forcesfree-body-diagramnewtonian-mechanicsreference framesvectors
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.
I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.
If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane so it accelerates at the same rate as the block, then the block won't move relative to the plane. What force on the plane is required to do this?
Assuming I've understood you correctly, the mistake you've made is to assume that the force you apply to the plane, $F$, is the same as the force the block exerts on the plane.
The force $f$ that the block exerts on the plane is not the same as the force $F$ that you exert on the plane.
When you apply a force F to the inclined plane it starts accelerating at some acceleration $a$ given by $a = F/(M + m)$. If you want the block to stay still relative to the plane the acceleration $a$ along the plane must be the same as the acceleration down the plane due to gravity:
$$ a \space \cos\theta = g \space \sin\theta $$
or:
$$ \frac{F}{M + m} \space \cos\theta = g \space \sin\theta $$
so:
$$ F = (M + m)g \space \tan\theta $$
Check out the solution in the image. It will be dependent on the inclination of the plane. Here, a is the horizontal acceleration of the inclined plane.
Best Answer
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: