# the angle of water surface in a moving car

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: 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:

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.

• 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. – Phil Sweet Jan 2 '19 at 1:09

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!