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?
(B) 2mg + Mg
(C) mg + Mg
(D) Mg + mg(sin α + sin β)
(E) Mg + mg(cos α + cos β)
A picture can be found here : http://www.aapt.org/physicsteam/2014/upload/exam1-2013-1-6-unlocked.pdf
Page 4 Problem 11
The Attempt at a Solution
I did a FBD getting a force of mg down on the cubic blocks and Mg on the triangle block and a normal force of mgcos(alpha) and mgcos(beta) up respectively on the cubic blocks and Mg up on the triangle block. I added up all the forces to get (E) but the right answer is (C). Where did I go wrong?