Measuring the Young modulus of a metal wire

By the end of this lesson

By the end you can describe how to measure the Young modulus of a metal wire, name the measuring instruments, explain each precaution, and turn the measurements into a value of $E$ .

The idea in one sentence

To find $E = \dfrac{F L}{A x}$ you need four quantities: the load $F$ , the original length $L$ , the cross-sectional area $A$ , and the extension $x$ .

$F$ and $L$ are easy. The two that need care are $A$ (a wire is thin — the diameter is under a millimetre) and $x$ (the extension is often less than a millimetre, too small to see against a metre rule).

So the experiment is really about measuring a very small extension accurately. That is why the wire is made long and thin, and why its extension is read with a vernier scale rather than a ruler.

The apparatus, and why it is shaped like that

Two wires, not one. The test wire carries the weights; the reference wire carries the same initial load but no extra weights. Both are in the same room, so both expand equally if the temperature changes, and both creep equally. Comparing one against the other removes those effects.
A long, thin wire. A long wire gives a bigger extension for the same strain; a thin wire gives a bigger strain for the same load. Both make the extension easier to measure. It also keeps the load needed comfortably small.
A vernier scale or travelling microscope. Extensions are often 0.5–2 mm, so a ruler is not good enough.
A micrometer screw gauge. The diameter is around 0.2–0.5 mm, and the area depends on its square — so an error in the diameter is doubled in the area.

The method, step by step

Measure the original length $L$ of the test wire between the support and the marker, with a metre rule.
Measure the diameter $d$ of the wire with a micrometer screw gauge in several places and in two perpendicular directions at each place. Average the readings.
Calculate the cross-sectional area, $A = \pi d^2/4$ .
Hang a small initial load on both wires to make them straight and taut, then record the vernier reading.
Add masses one at a time, recording the new vernier reading after each addition. Wait for the reading to settle before reading it.
Continue until you have 6–8 readings, all with small loads so the wire stays within its limit of proportionality.
Find each extension by subtracting the initial reading, then plot a graph of load $F$ against extension $x$ .

Why a graph, not one calculation

You could put one pair of readings into $E = FL/(Ax)$ and be done. A graph is better for three reasons:

It averages out random errors. One dodgy reading barely moves a best-fit line.
It shows whether Hooke's law held. A straight line through the origin confirms the wire was within its limit of proportionality.
The gradient is more precise than any single pair, because it uses the whole range.

The gradient of the $F$ – $x$ graph is $k = EA/L$ , so rearrange:

$E = \frac{k\,L}{A}$

where $k$ is the gradient in N m⁻¹, $L$ in m and $A$ in m².

Worked example: Worked example: from measurements to E

A student uses a steel wire of length 1.80 m and diameter 0.38 mm. A graph of load against extension is a straight line through the origin, and its gradient is $1.26 \times 10^{4}\ \text{N m}^{-1}$ .

Calculate the Young modulus of the steel.

Have a go first?

Worked example: Worked example: why the diameter is measured carefully

A student's diameter reading is 0.38 mm, but the true average diameter is 0.40 mm.

By what factor is the calculated Young modulus wrong, and in which direction?

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

Multiple choiceQuestion 2 · Apply

CalculateQuestion 3 · Stretch

Answer the 3 questions above to continue.