I have three kitchen scales with an identical box on each.
The first box is empty. Inside the second box is helium-filled balloon with a sticky weight on the balloon to achieve neutral buoyancy. The third box contains a well-fed, fat, and exceptionally well trained hummingbird hovering obediently.
We do this question pretty often. The answer is that they are all the same unless your boxes are actually sealed containers. The hummingbird would add to the weight if it is a closed system. Air is a fluid a a bird in it adds weight like a fish does in a tank.
The balloon and weight would increase the mass, but the scale’s reading wouldn’t change.
The hummingbird in the box would increase the scale’s reading (as well as the mass). I’m not sure the reading would change even if it’s sampling speed (correct term?) was faster than the wing beats.
Not exactly a good analogy. It would be like weighing two fish tanks at the bottom of a lake. They are both full of water and surrounded by water. When you add a fish to the tank, it would displace its weight in water out of the tank. It’s more like the balloon box
But the balloon filled with helium would displace the air that would have been there, decreasing the mass. Net result? Identical mass. You can’t say that the scale (weight) wouldn’t change but the mass would, because a scale simply measures the mass multiplied by the Earth’s force.
At sea level with average temperatures, the density of air is about 1.2 kg/m[sup]3[/sup], while helium is about 0.18kg/m[sup]3[/sup]. Let’s say the boxes are 1m[sup]3[/sup], and the balloon is a sphere 0.5m wide (0.25m radius).
That gives the contents of the empty box a mass of 1.2kg.
The balloon’s volume is 4/3π(0.25[sup]3[/sup] )= 0.0654m[sup]3[/sup] with a weight of 0.0118kg. This displaces the same volume of air, weighing 0.0785kg.
According to this site, helium will lift ~1.5 its mass, so 0.0118kg will lift 0.0177kg, so the balloon itself and the weight would have be 0.0177kg to achieve neutral bouyancy.
Thus the total mass of contents of the balloon-filled box is 0.0118kg+(1.2kg-0.0785kg)+0.0177kg = 1.151kg.
So apparently, we’re BOTH wrong
There is, of course, a very good possibility that my calculations are wrong is a variety of ways.
Unless the bird and added air taking up the equivalent volume of the balloon, weighs the same or less as the balloon with sticker, then it will be the heaviest. The balloon box and the air only box, weigh the same.
I think that tilde there is where your problem is. Approximations are fine, but if you take an approximation and an exact result and take the difference, you’re not going to get zero. Which is effectively what you’re doing here. Helium doesn’t have any inherent lifting property; it just has buoyancy, the same as anything else. A helium balloon weighted to neutral buoyancy has exactly the same mass as the same volume of air: That’s what makes it neutral buoyancy. So the box containing the balloon will have exactly the same total mass, weight, and reading on the scale as the “empty” box (containing air).
The hummingbird, meanwhile, is producing downwash, which is adding another force to its box. If the box is leaky, the force (or at least, the portion of it exerted inside the box) will be diminished, but it’ll still make some contribution. So the box with the hummingbird is sure to be heavier by some amount, and it’s just a matter of determining how much.
It will be heavier by an amount equivalent to the weight of the bird - the bird’s wings push on the air, the airt pushes on the inside of the box, the box pushes on the scales. If the bird is motionless in the middle of the box, there must be forces holding it there and they have nothing to act upon but the scales.
I disagree with this. The fish would displace it’s volume in water, not it’s weight (mass). A fish doesn’t necessarily have the same density as water. If the fish was made of lead, or if it was hollow, there would be a pretty drastic difference.