# Hydrostatic paradox one step further- weights of the different fluid amounts

1. Jul 23, 2011

### symphwar

1. The problem statement, all variables and given/known data

Two open-top containers, #1 on the left
and #2 on the right, with equal base area A
are placed on two scales. The #2 container
on the right has an lower diameter twice that
of its upper diameter and the height of its
lower (larger) diameter is half that of its water
height. Both containers are filled with water
to the same height H, as shown below.

http://img89.imageshack.us/img89/3197/pascalvases.th.png [Broken]

(http://imageshack.us/photo/my-images/89/pascalvases.png/)

What is the relationship between the weights
exerted by the flasks on the scales supporting
the containers?

1. Wleft = 8/5 Wright
2. Wleft = 6/5 Wright
3. Wleft = 5/3 Wright
4. Wleft = 7/4 Wright
5. cannot be determined
6. Wleft = 3/2 Wright
7. Wleft = Wright
8. Wleft = 2 Wright
9. Wleft = 7/5 Wright
10. Wleft = 4/3 Wright

2. Relevant equations

W = mg
m = V$\rho$
P = Patm + $\rho$gh

3. The attempt at a solution

My first try with this problem was to use the given dimensions of the container to determine that the flask on the right contains 3/4 as much water as the one on the left. However, 4/3 Wr was not correct. Looking at the differences between the flasks, I surmise that the flask on the right has the gravitational force from the mass of water, as well as downward forces from the horizontal "sides" of the container. These, added together, must equal the force from the bottom of the flask (upwards). This force from the bottom is equal to the pressure from the water on the bottom, and is, in classic hydrostatic paradox fashion, equivalent for both containers- F = 2A$\rho$gh. The weight of the left beaker, I think, would just be this force on the bottom. The sides of the right container are the additional factor I can't seem to weave in. I guessed that the normal-type force downwards from the flask "sides" would be F = PA = 1/4 A$\rho$gh per side, being h below the surface of the entire column of water and acting over 1/2 d of the bottom per side. But subtracting this, doubled, from the force acting on the bottom, just gives me the incorrect 4/3 ratio of weights again. What am I doing wrong? Thanks in advance for any input and sagely wisdom!

Last edited by a moderator: May 5, 2017
2. Jul 24, 2011

### ehild

Why 3/4? Remember, a flask exist is three-dimension, it is not a two-dimensional figure on a sheet of paper or on the display of a computer. Determine the volumes in terms of d and h. What do you get?

ehild

3. Jul 24, 2011

### symphwar

Well, knowing that the bottom area is the same for both, the volume contained in the left beaker would be A2h. Whereas, for the right flask, that would be....Ah + 1/2 Ah = 3/2 Ah. W,l = 2Ah vs. W,r = 3/2 Ah, so W,l / W,r = 4/3 again? I could be making a mistake assuming that the top, skinny part of the right container has a cross-sectional area of 1/2 A?

Last edited: Jul 24, 2011
4. Jul 24, 2011

### ehild

The diameter is half, what is the area?

ehild

5. Jul 24, 2011

### symphwar

D'oh. Lesson learned- never assume things are cubical (/forget what 'diameter' implies). Thanks so much! Obviously (argh), Wl = 8/5 Wr.

6. Jul 24, 2011

### ehild

Welcome in 3D world.

ehild

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