Force required when there is Friction between Pulley and String?

A massless rope is tossed over a wooden dowel of radius $$r$$ in order to lift a heavy object of weight $$W$$ off of the floor, as shown.

The coefficient of sliding friction between the rope and the dowel is $$μ$$. Show that the minimum down- ward pull on the rope necessary to lift the object is:

$$F=We^{\pi μ}$$

This was a question in Resnick Halliday Krane Ch-5. When I tried this problem, I was not able to make any progress whatsoever. Like I was not able to understand why the friction would act/how it would act as the rope is massless! So basically I still tried for a long time but I was not able to solve this.

So I looked at the solution, which unfortunately made no sense to me at all.

They somehow got the equation $$\frac{dT}{T}=μ \cdot d\theta$$ (where T=Tension).

Obviously I understand that once we set up this equation, the rest is trivial, as you just integrate this expression with the correct limits ($$0$$ to $$\pi$$ for $$d\theta$$ and $$F_{down}$$ to $$W$$ for $$dT$$). But I tried for a long time to understand how they got that equation but I absolutely do not understand it. Like I am not even sure how to begin with that.

This was the only problem in that chapter that I still don't understand the solution to (infact I had done this chapter 1 month back and I thought that if I try the problem again after some time, I might get it, but I tried it today and I still do not understand the solution). So I would be realllllly very grateful is someone could please explain the solution to this problem (or even give hints!), really anything would be appreciated. Thanks!

The $$dT = T d\theta$$ above is the normal contact force of the rope on the dowel, then if we include the $$\mu$$ we get the force that reduces the tension for that section, so $$dT = \mu T d\theta$$