Question

One end of a nylonrope of length 4.5m and diameter 6mm is fixed to a free limb.A monkey weighing 100 Newtonjumps to catch the free end and says there.Find the elongation of the rope and corresponding change in diameter. Given Young's modulus of nylon=4.8*10raised to 11

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{state}\mathrm{of}\mathrm{equilibrium},\mathrm{the}\mathrm{tension}\mathrm{in}\mathrm{the}\mathrm{rope}\mathrm{equals}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{monkey}. \\ \mathrm{Y}=\frac{\mathrm{Stress}}{\mathrm{strain}} \\ =\frac{\mathrm{T}/\mathrm{A}}{\mathrm{l}/\mathrm{L}}=\frac{\mathrm{TL}}{\mathrm{Al}} \\ \mathrm{or}\mathrm{l}=\frac{\mathrm{TL}}{\mathrm{AY}} \\ \mathrm{Elongation},\mathrm{l}=\frac{100\times 4.5}{{\mathrm{\pi r}}^2\times 4.8\times {10}^{11}} \\ =\frac{100\times 4.5}{\mathrm{\pi }({\displaystyle \frac{6\times {10}^{-3}}{2}}{)}^2\times 4.8\times {10}^{11}} \\ \mathbf{l}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{32}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{m} \\ \mathrm{Poisson}\mathrm{ratio}=\frac{∆\mathrm{d}/\mathrm{d}}{\mathrm{l}/\mathrm{L}}=0.2(\mathrm{For}\mathrm{nylon}) \\ 0.2=\frac{∆\mathrm{d}\times 4.5}{3.32\times {10}^{-5}\times 6\times {10}^{-3}} \\ \mathrm{or}∆\mathrm{d}=\frac{3.32\times {10}^{-5}\times 6\times {10}^{-3}\times 0.2}{4.5} \\ \mathbf{∆}\mathbf{d}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{m} \end{gathered}$$