How do I solve for the tension in attached cables?

Question

Hope that made sense.

My question asks "solve for the tension in each of the cables attached to the trafic light"

http://img.photobucket.com/albums/v371/MarkTivyn/lightpic.jpg

I can post what I got if anyone wants it, but I think I missed some steps.

Answer 1

The resolution vector of the two attached cables has to equal the opposite of the traffic light's downward vector of 980N. Draw the two vectors head to tail. They will make a 70 degree angle. Depending on which vector you move, the angle at the top where the resultant starts will be 50 or 40, and at the bottom will be 60 or 30 degrees. Use the Law of Sines to find the magnitude (tension) of each vector (cable).

sin 70 / 980 = sin 50 / x where x is the cable at 30 degrees; x = 798.9N

sin 70/980 = sin 60 / y where y is 40 degree cable:
x = 903.2 N

Is that what you got?

Answer 2

Hi, I retained the trigo notation just to keep rounding errors to a minimum.

Resultant X = 0 = -T1 cos 30 + T2 cos 40
Resultant Y = 0 = T1 sin 30 + T2 sin 40 - 980

Solve for T1 in the first equation:
T1 cos 30 = T2 cos 40
T1 = (T2 cos 40) / (cos 30)

Substitute T1 in the second equation and solve for T2:
0 = T1 sin 30 + T2 sin 40 - 980
0 = (T2 cos 40 / cos 30)(sin 30) + T2 sin 40 -980
980 = (T2 cos 40 sin 30 / cos 30) + (T2 sin 40)
980 = T2 [ (cos 40 sin 30 / cos 30) + (sin 40)]
T2 = 980 / [(cos 40 sin 30 / cos 30) + (sin 40)]
T2 = approximately 903

Substitute T2 = 903 back into the first equation and solve for T1
T1 = approximately 799

Answer 3

you did a good work. Keep going you will find the answer.