A 66.5 Kg chandelier is suspended 1.5 metres below a ceiling by three wires. Each wire has the same Tension and the same Length of 2 metres. What is the Tension in each wire?

The candelier generates a force F=mg=66.5kg \times 9.81m/s^2 due to gravity. The 3 wires
must balance this force in order to create an equilibrium. Thus the combined force on the 3 wires is
exactly F. As the problem tells you that the tension on each wire is the same, then you have that each wire must have tension F/3.

each wire must have tension F/3.

Not quite right. This would be true if all 3 wires were vertical, but they're not. Given that the chandelier is hanging 1.5m below the ceiling, and each wire is 2m long, this tells you that the wires are at an angle \theta to the vertical, where

cos\theta = \frac {1.5} 2.

The vertical component of force in each wire must be 1/3 the weight as you show, but the tension in them is greater than that, as there is both a horizontal and vertical component of tension in the wire. So:


T = \frac {W/3 } {\cos(\theta) }


where W = the weight of the chandelier, in Newtons.

Yes, ebaines is right, I have not considered wire length.

OK, thank you for your answers.
So, I hope I've got this right.
If the 3 wires were all vertical, the problem would be easier to solve, since all you would have to do would be to work out the weight of the Chandelier, which I calculate to be 652.37 Newtons. Then, in order for equilibrium to occur (i.e the Sum of the forces = 0), then each wire would have a tension of 217.46 Newtons.
However, as ebaines points out, each wire is at an angle to the vertical, which I calculate to be 41.4 degrees.
Thus, the tension in each wire would be greater than 217.46 Newtons. If 217.46 Newtons is the vertical, then:
cos 41.4 = 217.46 divided by T
(i.e. cos theta (41.4) = adjacent (217.46) divided by the hypotenuse (T)).
Therefore, T = 217.46 divided by 0.75 = 289.89 Newtons.
Hope I did this right!!