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.
Explanation
(a) Stree = (frac{F}{A}) = (frac{40}{3.14(1.0 times 10^{-3})}^2)
= (1.27 times 10^{7} times 10^7 NM ^{-2})
(b) Strain = (frac{e}{L} = frac{0.24 times 10^{-3}}{3} = 8.0 times 10^{-5})