The Balance: CR4 Challenge (03/31/09)

Apparently contradicting each other (one says total pressure the other volume...), but you claim both are spot on? Do we live in the same universe?

Whichever of those you mean is incorrect. What matters is the change in the mass of fluid (liquid or gas) that is displaced - that's why you can float on water but not on air air.

Hey now, Don't be a "fun sucker" LOL At least I stepped up and became a member! This has been a fun discussion, I tried to think it through and that's what I came up with. It makes sence to me now (or at least until we have the correct answer) . You are alway's right until proven wrong! Ha Ha. Well in my universe anyway.

OK, Now lets see the plastic ball drop!

J.

But if you think part time all other times you don't think do you ?

Are you like those famous cops (no names will be racist) who shoot first and then ask the name ?

And dont let the plastic ball drop. It is steel.

Check #1 - it was deliberate. .

The physicist wants the plastic to drop. And he being the theoritical and we being the practical (technology) will be on opposite sides. he wants plastic to drop ? we pull the steel side down and win

(What about putting a small magnet on bottom side ? it will pull the steel and not the plastic - will it be cheating ? of course not.

The challenge didn't say don't put magnet.

Also it the weighing is done on magnetic north pole (or south pole) then there is the attraction then let us calculate and prove physicist wrong.

Now what is the earths magnetic field and what will be the attraction of it on the ball ?

He he.

A minor spoiler*: I think the Earth's magnetic field is too uniform to have a measurable net effect on the balls. If the magnet is an essential part of the vacuum system, you should be in good shape; perhaps the magnets in the pump motor underneath the vacuum jar could serve? Or maybe the only available vacuum system is part of a NMR machine?

BTW, if there's a hard floor to drop onto, I'd rather the steel ball dropped - it'll bounce better; plus I far prefer the "ping" of steel to the "thud" of plastic.

*useful to get the Camaro to go further.

Welcome to the loony bin (and you have already seen the effect and looks like are one of us)

Ha Ha Nice! I guess I should have been "Thinking" when I came up with my handle. I leave the rest of my time for work! Te he he.

Maybe the questions should be multiple choice! A, B, or C If you pick "C" long enough you will eventually get one right!

"C" The plastic ball will drop.

Oh boy, lotsa problems with this one!

First of all,everyone is assuming a scale with pans or trays attached to the arms. Put a ball in a tray and what happens? It rolls around! If it does stop, it most likely will not stop in the center. Good luck getting two balls to stay still in this situation. What we need then is a scale equipped with slings.

The slings will have more contact with the surfaces of the balls, so then friction becomes an issue. The balls will have smooth surfaces, but the slings could be smooth or rough, but more likely rough, like canvas.

Being the resident weather nut, I am aware of a few things about atmospheric buoyancy. The buoyancy of moist air is the major factor in cloud formation and atmospheric instability. But when it comes to solids, buoyancy mainly affects microscopic particles. Otherwise, you have to use a lighter-than-air gas such as hydrogen or helium, or aerodynamic lift. I'm not doubting the effects of buoyancy on the balls, but I wonder if it will be enough to affect the outcome.

Does a helium filled balloon fall to earth if there was no air around it? Yes it does...its mass doesn't change nor arguably does it's weight, but it will still fall to earth.

The balloon will not just fall to earth. It will burst first. The remnants and any attached payload will fall to earth. That's why weather balloons and ARHAB experiments have parachutes between the balloon and the radiosonde or payload.

If the plastic ball is hollow and its walls are thin enough, it too will burst at or near vacuum conditions.

Now I wonder -- what if we conducted the experiment in reverse? We start with two solid balls, one plastic, the other metal (the OP doesn't say steel, it could be copper or aluminum for all we know), and a scale in a vacuum chamber. We use a remotely controlled mechanical arm to place the balls on the scale. They balance. Then we slowly introduce air to one atmosphere. What would happen?

OK, I'll put away my pot-stirring spoon now.

Actually gentlemen the real key to this question has nothing to do with balls, metal or plastic. It is purely that the object on one end displaces more fluid (air) than the object which balances it on the other side of the scales. Remove the fluid and you will remove more buoyancy from the more voluminous object increasing it's weight (but not it's mass of course).

A GA to the physicist for getting the first correct answer and SB should do penance and go to his children and admit his sin of making a muck up of such a simple challenge.

Talking of balls, I am amazed at how many members were not only unaware of the principles of buoyancy but are prepared to spout nonsense even when there are some clear explanations already posted. (A GA for the submarine explanation)

Can we drop it at the same time and see whether Newton is going to be happy?! I have no idea... just enjoying the thread.

In air both are partialy supported (floating) in proprotion to the air they displaced, when the air and thus that support is removed the plastic ball will be heavier since it displaced the most air.

The plastic ball will become "heavier" and the scale of the plastic ball will go down.

It is mentioned that the volume of the plastic ball is bigger than the metal ball.

In normal conditions (1 Atmosphere) the metal ball is getting only small "Floating" assistances

from the air around, but the big volume plastic ball gets more lift-up force from the air around it.

Therefore when the air is taken out, there is nothing to lift the balls that were submerge in air, and the big lift-up assistance that the larger ball received no longer exist. therefore the plastic ball will become heavier, and that will tip the scale.

For all the nonbelievers may I suggest a small experiment ? It can be done at home provided you have the necassary apparatus.

Take a balance (this is the only doubtful apparatus that I find, since I don't have it at home, but I can always make it as I used to do long back in my kid days - no not kid that was him but kid

Now put some thing heavier than water (but not steel) on one pan (say green potato)

Put this total scale (at least the pans) in Kris's bath tub (full of water) and carefully balance the balance by adding weight (steel) to the other side.

Now take it out of water and see whether they are still balanced ?

This will clarify all the doubts. (hopefully) - seeing is believing.

Can someone please clarify something for me:

If we take a beam balance, placed somewhere on this planet, not in a vacuum and assuming that the effect of gravity remains the same on either side of the beam balance and we put two objects of identical mass but different volumes on either side of the balance then the balance will tip in favour of the object with the smaller volume (since the effects due to buoyancy will not be equal). This is my understanding of what we are saying. Is this correct or have I completely misunderstood?

To be honest I would have said that they would remained balanced (I would have not included the buoyancy factor, I would just have looked at the mass)

Kind Regards

Mr. W.A Snow

You are right.

If two objects are of IDENTICAL mass, with different volumes, balance will tip in favour of smaller volume.

But problem is to make two objects of IDENTICAL mass in the air already around us.

In the current thread, two objects are of different mass, which weigh effectively same (gravity - buoyancy).

G1-B1= G2-B2 (G1 and B1 for Steel and G2 and B2 are for plastic)

B2> B1 in air, thus B2-B1 is +ve.

G1< G2.

Now, what you say for your example is

G1= G2

Still B1 < B2

G1- B1 > G2 - B2.

But what you say is make tw

Nobody said these are of equal mass, Only these are balanced on the either side of the balance. refer #68

sb,

Perhaps you would be so kind and go back and read what I actually wrote, not what you think I wrote. I never said that the original poster said that the masses were equal, I was asking a different question. You really should read the posts before responding.

Regards

Mr. W.A Snow

Yes, well put.

Fyz, in another post, talks about weighs (??!!!) in which you can tell if the two items are of nearly identical mass.

This is how I interperet it. "A metallic ball and a larger plastic ball are balanced on either end of a scale" Take that ballanced apparatus with the two objects from your current atmosphere and subject it to a condition void of atmosphere (vacuum). Will the two balls remain balanced?

It has been fun over analyzing this, but there is no need to make it complicated.

One simple answer is requested, and it has been answered a few times. There are three possible answers;

1. The Balls will remain balanced.

2. The metal ball will lower.

3. The plastic ball will lower.

A. It doesn't state that the balls weight the same, it only states that they are balanced.

My thought is that the apparatus will become unbalance in a vacuum in the favor of the larger object because of its larger volume (displacement). Once void of atmosphere it will "sink" relative to the object it is balanced with in it's initial atmosphere.

Josh.

The two spheres are affected by buoyancy in air. The plastic one must have more mass to have the same aparent weight in air. In a reduced pressure, the plastic one will have more 'weight', if they were perfectly balanced at 1.00 atmosphere. They will not remain balanced. The density of air, in grams per liter, of the plastic, and the metal ball, can be used to easily determine the actual change in apparent weight. If absolute pressure is reduced by 99%, then only 99% of the air density can be used in the calculation.

Each sphere is pushed up by the weight of the air that it displaces. In air the lift force on each sphere is equal to the volume of that sphere times the density of air which tends to zero in a vacuum. The balance will tip down on the side of the larger plastic ball as the air is removed.

The string is getting way too long. I should have read each one carefully. For any of us to not do so invites errors. Oh well.

One earlier poster asked how much difference buoyancy would make.

A 1 cc steel ball will weigh about 8 grams. A 4 cc plastic ball at SG=2, will weigh 8 grams. If they are balanced exactly at 1 atm, then absolute pressure is reduced to 0.01atm, the imbalance will be equal to the weight of 3 mL of air, which weighs about 29g per 22.4L or about 0.0039 grams. A good balance will show this imbalance of 4mg. We know that good garden-variety modern balances can meaure 1 part in 1 or 2 million. As of a century ago, an antique double pan should have been OK for 1 part in 500, which is 3.9mg in 8 grams. For SG=1, volume difference between the balls is 7 mL; the buoyant effect will be 9.1 mg.

I am too lazy to look any of this up, and use 4+ sig figs, but I do clearly recall these data from school 40 plus years ago.

We are still talking about flat pans. These are a modern invention. When one wants to use a row of modern standard weights, one needs a flat pan.

To compare goods sold to a standard that is not a finely machined metal cylinder, but perhaps a series of 'coins', the round bottom pan is not only easier to make, with a hammer and a bag of sand, but more useful. I can recall seeing nearly spherical pans in pictures. Look at an ancient balance diagram. With such a pan, a sphere will quickly settle to the bottom and not roll around at all.

After reading so many posts debating this question/challenge I'm just glad they aren't My Balls on the Scale.

Yes. There will be no air in the jar, therefore, the buoyancy of the balls do not apply.