Stupid trivia question of the day...

Here we have a stand.
On this stand there is a tank. The tank weighs 150 units.
In the tank there is water. The water weighs 100 units.
In the water there are rocks. The rocks weigh 50 units.
There are now 300 units of weight pressing down on the stand.
Here I have a fish. The fish weighs 5 units.
If I place the fish in the water, and the fish stays suspended off the bottom, how many units of weight are now pressing down on the stand?

305 units, I would say the fish's buoyancy is immaterial

Quote: Face

Here we have a stand.
On this stand there is a tank. The tank weighs 150 units.
In the tank there is water. The water weighs 100 units.
In the water there are rocks. The rocks weigh 50 units.
There are now 300 units of weight pressing down on the stand.
Here I have a fish. The fish weighs 5 units.
If I place the fish in the water, and the fish stays suspended off the bottom, how many units of weight are now pressing down on the stand?

It is impossible not to have the fish pushing down. The fish displaces water, adding weight to the tank.

305 was my original assumption. If 5 units of mass is added, then it has to be there, right?

But then I thought about my livewell. A cooler full of water has weight. Fill it with 20 crappie and I couldn't notice a difference. Of course a few pounds of fish might not be noticible when struggling with a 100lb box, but it made me second guess.

And then I thought about the 'pushing down' principle. Was it really? If it was pushing down, wouldn't it sink? Water would just move out of the way, would it not?

If a ballon, filled with just enough helium to float level, floats into your window at home, does your house get heavier?

Maybe I'm just having a dumb moment. =P I keep going back and forth on this.

Quote: Face

... If a ballon, filled with just enough helium to float level, floats into your window at home, does your house get heavier? ...

For most of your comments, I think you just need to hire some consulting services from Archimedes. For this specific question, I'll assume you are talking about the house and everything it contains at the moment. The entry of the helium balloon through the window would tend to make the house/contents heavier, except for the fact that I suspect an equal volume of air would be pushed back out of the window to make room for the balloon. Since the balloon floats level, the departing air has the same mass/weight as the balloon and helium, so there is no change in the total weight. Got that?

Quote: Doc

For most of your comments, I think you just need to hire some consulting services from Archimedes. For this specific question, I'll assume you are talking about the house and everything it contains at the moment. The entry of the helium balloon through the window would tend to make the house/contents heavier, except for the fact that I suspect an equal volume of air would be pushed back out of the window to make room for the balloon. Since the balloon floats level, the departing air has the same mass/weight as the balloon and helium, so there is no change in the total weight. Got that?

I think Archimedes would blow me off as a kook ;). You're explanation made me think, Doc. Maybe the weight of the air in the volume of said balloon would equal the weight of the balloon itself, and make your example true. But how about this...

We borrow Mythbuster's giant hanger and hypothetically place a scale under it. It weighs X
I walk inside. It now weighs X+200lbs.
I am handed a giant balloon with a helium tank attached.
A balloon of such size weighs 150lbs, the tank weighs 50lbs and the Helium has positive weight.
All told we have X+200+150+50+helium.
If I use the tank to inflate said balloon until the balloon, the tank and myself are now floating just off the ground, wouldn't the scale simply read X, with the air which was displaced not possibly weighing 400lbs+weight of helium?
This concept is not unlike a fish's swim bladder inflating with air and causing it's buoyancy, which is why I can't say 305 units for sure. Any thoughts?

Yes, the scale under the hanger would only indicate "X". The displacement of the rest of the weight by the helium filled balloon would not register (but the helium filled ballon would still have mass).

Quote: Ayecarumba

Yes, the scale under the hanger would only indicate "X". The displacement of the rest of the weight by the helium filled balloon would not register (but the helium filled ballon would still have mass).

So, if we were to apply this concept to the first post with the 'fish in a tank' example, then 300 units would be the force on the stand? =D

Only if the 5lb of water displaced by the fish is spilled out of the tank... (like the air in the balloon example is forced out of the house).

Quote: Face

If it was pushing down, wouldn't it sink? Water would just move out of the way, would it not?

Back to the tank, if you were to weigh the whole thing, it would be 305. Just because something floats in a medium, doesn't mean that its mass is neutral. Think about the a giant livewell, with a super sized crappie shaped airtight boxes made of depleted uranium. As you add each one to the livewell, you will note that an equal amount of water is displaced to keep them afloat. You would definitely notice the additional weight (unless the displaced water was allowed to flow away)

Quote: thecesspit

Only if the 5lb of water displaced by the fish is spilled out of the tank... (like the air in the balloon example is forced out of the house).

But by using this thinking in the warehouse example, that would mean that the air displaced out of the warehouse by me, my balloon and the tank would have to weigh 400lbs. Does that sound right? Doesn't to me.

See why this questions bugs me? lol =p

Quote: Face

... If I use the tank to inflate said balloon until the balloon, the tank and myself are now floating just off the ground, wouldn't the scale simply read X, with the air which was displaced not possibly weighing 400lbs+weight of helium? ...

Sorry -- ran out for a pizza and missed part of the discussion.

If the balloon is inflated sufficiently to float you, the tank, and the balloon itself, then it would indeed displace an amount of air weighing 400 lbs + weight of helium. That is Archimedes's principle. Typical atmosphere has a specific volume in the range of 13.8 cu. ft. per lb, so your balloon would need to be inflated to something in the order of 5,500 cubic feet.

And yes, in the original problem the scale would read 305 units, provided no water was lost.

Yes, it is 400 lbs. If it was less, you would not float.
That is why Doc was directing you to Archimedes - it's his law that says that the force, keeping you afloat is exactly equal to the weight of the displaced air. If you displace less than your weight, you don't float.

Edit: oops ... sorry, Doc :)

Quote: Ayecarumba

Back to the tank, if you were to weigh the whole thing, it would be 305. Just because something floats in a medium, doesn't mean that its mass is neutral.

With you so far....

Quote: Ayecarumba

Think about the a giant livewell, with a super sized crappie shaped airtight boxes made of depleted uranium. As you add each one to the livewell, you will note that an equal amount of water is displaced to keep them afloat. You would definitely notice the additional weight (unless the displaced water was allowed to flow away)

And then I lose you. I think displacement is a function of volume, not weight. This tank is not full to the brim, so displaced water does not exit. If I were to stand over it and dunk my hand in it, water would be displaced, none would be lost, and the weight on the stand would not change as the weight of my hand would transfer through my arm and eventually to my feet. I think this is the same concept as the giant balloon. I definately have weight and the balloon has weight, but we're supported by the balloon and not the floor of the warehouse. Which, in turn, I think extends to the fish. Until you ask me in 5 minutes at which point my veiw will have certainly changed. /insert manic cackle here.

Is this like trying to explain that 50 reds in a row doesn't means the next roll HAS to be black? Am I one of THOSE guys? ><

Quote: Face

This tank is not full to the brim, so displaced water does not exit.

It does not matter - the level goes up though, and the pressure at the bottom is level times density, the force (on the scale - i.e. weight) is pressure times area, and the weight of the displaced water - extra level times area - is the change in the force, which equals exactly to the weight of the body suspended in water.

Quote: Doc

Sorry -- ran out for a pizza and missed part of the discussion.

If the balloon is inflated sufficiently to float you, the tank, and the balloon itself, then it would indeed displace an amount of air weighing 400 lbs + weight of helium. That is Archimedes's principle. Typical atmosphere has a specific volume in the range of 13.8 cu. ft. per lb, so your balloon would need to be inflated to something in the order of 5,500 cubic feet.

And yes, in the original problem the scale would read 305 units, provided no water was lost.

Quote: weaselman

Yes, it is 400 lbs. If it was less, you would not float.
That is why Doc was directing you to Archimedes - it's his law that says that the force, keeping you afloat is exactly equal to the weight of the displaced air. If you displace less than your weight, you don't float.

Edit: oops ... sorry, Doc :)

Sorry to you both, was too busy losing my mind in a post and missed these comments. It appears those who stated 305 know what they're talking about. It makes sense, I suppose, I just thought I remember seeing this same 'balloons make you float' experiment actually done somewhere, and couldn't see how the volume of air displaced by such a contraption could weigh as much as the balloons and person themselves. Perhaps I underestimated the area, or more likely, the weight of air. I guess it is kind of elementary, once you stop listening to the b.s. in your head lol. Thanks for the lesson, sirs. I had fun =). On to other things, like seeing how many of my friends I can make lose their minds by posing this very question. Muahaha.

Imagine that you put 5 units of water in the tank, instead of a fish. Would the new water be suspended in the old water, or would it add to the weight of the stack?

(Don't answer this, it is asked rhetorically toward anyone who might choose the answer 300.)