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# Tension (cont.) ## Example A uniform wire of length $L$ and mass $M$ is hanging vertically. Calculate the elongation due to its own weight. ### Solution Let's consider a small element of length $dx$ at a distance $x$ from the lower end. * Tension at this point = weight of the wire below it...

# Tension (cont.) ## Example A uniform wire of length $L$ and mass $M$ is hanging vertically. Calculate the elongation due to its own weight. ### Solution Let's consider a small element of length $dx$ at a distance $x$ from the lower end. * Tension at this point = weight of the wire below it $T = \frac{M}{L}xg$ * Elongation of the small element $dl = \frac{Tdx}{AY} = \frac{Mg}{ALY}xdx$ * Total elongation $\Delta L = \int_{0}^{L}dl = \int_{0}^{L}\frac{Mg}{ALY}xdx = \frac{Mg}{ALY}\frac{L^2}{2} = \frac{MgL}{2AY}$ ## Example A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5} m^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5} m^2$ under a given load. What is the ratio of the Young's modulus of steel to that of copper? ### Solution $\Delta L = \frac{FL}{AY}$ Since $\Delta L$ and $F$ are same for both wires $\frac{L_1}{A_1Y_1} = \frac{L_2}{A_2Y_2}$ $\frac{Y_1}{Y_2} = \frac{L_1A_2}{L_2A_1} = \frac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} = 1.79$