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It is known that the force exerted by a static fluid on the walls of the containing vessel is perpendicular to the surface.

The general argument is that if the force were not perpendicular, the normal reaction on the fluid's surface would have a parallel component of force, thus producing shear and the fluid will flow.

But what if the fluid applies non-perpendicular force at two different points such that the net shear is zero? enter image description here

What I basically mean is that if the vessel applies a non-perpendicular normal force on the fluid, then in the configuration shown, the shear on the fluid elements cancel and we get a static condition, even though the force on the vessel is not perpendicular.

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When we make arguments with fluid elements, often we are implicitly taking a limit as the element volume goes to $0$.

In your diagram, two adjacent solid boxes could stay in that static configuration by pushing up on one another. But if the boxes contain a continuous fluid, then we could divide them up into even smaller boxes. Many of these would be along the surface and not have neighbors being pushed in an opposing way. They would thus try to flow in the direction they are pushed by the surface.

When a given small flowing element then reaches the point where the surface force changes, it would no longer be able to move in the same direction. However, as a result of being pushed from both behind and in front, it would be forced away from the surface. In the end, you would have small circulation cells along the surface, so at best the flow would be stationary but not static.

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  • $\begingroup$ How can one be sure that the liquids will move away from the surface? What if the forces acting on the fluid element from all the sides are equal. Then there would be no motion of the element. Much like water in a container open from the top - the pressure inside is trying to push the water up, but the force from the atmosphere keeps it in equilibrium. $\endgroup$
    – Sidd
    Sep 10, 2015 at 21:35

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