A thick metal wire of density rho and length | vedclass

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(B) Consider an element of length $dx$ at a distance $x$ from the free end of the wire.The weight of the portion of the wire below this element is $dw

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$\frac{1}{2} \frac{\rho g L^2}{Y}$

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(B) Consider an element of length $dx$ at a distance $x$ from the free end of the wire.The weight of the portion of the wire below this element is $dw = (A \cdot x \cdot \rho) g$,where $A$ is the cross-sectional area.The stress at this section is $\sigma = \frac{dw}{A} = \rho g x$.The strain is $\frac{d(\Delta l)}{dx} = \frac{\sigma}{Y} = \frac{\rho g x}{Y}$.Integrating from $x = 0$ to $x = L$ to find the total elongation $\Delta L$:$\Delta L = \int_{0}^{L} \frac{\rho g x}{Y} dx = \frac{\rho g}{Y} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{\rho g L^2}{2 Y}$.

$A$ thick metal wire of density $\rho$ and length $L$ is hung from a rigid support. The increase in length of the wire due to its own weight is ($Y =$ Young's modulus of the material of the wire).

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