Vertical component of $mg \sin θ$ From the 2013 F=MA exam:



*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?


(A) $2mg$
(B) $2mg + Mg$
(C) $mg + Mg$
(D) $Mg + mg(\sin{\alpha} + \sin{\beta})$
(E) $Mg + mg(\cos{\alpha} + \cos{\beta})$

So, the solution says, in order to solve this, you need to take the vertical components (relative to the table) of $mg \cos{\alpha}$ and $mg \cos{\beta}$ and add them to $Mg$. This gives you the magnitude of the total normal force the table needs to give.
However, I was wondering, don't $mg \sin{\alpha}$ and $mg \sin{\beta}$ both have vertical components too? So shouldn't you consider that when you are solving for the normal force?
 A: The tangential components of weight $mg\sin\alpha, mg\sin\beta$ do not act on 
the triangular block, because there is no friction. They act on the cubes, accelerating them down the inclines. The block does not resist this motion. So you do not include them in the forces acting on the block. 
Only the normal components $mg\cos\alpha, mg\cos\beta$ act on the block, because the block resists the motion of the cubes in this direction.  However, these normal reactions do not act vertically. Their horizontal components squeeze the block. Only the vertical components add to the force on the table. As you seem to be suggesting, these vertical components are $mg\cos^2\alpha, mg\cos^2\beta$. The options provided do not include the correct solution.
A: For the masses m you have the normal force perpendicular to the slope and acceleration parallel to the slope, and these are vector components of the gravitational acceleration. 
If the masses m were attached to the larger M then it would be simply $2mg+Mg$, but as you have noted, you take the components $mg\cos\alpha$ and $mg \cos\beta$ and add them to $Mg$. You add the normal forces on the masses m to the normal force on M. 
The complementary components $mg\sin\alpha$ and $mg\sin\beta$ gives you the acceleration down the slope for each. The green arrow is imaginary.

Approaching it this way, if $\alpha$ were 0° then $mg\cos\alpha=mg$ and the lateral force $mg\sin\alpha=0$. The normal forces present are thus $mg\cos\alpha + mg\cos\beta$ on the block and $Mg+?$ on the table.
As has been pointed out by sammy gerbil and others, the normal forces on the block are not downwards, but would be $mg\cos^2\alpha$ and $mg\cos^2\beta$ so the total normal force exerted by the table would be $Mg + mg\cos^2\alpha + mg\cos^2\beta$, which is not one of the offered solutions.

I invite further comments on this solution. 
A: The math by Sammy Gerbil is correct but he is missing the 'trick'  part of the question. The correct  answer is indeed provided and is $Mg + mg$. The thing to remember is $\alpha+\beta=\pi/2$.
