Unequal force at uneven attachment heights?

[derek88]

Unequal force at uneven attachment heights??

Friends:

My boss and I have been debating this physics problem:

Imagine that you have a cable suspended in between two poles (Pole 1 and Pole 2). The attachment height at Pole 1 is greater than that at Pole 2. As for the orientation of the x, y, and z-axis, let's say that the poles are set on the x-axis and they're pointing up in the direction as the y-axis. We both agree that the vertical tension (due to the cable) at Pole 1 is greater than the vertical tension at Pole 2, i.e. Vtension1 > Vtension2.

A wind blows on the cable. We will say that the wind blows along the z-axis. The cable swings outwards along the z-axis. This creates forces at the cable's pole attachments, in the z-direction (we can call the forces Ztension1 and Ztension2). I believe that since Vtension1 > Vtension2, it must be true that Ztension1 > Ztension2. My boss believes that Ztension1 = Ztension2. Which one of us is right?

I appreciate any thoughts on this problem! :):):)

 

[Simon Bridge]

This one is easy to set up in an experiment - rig two poles with a length of heavy string and a Newtonmeter at each end. This should be definitive, where any analysis can always be argued over.

You can also analyse a simplified picture by imagining now that the cable is a few masses joined by massless "string" and see how it plays out for different numbers of masses.

Formally:
http://www.aejm.ca/V4N1/Chatterjee.V4N1.pdf
... you'd have to modify this for crosswinds.

I'm thinking that if you sketch out the profile down the x axis, then the angle the cable makes to the tall pole will be less than the angle it makes to the short pole, which kinda suggests that the z-component of the tension against the tall pole is less than the short one.

Since the long side presents a larger surface area to the wind, it may get blown out more.
But I think the rule-of-thumb weight is leaning towards less z-tension for the tall pole. But it tells you what to watch for in the experiment.
Probably someone has a computer program for computing this.

 

[derek88]

Thanks for the reply!

I just assumed that the "blowout angle" (the angle from vertical) that the cable makes in a windy environment would be the same throughout the entire length. This is because I imagined that the cable was like a series of pendulums (just imagine an invisible string connecting each point of the cable to the axis of rotation). The angle of swing of a pendulum is Theta = arctan(Windforce/Cableweight). If we assume that both the wind force and the cable weight is evenly distributed, it seems that the blowout angle would be the same everywhere. Any thoughts?

 

[Simon Bridge]

Draw the picture you just proposed and you'll see you have to move one of the poles so it still connects with it's end of the cable.

But like I said - you can argue endlessly about this kind of thing. Either do the real math (see link) or do the experiment.

 

[derek88]

Well, I'll try to do the experiment by placing a stick on the ground and another on an elevated position, string a cable in between them, then turn on a big fan. I'll see if one stick falls down first or not. I'll post my results.

 

[Simon Bridge]

Sight along the two sticks and estimate the angles to each.