A block is kept on a frictionless inclined surface with angle of inclination $$\alpha$$. The incline is given an acceleration $$a$$ to keep the block stationary. Then $$a$$ is equal to?

To find the required horizontal acceleration $$a$$ of the wedge that keeps the block stationary relative to the incline, we can analyze the forces acting on the block from the **non-inertial frame of reference** of the accelerating wedge.

Equating these two opposing forces along the incline for relative equilibrium:
$$ma \cos\alpha = mg \sin\alpha$$

We can cancel the mass $$m$$ from both sides of the equation:
$$a \cos\alpha = g \sin\alpha$$

Solving for the acceleration $$a$$:
$$a = g \frac{\sin\alpha}{\cos\alpha}$$

Using the trigonometric identity $$\frac{\sin\alpha}{\cos\alpha} = \tan\alpha$$:
$$a = g \tan\alpha$$