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. enter image description here

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!


1 Answer 1


The dowel exerts a friction force on the rope that reduces the tension as we go around it. For example if the friction was high, someone could pull hard on the rope, but there would be a lower tension at the other side.

enter image description here

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 $dT = \mu T d\theta$


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