# Floating Table (With Buckets)

• Deployment
In summary: ...be in equilibrium if the weight of the table is equal to the weight of the buckets and the angle of the strings are equal.

#### Deployment

Hello Physics Forums, Deployment here. After browsing the internet I came across this photo..

I have been pondering and pondering on how this is actually possible. If you think about it, the buckets are actually holding down the table.

Does anybody have any idea how you would conduct calculations on doing this?

Thanks,
- Deployment​

I think what you have here is an optical delusion, a little piece of performance art made at the expense of physicists/engineers.

If you think about it, the buckets are actually holding down the table.

Are they?

Deployment said:
Does anybody have any idea how you would conduct calculations on doing this?
4 M_bucket >= M_table
where M, of course, is the mass.

By the way, this thread should be moved to General Physics.

Sumerion
Deployment said:
I have been pondering and pondering on how this is actually possible. If you think about it, the buckets are actually holding down the table.
Instead of buckets, imagine there were people pulling down on the ropes (if the table were bigger). Would that make more sense?

Does anybody have any idea how you would conduct calculations on doing this?
Treat it as any other static equilibrium problem.

SteamKing said:
I think what you have here is an optical delusion, a little piece of performance art made at the expense of physicists/engineers.
Why do you think that?

The table is either stuck to the wall or it's stuck to the buckets. It's impossible to tell which just by looking at this photo.

SteamKing said:
The table is either stuck to the wall or it's stuck to the buckets. It's impossible to tell which just by looking at this photo.

dont think so

I can imagine an upwards force from the table against the bottom of the buckets and the mass of the buckets ( including anything they contain ... maybe some water) producing an equal force down against the table top.

I don't see anything implausible in the scenario

Dave

SteamKing said:
The table is either stuck to the wall or it's stuck to the buckets. It's impossible to tell which just by looking at this photo.
Have you heard of a bosun's chair?

sophiecentaur, anorlunda and Varun Bhardwaj

I don't see anything odd at all. What did you expect to happen?

Firstly, sorry for placing this in the wrong section..I am just learning my way around here.

Secondly, this is 100% possible, with no tricks such as the table being attached to the wall, etc. We are currently doing this in physics class, and I was wondering if the geniuses at Physics Forums had any ideas as to where I should start.

I know that equilibrium is where I need to focus, and the idea that the same amount of static friction needs to be pulled on each rope. I am just trying to put together a road map of equations so I can figure out the answer

Today I will be given the values for..
* The weight of the table
* The weight of the buckets
* The angle of the strings

Then I will simply need to calculate how much water must be place into each bucket in order for the table to "float" ;)

- Deployment

Sumerion
Doc Al said:
Have you heard of a bosun's chair?

Yes, I've heard of a bosun's chair. But the chair doesn't work if it's not tied off or if someone isn't hoisting you up or down.

If you think the buckets are holding the tabletop down, how are you able to push on a rope? All I see are four sheaves, each with a single line running thru it, with one end tied to the table top and the other end tied to a bucket handle. If there is no mechanical connection between the buckets and the table top (bolts, glue, suction cups, whatever), the table top must be attached to the wall, which connection is not visible due to the viewpoint of the camera.

Unless you are claiming that gravity's hours have been cut back and it works only part time now.

Deployment said:
I know that equilibrium is where I need to focus, and the idea that the same amount of static friction needs to be pulled on each rope.
I think you mean tension in each rope, not static friction.

Im just trying to put together a road map of equations so I can figure out the answer

Today I will be given the values for..
* The weight of the table
* The weight of the buckets
* The angle of the strings
Sounds good.

As far as equations go, what are the conditions for equilibrium? (Hint: Consider the "buckets + table" as a single system. What forces act on it?)

Doc Al said:
I think you mean tension in each rope, not static friction.

Sounds good.

As far as equations go, what are the conditions for equilibrium? (Hint: Consider the "buckets + table" as a single system. What forces act on it?)

The buckets and table are all going to have to meet at the same point in the air. By putting the same amount of weight in each bucket, the buckets are taken care of and will all be equal. The question is how much weight.

I have the logic and can explain it to anyone, the equations are just what is holding me up at the moment.

Hopefully class teaches me a little bit more today..or someone on here kills it with knowledge!

- Deployment

SteamKing said:
Yes, I've heard of a bosun's chair. But the chair doesn't work if it's not tied off or if someone isn't hoisting you up or down.
The point of the bosun's chair is that you can hoist yourself up. You don't need someone else.

If you think the buckets are holding the tabletop down, how are you able to push on a rope?
? The ropes just pull, like always.

All I see are four sheaves, each with a single line running thru it, with one end tied to the table top and the other end tied to a bucket handle. If there is no mechanical connection between the buckets and the table top (bolts, glue, suction cups, whatever), the table top must be attached to the wall, which connection is not visible due to the viewpoint of the camera.
Is there a mechanical connection between you and the floor? Yet you can walk without floating into the air!

Unless you are claiming that gravity's hours have been cut back and it works only part time now.

Deployment said:
The buckets and table are all going to have to meet at the same point in the air.
Not sure what that means.

Answer my question: What forces act on the "buckets + table" system?

SteamKing said:
But the chair doesn't work if it's not tied off or if someone isn't hoisting you up or down.
The chair works as shown in the picture, as long as the man is heavier than the chair.

SteamKing said:
Yes, I've heard of a bosun's chair. But the chair doesn't work if it's not tied off or if someone isn't hoisting you up or down.

If you think the buckets are holding the tabletop down, how are you able to push on a rope? All I see are four sheaves, each with a single line running thru it, with one end tied to the table top and the other end tied to a bucket handle. If there is no mechanical connection between the buckets and the table top (bolts, glue, suction cups, whatever), the table top must be attached to the wall, which connection is not visible due to the viewpoint of the camera.

Unless you are claiming that gravity's hours have been cut back and it works only part time now.

It's perfectly possible to pull yourself up with a Bosun's Chair. You do need strong arms tho' and it's much easier with a beefy mate at the bottom of the mast. There is no difference between a Bosun's chair and a mass on the end of a loop which has been passed over a pulley. The passenger can change the length of the rope (i.e. height) by pulling in or letting out one of the ends of rope.

To be honest, there is a bit of cognitive dissonance in that picture. Despite my rational approach to it, which shows that it's perfectly possible, I still find it a bit magical. But isn't that what all conjuring tricks rely on?

In many ways, you could say that the buckets are 'holding the table down' against the tension in the rope. If there were four, bucket sized holes in the table, it would go up, wouldn't it??! (The tensions in the ropes would change when the holes appeared, of course.

There is also the matter of stability to consider. The arrangement even 'appears to be' in unstable equilibrium but there are those triangular structures with the strings under tension and a rigid table so there will be no change of shape even with a range of asymmetrical weights of bucket + water in the buckets. It could fail in the end, with big enough differences in weight.

With four buckets and four [pairs of] ropes, the system is over-constrained. The set-up process may involve careful measurement of ropes and positioning of buckets to ensure that none of the ropes go slack.

It might be simpler to analyze with just three buckets and three [pairs of] ropes.

jbriggs444 said:
With four buckets and four [pairs of] ropes, the system is over-constrained. The set-up process may involve careful measurement of ropes and positioning of buckets to ensure that none of the ropes go slack.

It might be simpler to analyze with just three buckets and three [pairs of] ropes.

That's an excellent idea - a la three legged stool problem. Doing it that way, you could assume unstretchable ropes and always get all three buckets in contact with the table.

You don't even need a chair.
You can pull yourself up with a rope thrown over a tree branch. You just make a little loop at one end so you put for foot through it and pull the other end of the rope.

Here is a more sophisticated, "modern" contraption:

She has some extra pulleys to reduce the force (by pulling more rope, in exchange, as can be seen).

Or even a bicycle:

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If you start off by imagining that the buckets are not in contact with the table, but static floating slightly over the table surface, then you know that the weight of the four buckets is close to the weight of the table... if you know the angle the ropes make from vertical as they run from the pulleys to the table, you could figure the difference.

If the buckets are too light, the table will fall, if the buckets are just the right weight, the table and buckets will hang without contact, and if the buckets are heavier than that, the whole thing will continue to hang with increasing tension in the ropes... until something breaks.

The critical thing looks like finding the rope tension that suspends the table statically without bucket contact, beyond which the whole thing will hang statically with increasing force between the buckets and table, and increasing rope tension constant throughout their lengths (Vanadium 50's scale under the buckets question).

To treat it mathematically, without motion and the buckets not in contact with the table, the ropes upward vertical component equals the downward vertical component of the weight of the table and buckets. Maybe what you want is a relationship of the force between the buckets and table pressing together, and the increase in tension in the ropes, as a function of bucket weight?

Something like:

Bucket weight = Wb
Table weight = Wt
Rope tension = T
Force between buckets and table = F
Trig adjustment for rope angles = ta

So
F= ta2T(4Wb+taWt) or something like that...?

Doc Al said:
The point of the bosun's chair is that you can hoist yourself up. You don't need someone else.

'Someone' is indefinite. It can mean the person in the bosun's chair or a second party.

Is there a mechanical connection between you and the floor? Yet you can walk without floating into the air!

Now, you are being a sophist. Yes, I can stand on the floor and walk without floating in the air. However, last time I checked, table tops did not float above the floor and have to be restrained by a contraption like is shown in the photo.

SteamKing said:
'Someone' is indefinite. It can mean the person in the bosun's chair or a second party.
The interesting (and relevant) option is when the person sitting in the chair pulls himself up.

Now, you are being a sophist. Yes, I can stand on the floor and walk without floating in the air.
My point was that as long are there is a normal force pressing you to the ground--or the buckets to the table--there is no need for any mechanical connection. And that is easily arranged.

However, last time I checked, table tops did not float above the floor and have to be restrained by a contraption like is shown in the photo.
I trust you don't really think the table is 'floating in the air'. It's hanging from ropes!

sophiecentaur
SteamKing said:
'Someone' is indefinite. It can mean the person in the bosun's chair or...
... a heavy bucket on the bosun's chair.

phinds
bahamagreen said:
If you start off by imagining that the buckets are not in contact with the table, but static floating slightly over the table surface, then you know that the weight of the four buckets is close to the weight of the table... if you know the angle the ropes make from vertical as they run from the pulleys to the table, you could figure the difference.

If the buckets are too light, the table will fall, if the buckets are just the right weight, the table and buckets will hang without contact, and if the buckets are heavier than that, the whole thing will continue to hang with increasing tension in the ropes... until something breaks.

The critical thing looks like finding the rope tension that suspends the table statically without bucket contact, beyond which the whole thing will hang statically with increasing force between the buckets and table, and increasing rope tension constant throughout their lengths (Vanadium 50's scale under the buckets question).

To treat it mathematically, without motion and the buckets not in contact with the table, the ropes upward vertical component equals the downward vertical component of the weight of the table and buckets. Maybe what you want is a relationship of the force between the buckets and table pressing together, and the increase in tension in the ropes, as a function of bucket weight?

Something like:

Bucket weight = Wb
Table weight = Wt
Rope tension = T
Force between buckets and table = F
Trig adjustment for rope angles = ta

So
F= ta2T(4Wb+taWt) or something like that...?

That would be the limiting case. The buckets can be as heavy as you like, as long as their combined weight is greater than the table's. You need a Cos(θ) factor in there but not if you assume long enough ropes. Bucket weights do not need to be equal for the table not to tip but the difference is limited by the moments on the table. If the tension in one or two of the ropes is greater than or equal to the weight of the corresponding bucket then you'd need to look more carefully at the actual values to see if the table would tip.

So we were coming to the conclusion today that what if you just took the weight of the table and divided it by 4...As long as each bucket is pulling with the same amount of force of course. The only concept that I keep over-thinking is that the ropes go through the pulley from the buckets to the table at an angle..and I keep asking myself if that angle will make any changes in our calculations.

I got some number for you guys today. Here they are,

Table Weight = 68.11N
Bucket 1 Weight = 165.9g | .1659kg
Bucket 2 Weight = 165.7g | .1657kg
Bucket 3 Weight = 167g | .1670kg
Bucket 4 Weight = 166.4g | .1664kg

The Table Dimensions are..
33.25 x 33.25 inches

The holes in the table obviously make a square, and are directly 14.25 inches from each edge. The square that they make is 10.25 x 10.25 cm

- Deployment

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So could it be this simple..

1.) The weight of the table is 68.11N, which can be converted into 6945.2871 grams.
2.) If you divide the weight of the table (In grams) by 4, you get the amount of weight that must be exerted on EACH bucket. Which is 1736.321775 grams.
3.) Now we have to factor in the weight that is already applied from the weight of the buckets, which can be seen in my last post, so after that factoring we can come to the conclusion that the following is how many grams that need to be added to each bucket..

Bucket 1 = 1570.421775g
Bucket 2 = 1570.621775g
Bucket 3 = 1569.321775g
Bucket 4 = 1569.921775g

Does this sound anywhere close to a reasonable prediction? Or do you also feel that the angle at which the ropes are tied and pull on the table matter in the factoring of this equation?

- Deployment

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I think it can be as simple as this, for the symmetrical case with long ropes.
Each rope supports 1/4 of the total weight, shared by both of its ends. The tension will be 1/2 that value - i.e. 1/8 of the total weight.

The force on each bucket from the table will be the difference between the weight of the bucket and the weight of the bucket plus its share of the table weight - i.e force on the bottom of the bucket would be 1/4 the weight of the table.

Without symmetry, you would need more calculations and, as mentioned earlier, the system is over-supported. So would you allow some finite stretch in the rope, to allow things to settle?

Surely the weight of all the buckets is balancing the weight of the table. The interesting questions arise: What will happen if the weight of the table exceeds the weight of all the 4 buckets? Obviously, the table will move down and the buckets upward due to difference in weights. However, what would happen if the weight of the buckets is increased? What finally will come out if you analyze it all would depend upon the tensile strength of the ropes- they might yield and break if the heavy loading finally !

Deployment said:
Hello Physics Forums, Deployment here. After browsing the internet I came across this photo..

I have been pondering and pondering on how this is actually possible. If you think about it, the buckets are actually holding down the table.

Does anybody have any idea how you would conduct calculations on doing this?

Thanks,
- Deployment​
Analyze each object separately. The forces on each bucket are: Weight of the bucket, tension in the rope, and the normal force from the table. These should add to zero.
The forces on the table are: The weight of the table, the normal force by each bucket down on the table (this is what you meant by "buckets are actually holding down the table", The vertical component of the tension in each string tied near the center of the table. These should add to zero. By symmetry the horizontal components of the tensions in the strings tied to the table will add to zero.
Write down these two equations, and you will find that with the given data, you can solve for everything.
The equation for the torques does not add anything. The symmetry of the arrangement guarantees that the torques will add to zero.

Weigh the table, divide it in 4, calculate the weight needed to equalize ¼ of the table weight using the string angle. Make sure the string angle is enough to position the table above the floor. Lift the table above your calculated height, add your necessary weight to the buckets then allow the table to lower and suspend at it’s equalized weight height.

By the way, a bosun's chair needs 2 men and a boy lifting, when anyone is sitting in it; you need to lift the weight of the chair as well as the person in it... Or! Normally you pull it up empty, tie it off then climb up and sit in it.

Tyler Worden said:
By the way, a bosun's chair needs 2 men and a boy lifting, when anyone is sitting in it; you need to lift the weight of the chair as well as the person in it... Or! Normally you pull it up empty, tie it off then climb up and sit in it.
HaHa. You have my sympathy there, and you are welcome to try to climb the mast on my boat, unaided and without mast steps.
But your comment demonstrates the vast difference between Mechanical Advantage and Velocity Ratio (=Efficiency), which is a point that only I seem to make, ever, on PF. The VR of a self-pulled bosun's chair is 2 and the MA is something less than 0.5 (assuming you can normally lift your own weight yet you need another person's help). So the efficiency is MA/VR = <25% ! Bosun's 'chairs' are normally canvas and webbing and will be no more than a couple of kg so the losses must be due to friction in the (usually pretty expensive) halyard pulleys and compression in the rope as it goes over. But I guess the pulleys are not designed to turn efficiently under a 100kg (1kN) load. I'd bet that a really good pulley (also large diameter) at the top would make things a lot easier - but who would think about that when their mast happens to be laying down in the yard?

This is simple. The weight of the table is equal to that of the buckets. A little push and you could make the buckets float 1 inch above the table. It would still be a static system and would not move.

1) First think about 1/4 of the table and 1 bucket. Then replace the bucket by 4 pounds of weight. Then replace the weight of the 1/4 table by another weight of 4 pounds. It will be a static system because the two weights are equal. If the weights are not equal things will move.

2) for the second step suspend 2 separate 4 pound weights over a table. As the string angle increases the weight of the table must be decreased to keep a constant tension in the string.

Tyler Worden said: