Q.A rectangular block rests on a horizontal table. A horizontal force is applied on the block at a height $h$ above the table to move the block. Does the line of action of the normal force $N$ exerted by the table on the block depend on $h$?

Step 1. Identify all the forces acting on the block.

Three (or four) forces act on the rectangular block resting on the table: its own weight $mg$, acting vertically downward through its center of mass $C$; the table's normal reaction $N$, acting vertically upward, distributed somewhere across the base of the block in contact with the table; the applied horizontal force $F$, acting at height $h$ above the table; and, if the block is not yet sliding, static friction $f$ opposing $F$ at the base.

Step 2. Apply translational equilibrium (vertical direction) — this fixes $N$'s magnitude, not its location.

Since there is no vertical acceleration, the net vertical force is zero:

This shows $N$'s MAGNITUDE is simply $mg$, entirely independent of $h$ — but this equation says nothing at all about WHERE, across the base, this net upward force effectively acts. That question is answered only by the separate, rotational condition.

Step 3. Apply rotational equilibrium (torques about the center of mass) — this is what fixes $N$'s line of action.

Take torques about the center of mass $C$ (a convenient, fixed reference point). The applied force $F$, acting horizontally at height $h$ above the table (hence at a vertical distance from $C$), produces a torque of magnitude $F\cdot(\text{vertical lever arm})$ about $C$, tending to rotate the block. For the block to remain in rotational equilibrium (not tip over), some other force must supply an exactly equal and opposite torque about $C$. Weight $mg$ passes straight through $C$ and contributes no torque about $C$ by definition. Friction $f$ acts at the base, essentially at the same height as $N$ (zero height, i.e. no vertical lever arm relative to the base line), so on its own it contributes a torque about $C$ set by its horizontal lever arm below $C$, but not one that can freely vary to balance an arbitrary $F$-torque. It falls to $N$ — by shifting where, across the base, it effectively acts — to supply the necessary balancing torque.

Step 4. See explicitly how $N$'s line of action must move. …

Separating translational equilibrium (fixes magnitude) from rotational equilibrium/t …

• Assuming that because $N=mg$ is independent of $h$, the ENTIRE behaviour of $N$ (including where it acts) must also be independent of $h$ — magnitude and line of action are governed by two DIFFERENT equilibrium conditions (force balance vs torque balance).
• Forgetting that $N$ is really the resultant of a distributed contact pressure across the whole base, which is free to shift its effective point of action even while its total magnitude stays fixed. …