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Boy Scout Problem

  • #1

Homework Statement


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?


Homework Equations


All the kinematics equations and Pythagorean Theorem.



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]

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
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.
 
  • #3
When you say y, are u talking about h?
 
  • #4
LowlyPion
Homework Helper
3,090
4
When you say y, are u talking about h?
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?
 
  • #5
LowlyPion
Homework Helper
3,090
4
So the first question resolves into

dy/dt = dHy/dt

And Hy is a function of t because x is a function of t.
 
  • #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.
 
  • #7
LowlyPion
Homework Helper
3,090
4
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.
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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