Question

A load of 10kg is suspended by a metal wire 3 m long and having a cross-sectional area $4m{m^2}$. Find
a) The stress
b) The strain
c) The elongation
(Young modulus of the metal is $2.0 \times {10^{11}}N{m^{ - 2}}$ )

Answer 1

Hint: Stress is the ratio of Force and Area. Find the force as force is the product of mass and acceleration due to gravity. Young modulus is the ratio of stress and strain. Substitute the value obtained stress and given young modulus to find the value of strain. Elongation is the product of strain and length of the wire. Substitute the values of obtained strain and length of the wire to calculate the elongation.

Complete step by step answer:
We are given a load of 10kg is suspended by a metal wire 3 m long and having a cross-sectional area $4m{m^2}$
We have to find its stress, strain and elongation of the wire.
a) Stress
$Stress = \dfrac{{Force}}{{Area}}$
According to Newton’s second law, Force is the product of mass and acceleration due to gravity.
$
  g = 10m/{s^2} \\
  m = 10kg \\
  F = m \times g \\
  \Rightarrow F = 10 \times 10 \\
  \Rightarrow F = 100N \\
 $
Cross sectional Area of the metallic wire is $4m{m^2}$
$
  Stress = \dfrac{{Force}}{{Area}} = \dfrac{F}{A} \\
  F = 100N, A = 4m{m^2} \\
  A = 4 \times {10^{ - 6}}{m^2} \\
  \Rightarrow Stress = \dfrac{{100N}}{{4 \times {{10}^{ - 6}}{m^2}}} \\
  \Rightarrow Stress = 25 \times {10^6}N/{m^2} = 2.5 \times {10^7}N/{m^2} \\
 $
b) Strain
Young modulus is a constant and is the ratio of stress and strain.
Given value of young modulus is $2.0 \times {10^{11}}N{m^{ - 2}}$
Stress is $2.5 \times {10^7}N/{m^2}$
$
  Y = \dfrac{{Stress}}{{Strain}} \\
  Strain = \dfrac{{Stress}}{Y} \\
  \Rightarrow Strain = \dfrac{{2.5 \times {{10}^7}N/{m^2}}}{{2.0 \times {{10}^{11}}N/{m^2}}} \\
  \Rightarrow Strain = 1.25 \times {10^{ - 4}} \\
 $
c) Elongation
Strain is the ratio of the elongation in the wire and actual length of the wire.
Elongation (E) will be the product of Strain and length of the wire.
$
  Strain = \dfrac{E}{L} \\
  E = Strain \times L \\
  L = 3m, Strain = 1.25 \times {10^{ - 4}} \\
  \Rightarrow E = 1.25 \times {10^{ - 4}} \times 3m \\
  \Rightarrow E = 3.75 \times {10^{ - 4}}m \\
 $

Note:Strain is defined as deformation of a solid due to stress. It is a ratio of length and length, so there are no units for strain. Elongation is a measure of deformation that occurs before a material eventually breaks when subjected to a stress.