Can the weight of a bird be measured, considering air friction? 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?
 A: 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.
A: 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. 
A: 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.
A: 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
