Solving for Tension in a Traffic Light: Fx and Fy Equations

[cowgiljl]

A traffic light weigh 235 Nt. It is supported by 2 wires cal the tension of each of the wires

T1 has an angle of 21 degrees
T2 has an angle of 38 degrees

There is no sag in the light

I have 2 equations

equ 1) Fx = T2 cos 38 - T1 cos 21 = 0
equ 2) Fy = T1 sin 21 + T2 sin 38 = 0

I set T2 = to T1

T2 = T1 (cos 21 / cos 38) = 1.2 T1

then subsitutedthat into equatin 2

T1 sin 21 + (1.2 T1)(sin 38) - 235 Nt = 0

T1 = 214 N

T2 = 1.2T1= 1.2(214)

T2 = 257 N

I think iI went about this in the right way i think

thanks

 

[Doc Al]

I didn't check your arithmetic, but it looks good to me (assuming angles are with respect to the x-axis). A few comments:

- I don't know what you mean by no sag: of course the wires "sag", that's why they are at an angle.
- Your equation #2 has a typo (you left out the weight), but you correct it later on.

 

[HallsofIvy]

It doesn't look so good to me.


What reason do you have for setting T1= T2?? It doesn't look to me like the tension in the two wires will be the same (the situation is not symmetric because the angles are different).

Oh, wait! You didn't set T1= T2! You solved equation 1 for T1. That's a whole different matter.

Also, as DocAl said, you second equation (for the vertical component of force) should be
equ 2) Fy = T1 sin 21 + T2 sin 38 - 235 = 0
so that the total vertical force is 0
or equ 2)
T1 sin 21 + T2 sin 38 = 235 so that the upward force of the wires balances the downward force of the light.

You did manage to get the right answer!