1. The problem statement, all variables and given/known data A weightless ladder at 7m long rests against a frictionless wall at an angle of 65degrees to the ground. A 72kg person is 1.2m from the top of the ladder. What horizontal force at the bottom of the ladder is required to keep it from slipping? d= 7m m(man) = 72kg r(man) = 1.2m Theta = 65 2. Relevant equations T = r(Fa)sin(theta) 3. The attempt at a solution I know that this is a torque question because of the nature of the ladder positioning. First off, I started by finding the distance of the man from the pivot point at the bottom of the ladder, which is: 7-1.2 =5.8m To get the angle of the force that the man is exerting down on the ladder, I used the Rule of Sum for the C-pattern angle. Since the bottom angle is 65degrees, I subtracted that from 90 to get 25degrees, the angle of the force acting on the ladder. Afterwards, I calculated his torque value that is going CW: T = r(Fa) sin(Theta) = 5.8(72*9.8) sin(25) = 1729.56N-m At this point, I was thinking that since the ladder has to be in staticity to prevent it from slipping, the Torque forces were all equal. Thus: T(CW) = T(CCW) From here, I plugged in the values for the equation and attempted to solve for (Fa) of the ladder. Since the force at the bottom is now in regards to the pivot point at the top of the ladder(?), the distance (r) is equal to 7m. The angle of the force would also then be 65degrees: T(CW) = T(CCW) 1729.56 = r(Fa) sin(theta) 1729.56 = 7(Fa) sin65 However, when I solve this equation for the (Fa) value, all I get is around 180N. Everyone in my class got 272N (approx.), and I am not sure how they did that. I am sure that I am missing something extremely simple, but I have spend over 45 minutes on this problem and I cannot figure out how to do it. Any hints on what to do after finding the Torque(CW) is much appreciated EDIT: Even after I fixed my answer (which was not 1279N-m), I still couldn't get 272N for the horizontal force. I am going to go crazy from this problem.