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Consider a liquid in a tub and this tub is accelerated. And so the liquid will have a slanted surface. I am clear about finding pressure difference between two points in the fluid having same height with respect to horizontal. But what if the two points have a height difference with respect to horizontal? How to account for pressure difference due to this difference in vertical height?

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Look on the water from the point of view of the accelerated reference frame oriented in such way that the surface of the water is parallel to plane $x'y'$ and depth below the water surface is measured by $z'$. In this frame, the total gravity (due to Earth's gravity and due to inertial force of acceleration) is directed perpendicular to the water surface and has intensity $\sqrt{g^2+a^2}$ (hypotenuse from the Pythagorean theorem). By the same argument as in usual circumstances, the pressure is function of depth $z'$: $$ p = z' \rho \sqrt{g^2+a^2}. $$

So to find out pressure at any point, find out its coordinate $z'$ and use the above formula.

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you will have to do this in two parts first of all take into account the horizontal difference and then the vertical distance

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    $\begingroup$ Is the vertical component of pressure just rho * g * h while horizontal part is rho * a * x where x is horizontal separation between points $\endgroup$ – user34304 Dec 3 '13 at 15:59
  • $\begingroup$ yes you are right but you will have to also pay attention to the slanted surface of water $\endgroup$ – Sahil Chadha Dec 4 '13 at 5:26

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