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Tension on a Rope Deflected by a Pulley: Differentials

  1. May 19, 2015 #1
    Hi all, first post here. I'm a junior Physics/Math double major at UMass Amherst, playing with some problems over the summer. I'll get right into it.

    A rope with constant tension T is deflected through the angle [itex] 2\theta_{0} [/itex] by a smooth, fixed pulley. What is the force on the pulley?

    It is easy to see the answer from geometrical considerations, but the exercise is to find the total force by integrating the force exerted by each small segment of the string. Below is a diagram, showing my use of variables:

    https://dl.dropboxusercontent.com/u/2223067/Tension-pulley-diagram-new.jpg [Broken]

    This shows our approach: since the rope is in equilibrium, the force on a small segment must be cancelled exactly by a force provided by the pulley. Let the force provided by the pulley on a small segment of rope between [itex] \theta[/itex] and [itex]\theta + \Delta\theta[/itex] be [itex]\Delta F[/itex].

    Then, from trigonometry, and an application of Newton's Second Law to the small segment, we have: [itex]\Delta F - 2T\sin\frac{\Delta\theta}{2}=0[/itex].

    Using [itex]\sin\theta \approx \theta[/itex] for small theta, we have: [itex]\Delta F = T \Delta\theta[/itex]. We conclude that a small rope segment exerts a force [itex]T \Delta\theta[/itex] inwards on the pulley.

    We need to sum these forces over each segment of rope in contact with the pulley. Since these forces are not parallel, we need to perform a vector sum. Symmetry convinces us that the resultant force will be in the positive x direction, or "to the right" in terms of the diagrams I've drawn.

    So, we write: [itex] \Delta F_{x} = T\Delta\theta\cos \theta [/itex], which gives us the x component of the force exerted by each rope segment.

    Now that you're in this deep, consider this: my book, K&K's An Introduction to Mechanics claims that if we take the limit [itex] \Delta\theta \to 0[/itex], we end up with [itex] dF_{x}=T\cos\theta \, d\theta [/itex].

    I have a problem with this. In my mind, [itex]dF_{x}[/itex] and [itex]d\theta[/itex] can be related in only one way: through the use of the symbol [tex] \frac{dF_{x}}{d\theta} \equiv \lim_{\Delta\theta \to 0} \frac{F_{x}(\theta + \Delta\theta) - F_{x}(\theta)}{\Delta\theta} [/tex]

    All of the dividing by differentials, or separation of variables nonsense must happen after this fractional differential symbol has been introduced according to its definition.

    This is a valid construction if and only if [itex] \Delta F_{x} = F_{x}(\theta + \Delta\theta) - F_{x}(\theta) [/itex], but I see no way in which this can be true. We defined [itex] \Delta F_{x} [/itex] to be the outward force on a segment of rope due to the pulley. I see no meaning to the outward force on a single point of rope due to the pulley. Those single points are [itex]\theta + \Delta\theta[/itex] and [itex] \theta[/itex], the difference of the forces exerted on which should give us [itex]\Delta F_{x}[/itex].

    I really doubt K&K is being sloppy, so I think I'm about to learn something about how differentials work. Can you see it?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 19, 2015 #2

    haruspex

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    ##\Delta F_x(\theta)## was defined as the net force in the X direction on a section of rope of arc ##\Delta \theta## starting at angle ##\theta## from the midline.
    Note that really that also makes it a function of ##\Delta \theta##, but that is left as implicit.
    Thus, ##\Delta F_x(\theta+\Delta \theta)## means the net force in the X direction on a section of rope of arc ##\Delta \theta## starting at angle ##\theta+\Delta \theta## from the midline.
     
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