# helium balloons

Oh, it is? Well, never mind. Whatever…

…by the water and my butt is supported by the chair. What difference does it make?"
the force exerted on a chair in which you sit and the force exerted by an object immersed in air or water is not the same. the chair is forced down (which is what weight is - a downward force) when you sit on it. this is by design so the chair will not break. if you put the chair on a scale and you sit on the scale, the resulting weight will be the chair + you. if you put a bucket of water on a scale and drop a bouyant object in, the resulting weight is less than the bucket of water + the object. why? because water does not support the object the way a chair support’s one’s butt. the object’s mass displaces the water in all directions. only a small portion is displaced in the down direction (the definitive ‘weight’ direction). the same goes for an object that is bouyant in air - the mass of the object does not direct all it’s force down. if it did, you wouldn’t want to stand under a hot air balloon if it hovered close to the ground.

the force exerted on a chair in which you sit and the force exerted by an object immersed in air or water is not the same. the chair is forced down (which is what weight is - a downward force) when you sit on it. this is by design so the chair will not break. if you put the chair on a scale and you sit on the scale, the resulting weight will be the chair + you. if you put a bucket of water on a scale and drop a bouyant object in, the resulting weight is less than the bucket of water + the object. why? because water does not support the object the way a chair support’s one’s butt. the object’s mass displaces the water in all directions. only a small portion is displaced in the down direction (the definitive ‘weight’ direction). the same goes for an object that is bouyant in air - the mass of the object does not direct all it’s force down. if it did, you wouldn’t want to stand under a hot air balloon if it hovered close to the ground.

didn’t mean to post twice - i kept getting error messages.

Quoth zwaldd:

Sorry, no, unless the bucket was full and some of the water spilled over. The portion of the force directed down due to the buoyant object is exactly equal to the weight. The vertical forces on the object (let’s say it’s a ball) are the weight pulling it down (F1), and the contact force of the water pushing it up (F2). Since the ball is not accelerating, and F=ma, the total force must be zero, so |F1| = |F2|. The forces on the bucket of water are a contact force from the scale, pushing up (F3), the weight of the water, pulling down (F4), and a contact force from the ball, pushing down (F5). Again, since the water is not accelerating, we can say that |F3| = |F4| + |F5|. Now, by Newton’s third law, that for every action there is an equal andopposite reaction, we can see that the magnitude of the force of the water on the ball equalls the magnitude of the force of the ball on the water, or |F2| = |F5|. This means that |F3|, the force of the scale on the system, is equal to |F4| + |F2|, or the sum of the weights of the water and the ball. Again, by Newton’s Third, this means that the force of the system pushing down on the scale, which is what the scale reads, is also equal to the sum of the individual weights.

If this doesn’t satisfy you, you can get out the bathroom scale and do the experiment. Make sure to use a heavy enough object that the scale can measure the weight easily.

don’t need a scale at all to test that, chronos. next time you go swimming, swim under a fat guy on a raft. you’ll be glad you were wrong.

I’ll add another attempt to answer the OP. Remember, the definition you gave there was, “weight is a function of force directed towards earth due to gravity.” Read the last three words there again.

In talking about ‘negative weight,’ you’re trying to take all forces into account, when your definition limits weight to only that part that’s due to gravity. So the forces of bouyancy are right out.

Consider this: What if you had an air-tight barrel, that contained a helium balloon? A precise scale placed under the barrel would read the sum of the weights of the the barrel itself, the air in the barrel, rubber of the balloon, and the helium in the balloon. Sure, the balloon would be at the top of the barrel, but it still contributes to the weight.

The confusion that makes it seem like it has negative weight is that the system you’re examining is not closed. The pressure of an atmosphere that you’re ignoring skews the results.

Here, again, you’re ignoring the rest of the pool. The bouyancy is provided by the whole pool, so there is no difference when you pass under a floating object. If you were instead to weigh the whole pool, you’d see the weight increase when the fat guy and his raft are included.

The bucket experiment is a closed system, so you will see what Chronos describes.

the air-tight barrel with the helium balloon in it would weigh less than the barrel without the balloon for the same reason that the balloon itself weighs less once you pump helium into it. but enough of my layman’s comments. here’s a message i received from a physics professor:

"The definition of weight as the force required to keep a body in equilibrium in its environment is not “my” definition, but one proposed by various physicists, and some textbook authors.

I agree with you that the helium balloon, under this definition, would have negative weight. Standard units of force would apply, as always.

Many textbooks speak of the “weight in a fluid” as the force required to suspend an object in the fluid. A standard freshman lab experiment measures the density of an object from its “weight in air” compared to its “weight in water”. The object is suspended by a string from a balance. It is “weighed” in force units. Then a container of water
is brought up from underneath the body, till it is hanging immersed in the water. Then the balance is read again, and this time the reading is less. This is usually done for denser-than-water objects.

But consider doing it with a cork. Now you’d have to have a string pulling the cork down into the water, perhaps with a pulley at the bottom of the beaker. The cork’s “weight in water” would be negative.

It just reminds us that “weight” is a property not only of the body, but of the body in its environment. Its weight on Earth is greater than its weight on the Moon. Mass, however, is a property of the body independent of its environemt, and would be the same even in space well removed from other objects. "

and i agree that the weight of the pool would increase, but not by the ‘land’ weight of the fat guy. just by that small portion of his mass that displaces water in the down direction.

correction - the helium ballon as a unit weighs less than the ballon without the helium.

You’re right, I was being imprecise. Imagine that the scale and barrel are in a vacuum. The bouyant force on the balloon still exists due to the air inside, but no bouyancy acts on the barrel itself. Can you see that you are now seeing a positive weight from the helium and the balloon? In other words, if you removed the balloon from the barrel (without adding air in its place), the barrel would weigh less?

As Chronos and others said, you can certainly change your definition of weight to change the answer to your question. But, in the OP, you said that weight is defined as “a function of force directed towards earth due to gravity.” That is, without any consideration of equilibrium or bouyancy. If you want to win your argument with your coworkers by shifting definitions, go ahead and try it. I prefer to hold you to what you said when you asked the question.

Where is the rest of the fat guy going? Are you proposing that being in water makes him thin?

The entire weight of the man and his raft must equal that of the displaced water in order for him to float. The water supports him, so it comes down to sailor’s chair comment. The weight of an object and the thing that holds it up is the sum of their two weights. Part of the total doesn’t disappear because bouyancy is involved.

You might as well say that because the scale is holding you up, you’re weightless.

the rest of the fat guy displaces water up and to the sides as well as down. in an open pool you could measure the displacement by seeing how much the water rises. in a hypothetical ‘closed’ environment, you could measure the increase in water pressure. neither of these is ‘weight’, though, by any definition.

the air pressure in the barrell would increase and you would be able to measure the weight increase only of that portion of the pressure directed towards the scale - very minimal, offset somewhat by the upwards force of the balloon.

i do want to win the argument and i have no problem using a definition provided by someone with more physics credentials then i have.

well, I got tired of this thread long ago but…

zwaldd says "if you put a bucket of water on a scale and drop a bouyant object in, the resulting weight is less than the bucket of water + the object. "

you are wrong but I am not going to argue. Just one question. How much are you willing to bet on that? I need to make some dough and I am willing to bet.

sailor i need to do this experiment myself before i put money on the line. i will get back to you on that.

darn it! I was hoping you would rush to take me up on this

zwaldd, make sure to do the experiment with the bucket of water, scale, etc. completely submerged in a big tank of mercury

Arjuna34

well i don’t have ready access to the materials necessary to perform this experiment - specifically, a scale accurate enough to weigh both a container of water, an object light enough to float in it, and the total combined weight. i have, however, submitted this scenario to my physics professor contact and am awaiting his reply.

i was able to perform Arjuna34’s version, but the whole thing took place inside a thermometer on a microscopic level. i’m not sure what was proven, since i don’t own a microscope.

I think you’re getting confused by how force and weight are transferred through water. In an open pool, the water indeed rises due to the fat guy, which increases the water pressure at the bottom of the pool, which, therefore, transfers more force to the ground, by an amount equal to the weight of the fat guy.

If that’s too confusing, ignore the “how” and look at it as an addition problem: The fat guy has weight and the water has weight. None of it disappears no matter how you combine them.

You need: a bathroom scale, a bucket, tap water, and a bag of ice. Ice floats, right? I can’t believe you can’t at least borrow each of these. I recommend you run the experiment; that should be a lot more convincing than verbal arguments.

Quoth zwaldd’s physics teacher:

Hmm, interesting definition… Do you know how much net force is required to keep something in equilibrium? Exactly zero. So, by this definition, the weight of anything in any environment is zero.

You could define weight as the sum total of the non-constraint environmental forces acting on an object (although you’d have to be careful how you defined “environment”), and that could be useful in some situations, but it also yields weird results like weightless submarines. As for the bucket, do you mind if I go in with you on that bet, sailor?

zut - no one here at work has a scale and i’m not really going to spend my free time on this. i’m sure one of the several hundred readers of this thread has a scale and can chime in with the results. by the way, my physics professor contact disagrees with my theory about the results of that experiment so i concede to his greater knowledge, although i don’t think he’s tried it himself either. he does however, believe in the negative weight concept (back to the original question) and i can’t line-item veto his opinions unless i hear from a more learned physicist.