A composite heavy rope made of two materials | vedclass

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(C) The stress in a rope at any point is given by $	ext{Stress} = \frac{F}{A}$,where $F$ is the tension at that point.For the lower rope,the maximum

Vedclass · Accepted Answer

$\frac{5}{6}$

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(C) The stress in a rope at any point is given by $	ext{Stress} = \frac{F}{A}$,where $F$ is the tension at that point.For the lower rope,the maximum stress occurs at its top,where it supports its own weight: $\sigma_l = \frac{m_l g}{A_l} = \frac{d_l A_l L_l g}{A_l} = d_l L_l g$.For the upper rope,the maximum stress occurs at its top,where it supports the weight of both the upper and lower ropes: $\sigma_u = \frac{(m_u + m_l) g}{A_u} = \frac{(d_u A_u L_u + d_l A_l L_l) g}{A_u} = (d_u L_u + d_l L_l \frac{A_l}{A_u}) g$.Now,calculating the ratio $\frac{\sigma_u}{\sigma_l}$:$\frac{\sigma_u}{\sigma_l} = \frac{d_u L_u + d_l L_l (A_l / A_u)}{d_l L_l} = \frac{d_u}{d_l} \cdot \frac{L_u}{L_l} + \frac{A_l}{A_u} = (\frac{2}{3} \cdot \frac{1}{2}) + \frac{1}{2} = \frac{1}{3} + \frac{1}{2} = \frac{2+3}{6} = \frac{5}{6}$.

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