Answer:3.51
Explanation:
Given
Coefficient of Friction [tex]\mu =0.4 [/tex]
Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]
Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]
[tex]N=T\cdot d\theta [/tex]
Friction [tex]f=\mu \times Normal\ Reaction[/tex]
[tex]f=\mu \cdot N[/tex]
and [tex]T+dT-T=f=\mu Td\theta [/tex]
[tex]dT=\mu Td\theta [/tex]
[tex]\frac{dT}{T}=\mu d\theta [/tex]
[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]
[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]
[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]
[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]
[tex]\frac{T_2}{T_1}=3.51[/tex]
