Physics Help Forum aceeleration im staraight path of fluid
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 Jan 15th 2016, 08:09 PM #1 Senior Member   Join Date: Jun 2014 Posts: 306 aceeleration im staraight path of fluid how to derive the formula of vertical rise of the surface ? Δz ? Attached Thumbnails
 Jan 19th 2016, 03:38 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,310 Consider a small volume of fluid at the surface - it will have a force acting on it from pressure of fluid below that equals its acceleration. The angle of that normal force is dependent on the magnitudes of the horizontal acceleration a_x to the sum of vertical acceleration a_z plus g - it has slope (a_z+g)/a_x. Snce the normal force is perpendicular to the surface, the surface has slope -a_x/(a_z+g). I assume you can take it from here.
Jan 28th 2016, 11:24 PM   #3
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Posts: 306
 Originally Posted by ChipB Consider a small volume of fluid at the surface - it will have a force acting on it from pressure of fluid below that equals its acceleration. The angle of that normal force is dependent on the magnitudes of the horizontal acceleration a_x to the sum of vertical acceleration a_z plus g - it has slope (a_z+g)/a_x. Snce the normal force is perpendicular to the surface, the surface has slope -a_x/(a_z+g). I assume you can take it from here.
vertical rise of liquid

can you explain why the formula of vertical rise of liquid is like in the notes? why not Δs =
(g +az ) Δx / ax ? In the post above , you just explain about the ax and az , how to derive the formula ?

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