Okay this is a bit of an old classic, but I don't think I've heard a good reasonable explanation for what happens: If you put a cage of canaries on a weigh scale and they take off does the scale still register the weight of the canaries? Does it matter if the cage is sealed or open to the air? What happens if there is no cage but a canary flies over the scale?
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2$\begingroup$ There is a good discussion about a similar situation in physics.stackexchange.com/q/519504/195949 $\endgroup$– Claudio SaspinskiCommented Dec 31, 2022 at 21:17
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1$\begingroup$ See youtube.com/watch?v=lVeP6oqH-Qo $\endgroup$– David WhiteCommented Dec 31, 2022 at 23:55
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1$\begingroup$ Can you Post either any research or failing that, your own argument? $\endgroup$– Robbie GoodwinCommented Jan 1, 2023 at 3:17
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/77626/2451 , physics.stackexchange.com/q/12756/2451 , physics.stackexchange.com/q/481711/2451 , and links therein. $\endgroup$– Qmechanic ♦Commented Jan 1, 2023 at 13:23
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1$\begingroup$ @Qmechanic Oh no, marking this question as a duplicate was unfortunate. Really took some effort into answering it! $\endgroup$– AlphaLifeCommented Jan 1, 2023 at 13:38
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4 Answers
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Assume a cage with solid walls, with the cage resting on the ground and the birds initially at rest on the bottom of the cage. Treating the cage and the birds as a system, the external forces on the system are: the weight of the cage downwards, the weight of the birds downwards, and the force of the ground on the cage upwards. The net external force determines the acceleration of the center of mass (CM). If the birds suddenly accelerate upwards, the CM acclerates upwards so the force from the ground exceeds the weights of the cage and birds; the force of the cage/birds on the ground (the weight shown on a scale) is the negative of the force of the ground on the cage/birds so the scale reads higher if the birds accelerate upwards. If the birds then move upwards at constant velocity, the acceleration of the CM is zero so the scale reading goes back to the initial value with the birds not moving.
If the cage is open to the outside air, the interaction with the surroundings must be considered. In this case it is easier to consider the cage alone as the system and the downward force on the cage is its weight plus the force from pressure of air on the bottom of the cage; the scales read the total of the weight of the cage and the force from the air pressure. For a bird flying over the ground, if the air pressure on the cage increases, the scale reads slightly higher.
answered Dec 31, 2022 at 19:34
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1$\begingroup$ Correct. You can even demonstrate this by standing on a scale and moving your arms up and down, and see the weight change. $\endgroup$– RC_23Commented Dec 31, 2022 at 19:56
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1$\begingroup$ One thing, in order to "move upward at a constant velocity" as you say, the bird would have to be exerting an upward force against the air to counter 9.8 m/s^2. Would this force not be reacted against the air and floor and ultimately be read by the scale? $\endgroup$– RC_23Commented Dec 31, 2022 at 19:58
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2$\begingroup$ For a solid-walled cage. Consider a bird as the system. Moving at constant velocity there is no net external force on the bird, so regardless of the velocity the net pressure up on the bird equals the weight of the bird. If the bird changes velocity (accelerates) upwards the net pressure up on the bird exceeds the bird's weight and this increases the pressure on the bottom of the cage. Considering the cage and bird as the system, the CM is not accelerating if the bird is not accelerating, but the CM is accelerating if the bird is accelerating. $\endgroup$ Commented Dec 31, 2022 at 20:07
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If the cage is totally closed and airtight, and the speed is independent of time (fixed flight), I think that we will measure the total weight, including that of the canaries.
You just have to make a balance of momentum on the whole. Since the total momentum $\vec{P}$ is a constant, $\frac{d\vec{P}}{dt}=M_{tot}\vec{g}+\vec{R}=\vec{0}$
The total weight and the reaction of the weighing pan compensate each other. So the reaction of the scale is the opposite of the total weight.
Sorry for my poor english !
answered Dec 31, 2022 at 10:59
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It depends on the size of the weighing scale.
A bird is very much like a rocket. To go up a little, it needs to displace an equal weight of air down. So even if a canary is standing on the scale or hovering in the air, the weighing scale wouldn't make any difference (No matter if the cage is open or closed).
Notice that I used the word "hovering." If the bird accelerates upwards, it would need to displace more air than its weight. So then the weighing scale would show a higher value than the bird's weight at rest.
However, this is only true if the weighing scale can capture all of the air flowing down from the bird. If the platform is small, the reading would be less because the weight is now distributed between the ground and the scale.
What happens if no cage but a canary flies over the scale?
The weight would fluctuate and would show a peak when the bird flies right above the scale.
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When people walk on the scale, the reading of the scale will fluctuate. Similarly, when the bird flies in the box, the reading of the weight scale under the box will fluctuate. A bird flapping its wings is like a man walking on a scale.
