A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young's modulus 7.0 x 10\(^{11}\) Nm\(^{-2}\). ...
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A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young's modulus 7.0 x 10\(^{11}\) Nm\(^{-2}\). Determine the extension produced. [g = 10 ms\(^{-2}\); \(\pi\) = 3.142]
Answer Details
The extension of a stretched wire is given by \(e = \dfrac{FL}{AY}\).
Data: mass \(= 11.0\,\text{kg}\), so \(F = mg = 11.0\times10 = 110\,\text{N}\); length \(L = 2.0\,\text{m}\); diameter \(= 2.0\,\text{mm}\), so radius \(r = 1.0\times10^{-3}\,\text{m}\); \(Y = 7.0\times10^{11}\,\text{Nm}^{-2}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.0\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \]Extension:
\[ e = \frac{FL}{AY} = \frac{110\times2.0}{(3.142\times10^{-6})(7.0\times10^{11})} \] \[ e = \frac{220}{2.199\times10^{6}} = 1.0\times10^{-4}\,\text{m} = 0.1\,\text{mm} \]Download The App On Google Playstore
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