- #1

Omar Wali

- 14

- 0

## Homework Statement

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/

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.

## Homework Equations

ΣF = 0 when searching for an equilibrium point.

## 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!