# The hummingbird glass box problems

Andrew Hamilton
Posers and Puzzles 12 Nov '08 14:40
1. 12 Nov '08 14:40
Here is some problems I thought up:
-they are really more of physics problems rather than ones of just logic:

A hummingbird sits at a bottom of an air-tight glass box and you then weigh the box and the hummingbird using some vary sensitive scales.

Then, while you are weighing this box with the hummingbird inside, you see the hummingbird suddenly takes off and starts to hover in the space exactly in the centre of the box so that it is no longer exerting weight down on the bottom of the glass box via its legs -do the scales momentarily show a different reading the moment when the hummingbird takes off but before the hummingbird reaches the centre of the box? -if so, very basically, how does the reading change with time?

-here’s an easier question:
once the hummingbird is hovering in a stationary position relative to the box, is the reading from the scales any different from what it was when the hummingbird was sitting down?

Now you repeat the measurements except this time with the top of the glass box removed so that the box is no longer air-tight. The hummingbird sits at a bottom of this air-tight glass box and you then weigh the box. Then, while you are weighing this box with the hummingbird inside, you see the hummingbird suddenly takes off and starts to hover in the space exactly in the centre of the box so that it is no longer exerting weight down on the bottom of the glass box via its legs. Once the hummingbird is hovering in a stationary position relative to the box, is the reading from the scales any different from what it was when the hummingbird was sitting down?

Now the hummingbird, while it is hovering, very slowly and at a constant speed starts to rise straight upward so that it goes from being at the centre of the box to then being where the lid of the box once was and then to being above and outside the box and continuous upwards until it is a mile above the box -do the reading from the scales change with time and, if so, very basically, how do they change with the hummingbirds position relative to the box?
2. 12 Nov '08 17:29
Most of the last part is irrelevant, the scales work by measuring the downforce applied by the Box onto its measuring surface.
When the hummingbird is sat on the bottom it will measure the downward force of the box and the Hummingbird. Once the hummingbird takes off it is exerting no downward force onto the scales thus the scales will only measure the weight of the box. Whether the Hummingbird is one inch or one mile from the bottom of the box is irrelevant as in all cases it is no longer exerting a force onto the scales
3. 12 Nov '08 19:367 edits
Originally posted by scarbelly
Most of the last part is irrelevant, the scales work by measuring the downforce applied by the Box onto its measuring surface.
When the hummingbird is sat on the bottom it will measure the downward force of the box and the Hummingbird. Once the hummingbird takes off it is exerting no downward force onto the scales thus the scales will only measure the we ...[text shortened]... bottom of the box is irrelevant as in all cases it is no longer exerting a force onto the scales
Wrong! ðŸ™‚
The only part that you got right there is “When the hummingbird is sat on the bottom it will measure the downward force of the box and the Hummingbird”
Anyone like to explain why all the rest is wrong?

Hints:

1, The weight of the bird is the same regardless of whether or not it is hovering.

2, when the bird flaps its wings to hover, it is exerting a downward force against the air below it
-think about how the forces are being transferred and how they relate to each other.

3, the bird has to first accelerate and then decelerate to move from the sitting position to the hovering position (and remember F=ma and every acceleration has a net force that causes that acceleration) and ever force has an equal but opposite force
-again, think about how the forces are being transferred and how they relate to each other.

I hope my hints don’t spoil it by giving too much away.
4. 13 Nov '08 02:00
The scale would probably change very, very slightly if it is very, very accurate. When the bird is hovering, its distance from the Earth increases, lessening the gravitational force between the bird and the Earth. However, I doubt this is significant enough to be taken in consideration, so for all intents and purposes, the bird's weight is the same wherever it is in the box.

While the bird is accelerating upward, the air pushes it upward at a GREATER magnitude than gravity is pulling it downward. For the air to push the bird, the bird must provide the equal and opposite force downward on the air. Thus, this downward force is greater than the force of gravity and the reading on the scale will increase. When the bird decelerates, the reading on the scale would decrease to the original reading for similar reasons.
5. 13 Nov '08 10:051 edit
Originally posted by twilight2007
The scale would probably change very, very slightly if it is very, very accurate. When the bird is hovering, its distance from the Earth increases, lessening the gravitational force between the bird and the Earth. However, I doubt this is significant enough to be taken in consideration, so for all intents and purposes, the bird's weight is the same wher ...[text shortened]... celerates, the reading on the scale would decrease to the original reading for similar reasons.
…The scale would probably change very, very slightly if it is very, very accurate. When the bird is hovering, its distance from the Earth increases, lessening the gravitational force between the bird and the Earth. However, I doubt this is significant enough to be taken in consideration, so for all intents and purposes, the bird's weight is the same wherever it is in the box.
. …

Correct -and thus the scales will measure both the box’s weight AND the bird’s weight regardless of whether it is hovering or sitting down inside the box.
As the bird hovers, the hovering bird will create a constant area of slightly higher air pressure below it compared to the slightly lower air pressure above it. In effect, the bird’s weight will be resting on the air below it which, in turn, will be resting on the bottom of the box thus the birds weight will still be exerted onto the bottom of the box (albeit indirectly) despite the fact it is hovering.

…While the bird is accelerating upward, the air pushes it upward at a GREATER magnitude than gravity is pulling it downward. For the air to push the bird, the bird must provide the equal and opposite force downward on the air. Thus, this downward force is greater than the force of gravity and the reading on the scale will increase. When the bird decelerates, the reading on the scale would decrease to the original reading for similar reasons.
...…

Correct.

-that just leaves only one question unanswered;

what happens to the scale readings if you take the lid off the top of the box and the hovering bird slowly rises up from inside the box to above it and continues upwards?
6. TheMaster37
Kupikupopo!
13 Nov '08 10:47
I Interpret "slowly rising" as "not accelerating".

No acceleration means no resultant force, so that doesn't change over time.

The only thing that is changing is the distance of the bird to the scale. In the posts above is mentioned that that doesn't affect the reading on the scale.

With that I have to conclude that nothing changes; a bird in a box weighs the same as an empty box ðŸ™‚
7. 13 Nov '08 11:144 edits
Originally posted by TheMaster37
I Interpret "slowly rising" as "not accelerating".

No acceleration means no resultant force, so that doesn't change over time.

The only thing that is changing is the distance of the bird to the scale. In the posts above is mentioned that that doesn't affect the reading on the scale.

With that I have to conclude that nothing changes; a bird in a box weighs the same as an empty box ðŸ™‚
…I Interpret "slowly rising" as "not accelerating".

No acceleration means no resultant force, so that doesn't change over time.
. …

Correct.

…The only thing that is changing is the distance of the bird to the scale. In the posts above is mentioned that that doesn't affect the reading on the scale.
...…

Not necessarily.

…With that I have to conclude that nothing changes; a bird in a box weighs the same as an empty box
….

How can that be true?

If the bird weights, say, 100g, and the box weights, say, 1000g, then the bird + the box weights 1100g and the empty box must surely simply weight just 1000g and not 1100g?

But if you try and imagine a bird hovering, say, a mile above the box, intuitively, do you think the scales would measure the weight of BOTH the box AND the bird!? -if not, shouldn’t there be some change in the reading of the scales at the bird moves upwards at a steady speed out of the box? -if so, would that change in the reading relate to how high the bird is above the box? -or exactly what that change in the reading would relate to if not to simply how high the bird is above the box?
8. divegeester
13 Nov '08 12:261 edit
After being arrested for keeping a bird in a box you could consider (from the confinement of your padded cell) that all forces are relevant and energy cannot be destroyed, only transfererd.

Therefore it is a matter of sensitivity of the scales and dissipation of the energy in the air which are the critical factors.

If the hummingbird was in a box the same width but a mile high with the bird hovering at the top, and the scale was sensitive enough, it would be able to measure the downward force of its wings through the increased air pressure on the scale.

Accepting: Minor energy loss from heat via molecule aggitation, depleted gravitational effects and the interference by the RSPCA!
9. 13 Nov '08 15:251 edit
If the bird is in the box, then the reading shouldn't change (barring changes in gravity with height).

If the brid flies out of the box, I suspect that the downward force is spread over a larger area of the ground so only a fraction of the bird's weight will be measured. By the time it gets any significant distance up, turbulence effects will probably mean that the effect of the bird on the scales will be very small.

Obviously, scales don't change their readings when a bird flies directly overhead!
10. 13 Nov '08 20:25
Originally posted by Schumi
If the bird is in the box, then the reading shouldn't change (barring changes in gravity with height).

If the brid flies out of the box, I suspect that the downward force is spread over a larger area of the ground so only a fraction of the bird's weight will be measured. By the time it gets any significant distance up, turbulence effects will probably mean ...[text shortened]... small.

Obviously, scales don't change their readings when a bird flies directly overhead!
…If the bird flies out of the box, I suspect that the downward force is spread over a larger area of the ground so only a fraction of the bird's weight will be measured.
. …

I think you are correct -I think you have hit it on the nail there.
11. 13 Nov '08 20:332 edits
Originally posted by divegeester
After being arrested for keeping a bird in a box you could consider (from the confinement of your padded cell) that all forces are relevant and energy cannot be destroyed, only transfererd.

Therefore it is a matter of sensitivity of the scales and dissipation of the energy in the air which are the critical factors.

If the hummingbird was in a ...[text shortened]... heat via molecule aggitation, depleted gravitational effects and the interference by the RSPCA!
…After being arrested for keeping a bird in a box. …

Perhaps not arrested for simply keeping a “bird” in a glass box? (unless it is airtight or you leave it in direct sunlight) but rather, specifically, arrested for keeping a hummingbird in a glass box, because deprive a hummingbird of the ability to travel to nectar-rich flowers while it is awake usually results in it dieing of starvation within only an hour or two! (although I really don’t know what the law says about this)

-but, I suppose you could provide it with an artificial “nectar-rich flower” in the box?
12. uzless
The So Fist
19 Nov '08 21:07
Originally posted by Andrew Hamilton
Here is some problems I thought up:
-they are really more of physics problems rather than ones of just logic:

A hummingbird sits at a bottom of an air-tight glass box and you then weigh the box and the hummingbird using some vary sensitive scales.

Then, while you are weighing this box with the hummingbird inside, you see the hummingbird sudden ...[text shortened]... d, if so, very basically, how do they change with the hummingbirds position relative to the box?
Each wing beat would produce a reading on the scale of larger than the resting weight, followed by less than, followed by greater than etc etc etc.

Simply because when the upswing beat occurs the bird is forcing air up, thus reducing the downward pressure on the scale.
13. AThousandYoung
All My Soldiers...
20 Nov '08 05:391 edit
Originally posted by Andrew Hamilton
Here is some problems I thought up:
-they are really more of physics problems rather than ones of just logic:

A hummingbird sits at a bottom of an air-tight glass box and you then weigh the box and the hummingbird using some vary sensitive scales.

Then, while you are weighing this box with the hummingbird inside, you see the hummingbird sudden ...[text shortened]... d, if so, very basically, how do they change with the hummingbirds position relative to the box?
do the scales momentarily show a different reading the moment when the hummingbird takes off but before the hummingbird reaches the centre of the box?

Yes. The hummingbird's upward motion is caused by a force, from it's wings and maybe legs as it hops up. This causes a downward force on the box, causing the scales to indicate a larger weight than you'd think.

if so, very basically, how does the reading change with time?

Afterwars the weight should fluctuate all over the place and eventually settle on the normal weight.

once the hummingbird is hovering in a stationary position relative to the box, is the reading from the scales any different from what it was when the hummingbird was sitting down?

No. The downward force of the wings on the air leads to the air pushing down on the scales exactly as much as the bird would while standing. We know this since the force to keep the bird up is exactly enough to counter it's weight, and the equal and opposite reaction is pushing down on the box.

Now you repeat the measurements except this time with the top of the glass box removed so that the box is no longer air-tight. The hummingbird sits at a bottom of this air-tight glass box and you then weigh the box. Then, while you are weighing this box with the hummingbird inside, you see the hummingbird suddenly takes off and starts to hover in the space exactly in the centre of the box so that it is no longer exerting weight down on the bottom of the glass box via its legs. Once the hummingbird is hovering in a stationary position relative to the box, is the reading from the scales any different from what it was when the hummingbird was sitting down?

My best guess is that the situation is unchanged with the top off.

Now the hummingbird, while it is hovering, very slowly and at a constant speed starts to rise straight upward so that it goes from being at the centre of the box to then being where the lid of the box once was and then to being above and outside the box and continuous upwards until it is a mile above the box - do the reading from the scales change with time and, if so, very basically, how do they change with the hummingbirds position relative to the box?

When the bird leaves the box, some of the downward force from it's wings will miss the box and flow outside of it. Thus as the bird gets farther from the box, it's weight will be removed slowly from the scales until they read only the weight of the box itself.
14. TheMaster37
Kupikupopo!
20 Nov '08 07:18
Originally posted by AThousandYoung

When the bird leaves the box, some of the downward force from it's wings will miss the box and flow outside of it. Thus as the bird gets farther from the box, it's weight will be removed slowly from the scales until they read only the weight of the box itself.
I think this is the correct way to look at it.

The bird is leaning on a cone of air below it. As the bird is flying low, the cone is small and entirely in the box. The higher the bird flies the bigger the cone becomes and more of it will be outside the box.
15. sonhouse
Fast and Curious
29 Nov '08 14:461 edit
Originally posted by TheMaster37
I think this is the correct way to look at it.

The bird is leaning on a cone of air below it. As the bird is flying low, the cone is small and entirely in the box. The higher the bird flies the bigger the cone becomes and more of it will be outside the box.
You really have to think outside the box for this one"ðŸ˜‰
One force overlooked here is the audio vibration the wings would produce, making a sine-ish wave that would wash over the sides of the container, increasing slightly the temperature of the overal box vs the outside temp, plus the temp of the bird adding to the temp of the air inside the lidded box, so the air pressure would also go up slightly due to the increase in temperature, but you would have to have a very sensitive pressure scale, not sure if your average capacitance manometer would measure it though.