1. The problem statement, all variables and given/known data A 30Kg 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 with the cats are in the positions shown. 2. Relevant equations xCM = m1x1 + m2x2 ... mtotal yCM = m1y1 + m2y2 ... mtotal F = mg 3. The attempt at a solution I found the centre of mass (CM) by assigning each cat a coordinate using the bottom left corner of the sign as the origin with (x,y). So: Cat1 = (0.2,0.5) Cat2 = (1.8,0.5) Cat3 = (2,0) Sign = (1,0.25) --> the centre of the sign Using the above two equations I found the centre of mass of the cats and sign to be (1.11m,0.278m) There should be no forces in the x axis and the forces in the y axis should all cancel out so: T1cable + T2cable = mg I know the position of each cable from the picture so: (using only y axis position) T1(0.5m) + T2(0.5m) = (45kg)(9.81)(0.278m) The tension in each cable should be equal? Or, because one cable is farther from the CM it should have less tension? Does this mean I should be taking into account the x axis? Please help!