1. The problem statement, all variables and given/known data A 5 meter long ladder, weighting 200 N rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.2 meters away from the wall. The center of mass of the ladder is 2.5 m from it's base, and the coefficient of friction is .20. How far up the ladder can a 600 N person climb before the ladder begins to tip. 2. Relevant equations Ffriction = uN F=ma torque = RF 3. The attempt at a solution So first I try identifying all of the forces acting on the ladder. In the X direction we have the Frictional force, and the wall acting on the ladder. In the vertical direction we have 200N from the ladder, the 600 N person, and the Normal force acting up on the base of the ladder. The angle it makes with the wall is 76.1 degrees. I use newtons Second law for each of these; Fy = Ffriction - Fwall = 0 Ffriction = Fwall Fx = N - 200 - 600 = 0 N = 800 Newtons Now hopefully I did that part right but when I sum the torques I get confused. So I guess I need to set an axis and then pick which directions are positive or not. I pick the Normal force as my axis. I'll pick counter clockwise as positive and clockwise negative. Sum of Torques = N(0) - 200 (2.5) - 600d + Fwall(5) = 0 The Fwall is equal to uN = (.2)(800)(cos 76)?? Yea I'm pretty lost here. Any help would be appreciated!