Object on a ledge I have tried to google this but don't know what to google. I know that normal force is the force exerted back on an object that opposes gravity, perpendicular to the surface. My question is, what are the physics of an object that isn't just flat on a surface, but only a fraction of it is. In other words, an object sitting partially off of a cliff. At what point does the force of gravity take over and make it fall as you slowly slide it towards the edge.
 A: If you have learned about free body diagrams, you need to create one, but instead of adding one normal force, add two forces, spanning the base of the object

Now you need to balance the weight $W$ with the two normal forces $N_1$, $N_2$ in the vertical direction, as well as balance the moments about the center of mass (the summation point is arbitrary here).
This leads to two equations, with two unknowns, the two normal  forces
$$\begin{aligned}
  N_1 + N_2 & = W \\
  b\,N_1 -a\,N_2 & = 0
\end{aligned}$$
The condition for stability is that $\boxed{N_2 \ge 0}$. Otherwise, you need to push down on the back of the object to keep it from going over.
A: Check out the Wikipedia article on center of mass. If the object is uniform, then the center of mass is at the center of the object. You can approximate the force of gravity as $mg$ acting only on the center of mass. From there, it should be clear that when the center of mass is over the ledge, the object will fall.
A: 
assume you have this  situation .
we put the coordinate system at the center of mass thus the distance a (centre of mass to the edge) can be positive or negative
you have two equations (static ):
$$\sum F_y=m\,g\,\cos(\alpha)-N=0\\
\sum \tau_A=m\,g\,\sin(\alpha)\frac h2-N\,a=0$$
from here you can obtain  the normal force $~N~$
$$N={\frac {m\,g\,h}{\sqrt {4\,{a}^{2}+{h}^{2}}}}$$
the torque about the center of mass is $~N\,a~$ .thus if a is grater then zero the block will rotate to the left. and if a is less the zero  this means that the center of mass passed the edge , the block will rotate to the right. for a equal zero the center of mass is above the edge  and the normal froce is equal to $~m\,g~$.
A: The required concepts
This is a question about the statics of rigid bodies (this term will help you when googling).
The fundamentals of statics is, that two conditions must be met for a rigid body to remain static:

*

*The sum of all forces acting on the body must be zero: $$\vec F_\text{total} = \sum_i \vec F_i = \vec 0$$


*The sum of all torques acting on the body must be zero ($\vec r_i$ is the point in the body that the force $\vec F_i$ acts on): $$\vec M_\text{total} =\sum_i \vec M_i = \sum_i \vec r_i \times \vec F_i = \vec 0$$
If the second condition is not met the body will begin to rotate.
The second condition seems to depend on the choice of the origin of the coordinate system, but careful analysis shows that while the individual torques change, the net torque is invariant to the choice of the coordinate system if the first condition is met:
$$ \vec M'_\text{total} = \sum_i \underbrace{\vec r_i'}_{=\vec r_i + \vec a} \times \underbrace{\vec F_i'}_{=\vec F_i} = \sum_i \vec r_i \times \vec F_i + \vec a \times \underbrace{\sum_i \vec F_i}_{\overset{\text{by 1.}}{=} \vec 0} = \vec M_\text{total}.$$
Application to the question
The sum of the forces has to be zero, this is given as long as the object does not begin to tilt (because the normal force is opposite to the weight force of the object).
So the answer to your question depends on the second condition $\vec M_\text{total} = \vec 0$, and in turn on the weight distribution of the object.
To do this properly one would have to write out the torque due to gravity and due to the normal force as an integral over the volume of the object resp. the contact area. We will not do this here, but rather use the result, that the torque due to gravity on a rigid body is equal to
$$ \vec M_g = m_\text{total} \vec r_\text{com} \times \vec g \tag{*}$$
(where $\vec r_\text{com}$ is the centre of mass). The proof is given below.
In the light of that result, we change our coordinate origin to the centre of mass of our object – the torque due to gravity will be zero in that frame!
So the torque being zero comes down to the torque due to the normal forces being zero. The normal forces can only point up (assuming a flat surface near the cliff edge). For simplicity we assume that the object only touches the cliff surface in three single points that are not collinear. (Note, that the normal force density in the footprint is not uniquely given by the static constraints for rigid bodies that touch on an extended surface – but the precise distribution doesn't matter for the result, so we choose three points to give us unique forces.)
Now, the object will tip when the three contact points don't enclose the centre of mass projected along $\vec g$. The forces all point of, and the "levers" all go to the same direction, so the tipping can't be prevented by any assignment.
Otherwise, and assignment can be found where the sum of the normal forces equals the weight and where the torque is zero. This is quite intuitive, but showing it strictly is tedious work. (You have to show that conditions (1) and (2) result in three linear equations whose lhs. has full rank, then there is a unique solution, then you have to show that the results are positive if the projected com is within the triangle spanned by the contact points).
So the result is that the object tips once its centre of mass passes out of the convex hull of the contact surface between the cliff edge and the object.
Once the object tips, it will at some point slide and fall. (Of course there are weird exceptions for weird cliff and object shapes, e.g. narrow bends where the object is then held by the other side of the cliff edge).
Additional material
Proof of (*)
\begin{align*}
\vec M_g = \int_V d^3r\, \vec r \times \vec g \rho(\vec r) = \left( \int_V d^3\, \vec r \rho(\vec r) \right) \times \vec g = m_\text{total} \vec r_\text{com} \times \vec g
\end{align*}
