Helium balloon question

I guess I'm getting (too) old but I just can't wrap my head around this.

Imagine a mylar party balloon that's filled with helium and which is attached to a 10 foot string. At the end of the string is a weight which is resting on the floor. At first of course the balloon will float at the end of the string, 10 feet in the air. Gradually however the helium will leak out and the balloon will over a period of days or weeks slowly sink towards the floor. I know that as the helium leaks out the balloon becomes heavier than the volume of air it displaces so it sinks. I'm probably going to feel silly asking this but why doesn't it fall all the way to the floor the moment there's not enough helium to keep it at 10 feet high?

As with a liquid, the pressure increases the deeper you go.

I've always assumed that as the balloon slowly loses buoyancy and drops, the tether (string, ribbon, etc.) load gradually reduces as more and more of it rests on the floor.

The string doesn't touch the floor until it sinks below 5 feet.

So a few feet of air makes that much difference?

I think it's about a few feet of tether, not air. Sorry if I misinterpreted you though, perhaps you were responding to the tongue-in-cheek pressure reply.

Surely the string touches the floor earlier, as the balloon starts to sink. After all it is tied to the weight, right?

We are only talking about a few molecules of Helium in any instant. If the Helium were leaking at any faster rate, the balloon would be down on the ground in seconds. As it is, you are losing tiny amounts of Helium.

From your description, the balloon is attached to a string on the ground. At some point, you lose enough of the Helium that it begins to sink. If it was not touching the ground, it will sink until the string touches the ground. In this case, the balloon begins to sink, still, but the string is already touching the ground. As soon as it sinks even a little, it takes a small amount of weight off the balloon. Sure, the amount that is removed might be the weight of a millimeter of string, but whatever that weight is, that's as much bouyancy as the balloon has lost. That SHOULD be a trivial amount of Helium. As the balloon leaves even a little more Helium, it will sink a bit more, which takes exactly the same amount of weight off as the loss of bouyancy.

And so on. It's not that it isn't sinking all the way, it is just that the amount of weight it has to lose to compensate for the loss of Helium is so utterly trivial because the loss is equally trivial.

There are two different effects at play, the weight of the string and the relative density of the atmosphere around the balloon.

As others have said, as the balloon sinks more string rests on the floor, taking weight off the balloon and allowing it to achieve a point of equalibrium before it hits the floor itself. Push down on the ballon from above and you will reduce the weight off the whole system to a negative so that, when you release it, the balloon will pop back up again, taking more string as it goes, until it once again achieves a point of equilibrium.

If the ballon starts off tethered to the floor then the string will be the larger of the two effects but what if the balloon has enough bouyancy to hold the weight a foot off the ground? As the helium leaks out it will still only shrink slowly. That's because air is denser at lower altitudes so it compresses the balloon further, reducing it's volume. Because bouyancy is a function of the volumne of air being displaced by the balloon and that volume is decreasing as the ballon descends that means the balloon is effectively becoming lighter relative to the atmosphere that surrounds it. So again, it will achieve a point of equilibrium and stop there. It's why hot air ballons don't just float off into space.


Edit>After writing that it occurred to me that it's probably worth talking about Archimedes. Archimedes' principle states that a body will displace its own mass when emersed in a liquid, not it's own volume. That might seem counter intuitive at first. After all, when you get in the bath it's your volume that's pushing water out, why would you mass have anything do with it? If you were made of lead why would that cause more water to be displaced? If you sat in a bath of liquid hydrogen a) you'd freeze to death in seconds but more importantly b) you would still only displace your own volume and, because hydrogen is much less dense that water (of which we're primarily made) that would be considerably less than your own mass.

The problem is that in the bath example, you reach the bottom before you reach a point of equilibrium. If you're allowed to sink until you reach a point of equilibrium then the liquid you're displacing is much denser than the liquid at the surface. That liquid is displacing the less dense liquid above it and so on up through the liquid until you hit the surface at which point your mass of liquid will be displaced, which will could potentially have a different volume to you.

I understand better what's happening now but let me amend the question since my description of the "thought experiment" unintentionally added a factor that is, I believe, adversely affecting some answers.

The real situation that got me thinking about this was a balloon tied to a string with a weight at the end just like I described above, but instead of the weight being on the floor (so that as the ballon sinks some of the weight of the string is removed) it was on the top of a bookcase so the ribbon never touched the floor (or the top of the bookcase) at all.

Having said that, I know it does have to do with the weight of the balloon+string+helium system because cutting off the string caused the partially "submerged" balloon to rise all the way to the ceiling. Had this been outdoors I know that the ballon would have continued to rise until it reached equilibrium at some altitude.

But even in that scenario such a "hanging loop" of string divides its load between the balloon and the weight. As the balloon "sinks" more and more load moves to the anchor weight.

Think of a system with a rope or cable anchored at one end with a pully and load in the middle. You have to expend effort to raise the weight higher by lifting/pulling on the loose end. Same thing, opposite direction: lowering the weight means less and less "pull" on the loose end.

Easy enough to show using a "pull scale" on the loose end to measure the load even without the pully and load if the cable is heavy enough to register on your scale.