Buoyant force

[Genady]

@Saber I have to admit that the answer which for years I thought solves the puzzle, I have doubts about it now  . I want to see where you and  @Lorentz Jr  go. If it turns to be the same answer, I'll share my doubts and we'll resolve them, hopefully.

[Saber]

7 minutes ago, Genady said:

@Saber I have to admit that the answer which for years I thought solves the puzzle, I have doubts about it now  . I want to see where you and  @Lorentz Jr  go. If it turns to be the same answer, I'll share my doubts and we'll resolve them, hopefully.

when  mr Lorentz   and  you   gave me hints  i  thought  that  your  hints  dont  line up......but i was  afraid to  ask......im really  confused  now......

[Lorentz Jr]

1 hour ago, Saber said:

Does  that  mean   weighs  on both  sides reach to the  bottom/tom  simultaneously ?

Yes, let's assume one cup and weight at each end goes around its nearest pulley at the same time.

Let's turn the belt until the tops of the cups on either side of the belt are lined up, keeping the weights locked in place. Forget about the depths of the cups.

There are N other cups on each side of the belt. Each cup on the right has a mass of water above it of volume hA, where h is the distance each weight can travel in its cup (assuming the weights are held in place by pistons instead of membranes), and A is the cross-sectional area of the cup. The corresponding volume above the weight on the left side is empty. Only air or a vacuum.

Each of the 2N other cups moves a distance D/N, where D is the total depth of the belt. So the work done by the weight of the unbalanced water above each cup on the right is [math]mg(D/N) = \rho hAgD/N[/math], and the total work done by all the other cups is [math]\rho hAgD[/math].

But each weight on the ends has moved up a distance h. So the work done on the end cups is [math]2Mgh[/math], but we said [math]W=Mg[/math] has to be greater than [math]\rho g DA[/math] for the weights to stay down when they're on the left side, so the work done on the end cups is at least [math]2 \rho h DAg[/math], which is twice the work done by the other cups. So the belt had to do work on the end weights, i.e. it lost [math]\rho h DAg[/math] of whatever kinetic energy it had before the turn.

Now we unlock the weights that just went around a pulley so they can fall into their new positions in their cups. Water pressure on the top/right weight will accelerate the weight, but some of the resulting kinetic energy will be lost when the weight stops moving (it's an inelastic collision with the cup), so it won't make up for the loss in raising the weight as it rounded the pulley.

Edited January 30, 2023 by Lorentz Jr

[Saber]

30 minutes ago, Lorentz Jr said:

Yes, let's assume one cup and weight at each end goes around its nearest pulley at the same time.

Let's turn the belt until the tops of the cups on either side of the belt are lined up, keeping the weights locked in place. Forget about the depths of the cups.

There are N other cups on each side of the belt. Each cup on the right has a mass of water above it of volume hA, where h is the distance each weight can travel in its cup, and A is the cross-sectional area of the cup. The corresponding volume above the weight on the left side is empty. Only air or a vacuum.

Each of the N other cups moves a distance D/N, where D is the total depth of the belt. So the work done by the weight of the unbalanced water above each cup is mg(D/N)=ρhAgD/N , and the total work done by all the other cups is ρhAgD .

But each weight on the ends has moved up a distance h. So the work done on the end cups is 2Mgh, but we said W=Mg has to be greater than rhogDA for the weights to stay down when they're on the left side, so the work done on the end cups is at least 2rhohDAg , which is twice the work done by the other cups. So we lost rho h DAg of energy.

Now we unlock the weights that just went around a pulley so they can fall into their new positions in their cups, but this doesn't help. The pressure on the bottom/left weight is greater than that on the top/right weight, so they won't help turn the belt.

Honestly thats  more  than  my  understanding capacity to  understand it @ once i   have to read it  a couple of  times and concentrate on it.....to  grasp it   i

[Lorentz Jr]

7 minutes ago, Saber said:

thats  more  than  my  understanding capacity to  understand it

The belt gets energy from the cups on the sides, but it has to give all the energy back in order to move the end weights from one side to the other.

Edited January 30, 2023 by Lorentz Jr

[Genady]

43 minutes ago, Lorentz Jr said:

Now we unlock the weights that just went around a pulley so they can fall into their new positions in their cups. Water pressure on the top/right weight will accelerate the weight, but some of the resulting kinetic energy will be lost when the weight stops moving (it's an inelastic collision with the cup), so it won't make up for the loss in raising the weight as it rounded the pulley.

Here is where I'm not convinced. Let's say the energy lost in the weights falling into their new position is K. The total energy is then,

ρhAgD - 2Mgh + 2Mgh - K    (?)

[Genady]

I think I know. The cups will never get over the turning points on the very top and on the very bottom of the upper and the lower wheels. This is because regardless of how the cups are attached to the belt, it takes h longer way for the upper weight to reach the top of the upper wheel than it takes for the lower weight to reach the bottom of the lower wheel. See, for example, this:

Thus, to get to the turning point the device needs at least Mgh extra energy. But, as @Lorentz Jr has derived above, this is equal to or more than the energy that all other cups can supply. So, the device will stop before reaching that configuration.      

QED  

[Saber]

I  admit im  dumber   than that  to  understand it.......haha ...........  i mean  if i  put  time and  effort....maybe  i  will    but   in the  country  i live  and  the  situation  i am   now........i  have to  work ......10 hours a day........really i  have  not that  much  of  mental  energy  to  put it  on that 

Ill save  it and try  to put  some  effort  on it  later......also  for the    post   @studiot     sent  for th  balloon buoyant  from  the book..............