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[itex]\rho[/itex] P = Patm + [itex]\rho[/itex]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[itex]\rho[/itex]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[itex]\rho[/itex]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!