A block on inclined plane 
A block of mass $m$ is placed on an inclined plane (a ramp). If a constant force $f$ is applied to the ramp so that it is accelerating horizontally at a proper rate, the block will remain at the same height. But what then is the force that cancels the component of the block weight parallel to the plane, (i.e. $mgsinθ$), and prevents the block from sliding along the inclined plane?
Note: all surfaces are frictionless.
 A: Notice that in-order for the block(of mass $m$) and the inclined plane(wedge of mass $M$) to move together, they must have a common horizontal acceleration given by: $$a=\frac{F}{M+m}$$
And thus for the block of mass $m$ it's horizontal acceleration must be equal to this, so there is a resultant force on the small block, acting horizontally(which I'll call $F_m$ which is given by $F_m=ma$ where a is the common horizontal acceleration of the block and wedge). 
Indeed there is no force opposing the component $mgsin(\theta)$ and you can see below that it is not required to be cancelled as it itself becomes a component of the resultant force $F_m$ which has components $N-mgcos(\theta)$ and as expected $mgsin(\theta)$:

Note: Diagram showing the forces on only the block of mass $m$. 
Another diagram requested to view the force diagram in another way which will give the same end result:

A: This question is correctly answered here, but it deserves an additional conceptual examination because of how clearly this problem illuminates the need to explicitly consider and include the Inertial Force in this force vector diagram.  
Newton's third law: All forces in the universe occur in equal but oppositely directed pairs.  A Force applied to any mass will produce the same force in reaction.  Note that the Inertial Force has a different character than the Field forces (e.g., gravity, electric, magnetic). The Inertial Force does not exert the sustained force of a potential field. Rather, the Inertial Force exerts a reactive opposing force only during the instant while accelerating and transferring kinetic energy.
The problem:


*

*What is the $F_{external}$ on the block required to accelerate the mass and wedge at the rate to suspend the block at a point on the wedge?


The Solution:


*

*Gravitational force $F_{gravity-downhill} = mg sin\theta$ accelerates the block down the frictionless surface of the wedge.

*Maintaining a static position on the block, requires balancing the downward acceleration of gravity with the upward acceleration of an external force.

*Horizontal acceleration of the wedge and block provides a vector component of upward Force: $F_{inertial-uphill}$.

*Solution requires determining the magnitude of $F_{external}$ which contributes the force vector component of equal magnitude and opposite direction to the downhill gravitational force.



A: The block does accelerate downward the inclined plane. This acceleration is caused by $mgsin( \theta )$. The block also accelerates along the normal. This acceleration is caused by the normal and $mgcos( \theta )$. 
There are no other relevant forces.
This seemingly peculiar view is a result of choosing a coordinate system along the plane and normal to the plane while we already know that the overall acceleration is along the horizontal axis.
If you do select the horizontal and vertical axis as your coordinate system, than only along the horizontal axis you obtain acceleration while there is zero acceleration along the vertical axis.
