calculating water pressure on inclined surface

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ketanco

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Ok basic question but I am a Geo guy and forgot this a little...

When calculating the water pressure Force on an inclined surface, all we do it is to calculate the height times gamma water times the area correct? because water acts equally in all directions, we do not have to multiply that force with the sine or cosine of the inclination right? i mean, the force we find, from gamma x h x area, means, it doesnt have to be horizontal, but acts perpendicular to that surface correct?

 
See here. http://engineeringregistration.tamu.edu/videos/PEDownloads/Fluids.pdf

There is a problem just like you are questioning on page 20. In its simplest form pressure=(density x gravity x height)/gravitational constant. In u.s. units gravity over grav. constant=1 and cancels, in SI units that's not the case (gravity remains in equation). The resultant force acts at the center of pressure, not the avg depth.

 
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