Cylinder constrained between two surfaces

Question

In the figure the system accelerates leftward. If acceleration increases, how does normal force exerted by sides $\overline{AB}$ and $\overline{BC}$ on cylinder change? The expected answer is $N_\overline{AB}$ remains constant and $N_\overline{BC}$ increases.

My reasoning is that $N _\overline{AB}$ must decrease because acceleration acts leftward so pseudo force acts rightward pushing it more firmly on $\overline{BC}$ increasing the normal reaction on $\overline{BC}$. By the same reasoning w.r.t. surface $\overline{AB}$ a component of pseudo force acts in an opposite direction thus reducing the normal reaction on $\overline{AB}$.

Another logic to justify my analysis is that sum of normal forces on the surfaces must remain constant but if one increases and another remains constant that is no longer the case.

Please help I'm confused.

Answer 1

Your cylinder is interacting with three parts of its surroundings:

1. There's a gravitational attraction towards the center of the Earth, which I'm assuming is "down" in your figure and so has only a vertical vector component.

2. There's a normal interaction with the wall $BC$, which has only a horizontal component because it is perpendicular to a vertical wall.

3. There's a normal interaction with the wall $AB$, which has both horizontal and vertical components.

The question is how these different interactions change depending on whether the cylinder undergoes horizontal acceleration or not. Horizontal acceleration doesn't change the weight of the cylinder. Vertical acceleration of the cylinder is not an option that's available to us, so the vertical component of the contact force that's normal to $AB$ must be a constant independent of the horizontal acceleration. But the direction of the force normal to $AB$ isn't adjustable either --- if it were, then we couldn't call it a "normal" (i.e. perpendicular) force.

So to set the horizontal acceleration, our only degree of freedom is the horizontal force that's normal to the vertical surface $BC$.