# Math Help - Find the tensions in the cables

1. ## Find the tensions in the cables

Question

A 250-lb traffic light supported by two flexible cables. The magnitude of the forces that the cables apply to the eye ring are called the cable tensions.
Find the tensions in the cables if the traffic light is in static equillibrium.

Thank you very much.

2. Originally Posted by Jenny20
Question

A 250-lb traffic light supported by two flexible cables. The magnitude of the forces that the cables apply to the eye ring are called the cable tensions.
Find the tensions in the cables if the traffic light is in static equillibrium.

Thank you very much.
Haven't you missed out something in the statement of this problem?

RonL

3. yes, a diagram.

30 degrees \ / 45 degrees
====================
Traffic light
is hanging
in the middle

The LHS is 30 degrees and the RHS is 45 degrees.

4. Also, I have the answer
ll Fsub1 ll approximately = 1183lbs,
and ll Fsub 2 ll approximately = 225lbs.

5. Originally Posted by Jenny20
Question

A 250-lb traffic light supported by two flexible cables. The magnitude of the forces that the cables apply to the eye ring are called the cable tensions.
Find the tensions in the cables if the traffic light is in static equillibrium.

Thank you very much.
NOTE: This was posted before I saw Jenny20's post giving specific angles. However there is still a bug in here somewhere, so I'm going to leave the post intact.

Something's bugging me here. Given the information in the problem, this has no solution. Let me explain that, then tell you my quandry.

First the weight of the traffic light automatically implies that the cable (which I will take to be massless for convenience) is forming an inverted triangle, with the eye ring at the apex. By a symmetry argument we can show that the not only the tension in the cable on both sides of the eye ring is the same, but also that the cables on either side make the same angle with the horizontal. (This angle, by the way, explains the fact that the cables are flexible as stated in the problem. If they were not then the cable is necessarily horizontal as it cannot stretch to form the triangle, and the tension in the cable becomes infinite. That is, the cable will snap no matter what the breaking tension is.)

So pick a coordinate system such that +x is horizontal and off to the right and +y is vertical. Call the angle the cables make with the horizontal $\theta$, the tension in the cable T, and the weight of the traffic light w. Newton's 2nd Law gives us:
$\sum F _x = 0 = Tcos(\theta) - Tcos(\theta)$ <-- This is always true so it doesn't give us any new information.
$\sum F _y = 0 = Tsin(\theta) + Tsin(\theta) - w = 0$

The second equation contains two unknowns, T and $\theta$. Thus (within the limits of the breaking tension of the cable) we may adjust the angle to be as small as we want by cranking up the tension in the cable. So there is no solution.

BUT...
Assume the eye ring is at the center of the cable. Call the length of the cable on one side of the eye ring to its support L. Then if we take the sum of the torques about an axis based where the cable is attached to its support (taking a CCW rotation to be in the positive sense) we should get 0 since we have static equilibrium. I'll take the axis to be the one on the left side of the diagram (the one on the right gives an identical result.)
$\sum \tau = 0 = TLcos(2 \theta) - wLcos(90 -\theta)$

$T(1 - 2sin^2(\theta)) - wsin(\theta) = 0$ <-- The L drops out!

From Newton's 2nd Law in the y direction we get that:
$sin(\theta) = \frac{w}{2T}$

Inserting this into the torque equation:
$T \left ( 1 - 2\frac{w^2}{4T^2} \right ) - w \frac{w}{2T} = 0$

$T - \frac{w^2}{2T} - \frac{w^2}{2T} = 0$

$T = \frac{w^2}{T}$

$T^2 = w^2$

$T = \pm w$

Which has a solution of T = w. Not only does this give a tension, it gives a specific angle ( $sin(\theta) = 1/2$), which is clearly a ridiculous answer given the situation. (We can always crank up the tension in the cable and change the angle as mentioned earlier.)

So what did I do wrong??

-Dan

6. Originally Posted by Jenny20
yes, a diagram.

30 degrees \ / 45 degrees
====================
Traffic light
is hanging
in the middle

The LHS is 30 degrees and the RHS is 45 degrees.
Ah! That explains it. (I still have a problem with my previous post, but this one I can explain! ) By the way, was there a typo in your post where you gave the answers? I don't agree with your F1.

Pick +x to be horizontal and to the right and +y to be vertical. Then:
$\sum F _x = 0 = - F_1 cos(30) + F_2 cos(45)$
$\sum F_y = F_1 sin(30) + F_2 sin(45) - w = 0$

The first equation says:
$F_2 = F_1 \left ( \frac{cos(30)}{cos(45)} \right )$

Inserting this into the second equation:
$F_1 sin(30) + F_1 \left ( \frac{cos(30)}{cos(45)} \right ) sin(45) - w = 0$

$F_1 \left ( sin(30) + \left ( \frac{cos(30)}{cos(45)} \right ) sin(45) \right ) = w$

$F_1 = \frac{w}{\left ( sin(30) + \left ( \frac{cos(30)}{cos(45)} \right ) sin(45) \right )}$

Putting w = 250 lb gives F1 = 183.013 N. This gives me F2 = 224.144 N.

-Dan

7. I am going to convert the lbs to Newtons. There are 0.225 lbs in a Newton.

The equilibrium equations are:

$\sum{F_{x}}=T_{2}cos(45)-T_{1}cos(30)=0$...[1]

$\sum{F_{y}}=T_{1}sin(30)+T_{2}sin(45)-56.25N=0$...[2]

The horizontal components must be equal in magnitude and from [2] the sum of the vertical components must balance the weight of the light.

Solve [1] for $T_{2}$ in terms of $T_{1}$

$T_{2}=\frac{T_{1}cos(30)}{cos(45)}=T_{1}\frac{\sqr t{6}}{2}$

The value for $T_{2}$ can be subbed into [2]:

$T_{1}sin(30)+(\frac{T_{1}cos(30)}{cos(45)})sin(45)-56.25=0$

Solve for $T_{1}$ and we find $T_{1}\approx{41.18} \;\ N$

This gives $T_{1}=41.18 \;\ N \;\ and \;\ T_{2}=50.43 \;\ N$

This gives $\boxed{T_{1}=183.02 \;\ lbs \;\ and \;\ T_{2}=224.133.... \;\ lbs}$

I hope I didn't mess up somewhere. Check it out. No doubt, Cap'n or topsquark will be along.

8. Yes, topsquark. This is a typo in my answer for fsub1. It should be 183lbs. =======================