A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the; (a) stress on the wir...
Question 1 Report
A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;
(a) stress on the wire;
(b) strain in, the wire. [\(\pi = 3.142\)]
Answer Details
Length \(L=5.0\,\text{m}\), diameter \(=2.0\,\text{mm}\Rightarrow r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}\), extension \(e=0.25\,\text{mm}=0.25\times10^{-3}\,\text{m}\), force \(F=50\,\text{N}\).
(a) Stress \(=\dfrac{F}{A}\), where \(A=\pi r^2\).
\[ A=3.142\times(1.0\times10^{-3})^2=3.142\times10^{-6}\,\text{m}^2 \]\[ \text{Stress}=\frac{50}{3.142\times10^{-6}}=1.59\times10^{7}\,\text{Nm}^{-2} \](b) Strain \(=\dfrac{\text{extension}}{\text{original length}}\).
\[ \text{Strain}=\frac{0.25\times10^{-3}}{5.0}=5.0\times10^{-5} \](Strain has no unit as it is a ratio of two lengths.)
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