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Homework Help: A Center of gravity problem

  1. Jan 16, 2010 #1

    M_G

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

    An 80 kg hiker carries a 20 kg pack. The center of gravity of the hiker is 1.1m above the ground when he is not wearing the pack. The center of gravity of the pack is 1.3m from the ground when it is worn. How far above the ground is the center of gravityof the hiker and the pack?

    I don't know how to solve such a problem, how to begin?
    Need help plz...
     
  2. jcsd
  3. Jan 16, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi M_G! Welcome to PF! :smile:

    What do you know about the centre of gravity of two different masses?
     
  4. Jan 16, 2010 #3

    M_G

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    Hi,
    If there is two weights on a weightless rod, the center of gravity will be at a distance X1 from W1 and X2 from W2, and the C.G will be closer to the heavier weight.......right?
     
  5. Jan 16, 2010 #4

    tiny-tim

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    Right. And if you hold it at that point, it will balance. :smile:

    But what's the formula, with W1 W2 X1 and X2, that tells you it will balance?
     
  6. Jan 16, 2010 #5

    M_G

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    The formula is:
    X=(x1*w1+x2*w2)/w

    Where:
    w=w1+w2
     
  7. Jan 16, 2010 #6

    tiny-tim

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    Actually, that's the slightly different formula, for x1 and x2 being the distances from a fixed point, and X being the distance of the centre of mass from the same point.

    ok … now apply that formula to the problem …

    what do you get? :smile:
     
  8. Jan 16, 2010 #7

    M_G

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    sorry, I can't get what you're pointing at...
     
  9. Jan 16, 2010 #8

    tiny-tim

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    The "balance" formula, that I originally asked about, is x1w1 = x2w2, where the w's are measured, in opposite directions, from the fulcrum.

    The formula you gave is where they're measured, in the same direction, from a completely general point (like the ground :wink:).
     
  10. Jan 16, 2010 #9

    M_G

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    got it,
    by applying that formula to the problem:
    x1=1.3
    x2=1.1
    X=1.1+Y (distance between the C.G & the person)
    Right???
     
  11. Jan 16, 2010 #10

    tiny-tim

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    (sorry … been out all evening :biggrin:)

    Yeeees … but what on earth are you doing? That doesn't get you anywhere, it's just stating the obvious.

    Use the formula (and use the 80 and the 20).
     
  12. Jan 17, 2010 #11

    M_G

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    Y+1.1 = (1.3*20+1.1*80)/20+80 ......?:uhh:

    I had a bad feeling that I'm proceeding wrong, very very wrong, I don't know why I can't get it?!
     
    Last edited: Jan 17, 2010
  13. Jan 17, 2010 #12

    tiny-tim

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    That's it! :smile:

    Are you happy now, with the general method for these centre of mass problems?
     
  14. Jan 17, 2010 #13

    M_G

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    Yes, I am:smile:...but, when I solved it for Y, I found Y=0.04m.....?I think it is a very small number :uhh: isn't it??
     
  15. Jan 17, 2010 #14

    tiny-tim

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    Well, it has to be very close to the centre of gravity of the hiker, and there's only 0.2m to play with anyway.

    (and of course, the question asks for the height above the ground, which is 1.1 + 0.04)
     
  16. Jan 17, 2010 #15

    M_G

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    Thanks !
     
  17. Jan 30, 2010 #16
    Don't forget that the hiker will lean forward if i carries a backpack!
    Sorry...
     
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