Bird in the box and other physics quandries

A coworker and I have an ongoing conversation about the following questions. Neither of us really know the answer, but we’ve developed many a wild conjecture. Without further ado, here they are:

Suppose you had a perfectly sealed box (the size doesn’t matter too much, but let’s say 10 feet cubed) with normal earth atmosphere inside it. Underneath it is a scale to measure the weight. Inside the box is a bird. If the bird starts flapping his wings and achieves liftoff, what would the effect be on the scale reading? Imagine bird takes off, flaps around a bit, lands and repeats.

Second question, same box. Instead of a bird, you have a 100 pound weight attached to the roof of the box (on the inside) and a nice fluffy pillow on the floor of the box. The weight has a release mechanism operated by remote control. Flipping a switch releases the weight and lets it fall to the pillow. Again, what would the scale show during the process of letting it hang, releasing it, watching it fall and then come to rest on the pillow.

Number three: Same box. This time, there is a fan affixed to the center of the box. The fan can be switched on and off by remote control. It is affixed such that it blows air towards the floor. Again, how would this affect the scale reading? (This ones a bit easy, but it helps set up the next one.)

Same box, also with a fan. But this fan is part of a “hovercraft” type device. It sits flat on the floor until the remote switch is thrown. Then it blows air enough to levitate a bit off the floor and drive around. Then the switch goes off and it comes to rest. For the last time, what would happen on the scale?
Okay, that’s pretty much it. It would be much appreciated if you labelled your responses complete guesses (that you aren’t really sure of) or as reasoned theories (that you think all the math works out on) or as answers (because people have asked and tested and answered these questions before). Or something like that.

Weasel, I’ll try Question #2

When the weight attached falls the scale registers a lightning of the weight of the box. As the falling weight passes through the “normal air” it just pushes the air aside without resting on it. when the weight lands on the bottom of the box the scale registers an increase in the weight of the box.

All reasoned theories, since experimentation would be (a) expensive as all-get-out, and (b) cruel to birds.

In all cases, the weight would remain the same, so long as the box is sealed and no air escapes. There will still be fluctuation up and down, however, just like when you “hop” a little on a scale, and your feet never leave the ground.

  1. Stays around the same weight, fluctuating a bit due to turbulence.
  2. Stays the same until it hits, then it goes WAY up, WAY down, etc, until it comes to rest at the same weight.
  3. Slight fluctuation
  4. Ditto.


Let’s assume for convenience that at the beginning of the experiments, the scale is “zeroed out” (i.e. it displays a weight of 0).

Bird question:
Of course, very few birds could achieve this, but this is a physics discussion, so that doesn’t really matter :slight_smile:

Of course a bird is very light, and so variations will be minor. When the bird lifts off, the scale will register greater than 0. While the bird stays in the air, the average weight will be 0. It may vary moment to moment depending on the way the bird flies. In fact the bird’s weight is still being supported by the floor, only now it is due to uneven air pressure. As the bird decends, the scale will show less than 0. When it lands, it will register greater than 0 for a moment, and then it will revert to 0.

Some of these variations might not show up if there is any substantial dampening in the scale’s support. I’m assuming that we are talking about a theoretical case with very low dampening.
The weight question:
I concur with Ishmael. As the weight falls, the scale will disply less than 0. As the weight hits, the weight will momentarily rise to greater than 0 and then return to 0. Again, this assumes low dampening. Greater dampening could cause the scale to never show a weight greater than 0. In fact, low dampening could result in a momentary oscillation between greater than and less than 0.
Question #3
No change in the scale. Pressure changes are immediately offset by the tension in the fan’s supports. The only change in the system that would be observable from the outside would be a change in rotational momentum. Of course that is off the subject.

Question #4
This is essentially the same as the bird question. With low dampening, there will be a momentary reading greater than 0 as the hovercraft takes off. The scale will return to 0 as the hovercraft maintains level flight. Again, this is because the weight of the hovercraft presses on the floor by means of uneven air pressure. As the hovercraft descends, the scale will show less than 0, and as it lands it will momentarily show greater than 0, and then return to 0 (with a possible oscillation).

My answers are based on fundamental principles of mechanics (action/reaction, conservation of momentum, gravitation, simple oscillators).

The weight of the whole object doesn’t change in any of the problems, from the definition of “weight”: the mass of the object times the accceleration constant of earth. Part of the statement of the problem is that the box is sealed, so we are assuming conservation of mass, right? So the weight doesn’t change.

But that doesn’t mean the scale won’t register changes. Scales work by measuring the force, not the weight. In the case of the flying bird, every flap of wings pushes air down to keep the bird up. This force must be transmitted through the air to the bottom of the box. The average weight the scale registers won’t change, but fluctuation around the average will increase for a flying bird over one sitting on the bottom of the box. The same reasoning works for question three, the hovercraft, although the fluctuations wouldn’t pulse with wing flaps. You’d just get the same weight as with the thing sitting on the floor of the box.

The dropping weight is different. The force the scale is initially offering against the box is equal and opposite to the pull of gravity downward on the box, because the box isn’t moving. If the earth’s acceleration is g, the scale is offering an acceleration of -g, because the whole business isn’t moving initially. When you release the weight, it is in free fall so that’s 100 lbs. less the scale has to offer to keep the box from moving toward the center of the earth (well, actually, a little bit of the 100 lbs. is transmitted to the bottom of the box from air resistance, but let’s ignore that small amount). When the weight hits bottom, though, the momentum changes very rapidly back to zero, and this requires a large force to stop it (recall that force is the rate of change of momentum). The scale is offering MORE force in order to accomplish this than it was before we dropped the weight. So in the weight problem, you start off with the scale reading some weight of the box; then when you drop the 100 lbs. the weight the scale registers go down; then when the 100 lbs. hits, the weight shoots way up for a moment before settling down again to the original weight. If you average over the whole process, you get the same weight you started with, but the scale’s reading changes during the process of dropping the weight.

Hmm…Undead beat me to it. I agree with him completely. I was working up a simplified (EXTREMELY simplified) explanation of some of the mechanics actually involved, and realized it’s too long to post here, so I’ve stuck it on a page and linked it. If you’re brave enough to withstand a little bit of math (just a tiny bit – I promise it won’t hurt :)) with an “English translation”, venture in!