Recent content by Richard Spiteri

(Δy.A)leaks out the right hand side tube Δy is the drop in level in the uncut tube

Figure ii cf Figure iii: I get the same result if we go with a/2. The height H and value Δx must be the same in both figures. Figure iii: In my solution I work out that if I cut either one of the two tubes on the right by Δy , then the whole system experiences a bulk shift of Δy resulting in...

I think what you are getting at is the conservation of mass that affects my H and Δx in Figure iii when compared to the results shown in Figure ii. If it makes it easier to remain consistent so we can compare, then let's change the areas a in each of the right-hand side tubes in Figure iii to...

In my mind, it is there initially at the same height as the one in the middle, the system equilibrates and then it is cut off by Δy. In my mind, it is there initially at the same height as the one in the middle, the system equilibrates and then it is cut off by Δy.

Thank you @Chestermiller and @Ibix for the threads above. I have worked it out using simpler and clearer variables and definitions and have come up with the following which should all be correct. Figure i above defines the variables and there is no question. Figure ii is what I postulated...

Also, what is the upthrust U (that the Red man must resist) on the Air Tank in #37? Personally, I think it is a constant equal to the volume of water that the Air Tank displaces U = LAir Tank . AAir Tank)ρwaterg if you assume that the Air tank dimensions are LAir Tank by AAir Tank. Or am I...

The figure above, is my original Fig 3 reposted for clarity In Fig 3, I isolated the hollow tubes from the outside water using the water bags. My question was "what force is the Green man exerting in each of the cases below" as the width a of the hollow tube increases from 0 to A? The figure...

@Chestermiller , This is your equation as relates to my original Fig 2. F = Mwater tankg + L(A-a)ρg (Note, @sophiecentaur , that I changed W to A and w to a so there is no confusion with upper case and lower case fonts and to reflect 'areas' rather than '2D widths'.) I "plotted" (more like...

@Ibix , my equation which equates pressures (that you say is correct) is: H = M/(ρW) - Δab This is still confusing to me as there is no dependency on lower case w (the width of the right hand tube). It says that H does not change regardless of lower case w ...unless w = f(Δab) I pressed the...

Please look at my thread #9 for the diagram. I cut a bit off the top (equal to x) and then defined the extra depth that the mass sinks as Δbc for a total of (Δab + Δbc) = Δac. It should be clear in the figure in #9.

This was my follow up question but I think I have enough feedback to try and figure it out for myself. I would say that: [p0 + ρg(H - x + Δac)]W - p0W = Mg and then solve for Δac .

Great, so my equation is correct. It was my assumption (thanks, @Ibix ) that the mass M displaces its own weight that was incorrect. Here is my follow up question:

This is good for me to hear. So, how should I treat the red mass if not floating? If it is sitting on the water and we assume no friction and no leaks between the manometer wall and the red mass, is that different from floating? Does it still displace its own weight or is the physics different...

@Ibix, you are right, I should have been more specific in stating this is a 2D (or a 2½D) model. To keep things simple assume the "volume" is always (area)*(unit depth). In that case the "volume" of water displaced must give you H = Δab(W/w) as shown in my figure. I cannot, however, see the...

Consider a manometer as shown above with different widths W and w. If we take a mass (red) that is frictionless and does not allow water to leak of mass M, then I would like to calculate the height of water H in the narrow tube of width w. I arrive at a non sensical answer of H = 0 in one...