A diving board of length 3.00 m is supported at a point a distance 1.00 m from the end, and a diver weighing 550 N stands at the free end. The diving board is of uniform cross section and weighs 255 N. Find the force at the end that is held down. In part a, I found the normal force at the support point, 1m from the end held down. N = [550N(3m) + 255N(1.5m)]/1.0m = 2030N Part b wants the force at the end held down. I don't really know how to approach this part, since the end held down is at my x=0m point. I found the center of mass to be 2.52m. I know the torque calculated should be about the support point. Sum torques = N(L/3) + N(0) - W_b(L) - W_d(L/2) where W_b = weight of the board W_d = weight of the diver But since the end held down is at zero, this term drops out and I'm left with what I got for the first part.