1. The problem statement, all variables and given/known data A uniform 15 kg ladder whose length is 5.0 m stands on the floor and leans against a vertical wall, making an angle of 25 degrees with the vertical. Assuming that the friction between the ladder and the wall is negligible, what is the minimum amount of friction between the ladder and the floor which will keep the ladder from slipping? 2. Relevant equations Fg = mg Torque = distance x force 3. The attempt at a solution I really am not sure about my approach. I figure I should try and find the force on the horizontal on the ground and that should be the friction needed. Fg = 15 x 9.8 at teh center of the ladder =147 N Torque also at the center = 147 x 2.5m = 367.5 The gravity should be going straight down and the torque perpendicular to the ladder. This gievs me the hypotenuse and one side of a triangle, left with the ground as the horizontal which solves as 337 N but the correct answer is 34 N. I'm guessing I would need to use the angle in there somewhere. Thanks for any help.