I need help on this problem. I thought I knew how to get it, but this is a multiple choice question, and my result wasn't any of the answers. An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 20 kg and it makes an angle of 60 degrees with the ground, find the force of friction of the ground on the foot of the ladder. Multiple choice answers: *7.8 x 10^2 N *2.0 x 10^2 N *50 N *1.7 x 10^2 N *100 N Here is the equation I thought was relevant, and I used: Tnet = (distance from ground to top of ladder where it meet wall)n - (distance from foot of ladder to man)mm - (1/2 x 10m)ml = 0 (ml = mass of ladder, mm = mass of man) Which for my attempt, came out like this: Tnet = 8.66n - 80 kg(2.5 m) - 20 kg(5 m) = 0 (To get 8.66, I did 10sin60) Which, since I solved for n, gave me: n = (80(2.5) + 20(5))/8.66 n = 34.642 N which was wrong. Any help showing me where I went wrong and what I need to do would be appreciated.