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This question already has an answer here:

Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?

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marked as duplicate by John Rennie, Emilio Pisanty, Qmechanic Oct 26 '13 at 0:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The weight will be the same on average over time.

The bird is supported by the air which in turn is supported by the box. To every action there is an equal and opposite reaction so the force supporting the bird must be transferred to the bottom of the box if no air can escape. However as the bird and air move around the overall centre of mass can move up and down so the force will only equal the total weight on average over time.

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The bird can hover because it is forcing air down with its wings at a force equal to its own weight. This force is continuously existing at the bottom of the box as the air decelerates on it. It's as if the bird was directly connected to the bottom--just sitting there minding it's birdly business.

If the bird is moving up or down in the box however...that's a different story. Any such movement indicates an acceleration, which will change the weight reading on the scale (it measures weight not mass).

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  • $\begingroup$ So as the bird is flying without moving up or down, the masses will be equal but if the bird changes its altitude, they will be different. Is it so? $\endgroup$ – moray95 Oct 23 '13 at 19:45
  • $\begingroup$ @moray95, yes it is so $\endgroup$ – Pranav Hosangadi Oct 24 '13 at 0:02

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