Chapter 9, Problem 46P

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

To determine

Mass of the steel wire.

Answer to Problem 46P

Solution: $\text{25}\text{.4 kg}$

Explanation of Solution

Given:

Diameter of the wire, $\text{d }=\text{ 2}.\text{3 mm }=\text{2}.\text{3 }\times \text{ 1}{0}^{-\text{3}}\text{m}$

Elongation of wire, $0.0\text{3}0\%\times \text{3 }\times \text{ 1}{0}^{-\text{4}}\text{mm }=\text{3 }\times \text{ 1}{0}^{-\text{7}}\text{m}$

Young’s modulus of the marble, $\text{2}00\times \text{ 1}{0}^{\text{9}}\text{N}/{\text{m}}^{\text{2}}$

Formula Used:

  $\frac{\sigma }{\epsilon }=\text{E}$

Where, $\text{ε}$ is the strain, E is modulus of elasticity

  $\sigma =\frac{F}{A}$

Where, $\sigma$ is the stress, F is the applied force, and A is the cross-sectional area.

Calculations:

  $\begin{array}{l}\frac{\sigma }{\epsilon }=E\\ \sigma =\epsilon E\\ \sigma =\frac{0.0003}{1}\times 200\times 10^9\\ \sigma =60\times 10^{6}N/m^2\end{array}$

To find the mass

  $\begin{array}{l}\sigma =\frac{F}{A}\\ \sigma =\frac{m\times g}{A}\\ m=\frac{\sigma A}{g}\\ m=\frac{60\times 10^6\times \pi {(2.3\times 10^{-3})}^2/4}{9.8}\end{array}$

  $m=\text{25}\text{.4 kg}$

Conclusion: Mass of the steel wire is $\text{25}\text{.4 kg}$ .