Super Physics Crisis!!!?

Question

A uniform rod of length 0.86 m and mass 11.5 kg is balanced on a sharp edge at 0.58 m from the left end. At the left end hangs a mass of 35.0 kg and the right end hangs an unknown mass so that the whole assembly is in equilibrium. Calculate the torque on the center of mass of the rod alone with respect to the pivot point.

Answer 1

If the rod is balanced 0.58 m from one end, it's 0.86 - 0.58 = 0.28 m from the other end. 0.58 m of the rod has a mass of (0.58/0.86)*11.5 = 7.76 kg, and 0.28 m has a mass of (0.28/0.86)*11.5 = 3.74 kg (and this checks out, because 7.76 + 3.74 = 11.5). And, naturally, the center of mass of 0.58 m of the rod is 0.58/2 = 0.29 m from the pivot, and the center of mass of 0.28 m of the rod is 0.28/2 = 0.14 m from the pivot point.

So on the left side we have 7.76 kg of rod at a distance of 0.29 m, and a 35 kg mass at a distance of 0.58 m. On the right is 3.74 kg of rod at a distance of 0.14 m and unknown mass m at a distance of 0.28 m. For this to be balanced, we must have 7.76*0.29 + 35*0.58 = 3.74*0.14 + m*0.28 ==> 2.25 + 20.3 = 0.52 + 0.28m ==> 22.03 = 0.28m ==> m = 78.68 kg. It makes sense that this mass is about twice as large as the other mass, because the shorter length of rod is about half as long as the longer portion.

So now we know the total mass of the rod and both masses. It's 11.5 + 35 + 78.68 = 125.18 kg. Therefore, the upwards force exhibited by the pivot must be (125.18 kg)*g = (125.18 kg)*(9.8 m/s^2) = 1,226 N. The pivot is 0.58 m from the left end of the rod, but the center of mass of the rod is 0.86/2 0.43 m from that end, meaning that the pivot is 0.58 - 0.43 = 0.15 m from the center of mass. As a result, the pivot exerts (1226 N)*(0.15 m) = 184 N-m of torque on the center of mass of the rod.