A piece of copper having a rectangular cross-section of $15.2 \;mm \times 19.1 \;mm$ is pulled in tenston with $44,500\; N$ force, productng only elastic deformation. Calculate the resulting strain?
Area of the copper piece:
$A=l \times b=19.1 \times 10^{-3} \times 15.2 \times 10^{-3}=2.9 \times 10^{-4} m ^{2}$
Tension force applied on the piece of copper, $F=44500 N$
Modulus of elasticity of copper, $\eta=42 \times 10^{9} N / m ^{2}$
Modulus of elasticity, $\eta=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}$
$\therefore$ Strain $=\frac{F}{A \eta}$
$=\frac{44500}{2.9 \times 10^{-4} \times 42 \times 10^{9}}$
$=3.65 \times 10^{-3}$
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A wire of cross-sectional area $3\,m{m^2}$ is first stretched between two fixed points at a temperature of $20°C$. Determine the tension when the temperature falls to $10°C$. Coefficient of linear expansion $\alpha = {10^{ - 5}} { ^\circ}{C^{ - 1}}$ and $Y = 2 \times {10^{11}}\,N/{m^2}$ ........ $N$
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The Young's modulus of a steel wire of length $6\,m$ and cross-sectional area $3\,mm ^2$, is $2 \times 11^{11}\,N / m ^2$. The wire is suspended from its support on a given planet. A block of mass $4\,kg$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth. The elongation of wire is (Take $g$ on the earth $=10$ $\left.m / s ^2\right):$
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