Finding Tension in Cables: Solving a Loudspeaker Suspension Problem

[sona1177]

Homework Statement

A 20 kg loudspeaker is suspended 2.0 m below the ceiling by two cables that are each 30 degrees from the vertical. What is the tension in the cables

Homework Equations

F=ma
Fx=max
Fy=may

The Attempt at a Solution

a=0
Here is how I solved it but I am not sure if it is right. Also, I don't understand how to use the 2.0 meters given in the problems statements. Anyways, here goes:

Fnetx=Wx + T1x + T2x =max=0
0 + T1 - T2=0
T1sinθ -T2sinθ=0
T1sinθ=T2sinθ
T1=T2

Fnety=wy + T1y + T2y=may=0
-w + T1 + T2=0
(20 * -9.8) + T1cosθ + T2cosθ=0
-196 + (T1 + T2)(cosθ)=0
(T1 + T2)(cosθ)=196
(T1 + T2)=196/cos(30)
(T1 + T2)=226.3

Since T1=T2, then 226.3/2=113 N. Therefore 113 N is the tension for both ropes?

Is this correct?

 

[sona1177]

somebody please help!

 

[PhanthomJay]

Yes! The 2 meters is not relevant to the solution of this problem.

 

[sona1177]

So the answer is 113 N for both cables? And my work is correct?

 

[PhanthomJay]

Yes, the tension is 113 N in each cable. In your work, though, you noted
Fnetx=Wx + T1x + T2x =max=0, correct, but then you said
0 + T1 - T2=0, instead of saying 0 +T1sintheta -T2 sin theta = 0, which you later corrected. You did the same thing in the y direction, looks like just a typo.

 

[sona1177]

Thank you. I'll make sure not to make that mistake again.