1. The problem statement, all variables and given/known data A 30 kg neon sign is suspended by two cables, as shown. Three neighbourhood cats (5.0 kg each) find the sign a comfortable place. Calculate the tension in each cable when the cats are in the positions shown. Ft1 = tension in left cable Ft2 = tension in right cable T=torque 2. Relevant equations Xcm=(M1*x1+M2*x2+...)/Mtotal ƩF(y) = 0 Ʃτ = 0 3. The attempt at a solution Xcm = (5.0kg)(0.2m) + (30kg)(1m) + (5.0kg)(1.8m) + (5.0kg)(2.0m) / 45kg = 1.11 m from the left edge of the sign. With left hand cable as reference point: ΣT = Ft1*(0) + Ft2(1.6 m) - (45kg)(9.80 m/s^2)(1.11m-0.2m)=0 Ft2 = (45kg)(9.80)(0.91m) / 1.6 m = 250.82 N , 2.5 x10^2 N ∑Fy = Ft1 + Ft2 = mg Ft1 = (45kg)(9.80m/s^2) - 250.82 N = 190.1 N 1.9 x10^2 N I feel as though though I should somehow be taking into about the vertical distribution of the weight on the sign using Ycm but I'm not sure if I need to or how to go about doing that. Thanks!