helium balloons

Chronos: I read that definition of weight as being the body force required to keep an object in equilibrium by balancing out all other forces such as buoyancy, reaction forces, etc. That interpretation jibes with your last paragraph, rather than your second.

However, I prefer the definition of weight (from my American Heritage dictionary) as: “The gravitational force exerted by the earth or another celestial body on an object, equal to the product of the object’s mass and the local value of the gravitational acceleration.”

This definition, zwaldd, precludes negative weight because both mass and gravitational acceleration are always positive. One reason I prefer this definition is because it’s less confusing: witness your own confusion as to what happens when a fat guy goes into a swimming pool. With the above definition, the weight of the fat guy stays the same whether he’s in or out of the pool, and it’s obvious what the total weight of the pool + fat guy is.

Also, tangentially, zwaldd, I’m glad you’re accepting the correct answer to the pail-on-a-scale problem, but I’m curious: no one where you work owns a bathroom scale? I thought everybody owned one of those! Is the bathroom scale a dying American icon?

zwaldd – let me put it this way…

You’ve got a bucket of water, say, 4 gallons. It’s half full, so no water will spill out if you add something small. Put it on a scale and it’ll weigh about 30 lbs (water weighs about 7.5 lbs/gallon). Let’s say that the water and the bucket weigh exactly 30 lbs.

You now add a volume of water weighing 10 lbs to the bucket (this is about 1-1/3 gallons). The scale now reads 40 lbs. Of course the water level in the bucket will rise.

Imagine that, instead of pouring 10 lbs of water into the bucket, you place a 10-lb water balloon into the bucket. The balloon is neutrally bouyant, but we can agree that it will displace a volume of water equal to its weight, thus water in the bucket will rise the same way it did above. And as above, the scale now will read 40 lbs.

Imagine that, instead of the 10-lb water balloon, you place into the bucket a 10-lb balloon filled with mineral oil (or any low-density fluid). This balloon will float at the top of the bucket, it will also displace a volume of water equal to its weight and the water line will rise exactly as it did in the previous scenarios. If you don’t believe that the weight on the scale now reads exactly 40 lbs, I want in on that bet also…

zut, i am laughing out loud at this line. i like the way you write. and yes, i’m sure someone at work owns a scale, they just don’t have it here and i’m not going to ask someone to bring one in.

sounds like a question for cecil.

Sailor, if you can’t cover all of zwaldd’s bet, I’ll take the rest of it.

I’ll see if i can perform the bucket experiment. If I do, I’ll post the results here.

I used a barrel with water in it, a Ziplock® bag filled with ice, and a bathroom scale.

[ul]
[li]Bucket of water — 32 lbs.[/li][li]Ziplog® bag filled with ice — 3.5 lbs.[/li][li]Ice in the bucket of water — 36 lbs.[/li][/ul]

Make of that what you will.

Perhaps this will help clarify things-- There are horizontal forces exerted by the man on the water, and likewise by the water on the man (action-reaction). However, these horizontal forces do absolutely nothing to help support the man. It’s only the vertical forces (or rather, the vertical components of the forces) that support him, and they support him completely. These vertical forces that the man exerts on the water must be counterbalanced by some other forces pushing up on the water, i.e., the bottom of the pool pushing up on the water. When the man gets in the water, the bottom must push that much harder to supprt him.

Well, sure. But what if you used helium ice in a vacuum-filled (vacuum-emptied?) ziplock bag.

:: flees ::

I make of that either a 0.5 error or a miracle.

I can see you have no scientific training because the first thing you learn is to fudge the numbers so they fit nicely the way they should. :wink:

Now what would happen if two buckets collided head-on? That’s what I want to know!

[QUOTE]
*Originally posted by sailor *
**

Ignore Manhattan’s post. He is obviously trying to confuse the issue. Let me simplify it instead: How fast are the buckets going?..Wait. Let’s use two singularities, each with infinite mass, approaching each other at the speed of light. Wait…Let’s make the Helium balloon out of anti-Helium. No…first we have to get Divine dispensation to explain the missing half pound of ice. Hold on…I’m losing track. Let me fire up some more meth to clear my head…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZ
OK, I ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZOK, I ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZZ…OK, I ZZZZZZZZZZZZZZZZZZZZZZZZ

tcburnett, don’t forget to take into account whether the experiment is done in the northern or southern hemispheres, whether it is a bucket filled with helium colliding with an antibucket filled with antihelium OR a bucket filled with antihelium against an antibucket filled with helium (this distinction is crucial) and, of course, barometric pressure because, I mean… where are we? in a cyclone or an anticyclone?

Manhattan, are you aware that ziplock bags do not block certain kinds of radiation which could affect the results of the experiment? huh? huh?