a 60-tooth gear. On the same shaft as the 60-tooth gear sits a 15-tooth gear, which drives a 45-tooth gear. With an input speed of 900 rpm, the final output speed is:
Work stage by stage. The first pair gives 900 × 20 / 60 = 300 rpm, and because the second gear shares its shaft, that speed becomes the input to the second pair: 300 × 15 / 45 = 100 rpm. Each stage reduces speed by three, so the compact two-stage train achieves an overall reduction of nine.
Q8 · 3.3.5 Cams and Followers
A cam follower stays at a constant height while the cam continues to rotate. This phase of the cam cycle is called:
Dwell occurs wherever the cam radius stays constant, so the follower is held still. Rise and fall correspond to increasing and decreasing radius. Designing the sequence of rise, dwell and fall into the profile is how a camshaft controls how far an engine valve opens, for how long, and at what moment.
Q9 · 3.3.6 Levers
A bicep curl, with the muscle attached to the forearm close to the elbow, is an example of:
The effort sits between the fulcrum and the load, so mechanical advantage is always below 1 and the muscle must exert far more force than the weight being lifted. What the body gains is range: a contraction of two or three centimetres sweeps the hand through some thirty-five, which is what makes throwing, writing and reaching possible.
Q10 · 3.3.6 Levers
A 50 N load is held 0.35 m from the elbow. The bicep attaches 0.04 m from the joint and pulls at 75° to the forearm. The effort force required is approximately:
Only the component perpendicular to the forearm creates a moment, so (E × sin 75°) × 0.04 = 50 × 0.35 = 17.5 N·m. That gives E = 17.5 / (0.966 × 0.04) ≈ 453 N. Omitting the sine term returns 437 N, which is the trap in this question.
Paper 2 requires extended written responses. Write your answer before revealing the example, then compare your approach, not just the content.
Question 1 · 3.3.2 Velocity Ratios · 4 marks
A belt drive has a drive pulley with a diameter of 320 mm rotating at 20 rpm, and a driven pulley with a diameter of 128 mm.

a) Calculate the velocity ratio (VR) of the belt drive.
b) Calculate the rotational speed of the driven pulley.
c) If the belt thickness is 5 mm, recalculate the speed of the driven pulley.
Show example answer

Given: d₁ = 320 mm, N₁ = 20 rpm, d₂ = 128 mm, belt thickness t = 5 mm

a) Velocity ratio:
VR = d₂ / d₁ = 128 / 320 = 0.4

b) Driven pulley speed (ignoring belt thickness):
Since VR = N₁ / N₂, rearrange to N₂ = N₁ / VR = 20 / 0.4 = 50 rpm. The same result comes from N₂ = N₁ × (d₁ / d₂) = 20 × 2.5 = 50 rpm. The smaller driven pulley turns faster than the drive pulley, so this arrangement gains speed and loses torque.

c) Driven pulley speed (including belt thickness):
When belt thickness is considered, the effective diameter becomes (diameter + thickness), as the belt's neutral axis sits at its mid-thickness:
N₂ = N₁ × (d₁ + t) / (d₂ + t) = 20 × (320 + 5) / (128 + 5) = 20 × 325 / 133 = 48.87 rpm

The belt thickness slightly increases both effective diameters, but affects the smaller pulley proportionally more, reducing the final output speed slightly below the no-thickness result.

Question 2 · 3.3.6 Levers · 6 marks
A third-class lever system represents a bicep curl. The load (weight held in the hand) is 50 N at a distance of 0.35 m from the elbow joint (fulcrum). The bicep muscle attaches to the radius bone at a distance of 0.04 m from the fulcrum. The force from the bicep is applied at an angle of 75° to the forearm.

a) Calculate the effective effort force required to hold the load in equilibrium (ignore the weight of the forearm).
b) Calculate the mechanical advantage (MA) of this lever system.
c) Explain why the body uses third-class levers despite their low mechanical advantage.
Show example answer

a) Effective effort force:

Only the component of the bicep force perpendicular to the forearm creates a turning moment. For equilibrium, the moments about the fulcrum must balance:

(E × sin 75°) × effort arm = Load × load arm
(E × 0.966) × 0.04 = 50 × 0.35
E × 0.03864 = 17.5
E = 17.5 / 0.03864 ≈ 452.9 N

b) Mechanical advantage:
MA = Load / Effort = 50 / 452.9 ≈ 0.11

c) Why the body uses third-class levers:

As Borelli proved in De Motu Animalium (1680), human body levers are primarily magnifiers of speed and distance, not force. Although the bicep must exert ~453 N to lift a 50 N load (MA = 0.11), the muscle only contracts approximately 2–3 cm to move the hand through ~35 cm. This large range of motion is essential for throwing, writing, reaching and fine motor skills. If the arm used a second-class lever (which would give MA > 1), movements would require less muscular force but would be much slower, making everyday actions impossible to perform at the speed required.

Question 3 · 3.3.3 Efficiency · 5 marks
A gear system has an input power of 15 W and an output power of 13.5 W.

a) Calculate the efficiency (η) of the gear system.
b) If the input torque is 1.2 Nm and the input rotational speed is 120 rpm, calculate the input power from these values.
c) State two reasons why real gear systems have efficiency less than 100%.
Show example answer

a) Efficiency:
η = (P_out / P_in) × 100% = (13.5 / 15) × 100% = 90%

b) Input power from torque and speed:
ω = N × (2π / 60) = 120 × (2π / 60) = 12.57 rad/s
P = τ × ω = 1.2 × 12.57 = 15.1 W

This agrees with the 15 W given in part (a), which is a useful check. Always convert rpm to rad/s before using P = τω, since this is where most marks are lost.

c) Two reasons for efficiency less than 100%:

  1. Friction between meshing gear teeth: as teeth slide and roll against each other under load, kinetic friction converts mechanical energy into heat. This is the primary loss mechanism in most gear trains.
  2. Churning losses: gears in a lubricated gearbox must push through oil or grease. The viscous resistance of the lubricant dissipates energy, particularly at high rotational speeds.
Question 4 · 3.3.4 Gear & Belt Systems · 4 marks
A compound belt drive system has two stages:
Stage 1: drive pulley d₁ = 400 mm, driven pulley d₂ = 200 mm
Stage 2: d₃ = 80 mm (coaxial with d₂), driven pulley d₄ = 50 mm
Drive speed N₁ = 60 rpm

a) Calculate the velocity ratio of each stage.
b) Calculate the overall velocity ratio.
c) Calculate the final output speed N₄.
Show example answer

a) Velocity ratio of each stage:
Stage 1: VR₁ = d₂ / d₁ = 200 / 400 = 0.5
Stage 2: VR₂ = d₄ / d₃ = 50 / 80 = 0.625

b) Overall velocity ratio:
VR_total = VR₁ × VR₂ = 0.5 × 0.625 = 0.3125

Note: You cannot calculate this as d₄/d₁ = 50/400 = 0.125: that formula only works for a simple two-pulley system. In a compound belt drive, each intermediate pulley pair contributes its own VR and all must be multiplied together.

c) Final output speed:
Stage by stage: N₂ = 60 × (400/200) = 120 rpm. Since d₃ is coaxial with d₂, N₃ = N₂ = 120 rpm.
N₄ = 120 × (80/50) = 192 rpm

Verification: N₄ = N₁ / VR_total = 60 / 0.3125 = 192 rpm ✓

Question 5 · 3.3.5 Cams and Followers · 6 marks
Analyse how cams and followers are used in modern mechanical systems. In your answer, refer to at least one historical development and two modern applications, and explain why cams remain relevant despite advances in electronic control systems.
Show example answer

A cam and follower system converts rotary motion (the rotating cam) into reciprocating motion (the back-and-forth movement of the follower pressed against it). The cam profile (its shape) encodes a specific motion pattern through its rising, falling and dwell phases.

Historical development: Trip hammers in Han Dynasty China (206 BCE–220 CE) used water-wheel-driven cams to produce a hammering action, one of the earliest mechanical applications of the rotary-to-reciprocating principle. In the 12th century, Ismail al-Jazari described programmable camshafts in his automata: changing the position of pegs and lobes on the shaft changed the sequence of movements. This represents an early form of mechanical programming, where the cam profile acted as the stored instruction.

Modern application 1 (Sewing machine stitch cams): Before computerised sewing machines, interchangeable cam discs produced different stitch patterns. As the cam rotated, a follower translated the motion into the needle bar's side-to-side movement. Changing the cam changed the stitch pattern without modifying any other part of the machine: a mechanically elegant solution to programmable output.

Modern application 2 (Internal combustion engine camshaft): The camshaft rotates at half crankshaft speed. Each cam lobe pushes a follower (tappet or rocker arm) to open an intake or exhaust valve at a precisely timed moment. The cam profile controls valve lift (how far the valve opens), duration (how long) and timing (when relative to piston position). A valve spring closes the valve once the lobe passes. Precision here is critical: incorrect valve timing reduces power, increases emissions and can cause engine damage.

Why cams remain relevant: Electronic control systems and servo motors can replicate cam motion, but mechanical cams offer reliability without sensors, controllers or software. In high-speed, high-temperature environments such as engines, cams require no electrical supply, tolerate oil and heat, and cannot suffer software failure. The internal combustion engine camshaft is one of the highest-volume precision mechanical components in manufacturing. "Camless" valvetrains exist but remain complex and costly by comparison.

Ismail al-Jazari, Wikipedia
en.wikipedia.org/wiki/Ismail_al-Jazari
The 12th century engineer whose camshaft driven automata predate European use of the camshaft by centuries. Includes the manuscript illustrations from the Book of Knowledge of Ingenious Mechanical Devices.
Giovanni Alfonso Borelli, Wikipedia
en.wikipedia.org/wiki/Giovanni_Alfonso_Borelli
The founder of biomechanics, whose 1680 De Motu Animalium showed that human joints work as third class levers that trade force for speed and range. Context for the lever calculations here.
The Engineering Mindset, YouTube channel
youtube.com/c/Theengineeringmindset
Animated coverage of belt drive calculations and gear train analysis, including velocity ratio, mechanical advantage and belt tension.
Lesics, YouTube channel
youtube.com/c/Lesics
3D animations of compound gear trains, overdrive gearboxes and camshafts. The animations make it clear why a compound train reaches a higher velocity ratio than a single pair of the same size.

Linking Questions

  • How does an understanding of the mechanical systems introduced in A3.3 inform the selection of components in real product design? (A3.3)
  • To what extent does the choice of material affect the efficiency and durability of mechanical systems such as gears and levers? (B3.1)
  • How do production methods and manufacturing tolerances influence the performance of precision mechanical components? (B4.1)
  • In what ways can the energy losses in a mechanical system contribute to the environmental impact of a product over its lifetime? (C2.1)
  • How might a user-centred approach change the selection of mechanical systems in consumer products such as power tools or assistive devices? (B1.1)