1. The problem statement, all variables and given/known data A 5.00-m long uniform ladder leans against a smooth wall and its base rests on a rough floor. The ladder has a mass of 18.0 kg and its base is a distance of 2.30 m from the wall. A person of mass 70.0 kg climbs 2.50 m up the ladder. If the ladder is to remain in place, what frictional force must be exerted by the floor on the ladder? 2. Relevant equations t = FR t =0 N=mgcosθ ΣF=0 Σt = 0 -mLgbg-mPgbP+Fn,w*a = 0 Fn,w= (mLgbg + mPgbP ) / a ΣFx = 0 Fs - Fn,w = 0 Fs = Fn,w = (mLgbg + mPgbP ) / a 3. The attempt at a solution wall (a)= sqrt(30.29)= 5.50m, bL=2.30m, bP=cos(67.3)*cP c= 5.00m, cP=2.50m θL= tan-1(5.50/2.30)= 67.3° Fs = ((18.0)(9.81)(2.30)+(70.0)(9.81)(.965)) / 2.5 = 194 N I'm not completely sure if I found all forces for this problem, did I find everything for the frictional force?