The surfaces between a 14 kg block, the 16 kg wedge and between the 16 kg wedge and the horizontal plane are smooth (without friction). the acceleration of gravity is 9.8 m/s^2. a block is released on the inclined plane (top side of the wedge).

what is the force f which must be exerted on the 16 kg block in order that the 14 kg block does not move up or down the plane?

If the wedge has an angle with the horizontal plane of α, the required force is F = 294N·tan(α)

Explanation:

To solve this problem you have to analyze the block and the wedge separately and like a single body. If there is no relative movement between them, that means that both have the same acceleration a*.

Knowing that we analyze the block first:

We can use a non-inertial system for the block. As seen in the image, the total acceleration in the NIS is 0 m/s². Therefore we applied the Newton equations:

Y) Ny - Wb = 0 ⇒ N cos(α) - 14kg·g=0  ⇒ N = 137.2N / cos(α)

X) F* - Nx = 0 ⇒ a*·14kg - 14kg·N sin(α) ⇒ a* = g·tan(α)

For the wedge, we can use an inertial system:

X) F - N´x = 16kg·a*  ⇒ F - 137.2N·tan(α) = 156.8N·tan(α)

F = 294N·tan(α)

seeing the graph would