Suspension cable statics calculus problem

  1. Mar 31, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    [​IMG]


    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    I find myself with 2 unknowns, 1 equation.

    [​IMG]
     
  2. jcsd
  3. Mar 31, 2012 #2

    I like Serena

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    You're on the right track. :)

    You have 1 equation for point A.
    Can you make another equation for point B?
     
  4. Mar 31, 2012 #3

    Femme_physics

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    I like you saying this lately :)

    Oohhh....

    Am I still on the right track?

    [​IMG]
     
  5. Mar 31, 2012 #4

    I like Serena

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    :cool:

    I'm afraid that in the 2nd line you lost ##F_H##.

    And btw, you're using 15000 [lb] for ##w_0##, but ##w_0## is given to be 600 [lb/ft].
     
  6. Apr 2, 2012 #5

    Femme_physics

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    You mean purely because of math?

    I see what you mean, since I don't know the length in each sectioning I can't use it like that. I'll jus use 600, with the units lb/ft. Yes?
     
  7. Apr 2, 2012 #6

    I like Serena

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    Yes.

    Yes.
     
  8. Apr 5, 2012 #7

    Femme_physics

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    I got it. My friend helped me with the math (the one who's registered as "niece of MD") :)

    [​IMG]


    [​IMG]

    We use X2 since the length can't be more than 25. Therefor we use FH2 as well..

    But my big problem is relating the distance, w0 and FH, to A and B.

    I end up with this diagram...

    [​IMG]
     
  9. Apr 5, 2012 #8

    I like Serena

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    Aha! So she does still do something every now and then! :)


    If you're interested, I have a shorter version:
    ##20 x^2 = 30 (25 - x)^2##
    ##2 x^2 = 3 (25 - x)^2##

    Since x and (25 - x) are both positive distances, we can take the square root and keep the positive versions:
    ##x \sqrt 2 = (25 - x) \sqrt 3##
    ##x \sqrt 2 = 25 \sqrt 3 - x \sqrt 3##
    ##x (\sqrt 2 + \sqrt 3) = 25 \sqrt 3##
    ##x = \frac {25 \sqrt 3} {\sqrt 2 + \sqrt 3} \approx 13.76##


    Looks good.
    But consider that the tensional force is not pointing down, but along the rope.

    Since they ask for the tension in the rope in A and in B, you need that
    ##F_V = {dy \over dx} \cdot F_H##
     
    Last edited: Apr 5, 2012
  10. Apr 8, 2012 #9

    Femme_physics

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    I think I got it, but I don't know what to make out of Fv as far as each of the reaction forces at A and B.

    [​IMG]
     
  11. Apr 8, 2012 #10

    I like Serena

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    You have the result Fv for point B here.
    Good.
    Oh, but the unit is lb, and not l/ft.

    On support B you have the horizontal force Fh and this vertical force Fv.
    So what's the total force?
     
    Last edited: Apr 8, 2012
  12. Apr 9, 2012 #11

    Femme_physics

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    Ahh....

    Each has their own force.

    So,

    [​IMG]

    Badabing badaboom?
     
  13. Apr 9, 2012 #12

    OldEngr63

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    Don't you need to add those vectorially? (or did I miss something as I skimmed down through the solution?)

    BTW, how did your gripper project turn out?
     
  14. Apr 9, 2012 #13

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    Ya know wee go n bada bing, bada boom n we ah atta thereah....day won't know that OldEngr63 hit um!

    (You're not supposed to simply add up forces that are perpendicular to each other. ;)
     
  15. Apr 10, 2012 #14

    Femme_physics

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    Ohhhhhhh Ok gotcha, now it makes perfect sense :)

    I need to find the resultant vector for each! *smacks forehead*

    So

    (CALCULATION ATTACHED)

    Rb = 9086 lb

    Ra = 7735 lb

    BADABING BADABOOM I said! :)



    Very, very, SLOWwwww... because of the teacher, not us. We're on our passover holiday right now. And I did eventually use a gripper's PDF guide to get ideas, and basically our main idea is a double-threaded spindle. But, right now, we're still awaiting orders and formulas.
     

    Attached Files:

  16. Apr 10, 2012 #15

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  17. Apr 10, 2012 #16

    OldEngr63

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    Thank you.
     
  18. Apr 11, 2012 #17

    Femme_physics

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    No, thank YOU! a lot. :)
     
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