Legacy mechanisms: transmission, bearings and efficiency
Compare the named drive and motion-conversion mechanisms, bearings and simple machine ratios in the older Systems & Control option.
Exam scope and assessment notes
StudyForA provides original teaching and scenario-based practice, not Cambridge past papers, official marks or an authenticated practical course. It cannot assess a candidate's actual workshop skills, finished product or Component 2 portfolio. Component 2 is school-based, internally assessed and externally moderated; the teacher authenticates candidate work. Any real making requires centre facilities, risk controls and direct qualified supervision. Candidate project work must remain their own. Follow the syllabus for the exam year.
Before you begin
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 20-tooth driver turns a 60-tooth driven gear. What fraction of the driver speed does the driven gear have?
What you will learn
- Choose among gear, belt, chain, bearing and motion-conversion systems using load, space, speed and maintenance needs.
- Calculate simple speed ratios, mechanical advantage, velocity ratio and efficiency from stated values.
Choose a transmission system for the task
The legacy option names bevel, worm and rack-and-pinion gears; flat, toothed, round and V-belts; sprockets and chains; and systems to maintain belt or chain tension. Compare the distance and direction between shafts, grip, slip, noise, load, alignment and maintenance.
Plain bearings use sliding contact; roller and ball bearings use rolling elements. Compare their suitability for load, speed, space, contamination and maintenance rather than choosing only by friction. Mechanisms can also convert movement through crankshafts, crank/slider systems, ratchet-and-pawl, eccentrics, cams, rack-and-pinion or screw threads.
Calculate ideal and actual performance
Mechanical advantage (MA) = output load ÷ input effort. Velocity ratio (VR) compares input movement or speed with output movement or speed. Efficiency = MA ÷ VR × 100% for consistent definitions and idealised school problems.
For a simple gear pair, speed is inversely related to tooth count: driven speed = driver speed × driver teeth ÷ driven teeth. State the direction change and use consistent units. Real losses from friction and deformation make actual output less than an ideal calculation.
Sources for this lesson
Pause and explain
If MA = 2 and VR = 4, what is the efficiency?
Worked example
An input effort of 30 N lifts a 90 N load. The machine's velocity ratio is 4. Find its MA and efficiency.
Show the worked solution
- MA = load ÷ effort = 90 ÷ 30 = 3.
- Efficiency = MA ÷ VR × 100%.
- Efficiency = 3 ÷ 4 × 100% = 75%.
- Check that the stated load, effort and velocity ratio use the same machine and operating condition.
Answer The mechanical advantage is 3 and the efficiency is 75%. The result describes the stated condition only; friction and load variation may change actual performance.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 12-tooth gear turns at 300 rpm and drives a 36-tooth gear. Find the driven speed and direction.
Give me a hint
Use driver speed × driver teeth ÷ driven teeth.
Compare my reasoning
- Write n2 = n1 × T1 ÷ T2.
- Substitute 300 × 12 ÷ 36.
- State the output direction for an external gear pair.
The driven gear turns at 100 rpm and in the opposite direction. This assumes a direct external gear pair with no intermediate gear and ideal transmission.
Look for these in your work
- Uses the correct ratio
- Calculates speed
- States direction and assumptions
A screw jack lifts a load of 600 N with an effort of 50 N. If its VR is 15, calculate MA and efficiency.
Give me a hint
Calculate MA first, then compare MA with VR.
Compare my reasoning
- MA = 600 ÷ 50.
- Efficiency = MA ÷ 15 × 100%.
- Interpret the result as actual performance for the given data.
MA = 12. Efficiency = 12 ÷ 15 × 100% = 80%. The remaining 20% represents losses under the stated conditions; do not assume the same efficiency for every load or jack.
Look for these in your work
- Calculates MA
- Uses stated VR
- Interprets a realistic efficiency
A drive must work in a dusty environment with occasional misalignment. Compare a belt drive and a chain drive and propose a test.
Give me a hint
Consider slip, lubrication, guarding, tension and tolerance to dirt.
Compare my reasoning
- Identify load, speed and environment.
- Compare belt and chain behavior and maintenance needs.
- Test the selected model under safe representative conditions.
A belt may slip under load and need tensioning; a chain can provide positive engagement but needs suitable lubrication and alignment and may be noisier. Dust can affect both. Test transmission and wear with representative load and contamination controls, guarded and supervised; then compare maintenance burden.
Look for these in your work
- Compares mechanisms in context
- Identifies maintenance issues
- Proposes a relevant safe test
Common mistakes
- Using driven/driver gear teeth in the wrong order
- Confusing mechanical advantage with velocity ratio
- Reporting efficiency above 100% without checking the data or definitions
What can you explain now?
If MA = 2 and VR = 4, what is the efficiency?
Compare with the explanation
50%.
Efficiency is MA ÷ VR × 100%, so 2 ÷ 4 × 100% = 50%.
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Make it stick
Recall the main decision in “Legacy mechanisms: transmission, bearings and efficiency”, name the evidence that supports it, and identify one limitation that still needs checking.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.