1. The problem statement, all variables and given/known data A ladder 8m long weighing 200N is resting against a rough vertical wall. A man of 720N climbs the ladder at point M. Take the coefficient of friction for both surfaces to be 0.25. The top of the ladder is point A and the bottom is B. The ladder makes an angle of 30 degress to the wall. http://postimg.org/image/syqq80ij7/ Here's the link to the image in case the picture isn't showing... http://s29.postimg.org/v3b393k5z/Untitled.png 2. Relevant equations How far should the man stand, from B, such that the ladder is at the point of slipping? 3. The attempt at a solution Taking upwards as positive... ΣForcesx = RA - FfA = 0 ∴ RA = μRB ⇒ 0.25RB Taking the rightward direction as positive... ΣForcesy = FfB - 200N - 720N + RB = 0 ∴ μRA + RB = 920N ∴ 0.25(0.25RB) + RB = 920N ∴ 0.0625RB + RB = 920N ∴ RB(0.0625 + 1) = 920N ∴ RB = 865.8823529N ∴RA = 0.25RB ∴RA = 0.25 x 865.8823529N ∴RA = 216.4705882N Taking clockwise as positive... ΣMomentsB = -(720sin30)(x) - (200sin30)(4m) + (RAcos30)(8m) + (FfAcos60)(8m) = 0 ∴ 360x - 400 + (216.4705882cos30)(8m) + (μRAcos60)(8m) = 0 ∴ 360x - 400 + 1499.752228 + ((0.25)(216.4705882cos60))(8m) = 0 ∴ 360x - 400 + 1499.752228 + 216.4705882 = 0 ∴ 360x = -1316.222816 ∴ x = -3.656174489m ∴x = 3.66m? The official final answer wasn't given and I'm quite certain that I made a mistake somewhere. On the internet, I've only seen ladder questions where only one of the surfaces is rough. Please help me! Many thanks in advanced.