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enter image description here in this picture, there is a container of water and it is moving on the slope. we want the acceleration of the container while the angle of water surface is parallel to the slope. so the book says the answer is that we let container move by its own weight so $\color{red}{a_x=g\,sin\alpha}$. but what i'm asking is that let us think the slop doesn't exist. so the accelerations would be like this: enter image description here and the formula of the angle of surface in fluid mechanics is: $$\color{green}{tan\,\theta= {\Sigma a_x\over \Sigma a_y}}$$ so in this case $tan \theta $ would be: $$tan \theta={gsin\alpha\over gcos \alpha}=tan\alpha$$ so $$\theta=\alpha$$ and this means: enter image description here

and this is wrong because we know if we let the container move by its own weight, the $\theta$ would be $0$ i don't know what am i missing here.

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  • $\begingroup$ Remove the water and replace it with a bowling ball in the middle. What would the bowling ball do? Water is just a bunch of small fictionless bowling balls. There is no net interaction needed from the container ends to accelerate the stuff in the container. If the bottom is parallel, the contents just sit there. The container just has to support the hydrostatic pressure. This is just a variation on the hammer and feather drop experiment. $\endgroup$ – Phil Sweet Jan 2 '19 at 1:09
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Let's assume the container is moving horizontally but under the forces $F_v =mg \cos(\alpha)$ as vertical force and $F_y = mg\sin(\alpha)$

The water surface angle will be $\arctan \left(\dfrac{\sin(\alpha)}{\cos(\alpha)}\right) = \alpha$

Therefor $\theta = \alpha$. So if you rotate the image by $\alpha$ the water will stand parallel with the incline!

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In your last picture you inverted the angle. It should be parallel with the ground. If you drew a line from the water surface, parallel to the ground, then the angle would be the same. You just didn't draw the geometry right. Let's say your water tank is stationary, your water surface will be parallel to the ground, but your angle will be the same if drawn from the opposite direction. The change in angle will reflect the change in velocity.

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The water will regain equilibrium position immediately upon placing the tank on the sloping ground, and the center of the mass will be shifting towards the toe of the hill, the combined effect is causing an additional force to the motion.

enter image description here

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