Toppling and Torque related to it When a body is said to be being just about to topple about one of its edge due to a force $F$ applied, why do we not consider the torque of normal about the edge point but about the centre of mass? Are we not trying to make the net torque zero at the edge? 
Why do we do $F\cdot b=N\cdot a/2$ (where $a/2$ is distance of box face from centre towards the direction where normal shifts and $b$ is the height at which the force is applied) and not $F\cdot b= N\cdot a$, to get the required force $F$ for toppling?
Also, how fast do normal forces adjust when they are required to?
 A: 
When a body is said to be being just about to topple about one of its edge due to a force F applied, why do we not consider the torque of normal about the edge point but about the centre of mass? 

We do consider the toppling torque about the edge (pivoting point):

The left hand box is perfectly balanced: the force $mg$ points perfectly to the pivot point $O$.
Now look at the right hand side. Assuming the box was immobile to start with, then to remain immobile Newton's Law tells us that no net forces or net moments may act on it.
The first condition is fulfilled if $F_N=mg$. But the fact is that the object is not balanced because $mg$ now provides a torque $\tau$ about $O$:
$$\tau=mg |OO'|$$
This torque will cause angular acceleration $\alpha$ (counter-clockwise, in this case):
$$I\alpha=\tau$$
Where I is the inertial moment of the object (about a horizontal axis running through $O$).
So the object will topple to the left (unless another external force is applied that counter-acts $\tau$).

Also, how fast do normal forces adjust when they are required to?

Instantly.
