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Old Jan 15th 2016, 08:09 PM   #1
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aceeleration im staraight path of fluid

how to derive the formula of vertical rise of the surface ? Δz ?
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Old Jan 19th 2016, 03:38 AM   #2
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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.
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Old Jan 28th 2016, 11:24 PM   #3
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Originally Posted by ChipB View Post
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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