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Dynamics - rope attached to an object

  1. Jul 13, 2010 #1
    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.
  2. jcsd
  3. Jul 13, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: Analyze forces acting on a small segment of the rope. You'll need to set up a (simple) differential equation and integrate to find your answer.
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