Physics 5.3- loudspeak suspended please help, final question

Question

A 13kg loudspeaker is suspended 2.0m below the ceiling by two cables that are each 30 degrees from vertical. What is the tension in the two cables??

I have tried many different ways and have come up with wrong answers. I really want to finish and this is the last one, please help me. I greatly appreciate it.

Answer 1

Draw a vector diagram. The wires are opposing the direction of gravity at angles. Their vertical components are what directly resist gravity, but their have horizontal components also.

They have no motion, so they both sum to zero.

horizontal:

T,·cosθ - T,,·cosθ = 0

vertical:

T,·sinθ + T,,·sinθ - mg = 0

----

T,·cosθ - T,,·cosθ = 0

The angle is the same for both, so the magnitudes of the velocities are identical.

|T,| = |T,,|

This gives:

T,·sinθ + T,,·sinθ - mg = 0

2·T·sinθ - mg = 0

2·T·sinθ = mg

T = mg/ [ 2·sinθ ]

Substitute in values:

T = ( 13 kg ) (9.8 m/s² )/ [ 2·sinθ ]

T = 127.4 kg·m/s²

T ≈ 130 N

in each

Answer 2

I think it's sin 30 = 0.5 as the increase in tension.

If the cables were straight, the tension would be 6.5 kg each. But because of the angle, add a half to that or 9.75 kg force for each cable. You can change to Newtons if you want.

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Answer 3

The total upward force is Tcos30+Tcos30

The total downward force is 13kg * 9.8m/s/s

If the speaker is not accelerating then the sum of the upward forces equals the sum of the downward forces, solve for T and you've got your answer.