ScotlandEngineering ScienceSyllabus dot point

How do levers and gear systems trade force for movement, and how do you calculate mechanical advantage and velocity ratio?

Levers and gear systems, calculating mechanical advantage, velocity ratio and efficiency, and the gear ratio of meshing gears.

An SQA Higher Engineering Science answer on levers and gear systems, calculating mechanical advantage, velocity ratio and efficiency, and finding the gear ratio and output speed of meshing gears.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Quick answer

A mechanism transmits motion and force, trading one for the other. Mechanical advantage is the load divided by the effort, $MA = \dfrac{\text{load}}{\text{effort}}$ : a value above 1 means the machine magnifies the effort force. Velocity ratio is the effort distance divided by the load distance, $VR = \dfrac{\text{effort distance}}{\text{load distance}}$ , set purely by the geometry. Efficiency is $\dfrac{MA}{VR} \times 100\%$ , always below 100% because of friction. For meshing gears, the gear ratio is the driven teeth over the driver teeth; the gear with more teeth turns slower (inverse proportion), so a reduction gear trades speed for torque. All three ratios are dimensionless because their units cancel.

Jump to a section

What this key area is asking
Levers and the principle of moments
Mechanical advantage, velocity ratio and efficiency
Gear systems
Putting it together
Examples in context
Try this

What this key area is asking

The SQA wants you to analyse levers and gear systems as mechanisms, calculating mechanical advantage, velocity ratio and efficiency, and finding the gear ratio and output speed of meshing gears. Mechanisms transmit and modify force and motion, and these ratios are the language for describing how.

Levers and the principle of moments

A lever is the simplest mechanism: a rigid bar pivoting about a fulcrum, with an effort applied at one point to move a load at another. It balances when the turning effects (moments) are equal, $\text{effort} \times \text{effort arm} = \text{load} \times \text{load arm}$ . A long effort arm and short load arm let a small effort move a large load, which is the source of a lever's mechanical advantage.

Mechanical advantage, velocity ratio and efficiency

Mechanical advantage (MA) measures force gain: how many times bigger the load force is than the effort force. Above 1 means the machine magnifies force.
Velocity ratio (VR) measures movement: how far the effort moves for each unit the load moves. It depends only on the geometry of the machine, not on friction.
Efficiency compares the two: a perfect (frictionless) machine has $MA = VR$ , so efficiency 100%. Real friction makes $MA < VR$ , so efficiency falls below 100%.

Gear systems

A gear train transmits rotation between shafts. When two gears mesh, the driver (input) turns the driven (output). Because the teeth must move past each other at the same rate, the gear with more teeth turns more slowly.

A gear ratio above 1 (more teeth on the driven gear) is a reduction: the output turns slower but with more torque. A ratio below 1 is a step-up: faster output, less torque. Meshing gears also reverse the direction of rotation; an idler gear between them restores the original direction without changing the ratio.

Gear train reduction

A motor runs at 1200 rev/min and drives a 15-tooth gear meshed with a 45-tooth gear on the output shaft. Find the gear ratio and the output speed, and state the effect on torque.

step 1 Gear ratio

$\text{gear ratio} = \dfrac{\text{driven teeth}}{\text{driver teeth}} = \dfrac{45}{15} = 3$ , that is 3:1.

step 2 Output speed

The driven gear has more teeth, so it turns slower in inverse proportion: $N_{\text{driven}} = 1200 \times \dfrac{15}{45} = 400\ \text{rev/min}$ .

step 3 Effect on torque

A 3:1 reduction lowers the speed to one third but raises the torque by about three times (ignoring friction). Trading speed for torque is exactly why a reduction gear is used to drive a heavy load slowly.

Putting it together

The three ratios describe any mechanism: MA for the force gain, VR for the movement, and efficiency for how much is lost. For a gear train the velocity ratio equals the gear ratio (the inverse speed ratio), so a 3:1 gear has a VR of 3. The same trade-off runs through every mechanism in this area: you can have force or speed, and the geometry fixes the exchange while friction sets the efficiency.

Examples in context

A bottle opener is a lever with a long effort arm and a short load arm, giving a high mechanical advantage so a small hand force lifts a stiff cap. A bicycle uses gears to match the rider's comfortable pedalling speed to the terrain: a low gear (large rear sprocket, high reduction) gives high torque for climbing at the cost of speed, while a high gear does the reverse on the flat. A hand drill or an electric screwdriver uses a reduction gear so the fast motor turns the bit more slowly but with the torque needed to drive a screw.

Try this

Q1. A machine lifts a 400 N load with a 100 N effort. Find the mechanical advantage. [1 mark]

Cue. $MA = \frac{\text{load}}{\text{effort}} = \frac{400}{100} = 4$ .

Q2. A 10-tooth driver meshes with a 40-tooth driven gear. State the gear ratio. [1 mark]

Cue. $\frac{40}{10} = 4$ , that is 4:1.

Q3. A machine has $MA = 3$ and $VR = 4$ . Find its efficiency. [2 marks]

Cue. $\text{efficiency} = \frac{MA}{VR} \times 100\% = \frac{3}{4} \times 100\% = 75\%$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher (specimen)4 marksA machine lifts a load of 600 N when an effort of 150 N is applied. The effort moves 800 mm while the load rises 150 mm. Calculate the mechanical advantage, the velocity ratio, and the efficiency of the machine.

Show worked answer →

Mechanical advantage compares the load to the effort.

$MA = \dfrac{\text{load}}{\text{effort}} = \dfrac{600}{150} = 4$ .

Velocity ratio compares the distances moved.

$VR = \dfrac{\text{effort distance}}{\text{load distance}} = \dfrac{800}{150} = 5.33$ .

Efficiency is the mechanical advantage over the velocity ratio.

$\text{efficiency} = \dfrac{MA}{VR} \times 100\% = \dfrac{4}{5.33} \times 100\% = 75\%$ .

Markers reward the load-over-effort ratio for MA, the distance ratio for VR (effort distance on top), and efficiency as MA over VR. Units cancel in each ratio, so all three are dimensionless.

SQA Higher (specimen)3 marksA driver gear with 20 teeth meshes with a driven gear of 60 teeth. The driver turns at 900 rev/min. Calculate the gear ratio and the output speed of the driven gear.

Show worked answer →

The gear ratio compares the teeth (or the speeds) of driven to driver.

$\text{gear ratio} = \dfrac{\text{teeth on driven}}{\text{teeth on driver}} = \dfrac{60}{20} = 3$ , often written 3:1.

The output (driven) speed: more teeth on the driven gear means it turns slower, in inverse proportion to the teeth.

$N_{\text{driven}} = N_{\text{driver}} \times \dfrac{\text{driver teeth}}{\text{driven teeth}} = 900 \times \dfrac{20}{60} = 300\ \text{rev/min}$ .

Markers reward the gear ratio as driven over driver teeth (3:1), and the output speed found by the inverse teeth ratio, giving 300 rev/min. A 3:1 reduction trades speed for torque.

Related dot points

Sources & how we know this

SQA Higher Engineering Science Course Specification — SQA (2019)
Higher Engineering Science Course Specification (PDF) — SQA (2019)