Where does normal force shift to? 
"A block of side length $a$  and mass $m$ resting on a ground with a coefficient of friction $\mu $. A force $F$ is applied on the upper edge of the block (think of the cross-section and one of the topmost vertices). How much force should be applied to topple the block?"

In the solution key, they showed that the normal reaction acts on one of the vertices rather than the center of mass. After reading a stack exchange article, I came to know that normal force does indeed shift. But I can not understand how to know where it will shift ( I need to know where exactly this is so I can take moments of force i.e: torque)
ref: Does the normal reaction shift when a force is applied? 
 A: For an ideal block standing on an ideal surface, each contacting point carries some load (counteracts a portion of the burdening weight). At each such point a normal force comes into existence to counteract the pressure at that point.
If horizontal and with a symmetric block with its centre-of-mass (CoM) right in the middle, then all the contact points carry equally much and experience the same small normal force. Regarding the torques they cause about the CoM, all those normal forces on the left side of the CoM cancel all those out on the right side. If the CoM is not in the centre, then some normal forces will have to be bigger than others (not evenly distributed).
You rarely need to care about this normal-force distribution over the contacting surface. Instead, how do we in an easier way note the normal forces' influence? We'll just sum them all up (so it still has the same influence in Newton's laws) and let them pass through the CoM (so it still has the same influence on the rotational version of Newton's laws). It has to be drawn at the contacting surface of course, since this is a contact force. And that's why we often only draw one normal force passing through the centre-of-mass. (Of this same reason, we also only draw one weight from the CoM.)
If you now push at the box from, say, the right side, then you are creating a torque which  pushes particles to the right up and particles to the left down. In other words, you are influencing the load and disturbing the normal-force balance. Surely, the contact points farthest to the left will experience a larger load and thus a larger normal force. This corresponds to the total normal force being shifted towards the left.
At the very moment where the box is about to tilt over, then all points are about to lift off from their contact with the surface - except for one point: the pivot point at the left corner. At this moment, the load has been lifted off from all those points, so their normal forces are zero, and only the corner pivot point carries any load - and it now carries the whole load. Thus the total sum of all normal forces now equals the normal force at this corner pivot point.
And this happens only at the tipping moment - before then, the total normal force is anywhere between the point under the box' CoM and the corner, depending on when in the pushing phase you are looking.
A: The magnitude of the normal force is sufficient to prevent the block from accelerating downward.  As long as the material does not yield, it will equal the weight of the object.
The location of the normal will shift to prevent rotation.  If you apply a torque $\tau$, the normal will shift sufficiently to oppose the rotation.  The limit of this process is when the normal has shifted to the edge of the object.  If the external torque were to increase at that point, the object would have an angular acceleration (topple).
