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Static cable car, approximate tensions

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Find approximate expressions for T1 and T2 that are
    valid when h/d << 1 and h/x << 1

    2. Relevant equations
    [tex]T_{1} =\dfrac{mg\left(d-x\right)}{dh}\sqrt{1+\dfrac{x^{2}}{h^{2}}}[/tex]
    [tex]T_{2} =\dfrac{mgx}{d}\sqrt{1+\dfrac{\left(d-x\right)^{2}}{h^{2}}}[/tex]
    [tex]h^{2} =\dfrac{\left(L^{2}-d^{2}+2dx\right)^{2}}{4L^{2}}-x^{2}=\dfrac{\left(L^{2}-d^{2}\right)\left(L^{2}-d^{2}+4x\left(d-x\right)\right)}{4L^{2}}[/tex]

    3. The attempt at a solution
    Not sure where to begin!
     

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  3. Nov 3, 2014 #2

    haruspex

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    There's a spurious h in the denominator in your first equation. I assume it's a typo.
    Since you want approximation for small h, rearrange things so that the surds take the form ##\sqrt{1 + \frac{h^2}Z}## for suitable Zs. Then you can use the binomial expansion.
     
  4. Nov 3, 2014 #3
    Hi, thanks, the h is spurious.
    I am looking at rearranging things now.
    James
     
  5. Nov 4, 2014 #4
    Having a bit of a block with the rearranging, any hints?
     
  6. Nov 4, 2014 #5

    haruspex

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    ##\sqrt{1+\dfrac{x^{2}}{h^{2}}} = \frac 1h \sqrt{x^{2}+h^{2}}##. Can you do the next step?
     
  7. Nov 5, 2014 #6
    I get that, thanks, and I get

    [tex]T_{1} =\frac{mg\left(d-x\right)}{h}\sqrt{\frac{x^2}{d^2}+\frac{h^2}{d^2}}[/tex]
     
  8. Nov 5, 2014 #7

    vela

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    That's not the form you want. You don't want ##x/d## inside the radical. You want a 1 in there. You have
    $$\frac 1h \sqrt{x^2+h^2}.$$ Factor something out of the radical so that one of the terms is 1 and the other is <<1.
     
  9. Nov 5, 2014 #8
    So is this right

    [tex]\frac{1}{hx}\sqrt{1+\frac{h^2}{x^2}}[/tex]
     
  10. Nov 6, 2014 #9

    haruspex

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    You have the right expression inside the square root, but you made a mistake in the extracted factor.
    When you've corrected that, do you understand what to do to get rid of the square root by making an approximation?
     
  11. Nov 6, 2014 #10
    Should it be this?

    [tex]\frac{x}{h}\sqrt{1+\frac{h^2}{x^2}}[/tex]

    The square root can be expanded by a binomial expansion

    [tex]\left(1+\frac{h^2}{x^2}\right)^{1/2}=1+\frac{1}{2}\,{\frac {{h}^{2}}{{x}^{2}}}-\frac{1}{8}\,{\frac {{h}^{4}}{{x}^{4}}}+\frac{1}{16}
    \,{\frac {{h}^{6}}{{x}^{6}}}-{\frac {5}{128}}\,{\frac {{h}^{8}}{{x}^{8
    }}}+{\frac {7}{256}}\,{\frac {{h}^{10}}{{x}^{10}}}[/tex]

    given that [itex]\frac{h}{x}\ll1[/itex] would it be

    [tex]\sqrt{1+\frac{h^2}{x^2}}\approx1[/tex]
     
  12. Nov 6, 2014 #11

    vela

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    If you wanted to do that, you could have gotten the same result much more simply:
    $$\sqrt{1+\frac{x^2}{h^2}} \cong \sqrt{\frac{x^2}{h^2}} = \frac{x}{h}.$$ Typically, you keep at least the first-order term of the expansion, i.e.,
    $$\sqrt{1+\frac{h^2}{x^2}} \cong 1 + \frac 12 \frac{h^2}{x^2}$$ so that
    $$\sqrt{1+\frac{x^2}{h^2}} = \frac xh \sqrt{1+\frac{h^2}{x^2}} \cong \frac xh \left(1 + \frac 12 \frac{h^2}{x^2}\right).$$
     
  13. Nov 7, 2014 #12
    Thanks for the help.
    I need to show that the cable can withstand a tension of

    [tex]\frac{mgd}{4h}[/tex]

    I have tried many things but cannot get the expression.
     
  14. Nov 7, 2014 #13

    haruspex

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    You need to post at least one updated attempt. Then we can see where you are going wrong.
     
  15. Nov 7, 2014 #14
    Hi

    I thought maybe adding the magnitudes of both [tex]T_1[/tex] and [tex]T_2[/tex] but that didn't work.
     
  16. Nov 7, 2014 #15

    haruspex

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    I don't think that's quite what you mean. Let me guess:
    The diagram represents a cable car moving across. As it moves, the angles, distances and tensions change. The car hangs from a wheel on the cable which is in rolling contact with the cable, so the tensions in the two parts of the cable need not be the same (because there is a static friction force).
    You are asked to show that the maximum tension on any part of the cable is [itex]\frac{mgd}{4h}[/itex].
    If so, there is no reason to be adding the two tensions. You only need consider one tension and its maximum value. By symmetry, the other will have the same maximum value.
    Please post your expression for T1 after making the approximation discussed. How will you find its maximum?
     
  17. Nov 8, 2014 #16
    Hi, after posting last night I was laying in bed thinking along those lines.
    At the half way point [itex]x=d/2[/itex], substituting into [itex]T_1[/itex]

    [tex]T_1=\frac{mg(d/2)(d-d/2)}{dh}\left( 1 + \frac{h^2}{2(d/2)^2} \right)=\frac{mgd}{4h}\left( 1 + \frac{2h^2}{d^2} \right)[/tex]

    so as [itex]\frac{h}{d}\ll1[/itex] then the expression in brackets is approx 1 so

    [tex]T_1=T_2=\frac{mgd}{4h}[/tex]

    Therefore the cable must be able to support this tension.
    Thanks
     
  18. Nov 8, 2014 #17

    haruspex

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    Yes, that's good. However, you've not shown that this is the maximum tension it will be subjected to. It's not clear to me whether you are supposed to show that.
     
  19. Nov 9, 2014 #18
    It doesn't, but showing why would be a good idea.

    For [itex]T_1[/itex] taking the approximation into account

    [tex]T_1=\frac{mg(d-x)}{dh}[/tex]

    taking the first derivative wrt [itex]x[/itex] and making this equal zero the value for [itex]x[/itex] is [itex]d/2[/itex] so half way along the cable where [itex]T_1=T_2[/itex]
     
  20. Nov 9, 2014 #19

    haruspex

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    Good. But does that show it is a maximum there?
     
  21. Nov 9, 2014 #20
    No, differentiating again gives a negative result, showing a maximum.
     
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