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Homework Help: Deriving the captstan equation

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A device called a capstan is used aboard ships in order to control a rope which is under great tension. The rope is wrapped around a fixed drum. The load on the rope pulls it with a force ##T_A## and the sailor holds it with a much smaller force ##T_B##. Show that ##T_A = T_B e^{\mu \theta}##, where ##\mu## is the coefficient of friction, and ##\theta## is the total angle subtended by the rope on the drum.

    2. Relevant equations
    Newton's second law

    3. The attempt at a solution

    So I think that I can get a solution. That is, I think that I can look at an infinitesimal section of the rope, apply Newton's second law in the horizontal and vertical directions, take the limit as ##\theta## goes to zero, and then integrate. However, I have some questions about which forces will be acting on the rope. My first question is, is the rope massless? If so, how would friction and a normal force be acting on the rope? Also, if the rope is massless, wouldn't the tension in the rope be uniform?
     
  2. jcsd
  3. Sep 18, 2016 #2

    kuruman

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    The rope is not massless. The normal force and friction increase as you move from the low tension to the high tension end.
     
  4. Sep 18, 2016 #3
    So if the rope is not massless, why don't we account for its weight?
     
  5. Sep 18, 2016 #4

    kuruman

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    The weight of the rope is negligible compared against the normal force.
     
  6. Sep 18, 2016 #5
    Is the normal force much greater than the weight because of the two tensions pulling down?
     
  7. Sep 18, 2016 #6

    kuruman

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    Yes, and the tensions increase as you go around the wrapped rope.
     
  8. Sep 18, 2016 #7
    Okay that makes sense. But how is the rope is equilibrium if the two forces on each side of the rope are so different? Shouldn't the rope accelerate in the direction of the tension caused by the load?
     
  9. Sep 18, 2016 #8

    kuruman

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    The maximum force of static friction increases as you move around the wrapped rope because the normal force increases. Derive the capstan equation including the dependence of the normal force on the angle and you will see how this works.
     
  10. Sep 18, 2016 #9
    Alright , that makes sense. Another question. After deriving that ##\Delta T = T \mu \Delta \theta##, we can have that ##\displaystyle \int \frac{1}{T} dT = \int_0^\theta \mu d \theta## However, I am not sure what the bounds of integration should be on the left integral. I think that it might look like ##\displaystyle \int_{T_B}^{T_A}##, but am not sure why this would be the case.
     
    Last edited: Sep 18, 2016
  11. Sep 18, 2016 #10

    kuruman

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    What is the tension when the angle is zero? That's the lower limit. What is the tension when the angle is θ? That's the upper limit.
     
  12. Sep 18, 2016 #11

    Charles Link

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    One thing you are needing, and it would help to show you with a diagram is the normal force per unit length from a rope around a circle is given by ## f_L=T/r ##. A derivation is done using a segment ## \Delta s ## where the angle changes by ## \Delta \theta/2 ## on each side. Meanwhile, the tension changes because of the frictional force which comes from the normal force. I see from a previous post that perhaps you already have the equation ## f_L=T/r ## or a variation of it in terms of ## \theta ##.
     
  13. Sep 18, 2016 #12
    Why would ##T_B## be the tension when the angle is zero, and ##T_A## be the tension when the angle is ##\theta##?
     
  14. Sep 18, 2016 #13

    kuruman

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    The capstan doesn't know the alphabet. Like I said, the lower limit of the integral on the left matches the lower limit of the integral on the right and similarly for the upper limits of the integrals.
     
  15. Sep 18, 2016 #14

    Charles Link

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    I think in post #9 your differential equation for ## \Delta T ## needs a minus sign. (The tension decreases due to the frictional force the more the thing gets wrapped around). Also you have an extra ## \theta ## next to the ## \mu ## that shouldn't be there. The ## T_B ## is the force from the sailor, (should be the upper limit of the ## T ## integral ), and ## T_A ## is the force from the sail. (needed to edit because I read the original problem more carefully.)(equation should read ## \Delta T=-\mu T \Delta \theta ##).
     
  16. Sep 18, 2016 #15
    I think I corrected the equation; also, I don't see why there would need to be a minus sign.

    However, I am just not seeing how the force on the sailor and the force that the sailor supplies would correspond to when ##\theta## is zero and when ##\theta## has a value.
     
  17. Sep 18, 2016 #16

    Charles Link

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    Please read my edited version of post #14.
     
  18. Sep 18, 2016 #17

    kuruman

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    If we agree that TA is the larger force, then TA = TBeμθ. This can be inverted to give TB = TAe-μθ, but the latter is not the traditional capstan equation. Either equation gives TA = TB when the angle is zero.
     
  19. Sep 18, 2016 #18

    Charles Link

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    I think the derivation will give (from the differential equation ) ## \ ## ## T_B=T_A e^{- \mu \theta} ##. A little algebra gives the textbook result.
     
  20. Sep 18, 2016 #19
    So why does ##T_B##, the smaller force, correspond to ##\theta = 0##, and ##T_A##, the larger force, correspond to the ##\theta## when it is nonzero? I'm just trying to understand why we use those limits of integration
     
  21. Sep 18, 2016 #20

    kuruman

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    Integration is a summation. You get a negative exponential if you add dT/T on one side and - μdθ on the other starting at the high tension end or add dT/T on one side and μdθ on the other starting at the low tension end. What I am saying here is that you can change the sign of the integral on the left and swap the limits of integration without changing anything.
     
  22. Sep 18, 2016 #21

    Charles Link

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    The sailor problem complicates the matter somewhat by calling it by the fancy names. Suppose instead that you lasso a horse. You would have a hard time hanging on to the horse, but if you could wrap the rope a couple of times around a telephone pole, you would have a much easier job of hanging onto the horse. ## T_A ## is the force from the horse, and ## T_B ## is the force you need to hold the horse.
     
  23. Sep 18, 2016 #22
    So physically, in the limit that ##\theta## approaches zero, we have low tension with rope being in equilibrium (which is ##T_B##), and in the limit where we have some higher angle ##\theta##, we have some greater tension ##T_A##?
     
  24. Sep 18, 2016 #23

    Charles Link

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    ## \theta ## is measured in radians. You get ## 2 \pi ## increase in ## \theta ## for every revolution. If you just pull the rope partly around the telephone pole, ## \theta ## will be less than ## 2 \pi ##. The problem can alternatively be worked using distance around the pole and force per unit length, etc. and tenstion as a function of distance around the pole.
     
  25. Sep 18, 2016 #24

    kuruman

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    Perhaps it should be clearer to you if you think of the capstan equation as a function of θ, namely
    Tsail(θ) = Tsailoreμθ.
    This is the condition for the rope to be at the threshold of sliding when the rope is wrapped around by angle θ. When θ = 0, i.e. no rope is wrapped around the capstan, the sailor needs to match the pull of the sail and nothing moves. If the sail pulls harder and the sailor is unable to provide the necessary force directly, the rope needs to be wrapped around the capstan to an appropriate angle θ in order to multiply the maximum force that the sailor can provide by a factor eμθ.
     
  26. Sep 18, 2016 #25

    Charles Link

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    It should be noted, the capstan holds the sail still. It does not provide extra force for moving it like a block and tackle (pulley ) system would.
     
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