Question

(II) A steel wire 2.3 mm in diameter stretches by 0.030% when a mass is suspended from it. How large is the mass?

Short answer

The mass of the steel wire is 25 kg.

Step-by-step solution

Young’s modulus

When a force (F) is applied to an object made of a certain material, the length of the object changes, this change in length is proportional to the original length (\({l_0}\)) and inversely proportional to the cross-sectional area (A), i.e., \(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\).

Here, E is the constant of proportionality, termed the elastic modulus or Young’s modulus of the material.

In this problem, the change in length of the steel wire is related to force F and Young’s modulus of the steel by the above expression. The value of Young’s modulus of steel is\(E{\bf{ = 200 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

Given information

The diameter of the steel wire is\(d = 2.3\;{\rm{mm}} = 2.3 \times {10^{ - 3}}\;{\rm{m}}\).

If\({l_0}\)is the original length of the iron wire and\(\Delta l\)is the change in length, the percentage change in the length of the wire is as follows:

\(\begin{array}{c}\frac{{\Delta l}}{{{l_0}}} \times 100 = 0.030\\\frac{{\Delta l}}{{{l_0}}} = 0.030 \times {10^{ - 2}}\end{array}\)

Since the wire gets stretched due to the mass suspended from it, the force acting on the wire is the weight of the mass. Thus, ifm is the suspended mass, then the force is

\(F = mg\).

Determination of cross-sectional area of the steel wire

The area of the cross-section of the steel wire is as follows:

\(\begin{array}{c}A = \pi {\left( {\frac{d}{2}} \right)^2}\\ = 3.14 \times {\left( {\frac{{2.3 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}}}{2}} \right)^2}\\ = 4.153 \times 1{{\rm{0}}^{ - 6}}\;{{\rm{m}}^2}\end{array}\)

Determination of the mass of the steel wire

The change in length of the steel wire due to mass suspended from it is given by the following:

\(\begin{array}{c}\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\\ = \frac{1}{E}\frac{{mg}}{A}{l_0}\end{array}\)

Thus, the expression for mass can be written as follows:

\(\begin{array}{c}m = \frac{{EA}}{g} \times \frac{{\Delta l}}{{{l_0}}}\\ = \frac{{\left( {200 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^2}} \right)\left( {4.153 \times 1{{\rm{0}}^{ - 6}}\;{{\rm{m}}^2}} \right)}}{{\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}} \times \left( {0.030 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)\\ = 25.43\;{\rm{kg}}\\ \approx 25\;{\rm{kg}}\end{array}\)

Thus, the mass of the steel wire is 25 kg.