Pascal's paradox

[saim_]

http://www.fas.harvard.edu/~scdiroff/lds/NewtonianMechanics/PascalsParadox/PascalsParadox002.gif
the picture is from http://www.fas.harvard.edu/~scdiroff/lds/NewtonianMechanics/PascalsParadox/PascalsParadox.html

The pressure on base of the first container (far left) and that of the second container (middle) is different, right? In the first container the pressure will be the same, as the pressure on the base of the second container, only exactly under the mouth of the container. On points not exactly under the mouth, the pressure on the base will be different in the first container. Is this correct? Otherwise, I cannot comprehend how there is a difference in the measured weights.

[Delta²]

According to the web page the pressure on the base of each container is equal and i think it is right.

The scales measures the force N that each container applies to the scale. Since the containers and the scales do not move at all this means that N=weight of container (conclude that from Newtwon laws of motion).

This force N is equal to force F that the water applies to the bottom of the container (and which is equal to F=pressure x bottom area ) only in the case of middle container. (because the A component -see below- is zero)

In left container N=F-2A1 where A1 is the vertical component of the force that the water applies to each side.

In right container N=F+2A2.

[saim_]

Hey, thanks alot! I think you are right. There will be a net force on the sides of the first container which is away from the containers and perpendicular to the surface of that side. Thus it will have a net upward component. Similarly in the third case, due to the way sides are tilted, the force on the sides will have a downward component. No vertical component for the second case. Thanks again.