Challenging question - forces

Ah. If you have different container shapes then different volumes then different masses then different weights therefore different forces.

The pressure of the water according to rho*g*h is a hydrostatic pressure of the water exerted on the inside of the container. That's an internal force that doesn't affect the total weight.

The pressure of the container on the scale is F/A, where F is the weight. Since they have the same area at the base, the pressure is higher for the cubic container since the weight is higher.

I really don't understand this. There is a force balance on every bit of container wall, balancing the water pressure inside with the air pressure outside and, where these are different, the container wall will slightly deform outwards and that deformation will generate a restoring force that balances the pressure difference.

I don't see how that affects the scales though.

Hmmmm. But that force is still an internal force, i.e. it doesn't actually act on the lever.

It's like if you set a mousetrap, there are a bunch of internal forces as the spring is loaded, but that wouldn't affect the weight when compared to a mousetrap that's not loaded.

Shouldn't make a difference. The downforce would change when the trap springs, but it will be the same after as it was before.

No that doesn't work because the force is not acting on the lever. The fact that some water pressure acts upwards on the inside of the container is irrelevant. There's a Newton's 3rd force pair from the container acting back into the fluid, which completely cancels the water pressure when summing. This is, in general, why internal forces don't affect the lever.