# Tension On Rope Ends

1. Jun 2, 2010

### mecheng2121

Hi,

My first post and I'm hoping someone can help/verify a problem I have. I've got a rope with a ball fastened to each end. The balls are secured inside a c-channel. There is a small amount of slack in the rope such that the balls are loose inside the channel (the rope is slightly longer than the distance between the channels).

I then want to apply a distributed load, Fw, across the rope. I can calculate the curavature of the rope and resulting angle, Q, of the rope at the ends from the geometry of the application (I'm going to assume no stretching of the rope or deflection in the channels).

I want to calculate the tension applied to the ball ends. What I come up with, based on my FBD, is the tension, T, on each ball end is equal to the distributed load, Fw / 2 sin(Q).

The problem, which I can't get my mind to accept, is that as angle Q gets very small (if I allow very little slack in the rope), the tension on the ball ends gets very large... much more than the input, distributed load Fw.

Then, if I play a little mind game & take it a step further by replacing the rope with say, a very stiff steel I-beam fastened to a wall at the ends, Q would be infitesimally small resulting in an incredibly large tension T pulling on the wall... how could a building ever be constructed to hold against an extreme tension even with a small distributed load?

Any comments would be greatly appreciated... Am I on the right track? Do I have to simply accept that my ball ends need to be ready to handle a ton of tension? Does my mind game make sense and if so, how can that be? Thanks!

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2. Jun 2, 2010

### gmax137

I think that since you have defined the problem in terms of a flexible rope, what you're seeing is, there must always be some 'droop' in the line (such that your angle theta is never quite equal to zero). If you define your problem in terms of a rigid beam laying between the two channels, you'd see all of the beam's weight carried by the channels (with the weight or force directed straight down, vertically). Just my 2 cents.

3. Jun 2, 2010

### mecheng2121

So... if I have a distributed load of say, 100 lbs and there is just enough slack to allow the rope to sag 0.50 degrees at the ends, then I have to make sure the ball ends can each withstand 5,730 lbs?!?

4. Jun 2, 2010

### Q_Goest

Hi mecheng, welcome to the board,
I think you're assuming here that the rope doesn't stretch, right? So what happens as the rope stretches?

Consider also what happens to the I-beam and the walls that support it when a load is applied. The beam begins to bow, so the ends come closer together, but the walls can also shift. There is no such thing as infinitely rigid. You can think of I-beams in buildings a bit like rubber. They will bend and stretch depending on their modulus of elasticity. That's true of strings, cables, chains or any other material.

5. Jun 3, 2010

### mecheng2121

Thanks,
I realize I'm assuming some things that in real-life might not be factual... like no stretch in the rope (say a steel cable instead), or rigid channels & walls. But I need to design the ball ends such that they can withstand the 100 lb distributed load.

The problem I'm having is before I actually started to do the calculations, I was thinking the ball ends needed to hold up to roughly 50 lbs ea. with some safety factor. However, I was assonished to find that they may actually need to be strong enough to hold 5,730 lbs ea... if my calculations and FBD are correct.

So I guess what I'm asking is - in general - would most people here agree with my FBD and calculations that show that with only the 100 lb load, each ball end will actually have to be strong enough to hold up to 5,730 lbs (assuming a very small amount of slack - 0.50 degrees).

Just kind of hard for me to believe I guess... but thanks again.