# Free Body Diagram for a Body on a Smooth Inclined Plane

If instead of decomposing into perpendicular and parallel force components, you use the 'regular' $$x$$ and $$y$$ axis as you did to get Eq. (2), then you will have a non-zero acceleration in both $$x$$ and $$y$$ directions. So, what you really should end up with for Eq. (2) would be $$ma_y=mg-N\cos\theta\tag2$$ because the forces are not balanced. Therein lies the reason why we decompose into parallel and perpendicular components -- because the block will only slide parallel to the incline, so we can nicely conclude that $$\Sigma F_\perp=0$$. However, in the usual $$xy$$ coordinate system, there will be an acceleration in both $$x$$ and $$y$$ directions. So you can't just assume one of them is zero.