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Homework Help: Tarzan! King of the Jungle

  1. Nov 17, 2004 #1
    I'm pretty sure that this question isn't as hard as I'm making it, but I'm having a brain block here.

    Tarzan, who weighs 700 N, swings from a cliff at the end of a convenient vine that is 17 m long. From the top of the cliff to the bottom of the swing, he descends by 5.3 m. The vine will break if the force on it exceeds 1580 N. What would the greatest force on the vine be during the swing?

    I'll show you guys how far I've gotton and then maybe you can help me from there, or show me where I am mistaken

    Emech(initial) = Ep + Ek
    Emech(initial) = mgh
    Emech(initial) = 700*5.3
    Emech(initial) = 3710

    Emech(final) = Ep + Ek
    Emech(final) = Es + Ek
    Emech(final) = (kx^2)/2 + (mv^2)/2

    *We can assume Emech is conserved

    3710 = (kx^2)/2 + (mv^2)/2

    kx^2 + mv^2 - 7420 = 0

    Also I know that

    F= m*a
    F=mv^2 / R

    kx^2 + mv^2 - 7420 = 0
    v^2 = (7420 -kx^2) / m

    F = 7420-kx^2 / R
    F = 7420-kx^2 / 17


    Fmax = kx
    1580 = kx
    1580/x = k

    Sub into F = 7420-kx^2 / 17

    F = (7420 - 1580x) / 12

    And thusly I am stuck.

    Too many unknowns, not enough equations.

    :surprised :confused:
  2. jcsd
  3. Nov 18, 2004 #2
    (kx^2)/2 ???

    Is this vine a spring?
  4. Nov 18, 2004 #3
    Its's not a spring, but it does unfortunatly have elastic properties. I really wish it didn't, the problem would be much easier.

    Took me about 20 minutes of getting the answer wrong and wondering why they told me Fmax before I figured that out though.

    When I realized it I thought I had solved the problem, only to realize I had gained two variables and only one equation.


    Now I am even more frustrated.
  5. Nov 18, 2004 #4


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    Just a question...how do you know that you are meant to assume that the vine has elastic properties? If the author intended for students to consider them, you would think he/she would give the elastic spring constant k, in the problem, as well as a statement that the elastic restoring force of the vine can be described by Hooke's law. It doesn't say so anywhere...???
  6. Nov 18, 2004 #5
    yea, I would also think that the book would specify that to you...
  7. Nov 18, 2004 #6
    Alright; perhaps you are right. I've done the problem for the case where the rope is not elastic however I get an incorrect answer.

    Maybe you can show me where I went wrong.

    [tex]Et_i = E_p +E_k[/tex]

    [tex]Et_i = E_p +0[/tex]

    [tex]Et_i = mgh[/tex]

    [tex]Et_i = 700(h)[/tex]

    [tex]Et_i = 3,710[/tex]

    *Assume no energy lost

    [tex]Et_f = E_p +E_k[/tex]

    [tex]Et_f = 0 +\frac{mv^2}{2}[/tex]

    [tex]3,710 = \frac{mv^2}{2}[/tex]

    [tex]\frac{7420}{m} = v^2[/tex]



    [tex]F = \frac {m \frac{7420}{m}}{R}[/tex]

    [tex]F = \frac {7420}{17}[/tex]

    [tex]F = 436.47N[/tex]
  8. Nov 19, 2004 #7
    It is sum of forces = ma

    I assume u have the free body diagram drawn for when tarzan is at the bottom

    then sum of the forces is not F, but :

    T - W

    where T is the tension and W is tarzans weight.
  9. Nov 19, 2004 #8
    Oh man. Thank you so much, alright so then the last step would be:

    [tex]\sum{F} = F_t - W[/tex]
    [tex]\sum{F} = F_t - 700[/tex]
    [tex]436.47 + 700 = F_t[/tex]
    [tex]F_t = 1136.47[/tex]

    Which...is so damn right.

    Thank you so much. I can't believe it was such a stupid mistake. :rofl: :surprised
    Last edited: Nov 19, 2004
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