Crate placed on cart on inclined plane Consider the following problem.

Why is the following approach wrong for crate A?
$$Mg\sin{30°} - N\sin{30°} = Ma$$
Solving this gives $$N = 58.1 \rm\space N$$ which is different from the answer given.
Isn't the principle/concept here the same in the both approaches: getting the forces in the same direction as the acceleration?
Also, why is the acceleration of the crate only vertical and not just $a$? I think this may be the reason that my approach for the first part is wrong.
 A: Let us first consider the things that will happen to the crate :
Observation 1 - The cart will move downward along with the crate.
Observation 2 - The crate will move backward along the smooth cart parallel to the ground. There will be a pseudo force acting on the crate. This case is almost like a lift moving downwards.
Observation 3- This is not a case of free fall. The system is not released from rest but instead it is given acceleration of $2\frac{m}{s^2}$.
Approach 1 :
The accelerating cart is a non-inertial frame, therefore, there will be a pseudo force acting on the crate whose direction will be exact opposite to the direction of acceleration of the cart. Since, the observer (you) is sitting on the cart (the non-inertial frame) the crate will not move downward with respect to you i,e, you will see the box at rest. This is a practical observation that the box won't move downward (i.e, into the crate), thus according to Newton's second law of motion,
$ma\sin{30^{\circ}}+N=mg$
$N=88.1 N$  
By this equation, we balance the forces acting on the crate in vertical direction in order to satisfy the practical observation that the crate wont move downwards wrt us (sitting on a non-inertial frame).
Approach 2 :
Now if you break the components of the normal force and the weight of the body along the direction of acceleration  and put them equal to $ma$, then this would mean that along the direction in which you have divided the components, the crate will be motionless wrt to you because you have equated these forces and cancelled them out. but the box actually moves towards you (if you are sitting towards the inclined plane, facing the crate).
This would only happen, if there is a component of acceleration along the line joining you and the crate. 
But while equating the forces, you cancelled them all out, leaving no component along the line joining you and the crate.  
And there are two other components, $mg\sin{60}$ and $N\sin{60}$, and you are not sure of which one is greater. But one thing you can be certain of is that the box wont move downwards wrt to you.
Conclusion :
Although it seems correct, we have no proof the you will get the right value of N by Approach 2 and apparently it gives the wrong answer.
But we know this for sure that the crate won't move downwards wrt to the cart. Thus, it is obvious that for this to happen all the forces in the vertical direction must give a net resultant of zero. 
I hope this gives you a reasonable reason why your method gives the wrong answer.
A: The reason your answer is wrong is that you are assuming the acceleration of the crate is the same as that of the cart.  In the $y$ direction both accelerations are the same, but there is no force on the crate in the $x$ direction, so it has zero acceleration in that direction, so it only moves down while the cart moves to the right and down.
The total acceleration of the crate is therefore not the same acceleration as the cart so your equations are wrong.
