Why Euler's rope friction formula cannot solve a massless string on a pulley

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Amin2014
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Consider a massless string which can rotate about a fixed pulley (first picture). The coefficient of static friction is μ. Assuming that the motion is impending, the goal is to find the equation that describes the variation in tension of the string.
( T2/T1 = eμΦ where Φ is the subtended angle.)

The usual method of solving this problem involves writing Newton's law for an infinitesimal element of the string and then integrating. In the second picture I've provided an extract from a different problem. Why can't we apply the solution provided in the second picture to the original problem stated above?
 
Attachments
  • Diagram of a fixed pulley hanging from a horizontal support, with a string passing over it and two vertical tension arrows labelled T1 on the left and T2 on the right, with the note T2 > T1 below.
    pulley.jpg
    5.7 KB · Views: 459
  • Solution extract from a textbook sample problem on a pulley of diameter 4 in. rotating about a shaft of diameter 2 in., showing a free-body diagram with load W = 500 lb and force P, and the moment equation giving P = 611 lb.
    Alternative Solution.png
    19.3 KB · Views: 528

Discussion

Here's my own solution:
N Cos β =μ N Sinβ therefore tan β = 1/μ
Writing Σ M = 0 about the point of application of R and taking the radius of the pulley to be r we have :
T2 ( r - rCos β) = T1 (r + r Cosβ)
dividing by r and rearranging:
T2 = T1 (1 + Cosβ) / ( 1 - Cos β)
Which yields a different relation between T1 and T2 than the method of integration.
 
Attachments
  • Free-body diagram of a pulley with a massless string: forces R upward, N and mu N at angle beta at the contact point, and tensions T1 and T2 pointing down, with the note T2 > T1.
    My solution.png
    4.6 KB · Views: 431
Amin2014 said:
Consider a massless string which can rotate about a fixed pulley (first picture). The coefficient of static friction is μ. Assuming that the motion is impending, the goal is to find the equation that describes the variation in tension of the string.
( T2/T1 = eμΦ where Φ is the subtended angle.)

The usual method of solving this problem involves writing Newton's law for an infinitesimal element of the string and then integrating. In the second picture I've provided an extract from a different problem. Why can't we apply the solution provided in the second picture to the original problem stated above?
Click on "alternative Solution.png", for better image quality.