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Assume that a bird of mass 1.5 kg is held in a box at height 500m and this setup is placed on a weighing machine. Now, the bird starts to hover in the box pushing the air downwards, and is at a height of around 450 m. If we measure the weight of the box, will we measure the bird's weight also. Consider air friction.

I believe that due to friction of air, the downward movement of the air may not touch the bottom of the box (the bird is hovering at around 450 m height) and is lost. Hence, we cannot measure the weight of bird.

Is this true?

bobie
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Santhi Kabir
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  • The box is completely closed. How do you know that there is a bird in there and not a stone of equivalent weight? Would knowing that there is a hovering bird in there change the result of a scale that keeps weighing the entire box? Would a person who knows about the bird see a different result than a person who believes that there is a stone in there? – CuriousOne Dec 15 '14 at 04:49
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    Does the friction of water mean that the fish swimming in the ocean don't contribute to the pressure at the ocean floor? – Peter Shor Dec 15 '14 at 05:14
  • Consider a sealed chamber filled with air and a bird in space, with a rocket exerting a steady force on the chamber, which will accelerate it at 1G if the bird is resting on the "floor" of the chamber. If there was suddenly less "downward" force on the floor of the chamber to oppose the "upward" force from the rocket when the bird took to the air, then the floor would have to accelerate more quickly, but by conservation of momentum the center of mass of the chamber cannot accelerate any more quickly than if the bird was on the floor, since the momentum of the exhaust is the same either way. So – Hypnosifl Dec 15 '14 at 06:01
  • (continued) if the floor accelerates more quickly then 1G the only way the center of mass of the chamber can continue to move at 1G is if the bird (or the air, I suppose) is suddenly rushing downward relative to the floor. So if the bird maintains a constant height on average and the density of air as a function of height is constant on average too, this implies that the time-average of the force on the floor must be the same when the bird is flying as when the bird is resting on the floor. – Hypnosifl Dec 15 '14 at 06:03

4 Answers4

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Let's assume that the box you have is perfectly closed and has a fixed amount of air in it. When the bird inside starts flapping it's wings it creates disturbances in the air present inside the box. The air molecules may start dancing in complex ways and it is difficult to completely describe this motion qualitatively.

Consider the whole system (box + air + bird). It's reasonable to assume that after some time the system will reach equilibrium (i.e when the bird reaches 450m and the air circulation stabilizes). The center of mass of this whole system will be stationary in equilibrium and hence the net external force acting on this system must be zero.

What are the external forces acting on this system?

  • Force of gravity acting on box
  • Force of gravity acting on bird
  • Force of gravity acting on air molecules (which is relatively negligible)
  • Normal reaction given by the weighing machine (i.e normal contact force)

(force of air on bird, air on box etc. are all internal to this system)

Hence, Normal reaction = weight of box + weight of bird and that is exactly what the weighing machine measures.

The air friction does lead to dissipation of energy. Where does this energy come from? As the bird is continually flapping it's wings, it is spending some biological energy in the process. It is this energy which is converted into mechanical energy (wing flapping) and transferred to the surrounding air molecules and is later partially dissipated due to air friction.

Quark
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  • Please clarify me on the normal reaction. How does the weight of bird is added to it, as the bird is not in contact with the floor, and the energy by flapping is dissipated due to air friction. – Santhi Kabir Dec 15 '14 at 07:42
  • @Santhi: It comes out as extra air pressure on the bottom of the box. The wings get lift by pushing air downward, which hits the bottom of the box, which pushes the air back up so it stays in the box. That means the bottom of the box is feeling more pressure. – Mike Dunlavey Dec 15 '14 at 13:04
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The bird is hovering in the box. The only way for it to hover is to increase the pressure underneath its wings and decrease the pressure above its wings. This pressure differential times the area of the bird will balance the exact weight of the bird.

The pressure differential may be thought of as a net downward impulse given to the air molecules by the bird. This net downward momentum will be conserved, as the air molecules bounce down towards the bottom of the box. Energy dissipation does not affect the net momentum of the air particles.

Eventually, some air molecules hit the bottom of the box, and the downward momentum of the molecules is transferred to the bottom of the box. In response, the bottom of the box pushes the air molecules up towards the bird, which the bird then pushes down again.

The compressed air below the bird may be thought of as a static compressed spring, which gives an identical upward force to the bird and downward force to the bottom of the box. The box then transfers this downward force to the scales, as surely as if the bird were simply standing on the box.

Further analysis would examine the low pressure above the bird which is like a stretched spring from the bird to the top of the box. The stretched spring pulls the top of the box down - a downward force which is also transferred to the scales.

Together the rarefied air above the bird (stretched spring pulling down the top of the box), and compressed air below the bird (compressed spring pushing down the bottom of the box) transfer the full weight of the bird to the box, which passes it on to the scales.

Richard
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  • I did some experiment like, I filled a balloon with air and released the air through the nozzle. I placed a paper near the end of the nozzle. The paper fluttered because of the momentum exchange. I moved certain distance(considering enough distance) away from the paper and did the same thing. This time, the paper did not flutter. Could you explain this to me. I thought that the above question is an analogy to the observation I made. But the answer gives different explanation. – Santhi Kabir Dec 15 '14 at 15:12
  • Will the Brownian motion of air have an effect on the motion of the particles and cause the momentum to dissipate? – Santhi Kabir Dec 15 '14 at 16:07
  • @SanthiKabir Under no circumstances can momentum ever dissipate. Momentum is always exactly conserved, even under relativity and quantum mechanics. Kinetic Energy can dissipate as it is transformed into other kinds of energy like thermal energy, but that can not affect the momentum. The difference with your first experiment is that the bird is in a closed box. The momentum from the balloon spreads out through the air, and continues until it hits the wall of the room you are in. As it is spread out over a greater mass of air, the velocity will be smaller, but the momentum will be conserved. – Richard Dec 15 '14 at 23:49
  • In the bird experiment, assuming the box has area 10x the surface area of bird or more, due to molecular collision between air molecules the momentum transfers and some molecules hits the side walls. So, the exact momentum will not transfer to the bottom of the box, and hence, I think we cannot measure the exact weight of the bird. Is this true? – Santhi Kabir Dec 16 '14 at 04:05
  • Momentum is always, always, always, conserved. The molecules can hit the side of the box, and momentum will be conserved. The molecules can spontaneously form a black hole and momentum will be conserved. The full gravity force from the bird hovering will always be transferred to the box. – Richard Dec 16 '14 at 04:11
  • The weighing machine can only measure the normal reaction, and the molecules that hit the bottom of the box, can only cause normal reaction. I just want to get clarity on this. – Santhi Kabir Dec 16 '14 at 04:17
  • What is inside the box is irrelevant. In this simple example, all you need to know about it is the combined mass of the box and contents, and the centre of mass. If the bird is in a stable hover, totally enclosed inside the box, you can effectively ignore the independence of the bird. It is part of the box. Any downward momentum must get transferred to the box, no matter what the shape or size of the box is. You sound smart enough to think of "what if"s forever, but every experiment for 100s of years has validated these conservation principles. – Richard Dec 16 '14 at 04:31
  • You are correct that you only need to consider normal forces. But air is complicated, and you also need to remember that the top of the box is pushed down by atmospheric pressure because of the rarefaction of the air above the bird. – Richard Dec 16 '14 at 04:36
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Assuming is a black box seen from the outside, consider the following situation: After a while, the bird tries to fly higher and pushes the ceiling of the box upwards. Will the weight of the box decrease? Then, after a while, the bird dies and falls to the floor. Will the weight of the box increase? If the box is sealed, it cannot change the time averaged weight, so the box will weight the same in the three situations.

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This is not true, when the box is sealed you are weighing the total mass of the box and everything in it, not the weight of the bird plus the weight of the box

Daz Hawley
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