Define and use the term 'limit of proportionality' for a load-extension graph and identify this point on the graph (an understanding of the elastic limit is not required)

Cambridge IGCSE Physics 0625 – 1.5 Effects of Forces

Learning Objectives (AO1‑AO3)

• AO1 – Knowledge: Define the limit of proportionality on a load‑extension graph; distinguish scalar and vector quantities; state the basic concepts of force, resultant (collinear), moment, principle of moments, centre of gravity and friction.
• AO2 – Application: Locate the limit of proportionality on a graph; solve simple problems involving forces, moments and stability; use the appropriate formulas.
• AO3 – Analysis & Evaluation: Plan a practical, record data with uncertainties, plot a graph, identify the limit of proportionality and evaluate sources of error.

1. Key Concepts for 1.5 Effects of Forces

2. Limit of Proportionality – Definition & Identification

Definition

The limit of proportionality is the highest load for which the extension of a material remains directly proportional to the load. In this region Hooke’s law holds:

\[

F = kx \qquad\text{(where }k\text{ is the spring constant)}

\]

It is a purely graphical concept; the material may still be elastic beyond the LOP, but Hooke’s law no longer applies.

How to locate the LOP on a load‑extension graph

1. Plot load (\(F\)) on the vertical axis and extension (\(x\)) on the horizontal axis.
2. Draw the best straight line through the first few points that appear linear (the “fit line”).
3. Find the last point that still lies on this straight‑line fit. The next point is the first that deviates.
4. Mark this last point as the limit of proportionality and record its coordinates \((F{\text{LOP}},\,x{\text{LOP}})\).

Worked Example

When the points are plotted, the first six points (0–30 N) lie on a straight line. The point \((30 \text{N},\,0.70 \text{mm})\) is therefore the LOP.

Check the slope (spring constant) in the linear region:

\[

k = \frac{\Delta F}{\Delta x}

= \frac{30\;\text{N} - 0\;\text{N}}{0.70\times10^{-3}\;\text{m}}

\approx 4.3\times10^{4}\;\text{N m}^{-1}.

\]

This value matches the one obtained from \(k = F/x\) for any point within the linear region, confirming that the fit line is correct.

Key Points to Remember

• The LOP is identified graphically – no need to know the elastic limit.
• All points up to and including the LOP obey Hooke’s law; points after the LOP do not.
• In exam questions the LOP is often described as “the point where the straight‑line portion of the graph ends”.

3. Related Core Topics in 1.5 Effects of Forces

3.1 Resultant of Collinear Forces & Translational Equilibrium

• Resultant: \(R = \sum F_i\) (add algebraically when forces act along the same line).
• Equilibrium condition: \(\sum R = 0\).
• Example: A hanging sign is supported by two upward ropes. The sum of the tensions equals the weight of the sign.

3.2 Moment of a Force & Principle of Moments

Moment about a pivot:

\[

M = F \times d \quad (\text{perpendicular distance } d)

\]

Rotational equilibrium:

\[

\sum M{\text{CW}} = \sum M{\text{ACW}}.

\]

Everyday illustration: Opening a door. The force applied at the handle (far from the hinges) produces a larger moment than the same force applied near the hinges.

Seesaw example:

• Child A (30 kg) sits 0.6 m from the pivot.
• Child B (20 kg) sits on the opposite side. Required distance \(d_B\):
  

  \(30g \times 0.6 = 20g \times dB \;\Rightarrow\; dB = 0.9\;\text{m}\).

3.3 Centre of Gravity (CG) & Equilibrium

• CG is the point where the weight of a body may be considered to act.
• For a uniform thin plate, CG coincides with the geometric centre.
• When the CG lies vertically below the base of support, the net moment about any point is zero → stable equilibrium.
• Practical method: Suspend a cut‑out shape from a string; draw a vertical line from the suspension point. Repeat from a different point; the intersection gives the CG.

3.4 Friction (solid & fluid)

• Solid friction: \(F_f = \mu R\) (μ = coefficient of friction, R = normal reaction).
• Fluid drag: \(F_d = \tfrac12 C\rho A v^2\) (C = drag coefficient, ρ = fluid density, A = cross‑sectional area, v = speed).
• Examples: A block sliding on a rough surface; a cyclist feeling air resistance.

4. Practical Skills Checklist (AO3)

1. Planning: State the aim (“To determine the limit of proportionality of a metal wire”), list materials, sketch the set‑up (fixed support, wire, set of masses, ruler).
2. Data collection: Record each load (N) and corresponding extension (mm) with uncertainties (e.g., ±0.01 mm for the ruler, ±0.1 N for the masses).
3. Plotting: Use graph paper or software; label axes with units, draw the best straight‑line fit through the initial linear points.
4. Identification: Locate the last point that still lies on the fit line; mark it clearly as the LOP and note its coordinates.
5. Evaluation: Discuss possible errors such as parallax, mass of the wire, non‑perpendicular loading, and suggest improvements (e.g., use a digital calliper, ensure the wire is vertical).

5. Summary Box

• The limit of proportionality is the highest load for which extension remains directly proportional to load (the linear region of a load‑extension graph).
• Identify it by drawing a straight‑line fit through the early data points and finding the last point that still lies on that line.
• It is a graphical concept distinct from the elastic limit; Hooke’s law ceases to apply beyond the LOP even if the material is still elastic.
• Understanding related ideas—scalar vs. vector, resultant forces, moments, centre of gravity, and friction—ensures you can answer any 1.5 question in the IGCSE exam.