Question

A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm². Find (a) the stress (b) the strain and (c) the elongation. Young modulus of the metal is 2.0 × 10¹¹ N m⁻².

Answer 1

Given:
Mass of the load (m) = 10 kg
Length of wire (L) = 3 m
Area of cross-section of the wire (A) = 4 mm² = 4.0 × 10⁻⁶ m²
Young's modulus of the metal Y = 2.0 × 10¹¹ N m⁻²

(a) Stress = F/A

F = mg
= $10\times 10$ = 100 N (g = 10 m/s²)
$$\begin{gathered} \therefore \frac{F}{A}=\frac{100}{4\times {10}^{-6}} \\ =2.5\times {10}^7\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Strain = $\frac{\Delta L}{L}$
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Or, $\mathrm{Strain}=\frac{\mathrm{Stress}}{Y}$

$$\begin{gathered} \mathrm{Strain}=\frac{2.5\times {10}^7}{2\times {10}^{11}} \\ =1.25\times {10}^{-4}\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(c) Let the elongation in the wire be $∆L$.
$$\begin{gathered} \mathrm{Strain}=\frac{\Delta L}{L} \\ \Rightarrow \mathrm{\Delta }L=\left(\mathrm{Strain}\right)\times L \\ =1.25\times {10}^{-4}\times 3 \\ =3.75\times {10}^{-4}\mathrm{m} \end{gathered}$$