Assume that you are holding a cage containing a bird. Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage?
Does it make a difference whether the cage is completely closed or it has rods to let air pass?
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3$\begingroup$ Are you asking if the weight of the cage + bird is reduced if the bird is hovering in the middle of the cage? $\endgroup$– John RennieCommented Sep 17, 2013 at 10:49
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2$\begingroup$ @DImension10 Abhimanyu PS: When OP is just asking for a conceptional qualitative (as opposed to a quantitative) understanding of a physics phenomenon, we usually don't tag it as 'homework' (unless there are other indications that it is indeed homework). That said, I'm pretty sure it is a duplicate. Ah yes, here we go: physics.stackexchange.com/q/12756/2451 $\endgroup$– Qmechanic ♦Commented Sep 17, 2013 at 11:59
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1$\begingroup$ @Qmechanic: I tagged it as such, because it seemed to be pretty common to get this as homework, even if it is qualitative. $\endgroup$– Abhimanyu Pallavi SudhirCommented Sep 17, 2013 at 12:29
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$\begingroup$ M. Tarun, see youtube.com/watch?v=lVeP6oqH-Qo $\endgroup$– David WhiteCommented Feb 2, 2019 at 2:43
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5 Answers
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If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. Since no additional external force is acting on the cage-bird-system when the bird is flying as compared to when it's not, the acceleration on the cage can be no different. The effect due to the flying bird only concerns the internal forces and since action=reaction, they cancel.
However, if the cage would not be closed, some of the 'wind' due to the bird could escape the cage and would become an external force, making the cage-bird-system lighter.
answered Sep 17, 2013 at 11:34
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$\begingroup$ Wouldn't air from the environment replace the vacuum created by the 'escaping wind' thus the keeping the mass of the air in the cage constant and not make the bird-cage-system ligter? $\endgroup$– zundarzCommented Sep 17, 2013 at 16:18
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1$\begingroup$ Well, of course air from the environment would go back into the cage, but it does not (substantially) alter its weight. Careful, you are confusing weight (the force pulling something down which may alter) with mass (an invariant). $\endgroup$ Commented Sep 17, 2013 at 16:22
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3$\begingroup$ Maybe this will make it more believable: If you sit on a skateboard and you have a hair dryer with you, it will cause you to accelerate. But the hair dryer is not just expelling air, it's also sucking air in. Still, you accelerate. $\endgroup$ Commented Sep 17, 2013 at 16:24
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To answer the second question first, a completely closed cage is a system which comprises the exterior cage, the air inside and the bird. So what all forces do we need to balance now? Gravity on air, bird and the cage. OK, So can the bird apply an additional force if it sits or swings or flap around the walls of the cage? And the answer might be against intuition but it is NO!
Let's imagine if it was a 'yes'. If the bird could apply a force on it's own closed system (or internal force) it could accelerate its own system as long it desires. Now this isn't right, is it? Otherwise space ships could just propel themselves by their travelers constantly kicking on the inside!
The air inside the cage does two things it expands and contracts (changes pressure) when the bird moves around or flaps its wings. So when the bird pushes some air towards the bottom of the cage the cage gets a force (a small impulse) which is immediately redrawn by another force now, on the top of the cage, by the elastic air that obeyed Newton's third law. In fact the bird itself will feel this recoil (that's why a bird can fly) and if the flap was strong enough it will hit the cage's top and does the same as above. So all you'll feel is an up and down movement of the cage! This is just a rearrangement of masses inside the cage, but no net forces are acting on the center of mass.
Now, for an open system external forces comes in to action. To put it simply, the bird will draw some air to the inside and this new air will apply new force that you will feel as a pull (since the air could now escape through the openings, the case above is not valid!)
Buoyant force has nothing much to do here except that it is what slightly reduces our weight (or the bird's) due to pure gravity by pushing us upwards since we are displacing the medium (air) we are in (Archimedes, eureka moment!) Thus the bird need to apply a force slightly less than it's gravitational force by earth in order to fly, but as I said no big deal here.
Now all this would get even more interesting by imagining a balloon instead of a bird. Think about it and tell me what you think.
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Some of the above discussions are incorrect; bird accelerating changes weight of cage/bird for bird in isolated cage sitting on a weight scale due to acceleration of center of mass. Assume closed cage. System is cage and bird. External forces are gravity (mass bird + mass cage)x g down and force of constraint holding cage up against gravity (e.g. cord holding cage off ground or weight on a scale with cage/bird on the scale on the ground). The force of constraint is the "weight" of the cage/bird system. Key is (basic physics) "Sum of external forces = mass of center of mass x acceleration of center of mass"; internal forces do not matter. As long as bird is not accelerating (e.g. stationary or hovering bird), then (mass bird + mass cage)x g down = force of constraint holding cage up and weight (force of constraint) is not changed. If bird is accelerating up or down the center of mass is accelerating up or down and there must be a net external force; this net external force is due to a change in force of constraint (increases if bird accelerating up so weight of system increases). See Halliday and Resnick/ Physics (best physics book ever written).
answered Feb 1, 2019 at 21:43
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$\begingroup$ This doesn't add much to existing answers. $\endgroup$ Commented Feb 1, 2019 at 22:19
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$\begingroup$ @John Darby I agree with you John. Just I do not understand when the bird is flying and accelerating upward (in a closed container), how this is transferred to the external force? $\endgroup$– EbiCommented Jul 19 at 21:49
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the weight of the cage will not change when a bird does anything inside it because it becomes an isolated system and all the force applied by the bird will be counted as internal force...................however when the cage is open air is free to come in and go thus affecting the weight of the cage ........in the second case momentum is not conserved....
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Whether the cage is enclosed or open, makes no difference. So long as the bird is not close to the top or bottom of the cage (when "ground effect" would come into play) the wings circulate air around the bird/cage. There is no net wind force either up or down - just recirculation. The bird doesn't need the ground to "push against"! It just needs to create a local pressure differential. Therefore, so long as ground effect is negated, the cage will get lighter when the bird flies. (the bird itself will have a cyclic "weight" according to whether its wings are going down or up)
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$\begingroup$ In a closed cage, what happens to the momentum of the molecules that your bird pushes down in order to stay midair? $\endgroup$– pelaCommented Apr 6, 2017 at 6:28
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$\begingroup$ Probably lost as heat due to the recirculation - obviously the energy has to go somewhere. $\endgroup$– StewartCommented Apr 7, 2017 at 7:02
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$\begingroup$ I mean the momentum, not the energy. The momentum must be conserved. It will be transferred to the bottom of the cage (via other molecules), and thus increase its weight by exactly the same as the bird weighs. So a closed cage will not decrease in weight because the bird is flying. You answer is only correct for the open cage, where the ait can escape. $\endgroup$– pelaCommented Apr 7, 2017 at 7:24