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Homework Help: I have no idea how to do this

  1. Jul 9, 2006 #1
    Hello, i have another question, sorry i didn't post them together. But i'm stuck again ... and this time i have no idea how to solve...

    Jane wishes to quicky scale a slender vine to vist Tarzan in his treetop hut. The wine is known to safely support the combined wieght of Tarzan, Jane, and Cheeta. Tarzan has twice the mass of Jane, who has twice the mass of Cheesta, if the wine if 60m long, what minimum time should Jane allow for the climb?

    Ok, well i only has this one idea that was name the mass of Cheeta x, the Tarzan is 4x, and Jane is 2x...and that's all i could think of... i feel dumb...

  2. jcsd
  3. Jul 9, 2006 #2

    Andrew Mason

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    What is it that would cause the force on the vine to be greater than Jane's weight?

    What average speed would Jane have if she completed the climb in t seconds? What is the minimum force required to do that?

  4. Jul 9, 2006 #3
    the net force? accleration times mass?? but wouldn't it stop at the top? or starting from rest? i need one variable to be 0m/s...
  5. Jul 9, 2006 #4

    Andrew Mason

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    Right. The trick is to minimize the force for a given time. Then find a time for which the force is just under the maximum strength of the vine.
    She starts at 0. She will end at a speed greater than 0 at the top. At that point she is not on the vine so presumably the tree branch (and gravity) will slow her down after she reaches the top.

    Plot speed vs. time on a graph. You want the path that has a slope that is always less than the maximum allowable acceleration.

    For a given time, t, what is the path that will result in the minimum slope? Keep in mind that the area under the graph (distance) has to be the same for all paths. When you have that figured out, it is a simple matter to find the minimum time.

  6. Jul 12, 2006 #5
    Finally found this post, thanks i did it like this

    (x+2x+4x) = (2x)(9.8) + (2x)(a)
    49x/2x = a
    a = 24.5

    then vf^2-Vi^2 = 2ad
    vf = 54.22m/s

    vf=vi + a*t
    t = 2.21s

    don't know if this is right. BUT THANK YOU
  7. Jul 12, 2006 #6

    Andrew Mason

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    This is the correct answer. Just to clarify the reasoning:

    The maximum force is 7/2 of the weight of Jane, or 3.5 mg where m is Jane's mass.

    For a given time, the peak force is minimized with constant acceleration. In terms of time and constant acceleration, the distance covered is:

    [tex]d = \frac{1}{2}at^2[/tex]


    [tex]F = ma + mg = m2d/t^2 + mg = 3.5mg[/tex]

    [tex]t^2 = 2d/2.5g \text{ So: } t = 2.21[/tex] seconds

  8. Oct 18, 2006 #7
    Well, this is interesting. This question came up on my physics test, but I didn't know how to answer it. So I started browsing, and it was my friend that had actually found this exact link. (We were working together)

    What I'm wondering is; How did you get 49x? We've been trying and trying, but we can't work it out so that a = 24.5

    Any help on this would be greatly appreciated!
  9. Oct 18, 2006 #8

    Andrew Mason

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    [tex]F = ma + mg = 7mg/2[/tex]

    letting m=2x:

    [tex]F = 2xa + 2xg = 7xg[/tex] so:

    [tex]a/g + 1 = 7/2[/tex]

    [tex]a/g = 5/2[/tex]

    [tex]a = 9.8*5/2 = 49/2 = 24.5 m/sec^2[/tex]

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