Before starting, the meaning of tendency needs to be clarified. I believe that by Tendency, your teacher means the obstructed motion of a body (or a fluid) under a force.
A liquid has the tendency to move down due to gravitational force (or under influence of gravitational force, same meaning). But the liquid’s downward motion has been obstructed by the container in which it is kept.
How can tendency of shifting away alone, change the pressure at point A ? Shouldn't the liquid at A move away from it in order to decrease the pressure there ?
This question can be answered both qualitatively and quantitatively. You have already derived the result quantitatively using <PERSON>’s laws of motion and a few assumptions. What remains is to understand the result qualitatively.
There are many-many ways to understand this. 927
One way is the way presented by your teacher. Another way presented in other answers is as follows:
* When at rest, the weight of the liquid in vertical column and the force exerted by atmosphere causes the liquid to apply force on the rightmost wall.
* When the tube is accelerated horizontally towards the right, a pseudo-force or fictitious force acts towards the left on the liquid in horizontal column with respect to the tube.
* This reduces the force exerted by the liquid on the rightmost wall, thus reducing the pressure at point $A$.
Below, I would like to present another way to derive the result. I have found this method to be much helpful while pondering over your question.
Assumptions
* The liquid is incompressible.
* The surface tension is ignored (I am not even sure what will happen to the meniscus under horizontal acceleration).
* The tube is thin (so that the horizontal pressure variation in vertical part and vertical pressure variation in horizontal part could be ignored).
Derivation
We use only <PERSON>’s laws of motion to derive the result. We take the ground as the frame of reference.
We choose a thin cylindrical part of the liquid with one end adjacent to the rightmost wall of the tube and another end directly at the end of vertical part of tube, and analyse the forces acting on it.
$F:$ The buoyant Force.
$\delta N:$ Normal Force applied by wall.
$\rho :$ Density of liquid.
$P_0 = P_{atm} + \rho gH$ : Force applied by liquid on left of the part of liquid.
We take a small circular cross-section ($\delta A$) to ensure negligible change in pressure across height of cylinder.
Let $P_i$ be the pressure exerted by liquid on the wall when tube is at rest. And, let $P_f$ be the pressure exerted by the liquid on the wall when the tube is accelerating.
Using <PERSON>’s third law: $$ P_i \delta A = \delta N_i \tag{1}$$ $$ P_f \delta A = \delta N_f \tag{2}$$ $N_i$ and $N_f$ are the Normal Forces in the two states.
When the part of liquid is in equilibrium: $$P_0 \delta A = \delta N_i$$ $$P_i = P_0 \tag{using 1}$$
Now, when the tube is accelerating horizontally towards the right with acceleration $a$: $$ ma = P_0 \delta A - \delta N_f$$ $$ \rho la = P_0 - \frac{\delta N_f}{\delta A} \tag{3}$$ Then: $$ \rho la = P_0 - P_f \tag{using 2 and 3} $$ $$ P_f = P_0 - \rho l a $$
$$ P_f < P_i $$
Note: If the presented set of assumptions is not sufficient then please comment. 439