Answer
$$F=(M+m)*g*tan(\alpha)$$
Work Step by Step
We begin by considering forces acting on the block:
Normal force points perpendicular to the surface of the wedge (at an angle of ($90^{\circ}-\alpha$) above the positive x-axis), and gravitational force $(mg)$ points downwards. Since the goal is for there to be no movement in the y-direction (the block must remain at a constant height), we can use Newton's Second Law to describe the sum of the forces on the block in the y-direction:
$$Nsin(90^{\circ}-\alpha) - mg = 0$$
The forces in the x-direction are defined by $$Ncos(90^{\circ}-\alpha) = ma$$
Solving for N for both equations gives
$$N=\frac{mg}{sin(90^{\circ}-\alpha)}$$
and
$$N=\frac{ma}{cos(90^{\circ}-\alpha)}$$.
Using the identities that $sin(90^{\circ}-\alpha)=cos(\alpha)$ and $cos(90^{\circ}-\alpha)=sin(\alpha)$, and solving for N, we get:
$$\frac{ma}{sin(\alpha)} = \frac{mg}{cos(\alpha)}$$
Solving for a and eliminating m from both sides gives
$$a=gtan(\alpha)$$
Since both the wedge and block will be moving together as one system, the acceleration will be equal for both objects. By Newton's Second Law, $F=(M+m)a$, so plugging in a into this equation gives us our final answer:
$$F = (M+m)gtan(\alpha)$$
In words, this means that the magnitude of the applied horizontal force (F) must be equal to the sum of the masses of the two objects times the magnitude of gravity times the tangent of the angle of the incline the block is on.
