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Quick Question

  1. Oct 25, 2004 #1

    hL

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    A child is lying on her back. The tension in the muscles of her neck is 55N as she raises her head to look past her toes and out the motel window. Ten minutes later she is screaming and sliding feet first down a water slide at a constant speed of 5.7 m/s in a horizontal curve of radius 2.40 m. She raises her head to look forward past her toes; find the tension in the muscles in her neck.

    Thanks in advance!
     
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  3. Oct 25, 2004 #2

    arildno

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    First case:
    The tension provides the necessary force to balance the weight of the lifted part of her body (for simplicity, let's call that "the head" in the following)
    Use this to determine the head's mass.
    For the second case:
    Note that when her head does NOT touch the slide, only the tension in her neck provides the force necessary for the centripetal acceleration the head experiences.
     
  4. Oct 25, 2004 #3

    hL

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    So I would do the second part of the problem focusing around the head? Gotcha.
    It'd be T=mv²/r, using the mass of her head I think. Thanks for responding.
     
  5. Oct 25, 2004 #4

    arildno

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    That's what I had in mind..
    Welcome to PF, BTW.
     
  6. Oct 25, 2004 #5

    hL

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    If you wouldn't mind helping me again, I have another question that I'm stumped on.

    A plumb bob does not hang exactly along a line directed to the center of the Earth because of the Earth's rotation. How much does the plumb bob deviate from a radial line @ 35 degrees north latitude? Assume the Earth is spherical.

    I combined the x and y components of the bob to get (4*pi²*r)/(T²g) = tan(x). What numbers would I use for r and T? I think I could use 24 hrs. for T, but r wouldn't be the Earth's radius.

    EDIT: Thanks for the welcoming. :smile:
     
  7. Oct 25, 2004 #6

    arildno

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    First of all:
    We both forgot that the tension in the first exercise b) ALSO must balance the weight, not only provide the centripetal acceleration..

    You're right, r is the planar radius at 35 degrees latitude
     
  8. Oct 25, 2004 #7

    hL

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    Since it's a horizontal curve, the weight, down, should be balanced by a frictional force, up. I think that's how it works in a certain amusement park ride whose name escapes me at the moment.
     
  9. Oct 25, 2004 #8

    arildno

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    Your head is FREE, is it not?
    (only connected to your body, not in touch with the slide, the only force which acts upon it other than the weight, is whatever your neck imparts to it..)
     
  10. Oct 25, 2004 #9

    hL

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    I see what your saying now. Thanks, I'll grind through this problem later tonight.
     
  11. Oct 27, 2004 #10
    Note for Plumb bob question if it helps( i got the right answer using this)
    It seems to work, since the angles aren't perpindicular

    C(Force of Tension)^2 =A(gravity)^2 +B(4pi^2 r(planar radius)/T^2 )^2 -2ABCOS(θ(which is 35º))

    Then use Sin Law to find the angle that the plumb bob deviates
     
    Last edited: Oct 27, 2004
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