(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(Halliday, Resnick, Krane - Physics, Fifth Edition. Chapter 5, Problem 11.)

A massless rope is tossed over a wooden dowel of radius [itex]r[/itex] in order to lift a heavy object of weight [itex]W[/itex] off of the floor. The coefficient of sliding friction between the rope and the dowel is [itex]\mu[/itex]. Show that the minimum downward pull on the rope necessary to lift the object is

[tex]F_{\text{down}} = We^{\pi\mu}[/tex].

2. Relevant equations

[tex]\Sigma \vec{F} = m \vec{a}[/tex]

[tex]f = \mu N[/tex]

3. The attempt at a solution

Note that in order to lift the object, the magnitude of the tension [itex]T[/itex] in the rope must be more than or equal to the weight of the object. (That is, [itex]T \geq W[/itex].) So the rope is pulled on one end by a force of magnitude [itex]T[/itex] and on the other by a force of magnitude [itex]F_{\text{down}}[/itex]. The dowel exerts a normal force [itex]\vec{N}[/itex] on the rope, and the magnitude of the friction between the rope and the dowel is given by [itex]f = \mu N[/itex].

The problem is that I have no idea how to deal with this normal force. If I draw a diagram where [itex]\vec{f}[/itex] opposes the motion of the rope, I end up with [itex]\vec{f}[/itex] and [itex]\vec{T}[/itex] pointing in the opposite direction as [itex]\vec{F}_{\text{down}}[/itex], but [itex]\vec{N}[/itex] is perpendicular to all of those forces. The rope is obviously not moving in the direction of [itex]\vec{N}[/itex], so it seems that some unknown force is balancing the normal force out.

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# Dynamics - rope attached to an object

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