If you have two jars, each on a set of weighing scales, one open and one sealed and there is bee in each, when each bee takes off what will be the changes in weight of the two jars?
Guessing that you take the lid off the second jar to let the bee fly out, no change at all.
Even if there was a bee in one jar and not in the other, as long as the bee wasn’t in contact with the jar, there shouldn’t be a difference in weight, I think!
The two jars will weigh the same until the bee from the open jar flies away from the scale. The weight of each bee-in-flight is registered on the scale because the thrust required to keep them aloft is displacing air which is hitting the scale.
AWB is right… this is a bit of a variation on the baseball on the airplane question that Cecil answered (I think)… The bee will exert a force to keep aloft that is the same as his weight… but the bee has to be in the jar.
OK… but! I was thinking (scary enough) that once the bee gets to a point high enough (somewhere above the neck of the jar, the force of the air under the wings would not only force air in the jar, but down along the sides, thereby decreasing the weight. Does that make sense? So in order for the weight to be the same, all of the loft force would have to be registered inside the jar.
“No, you got the wrong number. This is 9-1-2.”
I don’t buy AWB’s arguement. I believe that the jars will weigh the same until the bee in the open jar takes off. After that, the open jar will be lighter by the weight of the bee. (A portion of both bees mass will also be converted into energy but let’s assume our scale is not THAT accurate).
The downthrust should not apply in an open system. If it does, what happens to it after the bee clears the rim of the jar?
What if the experiment takes place outside and a bird flies overhead. Does the birds downward thrust affect the weight of the open jar? What if an airplane flys by?
In an open system (or a large enough closed system), the downward thrust is met by the air’s resistance to motion (inertia). Let’s substitute the bee with a helicopter. The downward thrust works primarily against the ground at liftoff. As the helicopter gains altitude, the downward thrust is met with resistance from the air below. The moving air tries to escape out the sides and is also met with resistance. So, you end up with an inverted ‘mushroom’ shaped area of increased pressure. When enough air is being compressed by the downward force and the accompanying resistance, the helicopter flies, without need for directly pushing against the ground.
well I’ll bee damned
Of course that’s just my opinion I could be wrong.
The helicopter flies via the lift created by the airfoils that are its rotor blades. The cross sectional shape of an airfoil is such that, when forced through the air, a pressure differential is created by virtue of forcing the same number of air molecules to traverse the greater distance over the top of the airfoil in the same time that the underflowing molecules traverse the shorter underwing (or airfoil) path. This creates a low pressure are above the airfoil. The pitch (angle) of the airfoil can accentuate the lifting force. The chopper does experience some “ground effect” when very close to the surface, but that has little to do with how they fly (and a lot to do with how they take off).
Nevertheless, I’ll WAG on into the question at hand. I’ve never contemplated how a fly achieves flight, but however he might, he attains relative weightlessness, so I’ll guess that while both flies are airborne, and assuming the lid for the open jar is also on the scale, and identical flies, etc…, the change in weight is the same.