A right-triangular wooden block of mass M is at rest on a table, as shown in figure. Two smaller wooden cubes, both with mass m, initially rest on the two sides of the larger block. As all contact surfaces are frictionless, the smaller cubes start sliding down the larger block while the block remains at rest. What is the normal force from the system to the table? $(\alpha \ne \beta )$ 
 

Consider the system of the block and the two cubes together.  The only external forces on this system are gravity  $F_g$ = Mg + 2mg, and the normal force N. The vertical acceleration a of this system’s centre of mass can be written as
 $(M+2m)a=F_g-N$
We will try  to find a, as this will then allow us to use the above equation to determine N. Let  $a_2$  be the acceleration of the cube on the angle ramp, and let   be the acceleration of the cube on the angle  $\beta$  ramp. Considering these cubes individually, the problem becomes a classic frictionless inclined plane scenario. Decomposing mg into 
perpendicular and parallel components yields  $a_1=g\sin \alpha and{a}_2=g\sin \beta$. 
To find a, however, we need the vertical components of  ${\overrightarrow{a}}_{1}and{\overrightarrow{a}}_2$. Using trigonometry, these are given by $a_1\sin \alpha and{a}_2\sin \beta$  respectively, so the acceleration of the centre of mass is  
 $a=\frac{m{a}_1\sin \alpha +m{a}_2\sin \beta }{M+2m}=\frac{mg({\sin }^2\alpha +{\sin }^2\beta )}{M+2m}$
Then, our expression for a simplifies to a = mg/(M + 2m), and combining this with the equation for N yields 
 $mg=F_g-N\Rightarrow N=F_g-mg=mg+Mg$