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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.
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Solution
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)
\[\therefore \frac{F}{A} = \frac{100}{4 \times {10}^{- 6}}\]
\[ = 2 . 5 \times {10}^7 \text{ N/ m }^2\]
(b) Strain = \[\frac{\Delta L}{L}\]
Or,
\[\text{ Strain } = \frac{\text{ Stress }}{Y}\]
\[\text{ Strain }= \frac{2 . 5 \times {10}^7}{2 \times {10}^{11}}\]
\[ = 1 . 25 \times {10}^{- 4} \text{ N/ m}^2\]
(c) Let the elongation in the wire be \[∆ L\] .
\[\text { Strain} = \frac{\Delta L}{L}\]
\[ \Rightarrow \Delta L = \left( \text{ Strain } \right) \times L\]
\[ = 1 . 25 \times {10}^{- 4} \times 3\]
\[ = 3 . 75 \times {10}^{- 4} \text{ m}\]
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