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Balloon in Truck.

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A balloon stands upright in a stationary truck. The truck then accelerates forwards. What direction does the balloon move in?

    2. Relevant equations
    Just Newton 1&2.

    3. The attempt at a solution

    1) A Free body diagram on the balloon. We know the balloon accelerates at the same rate as the truck in the x direction. There's no reason for there not to be equilibrium in the y, so (please see attached file) I choose the positive theta direction as shown (which happens to be the same direction as acceleration)

    $latex mg =Tcos\theta (F_{y}) $

    $latex ma=Tsin\theta = ma $

    $latex a=gtan\theta $

    Since a is in the positive theta direction, it is positive in the third equation. This implies that theta must be greater than 0, (i.e the balloon does actually move as shown.)

    The issue is, I'm not sure if this is acceptable as a solution! The method given in my book suggests that the approach is along the lines of

    (1) Newton's first law implies you would expect the balloon to move backwards while the truck accelerates.

    (2) But, the movement of all the air in the truck to the back (by Newton's first law) creates a pressure gradient that causes the balloon to move forward.

    So my question is, are these two approaches equivalent or not? If not, why, please!

    Many thanks for all your help.

    Attached Files:

  2. jcsd
  3. Dec 8, 2013 #2


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    hi bananabandana! :smile:
    i don't understand this

    if there was a pressure gradient, why would it have that effect on the balloon, and not on eg the furry dice hanging from the roof?

    start by drawing a free body diagram for the stationary car …

    how many forces are there on the balloon? which direction are they?

    ok, now draw the moving diagram, and use ∑F = ma :wink:
  4. Dec 8, 2013 #3
    I thought that was what I was doing in the first part! When stationary, T = mg.
    Then, as above??
  5. Dec 8, 2013 #4


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    i'm not really following what you've done :confused:

    and your T in the diagram is the wrong way round, it's the force acting on the balloon (along the string), so it has to be down :wink:
  6. Dec 8, 2013 #5


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    An approach that does not consider the (co-accelerating) air in the truck cannot work. You have to take buoyancy (or the pressure gradient, which is equivalent) into account.

    I don't see the direction of the acceleration in the truck in your sketch.
  7. Dec 8, 2013 #6
    Oh no! I've been very stupid. It's a string, so of course it can't be in compression. It is in tension though, because the helium is buoyant, so that the upthrust acting on the balloon is greater than the weight. However, the buoyant force always acts directly through the centre of mass, so won't be affected by the angle the ballon makes with the vertical.
    The tension though is different. I mean, as the balloon swings through an angle,we gain a horizontal component of Tension (acting in a direction opposite to acceleration) which we didn't have in the stationary case. This equates to the mass of the balloon times it's horizontal acceleration (the same acceleration as the truck.?)

    Overall though, this would suggest then suggest that tan(θ) >0 ? (-ve vertical component/ -ve horizontal component) of Tension =tan(θ) => tan(θ) is positive., so θ>0. So swings backwards! But that's not the right answer according to my book!!

    Help! :P
  8. Dec 8, 2013 #7
    Ah sorry mfb, so it's impossible to ignore the air in the truck as it affects the buoyancy of the balloon? Since it's the displacement of the air molecules which provides the upthrust, so it makes sense that upthrust depends on the denisty of the air....?
  9. Dec 8, 2013 #8
    What do you mean though that buoyancy is equivalent to the pressure gradient? I've not heard of this before!:P
  10. Dec 8, 2013 #9


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    The direction of buoyancy depends on the acceleration of the truck... and its magnitude is always larger than the gravitational force.
  11. Dec 8, 2013 #10
    How do we know the magnitude is always greater than gravity?
  12. Dec 8, 2013 #11


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    isn't the buoyant force always vertical (at least if you regard the weight as always vertical)?
    correct :smile:
    i'm really not following this without equations and a proper diagram

    how can you expect to find whether there's a plus or a minus if you don't do everything rigorously? :redface:
    archimedes' principle? :wink:
  13. Dec 8, 2013 #12
    Because I'm more interested in the principles of the situation than getting an exact number for the angle. It should be possible to find out intuitively whether or not it is positive or negative, I feel. The approach above was wrong though, I think because of what Mfb said. I'd ignored all the air in the container, hence incorrect conclusion. Though it probably was a bit lazy not to draw another diagram ;) Sorry.

    Yes to the archimedes too. But if we've got a pressure gradient going on, how do we know that there is always enough air to displace to support the balloon? (i.e the density of air changes due to the pressure gradient, so could there be a point where with sufficient acceleration, the ballon actually does not float?). I don't know! ;)
  14. Dec 8, 2013 #13


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    The balloon would not float otherwise.

    Vertical relative to the proper reference, which is not the direction of the center of earth here :wink:.

    You cannot calculate the angle anyway without more data.

    This is an issue if your truck accelerates with more than 10000 times the gravitational acceleration. It does not.
  15. Dec 8, 2013 #14

    Ray Vickson

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    This was a question on a PhD comprehensive exam at MIT back in the 1960s.

    The easiest solution is to use the "Principle of Equivalence": an acceleration is indistinguishable from gravitation. So, as the truck accelerates forward it "feels like" there is a slight backward component in the gravitational force; that is, gravity acts down and back. Since we know that helium-filled balloons float in a direction opposite to that of gravity, the balloon's string must point forward. You can even work out angles, etc., from this.
  16. Dec 9, 2013 #15


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    hi bananabandana! :smile:

    i've thought some more about what your book said …
    … i originally had the impression that it meant that there's only a front-to-back pressure gradient, which is incorrect (and gives the wrong answer for the furry dice)

    but now i think it means that there's an additional front-to-back pressure gradient, which has to be added to the "ordinary" top-to-bottom pressure gradient

    as mfb :smile: said …
    … it always follows the "adjusted gravity"

    to add to what Ray :smile: says …
    … when the car is stationary, an upright cube of air stays where it is because its weight ρgh3 is balance by the difference in force between the top and bottom surfaces, ie ∆pressure*area = ρgh*h2

    applying the principle of equivalence (in the non-inertial frame of reference of the moving car), an "adjusted upright" cube of air stays where it is, which can only happen if the pressure difference is exactly in the same direction as the "adjusted gravity" …

    ie (in the non-inertial frame of reference of the moving car) the pressure gradient and the adjusted gravity are both in the same direction, down and back
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