1. The problem statement, all variables and given/known data So the original problem goes as the following. A uniform ladder of mass (m) and length (L) leans against a frictionless wall, see figure. If the coefficient of static friction between the ladder and the ground is 0.41, what is the minimum angle (q) between the ladder and the floor at which the ladder will not slip? And I get the whole concept behind how to do it which is basically using sum of the forces and sum of the torques but my math seems to be wrong and different from the solutions I found online (my answer was 0.014 degrees which is obviously wrong). So can you tell me where I messed up in my math? 2. Relevant equations 3. The attempt at a solution Sum of the forces = 0 Nw = Normal force done on the ladder by the wall Nf = Normal force done on the ladder by the floor cg = Center of gravity (bad habit of mine that I use to distinguish between my other mg) x: Nw - Fs = ma (a = 0 thus ma = 0) y: Nf - cg = 0 Thus: Nf = cg Fs = Nw Sum of the Torques = 0 Set the axis of rotation at the base of the ladder Nw sin(theta) - (g sin(theta))/2 = 0 Substituting Nw for Friction static and then substituting Friction static with coefficient of static friction times cg I get the following: uk cg sin (theta) - (cg cos(theta))/2 = 0 Canceling out the cg I get the following: uk sin (theta) - (cos(theta))/2 = 0 Now this is where I screw up so can someone explain how I screwed up. Moved cos to the other side uk sin(theta) = (cos(theta))/2 Multiplied 2 to get rid of the 1/2 on the right side 2uk sin(theta) = cos(theta) Divided by sin (theta) 2 uk = cos(theta)/sin(theta) Which becomes 2 uk = arctan(theta) Thus theta = tan(2uk) Thanks for your help and sorry for asking for putting out so much detail for something so simple.