A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the; (a) stress on the wire; (b) ...
Question 1 Report
A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;
(a) stress on the wire;
(b) strain in the wire (\(\pi = 3.142\))
Answer Details
Given: F = 40 N, L = 4 m, diameter d = 2.00 mm = \(2.00\times10^{-3}\) m, extension e = 0.24 mm = \(0.24\times10^{-3}\) m, \(\pi = 3.142\).
(a) Stress on the wire
Cross-sectional area of the wire: \[ A = \frac{\pi d^2}{4} = \frac{3.142 \times (2.00\times10^{-3})^2}{4} = \frac{3.142 \times 4.00\times10^{-6}}{4} = 3.142\times10^{-6}\ \text{m}^2 \]
Stress: \[ \sigma = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \]
(b) Strain in the wire
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \]
Strain has no unit (it is a ratio).
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