How to measure stress/tension on a rope?

(a) Imagine an 5-100kg (we do not know exact weight) object is hanged with a piece of string/rope/wire and swings randomly.
(b) We do not have access to either end of this string. (c) How can we measure, electronically,em the stress/tension on the string?
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Study catenary curves. Apply some high school physics and algebra.
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Here's something that the OP may be interested in. http://www.squid-labs.com/projects/erope/index.html
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I think you need to say if you are allowed to cut the rope in order to insert a transducer (load cell etc).
Also the question appears a bit homework-like! So if it isn't, perhaps you should tell us a bit more about the application so that people are not put off replying.
--

John Devereux

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No, this is not homework, and I'm not a student.
Imagine a parachute. How can you measure stress/tension on a parachute line? You can not access the either end of the string/line. (One end connectected to parachute, the other is connected to carabiner.)
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Can we contact the string/line at all? Do we know what it is made of?
If so we could potentially measure stretch over a small sample if we can access it before the load is applied and can previously know the stretch characteristics.
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finger to keyboard and composed:

Would it be acceptable to fit a sensor to the harness and assume the load is equally distributed amongst the risers?
Otherwise, is this what you are looking for?
Collecting Parachute Test Drop Data: http://www.industrologic.com/cptdd.htm
-- Franc Zabkar
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wrote:

You probably could make a simple gizmo to measure the strain using inexpensive force sensors like below (search www.digikey.com for force sensors). You would just slip the parachute line in the gizmo sideways and when the string is pulled tight the gizmo flexes pinching the sensor in preportion to the pull on the line.
http://tinyurl.com/96srw
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You might try calling the U.S. Navy NAWS, China Lake, Ca. at 760-939-9011 and ask for the parachute department. I vaguely recall seeing something about the group being disbanded but it may still be in business and I would imagine they've instrumented a parachute at some time or another.
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Without access to the string, it can't be done in any way that could justifiably be called "electronically". You're talking about doing some serious physics here. Like: shoot a lot of x-ray intensity at it and have an expert interpret the diffraction pattern for you to determine the lattice length of the string, from that (assuming you at least know the material) the deformation and from that, in turn, the tension. Or shoot acoustic energy at it over a wide spectrum and try to find its resonance frequency.
--
Hans-Bernhard Broeker ( snipped-for-privacy@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.
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Tony Limson wrote:

Make a loop in the middle of the string and put an electronic scale or load cell in there.
Mitch
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Tony Limson wrote:

You have described a "pendulum" : oscillation frequency is proportional to pendulum length and weight on the end.
If you know the time for one oscillation, the gravitational acceleration, length of the rope, you should be able to solve for the mass of the pendulum (weight + rope).
Once you have the mass at the end and the velocity of it you should be able to calculate the force exerted on the rope through centrifugal force.
Sounds like a lot of mucking around.
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wrote:

Yes.
Umm, no. But thanks for playing.
Hint: where is the "weight" in 2*pi*sqrt(L/g)?
Regards,
-�ve
--
Change is inevitable, progress is not.

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Dave Hansen wrote:

Unless the rope is infinitely rigid (wonderful first year physics assumption), the period and the swing itself will vary with the mass, because the rope will stretch with the angular acceleration. I don't want to do the math, even if I could.
Not that this applies to this case, but just to be complete.
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[...re:2*pi*sqrt(L/g)...]

Assume a spherical horse moving through a vacuum... ;-)
Regards,
-�ve
--
Change is inevitable, progress is not.

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Dave Hansen wrote:

Try 2*pi*sqrt(L/g)= 2*pi*sqrt(m/k)
where m = mass

thanks for playing.
Regards,
-=Mark
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put finger to keyboard and composed:

Just to be pedantic, that's the oscillation period, not frequency.
And "proportional" implies a linear relationship, not sqrt.
And the formula strictly only holds for a "simple" pendulum, ie one where the oscillation angles are small.
See http://hyperphysics.phy-astr.gsu.edu/hbase/pendl.html
As for your interpretation of Hooke's Law, how long is a string when no force is applied to it? ;-)
Hint #1: F = k.dL, not F = k.L.
See http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/SHM/HookesLaw.html
Hint #2: Would a 100kg mass swing with 10 times the period of a 1kg mass, if released at the same angle?
-- Franc Zabkar
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I haven't seen all of this post, but the answer seems obvious.
Deflect the rope with a roller, bewtween two other rollers. Measure the force on the deflecting roller perpendicular to the rope. This could be fitted to a rope without cutting it.
Andrew M

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Andrew M wrote:

I didn't bother to read all of this post, but if the answer seems obvious, it's likely that someone has already proposed what you did. I guess we'll never know.
Mitch
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Exactly - The 3 sheave dynomometer. This method has been used in cranes for years to measure the weight on the end of the hook.
Andy
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