Question

One end of steel wire is fixed to a ceiling and a load of $2.5Kg$ is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of $2.0Kg$., is attached to the lower end of this wire, compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is ${10}^{-4}{m}^2$.

Answer 1

Here, $m_1=2.5kg$ and $m_2=2kg$

Young's modulus $Y=\frac{stress}{longitudinalstrain}$

[$Y_{steel}=20\times {10}^{10}N/m^2$]

For wire1, $stres{s}_1=\frac{F_1}{A}=\frac{(2.5+2)g}{{10}^{-4}}$

Thus, $Longitudinalstrai{n}_1=\frac{stres{s}_1}{Y}=\frac{4.5\times 9.8}{20\times {10}^{10}\times {10}^{-4}}=2.2\times {10}^{-6}$

For wire2, $stres{s}_2=\frac{F_2}{A}=\frac{2g}{{10}^{-4}}$

Thus, $Longitudinalstrai{n}_2=\frac{stres{s}_2}{Y}=\frac{2\times 9.8}{20\times {10}^{10}\times {10}^{-4}}=9.8\times {10}^{-7}$