A force of 40 N applied at the end of a wire of length 4m
PHYSICS
WAEC 2003
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 (\(\pi = 3.142\))
Explanation
Stress = \(\frac{F}{A} = \frac{40}{3.14(1 \times 10^{-3})^2} = 12.74 \times 10^{6} Nm^{-2}\)
(b) Strain = \(\frac{e}{I} = \frac{24 \times 10^{-5}}{4} = 6 \times 10^{-5}\)
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