1. The problem statement, all variables and given/known data problem is above 2. Relevant equations sum of torques = I * angular acceleration = 0 for rotational equilibrium 3. The attempt at a solution OK, here's what I tried to do. I wrote sigma Fy = Fleft pier up + F secondleftpierup - Fg,beam = 0 I assumed that the two piers have equal forces, so then I said that 2 * Fsinglepier = Fg,beam 2 * Fsinglepier = 100 kg * 9.8 m/sec^2. Fsinglepier = 50 * 9.8. Then I did my rotational equilibrium equation, and I assumed that there was no man in the pier initially. My axis of rotation is at the very left end of the pier: sum torques = +Fsecondpier [which is 50 * 9.8] * 3 meters - [ 100 kg * 9.8 * 5 meters] (for the center of mass of sphere and the force of gravity acting through it which gives = -3430 N-m. This obviously implies that the system is already experiencing a net torque without the man, so the man can not even be on the system to balance it out in any fashion. The solutions [which might be wrong, due to crummy question authors] say that 3 is the answer without providing any other guidance. Please help, and thanks in advance!