A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire ar...
Question 1 Report
A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.
Answer Details
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
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