Unequal force at uneven attachment heights?

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The discussion revolves around the physics of tension in a cable suspended between two poles at different heights. It is established that the vertical tension (Vtension1) at the higher pole (Pole 1) is greater than that at the lower pole (Pole 2), leading to a debate on the horizontal tension (Ztension1 and Ztension2) caused by wind blowing along the z-axis. One participant argues that Ztension1 must also be greater than Ztension2 due to the greater vertical tension, while the other believes they are equal. The conversation suggests conducting an experiment with a Newtonmeter to measure the tensions and validate the theoretical claims.

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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! :):):)
 
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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.
 
Last edited:


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?
 


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.
 


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.
 


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

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