Rise of Liquid from Horizontal Acceleration

[Ben2]

Homework Statement

"Consider the horizontal acceleration of a mass of liquid in an open container. Acceleration of this kind causes the liquid surface to drop at the front of the tank and to rise at the rear. Show that the liquid surface slopes at an angle \theta with the horizontal, where tan \theta = a/g, a being the horizontal acceleration."

Relevant Equations

p = p_0 + \rho gh

The only way I get this is to make a the vertical acceleration at the bottom corner and g the horizontal acceleration there. This is from Halliday & Resnick's Physics. I've been unable to find anything there or in REA's Physics Problem Solver. Thanks for any hints submitted.

 

[Lnewqban]

Please, see:
https://www.physicsforums.com/threads/acceleration-of-u-pipe-containing-water.999370/

 

[kuruman]

Suppose I attached an inverted L-shaped rod to the side of the accelerated container and suspended a string with a mass tied at the other end. Would the string hang vertically? If not, why not? Now consider the well known fact that masses at the ends of strings, a.k.a. plumb bobs are perpendicular to free surfaces of liquids. Would an observer moving as one with the liquid and the plumb bob see anything unusual? Do you see where this is going? If not, follow the link that was posted above just before I posted this.

 

[haruspex]

The only way I get this is to make a the vertical acceleration at the bottom corner and g the horizontal acceleration there.

Consider the case a=0.
Are you sure you are taking $\theta$ as the angle to the horizontal?