# Block on moving wedge (Maximum force for equilibrium problem

1. Dec 4, 2014

### Omar Wali

1. The problem statement, all variables and given/known data

Given: The friction between the block with mass 11kg and the wedge with mass 22kg is 0.48. The surface between the wedge with mass 22kg and the horizontal plane is smooth (without friction). The [A][/g] = 9.8m/[/2]. A Block is released on the incline plane (top side of wedge).

What is the maximum force F which can be exerted on the 22kg block in order that the 11kg block does not move up the plane? Answer in units of N.

2. Relevant equations

ΣF = 0 when searching for an equilibrium point.

3. The attempt at a solution

I setup a FBD of the 11kg block. I listed the force of static friction, the normal force, the reactionary force F opposite the direction of the applied force F on the wedge and the force of gravity.

The calculations went as follows:

ΣF = 0

(μ*[F][/N]) + ([F][/RF]/cos(θ)) - ([F][/g]sin(θ) = 0

I solved for [F][/RF] and got ≈ 17.395 N. This answer is wrong and I'm not too sure where my error was. Thank you for the help in advance!

2. Dec 4, 2014

### Omar Wali

OK I'm obviously very new at writing things out not sure why things are striked out and the subscripts did not go in properly. I'll correct them next time. I believe it is still easy to read, however.

3. Dec 5, 2014

### ehild

The force F acts on the wedge, and the reactionary force is exerted by the wedge on the object that applies F. The sum of the forces applied on the block will accelerate it together with the wedge.

If both the wedge and the block accelerate together, the acceleration, multiplied by the total mass of the system, is equal to the external force F. The acceleration is horizontal, so the vertical components of the forces must cancel.