Question

Suppose that we hang a heavy ball with a mass of 10 kg (about 22 lb ) from a steel wire 3m long that is 3 mm in diameter. Steel is very stiff, and Young’s modulus for steel is unusually large, $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{11}\mathbf{N}\mathbf{/}{\mathbf{m}}^2\mathbf{.}$Calculate the stretch$\mathbf{\Delta L}$ of the steel wire. This calculation shows why in many cases it is a very good approximation to pretend that the wire doesn’t stretch at all (“ideal non extensible wire”).

Short answer

$\mathrm{\Delta L}$of steel wire is$=2.079\times {10}^{-4}\mathrm{m}\approx 0.2\mathrm{mm}$, Stretch is very negligible.

Step-by-step solution

Given data

Mass = 10kg

Diameter = 3mm

Length = 3m

Step 2:Solution

Young’s modulus,

$\begin{array}{c}\mathrm{Y}=\frac{\mathrm{F}/\mathrm{A}}{\mathrm{\Delta L}/\mathrm{L}}\\ \Rightarrow \frac{\mathrm{\Delta L}}{\mathrm{L}}=\frac{\mathrm{F}}{\mathrm{AY}}\\ \Rightarrow \mathrm{\Delta L}=\frac{\mathrm{FL}}{\mathrm{AY}}\end{array}$

From equation,

$\begin{array}{l}\mathrm{F}=\mathrm{mg}=10\times 9.8=98\mathrm{N}\\ \mathrm{L}=3\mathrm{m}\\ \end{array}$

A =cross-sectional area =${\mathrm{\pi r}}^2$

$\begin{array}{l}=\mathrm{\pi }\times {(1.5\times {10}^{-3})}^2\\ =7.068\times {10}^{-6}{\mathrm{m}}^2\end{array}$

$\begin{array}{c}\mathrm{Y}=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\\ \mathrm{\Delta L}=\frac{98\times 3}{7.068\times {10}^{-6}\times 2\times {10}^{11}}\\ =2.079\times {10}^{-4}\mathrm{m}\approx 0.2\mathrm{mm}\end{array}$

Stretch is very negligible.