Quoted: I never stated whether the cage was open or closed.   Don't forget about turbulence and "frictional" losses in the turbulent airflow. That depends, is indeed correct and it opens the door to the necessary discussion regarding assumptions one is to make, the size of the parameter space to be investigated in the response,...all sorts of necessary things in order to more completely understand the question, its context and the considerations to be made in developing a technical approach and of course, the answer. Having not had this discussion, we (collectively) separately approached the problem in many different ways.  Only rarely is the one correct answer.  Usually only in the academic world.

errr...you're overthinking this.  it has nothing to do with airflow, or turbulence, or anything of that nature.  occam's razor--

by sealing the container, you have created a closed system.  the 'weight' of a closed system will not change unless there is a mass-energy conversion within it.

inertial effects may affect the observing system, but not the matter itself.

(and i'm a humanities guy!)  

sirensong,

I was the one who framed the question.  I'm telling you, I never stated whether the cage was open or closed.  That is in YOUR mind, not mine.

P.S. - scales don't measure mass, they measure force applied to the pan.

sorry--quoted the wrong guy.  the sealed container WAS STATED, and the conversation evolved from there.  i suppose i didn't register your non-sequitur.

(P.S.--if we're quibbling, i must point out that i never mentioned measuring mass.  that was in YOUR mind, not mine.)

Quoted: Quoted: Quoted: Exactly, here's another: A bird is resting on a perch inside a cage.  The cage is on a scale.  The bird leaves the perch and flies around inside the cage.  What happens to the scale's reading.  Why?  Please explain. The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage). What do I win? What if, instead of a cage, its a sealed container.  Will the weight change, then?

Quoted: Quoted: Quoted: Quoted: Quoted: Quoted: Exactly, here's another: A bird is resting on a perch inside a cage.  The cage is on a scale.  The bird leaves the perch and flies around inside the cage.  What happens to the scale's reading.  Why?  Please explain. The scale will read less because the bird is not being measured anymore (it's supported by the air, not the cage). What do I win? What if, instead of a cage, its a sealed container.  Will the weight change, then? No. It'll die very soon and won't be able to fly. Now gimme my prize from the first question damn it! Your answer was incorect, no prize given. The weight will be the same since the downforce of the bird's wings must equal at least the bird's weight in order for it to fly. The downforce will be transfered to the floor of the cage which is still on the scale. Yup, just like a helicopter presses downward on the helipad with equal force as it's own weight when it is hovering above it.  The problem we now face is that the spread of the weight is related to the height above the landing area, as well as the size of the target.  A bird flying around in the belfry of a really tall cage won't affect the weight on the scale nearly as much as in a squatty cage.  A helicopter exerts less and less weight on the helipad as it gains altitude above it.  At some point the air supporting the flying object does not translate significant force to the surface to measure.