A Fly in a Moving Car

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  • #1
In a car moving forward at 60 mph a fly is buzzing freely. If the car suddenly comes to a complete stop, does the fly crash into the windshield?
 

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  • #2
mbisCool
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Why don't you test it out?
 
  • #3
Stratosphere
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In a car moving forward at 60 mph a fly is buzzing freely. If the car suddenly comes to a complete stop, does the fly crash into the windshield?

Of course.
 
  • #4
Ivan Seeking
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Smoke, however, will move towards the back of the car. Why?
 
  • #5
NeoDevin
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Smoke, however, will move towards the back of the car. Why?

Same reason it floats upwards.
 
  • #6
humanino
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Smoke, however, will move towards the back of the car. Why?
Same direction as an inflatable balloon I would guess : because it's afraid of the windshield of course.
 
  • #7
humanino
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Same reason it floats upwards.
What a deep statement... indeed gravity is also an acceleration, and we're all afraid of falling.
 
  • #8
Ivan Seeking
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Okay, how about the classic: If a bird is sitting inside and on the bottom of a sealed box, and the bird begins to fly - hovers in place inside of the box - does the box lose weight?
 
  • #9
humanino
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Okay, how about the classic: If a bird is sitting inside and on the bottom of a sealed box, and the bird begins to fly - hovers in place inside of the box - does the box lose weight?
The flying bird will release heat, increasing the temperature of the box which will radiate energy away : the box does loose weight until the bird dies of hunger.
 
  • #10
junglebeast
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Okay, how about the classic: If a bird is sitting inside and on the bottom of a sealed box, and the bird begins to fly - hovers in place inside of the box - does the box lose weight?

I'm thinking that the wing beats create a downward air pressure that increases the weight of the box.
edit -- hmm...that's probably wrong since there is no source of new air...
 
  • #11
physics girl phd
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Okay, how about the classic: If a bird is sitting inside and on the bottom of a sealed box, and the bird begins to fly - hovers in place inside of the box - does the box lose weight?

The mass "m" of the box itself doesn't change, and the height certainly doesn't change significantly enough to change the value of little "g" (even if it's bouncing around), so the weight of the box doesn't change. Now if the box was originally sitting on a scale, and you wanted the net force of the box/bird system on the scale, that's a different question.
 
  • #12
Andre
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Smoke, however, will move towards the back of the car. Why?

That would only be true if smoke is heavier than air. But is it?

Test it! have somebody who smokes (not me) blow a bubble filled with smoke and compare the speed at which it sinks with that of a normal bubble.

Then why does smoke rise? Not because it is lighter than air.
 
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  • #13
Blenton
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Yes the box on the scale would see a decrease ~ the bird is no longer part of the boxes structure.
 
  • #14
NeoDevin
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Okay, how about the classic: If a bird is sitting inside and on the bottom of a sealed box, and the bird begins to fly - hovers in place inside of the box - does the box lose weight?

The net downward force of the air on the bottom of the box has to equal the net upward force of the air on the bird. There'll be slight fluctuations, as the bird flaps and falls, but on average no weight will be lost.
 
  • #15
tiny-tim
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Then why does smoke rise? Not because it is lighter than air.

I think it's because the smoke is hot, and heats the air, and the smoke is then carried on the updraft of hot air.

That's why smoke will be attracted to a chimney even if the chimney is not directly above … it follow the air current. :smile:
In a car moving forward at 60 mph a fly is buzzing freely. If the car suddenly comes to a complete stop, does the fly crash into the windshield?
Of course.

I don't know :confused: … won't the fly reach terminal velocity very quickly … alternatively, won't the air suddenly become very "viscous"?
 
  • #16
Pythagorean
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Yes the box on the scale would see a decrease ~ the bird is no longer part of the boxes structure.

actually, it is part of the boxes structure through the air in the box. If the bird lifts up off the box, he will have to be exerting force downwards (on the air) some of which will push down on the box, registering on the scale.

Of course, the full weight of the bird probably won't show up on the scale as some of the air molecules push against the sides of the box, which doesn't register on the scale. I don't know how they would relate mathematically, but I'm guessing you'd have to use navier-stokes or some what crap.
 
  • #17
Ivan Seeking
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That would only be true if smoke is heavier than air.

Really?

Btw, I have tried it.
 
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  • #18
Ivan Seeking
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Now if the box was originally sitting on a scale, and you wanted the net force of the box/bird system on the scale, that's a different question.

:rolleyes: Alright, the box is sitting on a scale and the bird begins to fly inside of the box... :tongue:

It was intended as a tricky question but not a trick question. :biggrin:
 
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  • #19
humanino
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:rolleyes: Alright, the box is sitting on a scale and the bird begins to fly inside of the box... :tongue:
Can you provide the accuracy of the scale, an order of magnitude for the proper mass of the box (not counting air inside), the inner volume of the box, the mass of the bird, and also the temperature and pressure if they are not standard. Then we could make an actual statement : this is not measurable...
 
  • #20
Pythagorean
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Can you provide the accuracy of the scale, an order of magnitude for the proper mass of the box (not counting air inside), the inner volume of the box, the mass of the bird, and also the temperature and pressure if they are not standard. Then we could make an actual statement : this is not measurable...

just keep it theoretical

V = volume
m = mass of bird
M = mass of box

etc...
 
  • #21
junglebeast
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actually, it is part of the boxes structure through the air in the box. If the bird lifts up off the box, he will have to be exerting force downwards (on the air) some of which will push down on the box, registering on the scale.

Of course, the full weight of the bird probably won't show up on the scale as some of the air molecules push against the sides of the box, which doesn't register on the scale. I don't know how they would relate mathematically, but I'm guessing you'd have to use navier-stokes or some what crap.

Although the air pressure directly the bird will be greater, if the box is large enough to hold a flying / flapping bird, then the air molecules (which can travel much faster than the bird's wing) will likely dissipate that downward pressure so that it is a uniform pressure in all directions, such that the added weight to the box is an imperceptible fraction of the bird's weight.
 
  • #22
humanino
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just keep it theoretical
Fine, but I already answered : the mass decreases due to loss of energy by electromagnetic radiation ! This is obviously negligible compared to even the thermal fluctuations of the scale, but theoretically, it is the only energy out of the box.
 
  • #23
Pythagorean
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Although the air pressure directly the bird will be greater, if the box is large enough to hold a flying / flapping bird, then the air molecules (which can travel much faster than the bird's wing) will likely dissipate that downward pressure so that it is a uniform pressure in all directions, such that the added weight to the box is an imperceptible fraction of the bird's weight.

yeah, I think that would be a specific case. While we're building the equation to describe this physical event, we should include the term and if epsilon goes to zero, so be it, but then we can find the critical point (box volume to bird's wing surface area or something) at which this is true.
 
  • #24
Pythagorean
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Fine, but I already answered : the mass decreases !

what a safe answer! :P
 
  • #25
junglebeast
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just keep it theoretical

V = volume
m = mass of bird
M = mass of box

etc...

This question is governed by the chaotic fluid dynamics of air currents. How can you write that as a simple algebraic equation without neglecting the factors that actually cause the real world result? If you only look at mass, volume, etc, you would conclude that an airplane cannot possibly fly because you ignored the effect of air resistance.
 
  • #26
Pythagorean
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This question is governed by the chaotic fluid dynamics of air currents. How can you write that as a simple algebraic equation without neglecting the factors that actually cause the real world result? If you only look at mass, volume, etc, you would conclude that an airplane cannot possibly fly because you ignored the effect of air resistance.

look up etc. sometime.
 
  • #27
humanino
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what a safe answer! :P
The thing is : I don't care of the hydrodynamical details of how the bird is still in the box. I have computers, I can get much more complicated stuff worked out provided I use enough power, but this is irrelevant, and not the right way to calculate it. If the question is "will the mass change" the only way for the mass of the box to change is by giving away energy, period. As I said, a much more relevant question is to get into actual numbers, to estimate the thermodynamical fluctuations and compare them to (1) the mass loss (2) the accuracy of the scale.
 
  • #28
tiny-tim
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Is this an unladen bird? :smile:
 
  • #29
Pythagorean
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The thing is : I don't care of the hydrodynamical details of how the bird is still in the box. I have computers, I can get much more complicated stuff worked out provided I use enough power, but this is irrelevant, and not the right way to calculate it. If the question is "will the mass change" the only way for the mass of the box to change is by giving away energy, period. As I said, a much more relevant question is to get into actual numbers, to estimate the thermodynamical fluctuations and compare them to (1) the mass loss (2) the accuracy of the scale.

Hrmm, I guess I didn't take it as the mass particularly changing. I was looking at the force on the scale changing.
 
  • #30
DaveC426913
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Although the air pressure directly the bird will be greater, if the box is large enough to hold a flying / flapping bird, then the air molecules (which can travel much faster than the bird's wing) will likely dissipate that downward pressure so that it is a uniform pressure in all directions, such that the added weight to the box is an imperceptible fraction of the bird's weight.
That would be an interesting trick. The implication is that, if you set something flapping in a closed environment, you can cause the whole environment to weigh less.

Commerical airliners would be all over this! If you set it up right, you could make a plane weightless just by putting enough birds in it.


No, it doesn't work. The downward pressure is not dissipated. Any air that moves towards the side or top of the box will have to displace air that's already there, pushing it down. The net pressure of all the air in the box will be downward and of a magnitude exactly equaling the weight of the bird.
 
  • #31
Pythagorean
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That would be an interesting trick. The implication is that, if you set something flapping in a closed environment, you can cause the whole environment to weigh less.

Commerical airliners would be all over this! If you set it up right, you could make a plane weightless just by putting enough birds in it.


No, it doesn't work. The downward pressure is not dissipated. Any air that moves towards the side or top of the box will have to displace air that's already there, pushing it down. The net pressure of all the air in the box will be downward and of a magnitude exactly equaling the weight of the bird.


I've been imagining the top off the box.

edit: I just realized it was a crappy cardboard box too and the bird is a hummingbird, as well. I'm taking a lot for granted here.
 
  • #32
DaveC426913
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I've been imagining the top off the box.
Then the bird would fly out... :confused:
 
  • #33
Pythagorean
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Then the bird would fly out... :confused:

hummingbirds can hover, and this particular hummingbird is very cooperative.
 
  • #34
humanino
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A truck carrying birds will be lighter if the birds fly around than if the birds sit/stand.
BUSTED
Adam and Jamie constructed a large box and placed it on top of scale and then filled it with captured pigeons. Then, the Mythbusters activated a special contraption that would force the pigeons to fly into the air, but they could not detect any discernible difference in the weight of the box. They then placed a model helicopter inside the box and had it hover above the ground, but this method also failed to produce any results. The Mythbusters theorized that the air being displaced by the birds’ wings and the helicopter rotors was pressing down the box, which is why there was no change in the overall weight.
It also works with an helicopter apparently :rofl:, although there is no mention of the fluctuations in the balance reading, which I guess with a helicopter should definitely be detectable !
 
  • #35
efekwulsemmay
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Don't you just love how A fly in a moving car goes to the bird in the box question.
:D
 

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