Fluids acceleration on inclined plane

[jonny23]

But if there is no friction between the box and the incline, then generally the box will either slide down or up the incline depending on the strength of the force F.

this is the general equation which i derived.
See following CASES :
if a=0 you see the equation gives angle of free surface as (alpha)
if a< g tan(alpha) where (alpha) is angle of inclined plane, then block moves downward wrt to inclined plane and free surface is given as per equation
if a =g tan(alpha) block is stationary wrt to the plane and the angle comes out to be zero which should be (as in post #14)
if a> g tan (alpha) block moves upward with respect to the plane and the angle comes out to be negative

 

[TSny]

If a represents the acceleration of the incline, then I agree with your conclusions about which way the block will slide along the incline. However, I get a simpler answer for the orientation of the surface of the water. I think you can get the answer with very little calculation. Go to the non-inertial reference frame of the block and consider a free body diagram of the block. In this frame there is an effective acceleration of gravity, $\small \vec{g}_{eff}$, that can be expressed in terms of the true acceleration of gravity $\small \vec{g}$ and the acceleration of the block relative to the earth, $\small \vec{a}_{block}$. From the free body diagram of the block, can you determine the direction of $\small \vec{g}_{eff}$ without first working out an expression for $\small \vec{a}_{block}$?

 

[jonny23]

then i am getting tan^-1 (a/g)
But i think its not right because if you put a =0 then the free surface would be horizontal but shouldn't it be paralllel to incline?

 

[TSny]

then i am getting tan^-1 (a/g)

But i think its not right because if you put a =0 then the free surface would be horizontal but shouldn't it be paralllel to incline?

Right, we know that the answer for a = 0 should be that the surface is parallel to the incline. Not sure how you got tan^-1 (a/g).

 

[256bits]

Up to about post #13, the thread was mostly in relation to the original problem. Nobody could come up with an explanation for the supposed correct answer. Instead, the consensus is that the body would not move relative ro the liquid.
After that, the thread broadened into interesting discussion of how the liquid would behave if it does not fill the box, under various conditions of friction and applied forces.

Problem states that the closed box is allowed to slide down, not that it is sliding down the ramp. I interpret this to mean it eventually ends up moving horizontally.

At the curve the box will experience a dw/dt, a change in dy/dt and a change in dx/dt.
It could be that with different angles of incline, the differences in these values will cause the object to shift its position within the box.

 

[haruspex]

Problem states that the closed box is allowed to slide down, not that it is sliding down the ramp. I interpret this to mean it eventually ends up moving horizontally.

I read it that it is sliding down the ramp. I see no suggestion that we are to consider a later stage in which it is moving horizontally.

At the curve the box will experience a dw/dt, a change in dy/dt and a change in dx/dt.
It could be that with different angles of incline, the differences in these values will cause the object to shift its position within the box.

As I posted before, there's a simple thought experiment. If the object has the same density as the water then you can replace by 'block' of water. There need be no physical boundary on this block, it is just a region of the water we choose to identify as a distinct object. Why should that move around within the body of water generally (other than rotate, maybe)?