How do levers and linkages change the size, direction or type of a force or motion?
Mechanisms based on levers and linkages: the three classes of lever, mechanical advantage and velocity ratio, the principle of moments applied to levers, and linkages (reverse motion, parallel motion, bell crank) that change the direction or type of motion.
A focused answer to OCR A-Level Product Design on mechanisms based on levers and linkages: the three classes of lever, mechanical advantage and velocity ratio with worked calculations, the principle of moments applied to levers, and the common linkages that change the direction or type of motion.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
Jump to a section
What this dot point is asking
OCR wants you to know the three classes of lever, calculate mechanical advantage and velocity ratio, apply the principle of moments to levers, and describe the common linkages. Mechanisms change the size, direction or type of a force or motion, and they carry calculation marks.
The three classes of lever
The single most-tested point is identifying the class by which component is in the middle: fulcrum (first), load (second), effort (third).
Mechanical advantage and velocity ratio
The principle of moments applied to levers
Linkages
Exam-style practice questions
Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
OCR 20194 marksA crowbar is used as a lever. The effort is applied 0.60 m from the pivot and the load sits 0.10 m from the pivot. Calculate the mechanical advantage and the effort needed to lift a 450 N load.Show worked answer →
A Component 01 lever calculation. Marks for the mechanical advantage, the method and the effort with units.
For an ideal lever, mechanical advantage equals the effort arm divided by the load arm: . Using the principle of moments, , so N. (Equivalently, effort = load divided by MA = N.)
A common dropped mark is inverting the mechanical advantage (load arm over effort arm) or omitting the unit; MA is effort arm over load arm, and a value above 1 means a small effort moves a large load.
OCR 20214 marksIdentify the class of lever for a pair of scissors and for a wheelbarrow, and explain the difference in where the fulcrum, effort and load lie.Show worked answer →
A Component 01 application question. Marks for each class and the explanation of the arrangement.
A pair of scissors is a first-class lever: the fulcrum (the pivot/rivet) is between the effort (the hand on the handles) and the load (the material being cut). A wheelbarrow is a second-class lever: the load (the contents) is between the fulcrum (the wheel) and the effort (the hands lifting the handles), which gives a large effort arm and a mechanical advantage greater than 1, so a small effort lifts a heavy load. The difference is the order: first class has the fulcrum in the middle; second class has the load in the middle (and third class has the effort in the middle, as in tweezers, favouring speed and range over force).
A common dropped mark is confusing second and third class; the middle component names the class only for first (fulcrum), and for second it is the load, for third the effort.
Related dot points
- Structures and forces: types of structure (frame, shell, monocoque), the forces of tension, compression, shear, bending and torsion, the principle of moments and equilibrium, and methods of reinforcing and stiffening to improve strength and rigidity.
A focused answer to OCR A-Level Product Design on structures and forces: frame, shell and monocoque structures, the forces of tension, compression, shear, bending and torsion, the principle of moments and equilibrium with worked calculations, and methods of reinforcing and stiffening.
- Rotary mechanisms: gears and gear trains (gear ratio and output speed), cams and followers (converting rotary to reciprocating motion), and pulleys and belt drives (velocity ratio and speed), with calculations of ratio and speed.
A focused answer to OCR A-Level Product Design on rotary mechanisms: gears and gear trains with gear-ratio and output-speed calculations, cams and followers converting rotary to reciprocating motion, and pulleys and belt drives with velocity-ratio and speed calculations.
- Electronic systems as input, process and output blocks: sensors and switches as inputs, processing devices, and output transducers, with Ohm's law, series and parallel resistors, and the potential divider used to sense light and temperature, including calculations.
A focused answer to OCR A-Level Product Design on electronic systems: the input, process, output model, sensors and output transducers, Ohm's law, series and parallel resistors, and the potential divider used with an LDR or thermistor, with worked calculations.
- Applied electrical and mechanical calculations: electrical power and energy, Ohm's law in context, mechanical advantage, velocity ratio, efficiency and the moment of a force, and selecting and applying the right formula to a design problem.
A focused answer to OCR A-Level Product Design on applied electrical and mechanical calculations: electrical power and energy, Ohm's law in context, mechanical advantage, velocity ratio, efficiency and moments, and how to select and apply the right formula to a design problem.
- Stress, strain and Young's modulus: the definitions and formulae, their units, the stress-strain relationship and the meaning of stiffness, with worked calculations applied to product components.
A focused answer to OCR A-Level Product Design on stress, strain and Young's modulus: the definitions, formulae and units, the stress-strain relationship and the meaning of stiffness, with worked calculations of stress, strain and Young's modulus applied to product components.