Forces on the slope of a triangular block

[caspeerrr]

Homework Statement

I uploaded the question as an attached file.

Homework Equations

Fz = mg , F = ma ,

The Attempt at a Solution

The slope : x
Perpendicular to the slope: y
I thought the Y forces can be neglected because the normal force counters the y component of the gravitational force.
Fz,x = mg sinθ
Without an external force, Fz,x is the only force along the x-axis. So the acceleration a=F/m. This acceleration must be equal to the acceleration of the big block, so F=Ma = (mgsinθ)/m * M = Mgsinθ.
This seems logical to me but the answer book states a totally different answer: (m+M)gtanθ.
What did I do wrong?
Thanks in advance!

 

[MarkFL]

What we require here is for the acceleration of the block down the incline plane to be equal to the acceleration of the system along the plane:

$a\cos(\theta)=g\sin(\theta)$

Now, use Newton's 2nd law of motion to rewrite $a$, and solve for $\vec{F}$.

 

[haruspex]

I thought the Y forces can be neglected

The acceleration of the block has a component in the y direction, so the y direction forces contribute to it.

The slope : x
Perpendicular to the slope: y

The hint said to use vertical and horizontal axes.