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Force at bottom of hill

  1. Oct 29, 2013 #1
    [Solved] bottom of hill

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

    No friction, starts at rest.
    What is the force factor at the bottom of the circular loop?
    (see attachment)

    2. Relevant equations

    Conservation of Energy Equations

    3. The attempt at a solution

    EDIT: Solved below (incorrect solution was here)
     
    Last edited: Oct 29, 2013
  2. jcsd
  3. Oct 29, 2013 #2

    tiny-tim

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    hi darksyesider! :smile:
    yes, you certainly need to know the radius :confused:

    i'll guess it's supposed to be 2h
     
  4. Oct 29, 2013 #3
    I think this problem is missing the radius. I don't see how else to solve it.
     
  5. Oct 29, 2013 #4

    tiny-tim

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    yes :smile: try 2h
     
  6. Oct 29, 2013 #5
    So is the answer 5 times?

    I got:

    [tex] F_N = \dfrac{m(2gy)}{r}+mg [/tex]

    [tex] = mg(\dfrac{2y}{r}+1) [/tex]

    Since force factor = Fa/Fg

    [tex] ff = \dfrac{mg(\dfrac{2y}{r} + 1 )}{mg} [/tex]

    Substituting in y = 2r we get 5 times.
     
  7. Oct 29, 2013 #6

    tiny-tim

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    if force factor = Fa/Fg, yes :smile:
     
  8. Oct 29, 2013 #7
    Sorry, but is that the incorrect definition? :(
    I can't find any other definition of it…
    And thanks a lot for the help!!
     
  9. Oct 29, 2013 #8

    tiny-tim

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    i've never heard of it before :redface:

    but i'm happy to take your word for it … and if it is, your answer looks fine :smile:
     
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