Can a Closed Jar Measure the Weight of Flying Bees Inside?

[Dr. Nitrus Brio]

We have bees flying in a jar, not touching it. Jar is closed and air in it has standard atmospheric pressure. Here is the question: if we put that jar on a weighing-machine, is ti going to show the mass of the bees?

I had some theories about it, but many people thought differently, trying to convince me that they are right. I don't believe them - I want to hear your opinion.

I apologize if I'm spamming. I'm new here and I searched through many different threads, but I didn't find anything similar.

 

[jim mcnamara]

F=ma

Force on the scale is weight. - mass of bees times g, the acceleration due to gravity.

While bees are flying around inside, the their wings are pushing air downward, equal to their acceleration due to gravity. This means the air is likewise "pushing" against the bottom of the container.

If the bees were parked on top of a vertical thin rod, the answer would be obvious. Because we cannot see air movement we don't make a connection between the bees and the container walls.

Yes, the mass will show.

 

[Sojourner01]

My thought is yes.

The bee must exert a continuous force downwards in order to maintain itself in midair. It might fly higher or lower, but it's confined within a fixed distance - between the bottom of the jar and the top - so the time-average of the force it exerts is constant. Now technically, the bee is pushing the air downwards, but because the air is itself confined in the jar it must at some point transfer the momentum given to it by the bee to the bottom of the jar. This exerts a force on the bottom of the jar, on average equal to the weight of the bee.

So yes, you can weigh the bee.

 

[Dr. Nitrus Brio]

Ok. That sounds pretty logical. But, when bee moves her wings, she makes the air to flow. I was thinking that this flow of the fluid might be turbulent. Does this have anything to do with it? Maybe the bee keeps herself in the air thanks to Bernoulli's principle. Helicopters can fly above us, but we are not going to feel their weight.

 

[cesiumfrog]

Brio, never seen helicopter-induced crop-circles on tv?

 

[DaveC426913]

This is identical to a classic riddle involving a truck filled with birds crossing a bridge. (I wish I could find an example.) The owner bangs on the side of the truck to get the birds flying, hoping to reduce the weight of the truck. The answer is: no it doesn't work.

 

[Dr. Nitrus Brio]

Brio, never seen helicopter-induced crop-circles on tv?

Of course I did (although I don't watch the TV very much).

This is identical to a classic riddle involving a truck filled with birds crossing a bridge. (I wish I could find an example.) The owner bangs on the side of the truck to get the birds flying, hoping to reduce the weight of the truck. The answer is: no it doesn't work.

Nice example. I have something more to ask. When bee moves her wings, she transfers kinetic energy of the wings and impulse of force to the particles in air. This force impulse then transfers to the other particles, if the collisions between particles are perfectly elastic.
If the particles have same mass then we get: mv=mv1+mv2 => v=v1+v2 The more collisions happen, the particles will have smaller velocity, but there will be more and more of them and the starting impulse will be preserved. Correct me if I'm wrong. But what happens with the particles which don't hit the bottom of the jar?

 

[Sojourner01]

But what happens with the particles which don't hit the bottom of the jar?

Many of them won't - I don't believe the mean free path in air is that long. However, a volume of air molecules directly underneath the bee will have a net momentum downwards, which must be conserved in some fashion. If they don't reach the bottom of the jar, they will transfer some of their momentum to whichever air molecule they strike, which will then have a momentum with a statistically greater probability of being downwards. The downward momentum in this way spreads out, but it is definitely conserved and must eventually reach the bottom.

 

[pixel01]

I like the explanation of Jim Mc and Sojourn.
My way to explain this is just as simple. The jar and the bee including all the gas inside is an isolated system. For an isolated system, the sum of momentum is zero. So no matter the bee is flying or standing, the jar will not exert any extra force on the scale.
In your example of momentum of the air particles, not all the particles will touch the jar's bottom wall, but those are impacted will transfer the impulses to the bellow and so on. Finally the momentum will get to the bottom.

 

[Dr. Nitrus Brio]

But, no matter if the impulse is preserved, if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be "transfered" to scale. Of course, they eventually hit the bottom, but not at the same time.

 

[Doc Al]

But, no matter if the impulse is preserved, if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be "transfered" to scale.

Why not? What holds up the bees? The air! What holds up the air? The jar. What holds up the jar? The scale.

Even if the air particles being smacked down by the bee's wings do not hit the bottom, they will certainly increase the pressure underneath the bee enough to support the bee's weight. And that increased pressure will be transmitted to the bottom of the jar.

 

[pixel01]

But, no matter if the impulse is preserved, if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be "transfered" to scale. Of course, they eventually hit the bottom, but not at the same time.

I think DocAl explain clearly enough, but I would like to show you the weird point in your reasoning. You said: "if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be transferd to scale". That's not true, even if they don't hit the bottom wall, they transfer the impulse to the layer bellow and so on. The force will immediately exert onto jar bottom and onto the scale in no time at all. Why don't you ask youself : what make some of the air particles float inside the jar?. Afterall, they must rely on each other and finally onto the bottom wall. The bee is just one special air particle that is so big that it has to flap itself to be float among other air particles.

 

[Dr. Nitrus Brio]

Even if the air particles being smacked down by the bee's wings do not hit the bottom, they will certainly increase the pressure underneath the bee enough to support the bee's weight. And that increased pressure will be transmitted to the bottom of the jar.

Of course that the pressure will be transmitted to the particles underneath, but the bottom of the jar can only be hit by approximately same number of particles (pressure dependant). So, all particles that have the "extra" impulse cannot hit the bottom wall at the same time. Some of them must hit the side walls of jar before transferring their impulse to those below.

 

[Doc Al]

Not quite getting your point. Air will hit the sides as well as the bottom, but it's the bottom that will do the pushing up needed to support the bee.

 

[Dr. Nitrus Brio]

...but I would like to show you the weird point in your reasoning. You said: "if all of the particles in air don't hit the bottom wall, then the whole weight of the bee(s) is not going to be transferd to scale". That's not true, even if they don't hit the bottom wall, they transfer the impulse to the layer bellow and so on.

Maybe I expressed myself a bit wrong. I realize that the impulse will be transferred to the layers below, but the direction is not only down. They also hit the side walls. What happens then? Do they lose their velocity if they hit the wall? They transfer the impulse they have to the walls, not only to each other. Am I right?

 

[Dr. Nitrus Brio]

Air will hit the sides as well as the bottom, but it's the bottom that will do the pushing up needed to support the bee.

Doesn't the higher pressure below the bee cause the force which "neutralizes" her weight? I don't see what the bottom has with it. It's just being hit by these particles later.

 

[Doc Al]

Doesn't the higher pressure below the bee cause the force which "neutralizes" her weight?

Sure. And now that higher pressure must be supported by the layers below, as pixel01 explained.

I don't see what the bottom has with it. It's just being hit by these particles later.

I am assuming a steady state situation: the bee is hovering. The air is continually being pushed down by the flapping wings, creating a greater pressure beneath it, which in turn gets transmitted to the bottom of the jar.

 

[Sojourner01]

They also hit the side walls. What happens then? Do they lose their velocity if they hit the wall? They transfer the impulse they have to the walls, not only to each other. Am I right?

That's right, but momentum is a vector quantity, so it must be conserved in direction and not just in magnitude. An air molecule hitting the side may or may not transfer a portion of its momentum downward; whether it does or does not, the quantity is conserved. The average momentum of the air in the jar is not zero - it has a net momentum downward, which since we know no air is escaping from the jar must be transferred away - hence a force downward.

 

[Dr. Nitrus Brio]

I FINALLY realized. I wasn't sure about all of this by now. Particles don't lose anything when they hit the side walls. The walls are being hit approximately with the same force from all sides. So, the impulse from the bee goes straight to the bottom and scale shows the mass of the bee (of course, scale cannot measure mass, it measures weight). Maybe some of them can escape from the bottom, but I think that it itsn't the crucial thing.

 

[Ivan Seeking]

Just for fun, we tried to do this experiment after a chem lab class one day. The trouble was that insects don't like to fly when you put them in something. They would never stay aloft long enough for us to see anything but the impulse created when they took off and landed.

 

[peaceharris]

We have bees flying in a jar, not touching it. Jar is closed and air in it has standard atmospheric pressure. Here is the question: if we put that jar on a weighing-machine, is ti going to show the mass of the bees?

I had some theories about it, but many people thought differently, trying to convince me that they are right. I don't believe them - I want to hear your opinion.

I apologize if I'm spamming. I'm new here and I searched through many different threads, but I didn't find anything similar.

I think it will show a weight less than the combined weight of the jar and the bees. Using Bernoulli's principle, where the velocity of the air is more, the pressure will drop there.

If you have a aeroplane flying just above you, not touching you, obviously you wouldn't be squashed by the weight of the plane. Same thing here, there will be a local pressure drop due to the increased velocity of air just around the bee in accordance to Bernoulli's principle.

 

[Dr. Nitrus Brio]

I think it will show a weight less than the combined weight of the jar and the bees. Using Bernoulli's principle, where the velocity of the air is more, the pressure will drop there.

Yes, the static pressure will drop, the overall pressure is always the same. But what happens with the dynamic pressure? It increases. Does it have anything to do with this?

If you have a aeroplane flying just above you, not touching you, obviously you wouldn't be squashed by the weight of the plane.

Maybe because the pressure is transferred all over the air, not just downward like in this bee case. And smaller number of particles will hit us. The bees may fly thanks to Bernoulli's principle, but not in the same way like planes.

What if I put some body in the water? Am I going to feel its weight, despite the lift force (if I put all of the water on the scale also)?

 

[peaceharris]

Yes, the static pressure will drop, the overall pressure is always the same. But what happens with the dynamic pressure? It increases. Does it have anything to do with this?



Maybe because the pressure is transferred all over the air, not just downward like in this bee case. And smaller number of particles will hit us. The bees may fly thanks to Bernoulli's principle, but not in the same way like planes.

What if I put some body in the water? Am I going to feel its weight, despite the lift force (if I put all of the water on the scale also)?

I think you might be able to experimentally verify this if you have a sensitive weighing scale. Get some swimming fish, record the weight of the tank. Then kill the fish and reweigh the tank. I suspect that the weight of the tank with the dead fish should be more than with the swimming fish.

I did a internet search for someone who has done a similar experiment, but haven't found any.

 

[Sojourner01]

I suspect that the weight of the tank with the dead fish should be more than with the swimming fish.

Wrong. We've already 'proved' that this thought experiment has the opposite conclusion.

 

[peaceharris]

Wrong. We've already 'proved' that this thought experiment has the opposite conclusion.

I disagree with the 'proofs'. The proofs have all neglected that pressure will decrease when velocity increases in accordance with Bernoulli's principle.

Consider a huge box with an airplane flying inside, and a scale underneath the box. Will the weighing scale show the same reading when the plane has landed inside the box, and when the plane is flying inside the box?

 

[Doc Al]

I think you might be able to experimentally verify this if you have a sensitive weighing scale. Get some swimming fish, record the weight of the tank. Then kill the fish and reweigh the tank. I suspect that the weight of the tank with the dead fish should be more than with the swimming fish.

Whether the fish are swimming or dead, the tank must support the weight of the fish. When swimming or floating, the water directly supports the fish--but that just increases the pressure that the water exerts on the bottom of the tank. No magic here: The scale will read the weight of the tank and everything in it--fish included.

I disagree with the 'proofs'. The proofs have all neglected that pressure will decrease when velocity increases in accordance with Bernoulli's principle.

What are you talking about with "Bernoulli's principle"? Do you not agree that is it the interaction of plane with air that supports the plane? In effect, the air pushes up the plane--and the plane pushes down on the air.

Consider a huge box with an airplane flying inside, and a scale underneath the box. Will the weighing scale show the same reading when the plane has landed inside the box, and when the plane is flying inside the box?

Of course it will! Do you think the plane is suddenly "weightless" just because it's flying? Just like with the bees, the weight of the plane is supported by the air, which increases the force acting on the bottom of the box. The scale supports the weight of the box and everything inside it--including air and plane.

 

[russ_watters]

Sit under a helicopter hovering just above the ground and see if you can feel the force, peaceharris...

 

[jambaugh]

BTW This was tested on an episode of Myth Busters. They used pigeons and a RC Helicopter inside a trailer. It confirmed the fact that indeed the weight of the flying object still gets transmitted to the floor of the container.

If this where not the case then you could in principle fly up to the top of the container and lift it off the ground from the inside ... or similarly put a propeller inside the cabin a space-ship and push it without rockets. From the outside you are attempting to violate the law conservation of momentum. In practice if you try this you will be "arrested"! ;-D
Regards,
James Baugh

 

[peaceharris]

Whether the fish are swimming or dead, the tank must support the weight of the fish. When swimming or floating, the water directly supports the fish--but that just increases the pressure that the water exerts on the bottom of the tank. No magic here: The scale will read the weight of the tank and everything in it--fish included.

What are you talking about with "Bernoulli's principle"? Do you not agree that is it the interaction of plane with air that supports the plane? In effect, the air pushes up the plane--and the plane pushes down on the air.

The air underneath the plane doesn't support the weight of the plane plus the atmospheric weight. Since there is air velocity difference above and below the wings, you must subtract a pressure term that is proportional to the difference of velocities squared. Why do you keep neglecting this dynamic pressure?

Of course it will! Do you think the plane is suddenly "weightless" just because it's flying? Just like with the bees, the weight of the plane is supported by the air, which increases the force acting on the bottom of the box. The scale supports the weight of the box and everything inside it--including air and plane.

In an earlier post I said "I think it will show a weight less than the combined weight". This is quite different from 'weightless'.

BTW This was tested on an episode of Myth Busters. They used pigeons and a RC Helicopter inside a trailer. It confirmed the fact that indeed the weight of the flying object still gets transmitted to the floor of the container.

If this where not the case then you could in principle fly up to the top of the container and lift it off the ground from the inside ... or similarly put a propeller inside the cabin a space-ship and push it without rockets. From the outside you are attempting to violate the law conservation of momentum. In practice if you try this you will be "arrested"! ;-D
Regards,
James Baugh

What was the weight of the trailer, the RC (does RC mean remote control?) helicopter, and how accurate was the scale?

Here is a numerical value from the internet:
http://www.boysstuff.co.uk/product.asp?id=13565
RC helicopter = 10g
Assuming the trailer weight is 100kg, you need a scale with 5 significant digits at least to detect a small change in weight when the helicopter takes off. Why didn't they do this experiment using a light box instead? Was a trailer necessary? Did they use a sensitive scale?

 

[peaceharris]

Sit under a helicopter hovering just above the ground and see if you can feel the force, peaceharris...

http://star.tau.ac.il/QUIZ/05/Bernoulli.jpg

Consider the photo above. If the water 'felt' the weight of the plane, wouldn't the water be pushed downward?

 

[russ_watters]

Uh, it is - don't you see the rooster-tail? The plane goes fast enough that the pressure wave isn't directly under the plane.

Anyway, I'm pretty sure that's a painting.

[edit: There is a good actual photo of a B-1 doing this and you can see in the photo that the rooster-tail is actually diamond or wedge shaped because it is caused by the wingtip vortices, not the engines. It is difficult to see in this version of the pic, but there is a noticeable area of disturbed water just behind the plane and in front of the rooster-tail: http://www.freerepublic.com/focus/f-news/1496427/posts ]

edit2: apparently the B1 "photo" is a painting too. Anyway, they did a better job with it...

 

[russ_watters]

There is no need to look for complex examples of this - everyone has certainly seen pictures of helicopters hovering just over the water, blasting-out a circular disturbance below them. It is a very straghtforward calculation to figure out just how much water needs to be displaced to hold aloft a helicopter hovering just above the surface of a lake or ocean:

An SH-60 has a gross weight of 21,000 lb and a rotor diameter of 53'.

That's 2206 sq ft of area.
At 62.4 lb/ft^3 weight density for water, 21000 lb displaces 336 ft^3.

336 ft^3 / 2206 sq ft is 1.8 inches deep.

 

[peaceharris]

Uh, it is - don't you see the rooster-tail? The plane goes fast enough that the pressure wave isn't directly under the plane.

Anyway, I'm pretty sure that's a painting.

It's not a painting. The air pressure drops due to the increased velocity of air, water is sucked up. Just like drinking from a straw, you reduce the pressure, water is pulled up.

 

[russ_watters]

No, peaceharris, that isn't what happens at all. The air pressure on the bottom of the wing is higher and that air is pushed down. The air pressure above the wing is lower and that air is pulled down. Behind the plane, air is moving downward. Again, it has to - otherwise we'd have a violation of Newton's laws: http://www.mansfieldct.org/schools/MMS/staff/hand/flightbernoulliandNewton.htm

Googling for that painting comes up with discussions of it along with real pictures and video of what a plane in ground-effect looks like. For example: http://alexisparkinn.com/photogallery/Videos/Supersonic%20F14%20flyby.mpeg

Notice how the disturbance under/behind the plane looks a lot like what you see under a hovering helicopter. You can even see in that distrubance that he's accelerating past mach 1. It start's off in front of the plane and moves back and becomes wedge-shaped as he nears the ships.

Anyway, you can tell that painting is a painting because of the coloration, the weird "camera" angle, and the fact that they got the plane's effect on the water wrong!

 

[russ_watters]

Ok - here's a link to the artis'ts website, showing both paintings: http://www.drublair.com/comersus/store/comersus_listCategoriesAndProducts.asp?idCategory=55