Suppose we consider friction between string and the pulley while still considering the string to be massless. So, in such a theoretical case, the tension in the string in contant with the pulley will not be constant and its variation is calculated as follows.
If the string is just on the verge of slipping on the pulley, then friction acting is limiting and if `T_(2) gt T_(1)` then friction will act towards `T_(1)` in tangential direction as shown.

`d N = (T + df) sin (d theta//2) + (T + dT) sin (d theta//2)`
`dN = T sin (d theta//2) + df sin (d del//2) + T sin (d theta//2) + d T sin (d theta//2)`
`dN ~~ 2T sin (d theta//2)`
`d N ~~ 2T (d theta)/(2)`
`d N = T d theta` ........(1)
Alsso, `(T + df) cos ((d theta)/(2)) = (T + d T) cos ((d theta)/(2))`
`T cos ((d theta)/(2)) + df cos ((d theta)/(2)) = T cos ((d theta)/(2)) 1 dT cos ((d theta)/(2))`
`df = dT = mu d N`.........(2)
from (1) & (2)
`dT = mu T d theta`
`int_(T_(1))^(T_(2)) (dT)/(T) = mu int_(0)^(theta) d theta`
`ln ((T_(2))/(T_(1))) = mu theta`
`T_(2) = T_(1) e^(mu theta)`, where `theta` is the angle of contact between the string and the pulley Based on above information, answer the following questions.
Two masses `m_(1) kg` and `m_(2) kg` passes over an atwood machine. Find the ratio of masses `m_(1)` and `m_(2)` so that string passing over the pulley will just start slipping over its surface. The friction coefficient between the string and pulley surface is 0.2.

Options

A. `e^(0.2 pi)`
B. `e^(0.2)`
C. `0.2`
D. `e^(0.4 pi)`

Answer

A

Explanation

For a simple at wood's machine we can write for the two masses
`m_(1) g - T_(1) = m_(1)a`
and `T_(2) - m_(2) g = g = m_(2) a`
and if string starts slipping, we have
`T_(1) = T_(2) e^(mu pi)`
or `m_(1) (g - a) = m_(2) (g + a) e^(mu pi)`
If sting just slips. we can use `a = 0`, thus we have
`(m_(1))/(m_(2)) = e^(0.2 pi)`