Question

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?

Answer 1

Answer:

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)