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Homework Help: Me Tarzan, You Jane. Force Problem.

  1. Jul 12, 2008 #1
    holy physics! i can't wrap my head around this.

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

    Jane wishes to quickly scale a slender vine to visit Tarzan in his treetop hut. The vine is known to safely support the combined weight of Tarzan, Jane, and Cheeta. Tarzan has twice the mass of Jane, who has twice the mass of Cheeta. If the vine is 60 m long, what minimum time should Jane allow for the climb?

    3. The attempt at a solution

    this is what i've scrounged so far:

    d= 60 m
    V1 = 0

    jane wants to quickly scale the vine - let's find a, then t

    cheetah's weight is x, jane is 2x, tarzan is 4x. the vine can hold 7x, so jane can apply 3.5x her own weight.

    Fnet = ma
    Fg = mg = ma
    3.5mg = ma [cancel masses]
    3.5 (9.81) = a
    a = 34.3

    then plug a into one of the kinematics to find t. is this correct?

    edit: i got t = 1.87 s

    i guess what i am trying to say is that it seems too easy now compared to the amount of time i just stared at the question so maybe i have made some terrible mistake
    Last edited: Jul 12, 2008
  2. jcsd
  3. Jul 12, 2008 #2
    Interesting introduction...what you did seems right considering that her net acceleration would be constant.
  4. Jul 12, 2008 #3
    ooo tambourine you were so close. You forgot a basic detail in your last equation. I am guessing that you used Ke=1/2kg*v^2 and then t=d/v

    ok, when you use the Kenectic energy formula, you go a velocity of 34.31m/s correct?
    than you probably did:

    t=34.31m/s*60m, well thats where you were slightly wrong. Remember that the velocity changes from 0m/s to 34.31m/s. That means that you had to find the average velocity witch is 17.16m/s

    now you can do this:
    t= 3.5s

    You were so close :(. Beautiful work though
  5. Jul 12, 2008 #4


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    Hi tiale11,

    I don't believe your velocity of 34.31m/s is correct. How did you get that number?

    It looks like you just set the change in kinetic energy equal to the change in potential energy, but that is not right for this problem. Energy is not conserved here. Or did you do something else?
    Last edited: Jul 12, 2008
  6. Jul 12, 2008 #5
    i am 100% sure that i am correct.
    to find the velocity, I used Ke= 1/2*m*v^2 and soleved for v
    and yes, you can use the potential energy as the kenetic energy. Think of droping Jane with the same acceleation(replaces gravity) that she went up. At the top she will have no kenetic energy but have potential energy. As she hits the floor she will have TRANSFERED all of that potential energy to make it become kenetic energy. The same concept for bottom up. Again i am sure my answer is correct. You have to remember CONSERVATION of ENERGY!!!!!!!
    How is the energy not conserved here?

    I dont know if you notice but your time is half of mine. your error is because you did not use the average velocity to find the time.
    Show me how you found your velocity and time.
  7. Jul 12, 2008 #6


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    I did not tell you what my time I got for the problem. I think the answer the original poster got and you got are both incorrect.

    Since she is pulling herself up, work is being done on her, and energy is not conserved here.
  8. Jul 12, 2008 #7


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    Hi tambourine,

    I don't think your answer is quite right.

    I believe that here is your error. When you consider the force diagram acting on Jane, there are two forces acting on here--the tension and the weight force. But here you only have the tension acting on her. Once you put the weight force in the equation you should get the right answer.

    In other words, you found the maximum tension was 3.5 times here weight; but part of that tension has to counteract the weight force. Only what's left over actually accelerates her upwards.
  9. Jul 12, 2008 #8
    thanks for the responses!

    tiale11, i don't quite understand what you did. how did you solve using the kinetic energy formula if mass wasn't given?

    i got t=1.87 by plugging d=60 m, V1=0, and a=34.3 into d=V1t + 1/2at^2

    alphysicist, this is the concept i'm having trouble with. would the tension and her weight both point in the same direction, downwards? and if so, what is propelling her up? she's applying tension on the vine, but that force points down. when she is at rest, hanging on the vine, would the tension force equal her weight? since without her, there is no tension.

    can you please help me understand this? heck, anyone, feel free to jump in! we could talk in circles together. it would be fun.
  10. Jul 12, 2008 #9


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    tiale11's approach was based on the belief that energy was conserved, which is not true here.

    In the force equation (from her force diagram), we need the force that the rope puts on her. If she pulls down on the rope, which way does the rope pull on her? That is the way I would think about it. So which way is the tension? And what equation would you get?
  11. Jul 13, 2008 #10
    so.. the rope pulls up at her, equal and simultaneous-

    Fnet = ma

    if down is positive:

    Fg - Ft = ma
    3mg - Ft = ma ?
  12. Jul 13, 2008 #11


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    No, I don't think that is right. The rope pulls up, and the weight pulls down, so:

    T - W = m a

    where T is tension and W is the weight. So what is W? And what is T? Remember that you had the tension in your first post. What do you get for a?
  13. Jul 13, 2008 #12
    3mg - mg = ma
    so 2mg = ma ?
  14. Jul 13, 2008 #13


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    In your original post, you found that the maximum tension allowed was 3.5 times her weight, which I think is the correct value.
  15. Jul 13, 2008 #14
    oh, i can't see numbers straight anymore. meant 3.5. thank you alphysicist :D
  16. Jul 13, 2008 #15


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    Sure, glad to help! And I think these long hours are starting to get to me, too.
  17. Jul 13, 2008 #16
    dudes, i am so sorry. I thought Jane did not climb but hung on onto the vine which acted like a spring :(. lol Now i see why you folks said that energy was not conserved. I was on a totaly different topic.
    Ok so now let me get this right. the question asks to find the time that Jane will climb the 60m vine. Is it like a pulley problem with 2 directions? please explain so i can have a crack at it
  18. Jul 13, 2008 #17


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    No, there's only one direction involved. The vine would be attached to the tree at the top.

    The idea is that the problem gives you a way to find the upper limit of tension that the vine can withstand before breaking. The tension is the only upward force acting on Jane, so that allows you to find the maximum upward acceleration.

    It's different because the normal thing is for ropes to be tied to the object, but that is not necessary; there just has to be some way for the rope and object to apply forces to each other.

    Also, this doesn't help in finding the answer, but if you want to think more about the problem, you could say that it's not really the tension force that pulls her up; it's really the frictional force between the vine and her hand. You would then need to draw a force diagram for the vine, which would show that the tension force and frictional force are equal in magnitude.
  19. Jul 13, 2008 #18
    ok so, the vine is like a rope, it is not moving at all. Jane does all the work. She does not hold on to a certain spot on the vine and it pulls her up. is that correct?
  20. Jul 13, 2008 #19


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    Right, that's how I visualized it; she's climbing up, hand over hand. (The problem would be the same, though, if she was just holding on and someone was pulling the vine up--assuming you can ignore the mass of the vine.)
  21. Jul 13, 2008 #20
    o0o ok. now i am gonna solve this using what alphysicist said about another person pulling Jane up a weigthless vine.
    I am now going to assume that Tarzan, Jane and cheeta are the maximum weight on the vine that the other "person" can pull up. ****I am doing this like a pulley problem****
    the tension of all of them will equal T=7x*a. Now, the acceleration that will be produced on Jane alone on the vine will be : a=(7x*a-2x*a)/2x=24.525m/s^2. Now, using d=vi*t+1/2*a*t^2 i solved for t and calculated it to be 2.212s. Well considering what i assumed for this problem, this should be right
  22. Jul 13, 2008 #21


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    Your answer looks right to me.
  23. Jul 14, 2008 #22
    man, finally! :). Well next time ill make sure i understand what the question is asking first before answering
    thanks for clearing stuff out
  24. Jul 14, 2008 #23


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    Staff: Mentor

    Assuming 2.1s and 25 m/s^2 she ends at about 200 km/h, how does she stop? :rofl:
  25. Jul 14, 2008 #24


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    There's a big branch that stops her, so all she needs is something for her head. That's where all of those egg drop experiments come in. Perhaps the old "insert her head in pantyhose and staple the ends to the box"?
  26. Jul 14, 2008 #25
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