Why is the normal force $(M+m)g$? I am trying to understand the solution to this problem. The problem asks to find F such that m stays fixed relative to M. In the solution, it is mentioned that the normal force for block M is (M+m)g, I don't understand that. I thought it is supposed to be only Mg. 
The solution states - The normal force on the first block, M is Mg + u_2*F_bb = (M+m)g.
Normal Def from wiki - is that component of the contact force that is perpendicular to the surface that an object contacts. 
Since block M is in contact with 2 surfaces, is that why they are adding the the Mg+ u_2*F_bb?
I think I am just confused about the definition of Normal Force and it's application in this problem. 


 A: The normal force is the force the table (or surface) must exert on the block $M$ in order to keep it stationary in the vertical $y$ direction. This means that the normal force must be equal and opposite to the net downward force that is being applied on M. The net downward force being applied on $M$ in this question is composed of two parts. One part is the weight of $M$ which is given by $F_w=Mg$, the other is the force that the little block $m$ exerts on $M$ in the downward direction. By Newton's third law, this force must equal the force that $M$ exerts on $m$ in the upward direction. This force, as was argued earlier in the text, must have magnitude $mg$ in order for little $m$ to remain stationary in the vertical direction. Thus, the total force that is pulling down on $M$ is $F=F_w+mg=(M+m)g$. The normal force must equal this in order to counteract it so that $M$ does not move in the vertical direction. 
A: Remember that a normal force $n$ is just a "holding back" force. No law says it must equal $Mg$. It might. But that would just be a coincidence. The only rule is that $n$ equals whatever it must to hold back against something.


*

*Put an apple on a table. The normal force holds back against it's weight:
$$n=mg$$

*Push down on top of this apple. The normal force must now hold back against both the apple's weight and this push:
$$n=mg+F_{push}$$

*Push a book sideways onto a wall. The wall exerts a normal force sideways to hold back against this push, but not against the weight. The book's weight does not push towards the wall:
$$n=F_{push}$$


These examples show that a normal force has got absolutely nothing to do with $mg$. It might equal $mg$ in some situations - that would just be a coincidence.

In your specific situation, the normal force from the floor is holding back against block $M$ including whatever that pushes down in it.
The little block $m$ is pulling down in $M$. It tries to fall down but is carried by the friction force, so it is now "fixed" onto $M$ -  it doesn't really matter how it is fixed, with a screw, with glue, or with friction from being squeezed together. In any case, this corresponds to being fixed and thus it is burdening $M$ by pushing down in $M$ with it's full weight $mg$. 
The normal force therefor has to hold back against both weights, $$n=Mg+mg=(M+m) g$$ 
A: I wanted to write this as a comment but it seems I don't have to reputation to do so, so here goes. Since it is given that the masses should be stationary relative to each other, to understand why the vertical normal force exerted on $M$ by the surface is $(M+m)g$, it is probably easier to treat the combination of the two masses as a single object of mass $M+m$, and draw a free body diagram for this object. Gravitational force acting downward is clearly $(M+m)g$, and the only other force acting in the vertical direction is the upward normal force exerted by the surface. Since there is no motion along the vertical, the normal force must equal $(M+m)g$.
As for why $m$ exerts a downward force on $M$, note that $M$ must exert an upward force of $mg$ on $m$ (due to friction) to keep it from sliding downward. This means that, by Newton's third law, $m$ exerts a force on $M$ of the same magnitude $mg$, but downward. You can also arrive at the conclusion that the normal force on $M$ must be $(M+m)g$ by using this fact and drawing a free body diagram for $M$, using the same reasoning as enumaris has described.
A: First thing is friction occurss in pair , so when small box is stationary in vertical direction that means it's weight mg  is balanced by the force of friction on it in upward direction that's is u_2*F_bb so, equal and opposite force must act on big block M in downward direction so total force
On M is (m+M)g from equations so formed
