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Force and Newton's Laws

  1. Sep 24, 2007 #1
    A research balloon of total mass M is descending vertically with downward acceleration a. how much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward life of the air balloon does not change.

    I figured that
    let Fu = upward lift force
    Fnet = Fg - Fu = mg - ma
    and Fg must be greater than Fu since there is downward acceleration
    I'm not sure if I want to use big M to represent the mass I throw out
    or use little m and it be a seperate quantity from the balloon.
    I'm unsure of how to proceed from here,
    any help is much appreciated

    Thanks,
     
  2. jcsd
  3. Sep 24, 2007 #2

    learningphysics

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    Fnet = Ma

    Fg - Fu = Ma

    Mg - Fu = Ma etc...

    Once you calculate Fu... you know it remains the same.

    You want the upward acceleration = a

    Fu - Fg = Mnew*a
     
  4. Sep 25, 2007 #3
    so if
    upward acceleration = a
    Fu - Fg = Mnew*a
    a = (Fu - Fg) / Mnew

    I would have to throw up M - Mnew ballast

    how could I show this?
     
  5. Sep 25, 2007 #4

    learningphysics

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    substitute in the Fu calculated in the first part... substitute Fg = Mnew*g. Then solve for Mnew.

    Then get the difference in masses between M and Mnew.
     
  6. Sep 25, 2007 #5
    the Fu = -Ma + Mg ?

    Fu - Fg = Mnew*a
    Fu - Mg = Mnew*a

    -Ma + Mg - Mg = Mnew*a
    -Ma = Mnew*a
    Mnew = -Ma/a
    Mnew = -M

    Mnew - M= Mballast
    -M - M = Mballast
    -2M = Mballast

    would this work?
    Thanks
     
  7. Sep 25, 2007 #6

    learningphysics

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    You should use Mnew*g not Mg.
     
  8. Sep 25, 2007 #7
    Fu = -Ma + Mnew*g
    Fu - Fg = Mnew*a
    Fu - Mnew*g = Mnew*a

    will it change the rest?
    If so, how can I cancel out the other terms
     
  9. Sep 25, 2007 #8

    learningphysics

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    don't cancel anything... substitute in Fu from the first part (Fu does not change)... solve for Mnew.
     
  10. Sep 25, 2007 #9
    Fu = -Ma + Mnew*g
    Fu - Fg = Mnew*a
    Fu - Mnew*g = Mnew*a
    -Ma + Mg - Mnew*g = Mnew*a
    -Ma + Mg = Mnew*a + Mnew*g
    M(-a+g) = Mnew(a+g)
    Mnew = M(-a+g)/(a+g)
     
  11. Sep 25, 2007 #10

    learningphysics

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    looks right. now get the difference... is M - Mnew.
     
  12. Sep 25, 2007 #11
    Mballast = M - Mnew
    Mballast = M - M(-a+g)/(a+g)
    Mballast = M[1 - (-a+g)/(a+g)]

    Is there any more I can do, or is this the mass I have to toss off?

    Thanks
     
  13. Sep 25, 2007 #12

    learningphysics

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    you can simplify a little... use a common denominator of a+g.
     
  14. Sep 25, 2007 #13
    how can I divide -a+g by a+g
    would it be -1 + 1 = 0

    so leaving Mballast = M ?
     
  15. Sep 25, 2007 #14

    learningphysics

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    [tex]M(1 - \frac{-a+g}{a+g}) = M(\frac{a+g + a - g}{a+g}) = M(\frac{2a}{a+g})[/tex]
     
  16. Sep 25, 2007 #15
    beautiful
    that makes sense
    thanks very much
     
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