Question

A wire of length 5.0m and diameter 2.0mm extends by 0.25 mm when a force of 50N was used to stretch it from its end Calculate a. the stress on the wire b. strain in the wire take pi = 3.142

Answer 1

Answer:

L = 5

d = 2*10^-3

r = 10^-3

A = pi r^2 = pi (10^-6)

stress = 50 N / 3.14*10^-6 = 15.9 * 10^6 Newtons/ m^2

strain = delta L/L = .25*10^-3 / 5 = .05 *10^-3 = 50 * 10^-6 m

E = stress / strain

Explanation: