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.
xCM = m1x1 + m2x2 ...
yCM = m1y1 + m2y2 ...
F = mg
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!
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