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 (\(\pi = 3.142\))
Verified Correct Answer: .
Physical Principles & Numerical Solution
Stress = \(\frac{F}{A} = \frac{40}{3.14(1 \times 10^{-3})^2} = 12.74 \times 10^{6} Nm^{-2}\)
(b) Strain = \(\frac{e}{I} = \frac{24 \times 10^{-5}}{4} = 6 \times 10^{-5}\)
Source: Myschool Past Questions Archive (Record #54029)