Cylinder Load: Newsletter Challenge (07/01/08)

p - pressure

s1 - cross section first cyl.

s2 - cross section second cyl

g1 - load first cyl

g2 - load second cyl

p=g1/s1; p=g2/s2

so g1/s1=g2/s2

s1=s2/2

so when the additional load is added to the system piston one is going from height 0.6m to 0. piston two is going up displaced by same volume as piston one becouse the pressure will be the same (if we change the load on second piston also the pressure will change and the volume of gas also, but the case is not like that) so.

s1*h1=v; s2*h2=v and s1=s2/2 so

s2/2*h1=s2*h2

h2= h1/2=0.6/2=0.3 m

so the diference is 0.6 + 0.3 = 0.9 m total.

The smaller piston will fall to zero and the larger will rise to 1.2m. Assuming that the two were in mass - pressure equilibrium before the change in state.

Mr. Gee

In the statement of the problem, in my humble opinion, one would assume that the two pistons are in equilibrium which they cannot be. In force balance problems equilibrium is reached when the forces are equal by computing pressure working across an area. After practice one begins to understand that it takes a lot of force acting on a smaller piston to counteract a small force acting on a large piston. The problem stated that the smaller force piston also had the smaller of the two areas and was working against a piston with a larger force working across a larger area. The only way this system can be in equilibrium is if one of the areas is negative.

Based on this hypothesis, the smaller pistons has to be at it's upper stop or the larger piston is at it's lower stop. Either way, having equal weight on each of the two pistons will not begin to bring the system into equilibrium either. Therefore, after adding the weight to the smaller piston, the change in height of the two pistons will be unchanged.

It the wording it says initial the pistons are at the same height but since one is heavier they would end up in different heights, or unbalanced however when you double the mass of the smaller piston is would be 4kg. This would cause the two pistons to be balanced. The height difference to be 0 or equal.

Dear frnds,

With the phrase " The system is place in vacuum ", the poster of this question wants to prove that the solutions will be under constant pressure of gas , by my opinion.

This yields that P₁= P₂ in the P₁V₁=P₂V₂ equation.

So, comments that talking about the small piston will drop to bottom of the cylinder and finalised with 0,9m height, are all true answers.

I don' think that it is necessary to calculate the work done by gas in the thin pipe.

use this point of view at this time, hopefully we can arrive at a common solution.

regards

As the "contributor" most responsible for attacks on an alternative problem, I'd like to reassure you: there was already agreement among most of us playing with alternatives that the proper answer to the problem as posed is "very slightly less than 0.9=metres"*. Sorry if that was not clear.
*The value of "very slightly less" depends on the density and ideality of the gas, and on the total volume and deposition of gas in the system.

On the other hand, the vacuum constraint is unnecessary. I don't think it even helps reduce the effect of the mass of the gas in the cylinders, unless you choose a fill gas with vastly different molecular weight to the surrounding atmosphere (you could even choose a fill-gas to make them compensate)

Thanks Physicist?

You have told the point quite well. I also egree with you that there would exist any other gas instead of vacuum and this would not affect the behaviour of the pistons.

If the fill gas is not ideal, the equation P₁V₁=P₂V₂ will not true of course.

I want to emphasize is that the question itself tends to " finding out the probabilities ", instead of a physical problem that can be solved with math, because of the absence of some parameters in the question.

The answer "very slightly less than 0.9=metres" is not a mathematical result, it is true though.

One could say " how much less than 0.9m? " and introduce a new problem. I mean this question could not arrive a clear solution.

I hope i can tell my stressed point.

'The system in place is a vacuum.' Presumably you mean that the system is placed in a vacuum? The checking of your text is somewhat lacking. How many other messages have you got regarding this?

Chanter 'n spanner

Amen!

As a layman with only logic to guide me The answer is 1.2m . The smaller piston has doubled in weight therefore will "sink" to bottom of cylinder ie. .6m The larger cylinder is connected to the smaller cylinder by a gas medium therefore it will rise accordingly the same amount, .6m It is in vacuo therefore no resistance.

Simple math: .6m + .6m = 1.2m

This is the principal of the Hydraulic lock for raising boats at a city called Peterborough , Ontario, Canada. Look it up !

Problem cannot be truly solved without more information.If we Assume that the pistons are made of the same material, and have the same density, therefore the same ratio of area/mass, the info is adequate, but these type questions should not leave room for assumptions. Since the author of the question expressed the variables as mass,and load, rather than weight, and the environment as a vacuum, could he be talking about in space, free of gravity's influence?If so, then no difference at all.

I know this is splitting hairs, but assuming unstated info can lead to errors.

I believe the answer is 0.6 m.

The way I've looked at it is to use Bernoulli's Theorem [(v²/2g) + (P/ρg) + h = 0]. Assume the smaller cylinder has diameter 'd' and the larger one, 'd₁'. Thus, considering the base of the cylinders (at the thin connector pipe), we initially have:

For the smaller cylinder: v²/2g (=0) + P/ρg (=2 /((πd²/4) ρg) + h (=0.6) = 0 = for the larger cylinder= v²/2g (=0) + P/ρg (=4 /((πd₁²/4) ρg) + h (=0.6)

Hence: d₁ = √2 d

Also, since the Bernoulli term is a constant, then we obtain, 2 /((πd²/4) ρg) + h (=0.6) = 0 → 8 /(πd²ρg) = -0.6

Now, by adding a 2 kg additional mass to the smaller piston and applying the above theorem again, we have:

For the smaller cylinder: v²/2g (=0) + P/ρg (=4 /((πd²/4) ρg) + h₁ (unknown) = constant = for the larger cylinder v²/2g (=0) + P/ρg (=4 /((πd₁²/4) ρg) + h₂ (unknown).

By substituting d₁ = √2 d into the right hand side above, we obtain:

h₂ – h₁ = Difference between the height of the two pistons) = 8 /(πd²ρg) = (-0.6 m from above).

When the small piston comes to rest at the bottom, it is no longer entirely supported by the gas. So P/(ρ.g) in the small cylinder is no longer equal to
weight(small-piston+load)/(area_of_small-cylinder)
[This is now just an upper bound on the pressure].

That invalidates your statement:
For the smaller cylinder: v²/2g (=0) + P/ρg (=4 /((πd²/4) ρg) + h₁ (unknown) = constant = for the larger cylinder v²/2g (=0) + P/ρg (=4 /((πd₁²/4) ρg) + h₂ (unknown).