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Homework Help: Kinematics problem

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data
    To protect his food from hungry bears, a boy scout rasies his food pack with a rope that is thrown over a tree limb at height h above his hands. He walks away from the vertical rope with constant velocity (v boy), holding the free end of the rope in his hands.
    (a) Show that the speed v of the food pack is given by (x)(x^2 + h^2)^(-1/2)(v boy) where x is the distance he has walked away from the vertical rope

    The diagram given is something like this,

    | \
    | \
    | \
    | \
    |___\ ------------->

    * The diagram is not coming out properly, anyhow its a right angled triangle. with the below characteristics.
    *h is the constant distance between the tree top and the hand of the boy (the vertical line)
    *x is the variable horizontal distance (the horizontal line)
    *The point where the diagonal and vertical line meet is like the "pivot". That is where the rope goes one round around a twig.
    * ----> is v boy

    2. Relevant equations

    3. The attempt at a solution
    V = distance displaced / time
    Distance displaced = (h^2 + x^2)^(1/2) - h
    Assumption ? < The change in diagonal length when the boy moves = distance displaced >

    According to Pythagoras theorem the above (h^2 + x^2)^(1/2) is derived. And the change being the ( final diagonal distance - initial diagonal distance (when x=0) ).

    So I have come until this point, but I am unable to simplify it further so that its similar to the equation I'm supposed to derive. Can anyone hint to me or guide me along on which part I have made a mistake or whats the possible next step? Thanks

    P.s: hope the drawing helps. If any additional information is needed do tell me so that i can take a look and see whether its provided. A BIG THANKS again.
  2. jcsd
  3. Sep 6, 2010 #2

    When the boy is close to the tree the boys walking with the rope in his hands hardly lifts the basket at all. However, when he is far away every step he takes the basket moves almost an equivalent step/amount.

    As you posted this in Advanced Physics you may want to consider which trigonometric identity would express exactly this. You can then see that this identity can be expanded to be the expression they require.

    Does this help?
  4. Sep 6, 2010 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Good. Now express the relationship between diagonal length and the horizontal distance x. (Use Pythagoras.) Use a bit of calculus to relate their rates of change.
  5. Sep 8, 2010 #4
    I understand that change in distance is related to the change in diagonal length which inturn can be calculated through the Pythagoras theorem.

    <Assumption: Change in diagonal length = change in vertical length, since the rope is getting reeled over>

    So as such, let the diagonal length at the start be y = h (given)

    h^2 + x^2 = y^2

    y = (h^2 + x^2)^(1/2)

    Change in y = (h^2 + x^2)^(1/2) - h

    Velocity of the mass should be = change in y/ time taken

    Since time taken = x/v boy

    Velocity of the mass should be = ((h^2 + x^2) ^ (1/2) - h )/(x/ v boy)

    = (v boy)(h^2 + x^2) ^ (1/2) - h )/x
    = not the right answer =(

    So I get the general gist of the question, seems like I know how to solve it but I'm missing something. Can some one help me point out or better still tell me I'm correct and that all I am doing wrong is simplifying the equation. A big thanks for the help from Doc al and h man. =)
  6. Sep 8, 2010 #5

    Doc Al

    User Avatar

    Staff: Mentor

    I'd express it like this: The change in the diagonal = change in the height of the package. So all you need to do is relate the instantaneous rate of change of the diagonal to the instantaneous rate of change of the boy's horizontal position. And you know that the diagonal and boy's positoin are related by Pythagoras.

    OK. I would immediately begin to find the rate of change. How do you do that?*

    OK, but not needed.

    Careful. You want the instantaneous speed as a function of position, not the average speed.

    *Hint: Calculus!
  7. Sep 8, 2010 #6
    Dam! Thanks! Has been staring at me all along! Laid off physics and other stuff for a long time so am a bit rusty, guess that explains this. Anyhow thanks Doc Al for the help. Cheers to you. =)
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