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Boy Scout Problem

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data
    To protect his food from hungry bears, a boy scout raises 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 vboy, holding the free end of the rope in his hands (Fig. P2.71). (a) Show
    that the speed v of the food pack is given by x(x2 + h2)–1/2 vboy where x is the distance he has
    walked away from the vertical rope. (b) Show that the acceleration a of the food pack is
    h2(x2 + h2)–3/2 v2
    boy. (c) What values do the acceleration and velocity v have shortly after he
    leaves the point under the pack (x = 0)? (d) What values do the pack’s velocity and acceleration
    approach as the distance x continues to increase?


    2. Relevant equations
    All the kinematics equations and Pythagorean Theorem.



    3. The attempt at a solution
    I have been working at this for awhile now and can't seem to get it. I know that x=vboyt. I know that there is some trig here along the lines, but I can't get how the food box moves. Ok if the boy moves x then l will grow also, but at what rate? However much l grows is how high up the box goes. I can't seem to figure it out.


    http://img443.imageshack.us/img443/9238/physicsquestionku9.jpg [Broken]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 20, 2009 #2

    LowlyPion

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    Since you are interested in dy/dt (velocity of the pack) then express y in terms of x.

    Pythagoras was good for more than building pyramids.
     
  4. May 20, 2009 #3
    When you say y, are u talking about h?
     
  5. May 20, 2009 #4

    LowlyPion

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    Y is a direction.

    h is a height above the boy's head. That makes h a scalar constant doesn't it?

    Using the Pythagorean Theorem express the hypotenuse Hy as a function of x.

    You know as in

    Hy2 = x2 + h2

    Don't changes in the Hy represent 1:1 changes in y?
     
  6. May 20, 2009 #5

    LowlyPion

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    So the first question resolves into

    dy/dt = dHy/dt

    And Hy is a function of t because x is a function of t.
     
  7. May 20, 2009 #6
    Ok so Hy=sqrt(x^2+h^2). What does that do? I'm sorry I clearly don't understand this concept... Let me review it and I'll come back to it. Also I haven't learned too much calculus, only knowing basics, so if there is an algebraic approach to this I might I understand it better. Sorry for your time LowlyPion.
     
  8. May 20, 2009 #7

    LowlyPion

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    Just think about it conceptually. Boy runs with a velocity (dx/dt).

    How does Hy change with x ? You have that equation.

    So maybe take the derivative of Hy with respect to x? What does that give you? Since the rate of change of Hy = rate of change of y ... as it must since the rope doesn't change length ...
     
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