I think it's because water moves on a curved surface with centripetal force. Because the gravity of water provides centripetal force, the pressure of water on the curved surface is reduced. When the velocity of water increases, the pressure of water on the curved surface will be more reduced. When the velocity of water is large enough, it separates from the curved surface, because gravity can no longer provide centripetal force. But some people think that viscosity reduces the pressure of water on curved surfaces. Are they right? enter image description here

This surface is a convex surface.

  • $\begingroup$ I think it is wrong to say to say gravity provides a centripetal force. Sure, you have to take in the normal force of the surface, but think about the case where the slope is either 0 degrees is or 90 degrees. There is no net force pointing toward the "center". $\endgroup$ – wcc May 14 at 21:56
  • $\begingroup$ @AmIAStudent I think gravity has a normal component. $\endgroup$ – enbin zheng May 14 at 22:11
  • $\begingroup$ perhaps you can draw a free body diagram to convince yourself that it is true or not. $\endgroup$ – wcc May 14 at 22:12
  • $\begingroup$ @AmIAStudent Imagine that there must be pressure when water moves on a horizontal plane because of the gravity of the water. When water moves on an inclined plane, does it have no pressure on the inclined plane? $\endgroup$ – enbin zheng May 14 at 22:29
  • $\begingroup$ I agree there is pressure in the water, but pressure is a scalar and has no direction. I was pointing out that the net force from gravity of water and normal force from the surface seems unlikely to produce a centripetal force. Centripetal force means there is a force pointing to the center to which the radius of the curvature is defined. $\endgroup$ – wcc May 14 at 22:45

You can see this by applying Euler's equation across streamlines on the curved flow (ignoring gravity if it's a thin flow):

$$ \frac{dP}{dr} = \rho \frac{V^2}{R}$$

where $R$ is the radius of curvature of the bend, and $V$ is the streamline flow velocity. Clearly, pressure must increase radially outward. However, pressure at the surface is atmospheric; pressure below the surface is therefore below atmospheric (again, ignoring gravity).

  • $\begingroup$ If atmospheric pressure is neglected, centripetal force is provided by the gravity of water. $\endgroup$ – enbin zheng May 15 at 4:12

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