This set of notes covers the mechanical part of the Cambridge IGCSE 0445 syllabus, linking each topic to the relevant assessment objectives (AO1 – knowledge, AO2 – application, AO3 – analysis). It also includes the required background on materials, structures, loads, friction, and basic control concepts so that the notes form a complete revision resource.
Understanding the properties of materials and how they behave in simple structures is essential for designing reliable mechanisms.
| Material property | Typical IGCSE material | Common use in mechanisms |
|---|---|---|
| Density (ρ) | Steel ≈ 7850 kg m⁻³, Aluminium ≈ 2700 kg m⁻³ | Weight of moving parts, inertia of flywheels |
| Young’s modulus (E) | Steel ≈ 200 GPa, PVC ≈ 3 GPa | Stiffness of levers, gear shafts, cam bases |
| Yield strength (σy) | Steel ≈ 250 MPa, Brass ≈ 100 MPa | Maximum allowable stress in gears, shafts |
| Hardness | HRC ≈ 60 (hardened steel) | Wear resistance of gear teeth, cam followers |
Simple structural forms
Example (AO2): A simply‑supported wooden beam 1.2 m long carries a centred load of 200 N. The maximum bending moment is M = F L/4 = 200 × 1.2 / 4 = 60 Nm. Using a pine beam with section modulus Z = 2.5 × 10⁻⁶ m³, the bending stress is σ = M/Z = 24 MPa, well below the typical yield stress of pine (≈ 40 MPa).
Loads are classified as:
For static equilibrium (AO1):
\[ \sum F_x = 0,\qquad \sum F_y = 0,\qquad \sum M = 0 \]Reaction forces at supports are found by resolving the above equations. In mechanism design these reactions dictate the required bearing sizes and mounting points.
Friction reduces the ideal mechanical advantage and generates heat.
| Contact type | Typical coefficient of friction (μ) | Recommended lubrication |
|---|---|---|
| Steel on steel (dry) | 0.15 – 0.20 | Oil or grease |
| Steel on steel (oiled) | 0.05 – 0.08 | Light oil |
| Bronze on steel | 0.02 – 0.04 | Grease |
| Plastic on plastic | 0.10 – 0.20 | Silicone spray |
Efficiency (η) of a simple machine can be approximated by
\[ \eta \approx \frac{1}{1+\mu} \]When solving AO2 problems, always state whether the calculation is “ideal” (μ = 0) or “real” (include a reasonable μ).
A lever is a rigid bar rotating about a fixed fulcrum.
Mechanical advantage (MA)
\[ \text{MA} = \frac{F_{\text{output}}}{F_{\text{input}}}= \frac{d_{\text{effort}}}{d_{\text{resistance}}} \]Classes of levers
Worked example (AO2)
A Class II lever lifts a 120 N load. Effort distance = 0.30 m, resistance distance = 0.10 m. Find the effort.
\[ \text{MA}= \frac{0.30}{0.10}=3,\qquad F_{\text{effort}}= \frac{120}{3}=40\ \text{N} \]Linkages are assemblies of rigid links joined by pivots (pin joints) to produce a prescribed motion.
Common planar linkages
Linkage terminology
| Term | Definition |
|---|---|
| Ground link | Fixed to the frame; reference point. |
| Input link (driver) | Directly driven by a motor or hand. |
| Coupler link | Transfers motion between input and output. |
| Output link (driven) | Performs the useful work. |
Gears transmit torque between shafts and can change speed, direction, and axis of rotation.
Gear ratio (i)
\[ i = \frac{N_{\text{driven}}}{N_{\text{driver}}}= \frac{D_{\text{driven}}}{D_{\text{driver}}} \]Speed–torque relationship
\[ \omega_{\text{driver}} N_{\text{driver}} = \omega_{\text{driven}} N_{\text{driven}},\qquad \frac{\omega_{\text{driven}}}{\omega_{\text{driver}}}= \frac{1}{i},\qquad \frac{T_{\text{driven}}}{T_{\text{driver}}}= i \]Common gear types
| Gear type | Purpose | Typical application |
|---|---|---|
| Spur | Parallel shafts, simple | Clocks, simple drives |
| Helical | Smoother, higher load | Automotive transmissions |
| Bevel | Change shaft direction (usually 90°) | Differentials |
| Worm | Very high reduction, self‑locking | Conveyor lifts, hoists |
| Rack‑and‑pinion | Rotary → linear | Steering, CNC axes |
Worked example (AO2)
A driver gear (20 teeth) drives a 60‑tooth gear. Determine i, speed reduction and torque multiplication (ignore losses).
\[ i = \frac{60}{20}=3,\qquad \frac{\omega_{\text{driven}}}{\omega_{\text{driver}}}= \frac{1}{3},\qquad \frac{T_{\text{driven}}}{T_{\text{driver}}}=3 \]Pulleys change the direction of a force and, when combined, can multiply force.
Ideal mechanical advantage
\[ \text{MA}= n \quad (n = \text{supporting rope sections}) \]Speed relationship (ideal)
\[ v_{\text{load}} = \frac{v_{\text{effort}}}{\text{MA}} \]Worked example (AO2)
A block & tackle has three supporting rope sections. An effort of 150 N is applied. Find the load that can be lifted (ideal).
\[ \text{MA}=3,\qquad F_{\text{load}} = \text{MA}\times F_{\text{effort}} = 3\times150 = 450\ \text{N} \]A cam converts continuous rotary motion into a prescribed linear motion of a follower.
Key parameters (radial follower)
Displacement equation (simple circular cam)
\[ s(\theta)=r(\theta)-r_b \]Design tip (AO3): Sketch the required lift‑dwell‑return diagram first, then use a cam‑generating template or CAD to obtain the profile that satisfies the required velocity and acceleration limits.
A crank is a lever arm attached to a rotating shaft; together with a connecting rod it produces reciprocating motion.
Definitions
Slider position as a function of crank angle \(\theta\) (assuming a simple slider‑crank)
\[ x = r\cos\theta + \sqrt{L^{2} - r^{2}\sin^{2}\theta} \]Worked example (AO2)
Given \(r = 40\ \text{mm}\) and \(L = 150\ \text{mm}\), find the slider position at \(\theta = 60^{\circ}\).
\[ x = 40\cos60^{\circ} + \sqrt{150^{2} - 40^{2}\sin^{2}60^{\circ}} = 20 + \sqrt{22500 - 1200} \approx 20 + 146 = 166\ \text{mm} \]Although the focus of this document is mechanical, the IGCSE 0445 syllabus also requires knowledge of simple control elements.
Example (AO2): A motor is started by a push‑button (normally open) and stopped by a second push‑button (normally closed). Draw the circuit using a relay coil to keep the motor running after the start button is released.
| System | MA formula (ideal) | Typical range (ideal) |
|---|---|---|
| Lever (Class I & II) | \(\displaystyle \frac{d_{\text{effort}}}{d_{\text{resistance}}}\) | 1 – 10 |
| Gear train | \(\displaystyle \frac{N_{\text{driven}}}{N_{\text{driver}}}\) | 0.1 – 20 |
| Pulley (block & tackle) | \(\displaystyle n\) (supporting rope sections) | 2 – 10+ |
| Cam (radial follower) | \(\displaystyle \frac{\text{Lift}}{r_b}\) (approx.) | 0.5 – 5 |
| Crank (slider‑crank) | \(\displaystyle \frac{2r}{S}=1\) (force conversion depends on dynamics) | 1 (force) – varies with speed |
Question: A Class III lever lifts an 80 N weight. Effort distance = 0.05 m, resistance distance = 0.15 m. Find the effort and the MA.
Solution:
\[ \text{MA}= \frac{0.05}{0.15}= \frac{1}{3},\qquad F_{\text{effort}}= \frac{80}{1/3}=240\ \text{N} \]Question: A motor drives a 12‑tooth pinion which meshes with a 36‑tooth gear. The 36‑tooth gear is fixed to a second 12‑tooth pinion that drives a 48‑tooth output gear. Determine the overall gear ratio, speed reduction, torque multiplication and comment on the suitability for a low‑speed, high‑torque winch.
Solution:
\[ i_1 = \frac{36}{12}=3,\qquad i_2 = \frac{48}{12}=4,\qquad i_{\text{total}} = i_1 i_2 = 12 \] \[ \frac{\omega_{\text{out}}}{\omega_{\text{motor}}}= \frac{1}{12},\qquad \frac{T_{\text{out}}}{T_{\text{motor}}}=12 \]Because the overall reduction is 12:1, the output speed is low and torque is increased twelvefold – ideal for a winch where high pulling force is required at modest speed.
Question: A block & tackle has four supporting rope sections. The operator can apply a maximum effort of 180 N. Assuming a 10 % loss due to rope‑slip, calculate the maximum load that can be lifted.
Solution:
\[ \text{Ideal MA}=4,\qquad \eta = 0.90,\qquad \text{Real MA}=4\times0.90 = 3.6 \] \[ F_{\text{load}} = \text{Real MA}\times F_{\text{effort}} = 3.6\times180 = 648\ \text{N} \]Question: A radial cam must raise a follower 15 mm in 90° of rotation, hold it for a dwell of 180°, then return to the base position in the remaining 90°. Sketch the displacement diagram and describe a simple cam profile that would achieve this motion.
Answer outline (AO3):
Question: A slider‑crank mechanism has a crank radius of 30 mm and a connecting‑rod length of 120 mm. Calculate the slider position when the crank angle is 120°.
Solution:
\[ x = 30\cos120^{\circ} + \sqrt{120^{2} - 30^{2}\sin^{2}120^{\circ}} = 30(-0.5) + \sqrt{14400 - 900(0.75)} = -15 + \sqrt{14400 - 675} = -15 + \sqrt{13725} \approx -15 + 117 = 102\ \text{mm} \]Question: A motor must run in the forward direction when Switch A is pressed, and in reverse when Switch B is pressed. Both switches are normally open. Sketch a simple relay‑based control circuit.
Answer (AO2):
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