Be careful - you are applying the reasoning about a static situation to a situation that is not static. This rod will be rotationally and linearly accelerating, so you can no longer assume the net force or net torque is zero - $\Sigma F_y = m a_y$, and $a_y \ne 0$.
If the rod is instantaneously horizontal and at rest, it's not too hard to find $F_n$ at that moment:
$$\Sigma F_y = M a_y$$
$$F_N - M g = M a$$
We can get more information about the rod by using the rotational version of Newton's 2nd law:
$$\Sigma \tau = I \ddot \theta$$
Where $\tau$ is a torque, $I$ is the moment of inertia of the rod, and $\theta$ is the angle between the rod and the horizontal. Let $l$ be the distance between the center of mass and the fulcrum:
$$M g l = I \ddot \theta$$
We can find a geometrical relationship between the angular and linear accelerations:
$$y_{CM} = - l sin(\theta), ~~x_{CM} = l cos(\theta)$$
Using the simplifying fact that we are only concerned with the instant when the rod is at rest and horizontal, we can differentiate with respect to time (what follows is only valid at that instant):
$$\ddot y_{CM} = a_y = - l cos(\theta) \ddot \theta + l sin(\theta) \dot \theta^2$$
$$\ddot y_{CM} = a_x = -l sin(\theta) \ddot \theta - l cos(\theta) \dot \theta^2$$
Since $\theta = 0$ and $\dot \theta = 0$ at this instant,
$$a_y = - l \ddot \theta, ~~a_x = 0$$
$$\ddot \theta = - \frac{a_y}{l}$$
Substituting into the rotational version of Newton's 2nd law:
$$M g l = - I \frac{a_y}{l}$$
$$a_y = - \frac{M g l^2}{I}$$
Substituting into the linear Newton's 2nd law:
$$F_n - M g = - \frac{M^2 g l^2}{I}$$
Substituting the moment of inertia about the fulcrum (which can be obtained with the parallel axis theorem) $I = \frac{1}{12} M L^2 + M l^2$$
$$F_n - M g = - \frac{M^2 g l^2}{\frac{1}{12} M L^2 + M l^2}$$
Simplifying and solving for $F_n$...
$$F_n = M g \left(1 - \frac{l^2}{\frac{1}{12} L^2 + l^2}\right)$$
To sanity-check this answer, we can return to the balanced case where the center of mass is directly over the fulcrum ($l = 0$), which gives us the familiar result $F_n = M g$. Additionally, as $L \rightarrow \infty$, the moment of inertia increases, which should cause the rod to rotate more slowly, and indeed, $F_n$ approaches its value for the static situation in that limit.
If you had a very squishy object, it will exert a small force on the ground, whereas if you have a very hard object there will be a large force. From this it's clear that asking for the force is ambiguous, we're going to need to introduce some other variable.
As the question suggests, one thing we can do is include a variable, lets call it $\Delta t$, which tells us the duration of the collision between object and ground. The nicest way to do this is to write down the force equation you had:
$F(t) = \frac{dp}{dt}$
I've included the time on the left hand side to remind us that the force will change as a function of time over the course of the collision.What we can do is integrate this equation from $t=0$, the time of contact, to $t=\Delta t$, the time at which the object comes to rest. Then
$\int_0^{\Delta t} F(t) dt = \int_0^{\Delta t} \frac{dp}{dt} dt = \Delta p$
We can multiply and divide by $\Delta t$ to see that
$\Delta p = \Delta t \left(\frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt \right) = \Delta t \,\, F_{avg,t}$
Presumably this is how you found $F = M \sqrt{2g h} / t$, but I wanted to be clear what $F$ meant in that equation.
As an alternative which doesn't use the time, we can exploit the work energy theorem:
$F(x) = \frac{dW}{dx}$
Again, integrate both sides, this time from $x=0$ to $x=\Delta x$, the total distance over which the collision occurs. This time we find
$\Delta x F_{avg,x} = W$
By the work energy theorem $W = \Delta E$ where $\Delta E$ is the change in energy of the object, so
$F_{avg,x} = \frac{1}{2\Delta x} M v^2$
This gives us a way to write down an average force without reference to the time. The tradeoff is that now we have the distance over which the collision occurs, and we find the force averaged over position rather than over time.
Best Answer
I'm not 100% sure about this but here goes. If your cupboard is in equilibrium, or not undergoing any changes in motion, the wall is pushing up on it with the same force it is pushing down on the screws/nails holding up the cupboard. That is to say $$ΣF = 0$$
Using: $$F = ma$$ and assuming you are near sea level, we can use the acceleration due to gravity at sea level in place of a. $$F=m(9.8m/s^2)$$ We can use the mass you mentioned $$F=(15Kg)(9.8m/s^2)$$ And since a Newton is 1Kg * m/s^2, we get $$F=147N$$ If it is already supported in two places then each one has about 73.5N on it. If you add another bracket(for a safety margin), it would be 49N on each one.
$$147N / 3\text{ Total brackets} = 49N/\text{Bracket}$$ Since the load and the cupboard and the books have the same mass, you can just double our force calculation. You would also double the number of brackets. $$(147N ⋅ 2) = (3\text{ Total brackets}) ⋅2$$ $$294N = 6\text{ Total brackets}$$ $$\text{Because:}$$ $$\frac{249N}{49N/\text{Bracket}}$$ $$\text{Cancel the Newtons, get the Brackets out of the denominator and divide the fraction:}$$ $$6\text{ Brackets}$$ 6 Brackets minus the two already in your cupboard, leaves you with 4 brackets to be evenly spaced along the bottom.
I hope this will help you decide how to mount your cupboard.
Edit: Reading the comments above reminded me about the torque. If nothing else, I think a bracket at the top might stop the cupboard from leaning out from the wall/bending the brackets on the bottom.
Edit: You could also technically just eyeball it, and say "Well, I add another 30Kg, I have brackets rated at 15Kg, so I'll add two more, plus one at the top to stop it from falling forward." However, I don't think this is a very safe approach, and would recommend adding another bracket on the bottom to your eyeball calculation as a safety margin, if you were to do this.