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The question asked was "what should be the acceleration such that the pressure at both the points marked by thick dots be equal? the vessel is open and cubic with side 5m?"

Initially i considered the diagram to be resembling the figure B where $$\tan(\theta)=a/g$$ but then i realized since the both points are the midpoints of the respective sides, the height of the liquid above the lower point must be half of the height above the point on the side. I took the diagram then to resemble D. Then i got the answer as $a=2g$. Is this CORRECT?
If this had been a closed vessel what would the answer have been? Figure C That is my question.

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I am sorry to rush, but the answer to these questions are required quite urgently. thank you in advance. –  Satwik Pasani Jun 22 '13 at 16:21

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Actually the pressure at those two points doesn't only depend on the height from the surface but also the distance of those points from the right wall.
This is because there is an acceleration towards the right, and if you move from right to left in the same horizontal level in the container, you would not get a pressure difference corresponding to gravity, but you would get a pressure difference corresponding to the rightward acceleration.

A better way to look at this model is to view from the frame of the container and apply a pseudo-force on the liquid corresponding to the acceleration.
It can be said that a net force(vector sum on gravity and pseudo-force) will act on the liquid, and the surface of the liquid will align itself perpendicular to the direction of the force. Now the condition of equal pressure will be found when the perpendicular distance of those two points is same from the surface(can be regarded as the apparent depth of the point).

Applying that condition you should get your answer. P.S. i haven't calculated the answer but i don't think $a=2g$ is right...

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I got the intuition behind the method that I have to apply. But what if the vessel is closed and the net amount of water remains same therefore preventing any change in surface profile? –  Satwik Pasani Jun 28 '13 at 5:29
    
you'll have to find the force the walls exert on the fluid, and treat it just like external atmospheric pressure. But the force will be different at each point, and i'm not really sure how to calculate that. that's why i didn't post it. –  udiboy1209 Jun 28 '13 at 12:46

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