Suppose I had 2 solid anchor points say 20 ft apart. I pulled a non stretch string with a breaking strength of say 1,000# to 90% of it capacity and then tied it off without loosing any tension. Now I walk to the center of the string and push sideways with say 100# force. How much force would be exerted on the anchor points and how is it figured?
You’re assuming an impossibility, and hence the literal answer to your question would also be nonsensical. There is no such thing as a non-stretch string, and the real-world answer to this question would depend on just how stretchy the string actually is.
Under the (obviously unrealistic) assumption that the system (string + anchor points) is perfectly rigid, ISTM the 100# side force on the string produces only a side force on the anchor points - there is no additional force pulling them together.
To get a feel for this, mentally replace the string with a steel bar 8" in diameter.
That makes sense, too easy!
I don’t understand the math behind it but you should look at the answers about this when the question is rock climbing anchor points.
(IIUIC your looking at significantly more than 100# on both anchors.)
CMC fnord!
You are defining the tension force as 900#. You are adding another force of 100# perpendicular to the string and assuming it’s a rigid body, which is not an uncommon assumption. You have 900#s pulling on each anchor in an axis. And 50# on each anchor perpendicular to that axis. The resultant can be determined with vector addition. The magnitude being (900#^2+50#^2)^.5. The direction would require a tiny bit of trig.
It’s just a basic statics question.
Things can be under tension and impact anchor points with negligible strain for this type of analysis. Think of music instruments and their strings or pre-tensioned steel rods in concrete. The degree of elongation is just superfluous for determining the resultant forces.
I should revise the question to a low stretch 1% before breaking stretch. It would have been more pertinent.
All that matters is if it’s in equilibrium and not significantly deformable if you care about static forces. Now, if it’s deforming under load you got a more challenging problem.
I going to give the real life example instead of drawing very poor comparison. An archery has a normal brace height of about 6". This is the distance the string is from the handle when the bow has a string on it. I will use a normal weight 50# straight bow for example. Now if I were to increase the length of that string until the string in its fully stretched out condition was maybe 1/4" shorter than the bow. So now we have zero brace height. The string is resting firmly across the belly of the bow. At this point the tension of this string is in the thousands of pounds, but only in one very small position and once past that position the bow will either bend or buckle under the tension. Yet I can easily pull the string arrow from the bow in spite of the huge tension force currently on the string. This is the common assumption, I don’t think that very small perfect condition for this could ever actually be achieved and if it was ever achieved it would have to be a much fatter bow sting than normal and I doubt we could make that initial move with our hands. I was just looking for something to compare to while playing around.
Hmm. That’s due to the strength of the material, aka the bow, that you are actually deforming when you pull the string. You aren’t fighting against the tension of string when you pull it in that case.
Actually, if the string doesn’t stretch, then the force at the anchor points is infinite. Or at least, at the breaking tension of the string, at which point the string breaks. If the string stretches by only a small amount, then the tension at the anchors is very large. To see this, look at the force diagram at the point on the string where you’re pushing it.
You obviously can’t assume that the string is perfectly rigid, because even aside from the impossibility of perfectly-rigid objects, it’s a string. Strings are noted, defined even, by their complete lack of rigidity.
But in problems of this nature you tend to make simplifying assumptions so that the problem is solvable. And one of the simplifying assumptions is that of rigidity. Another is instantaneous force transmission. Now it sounds silly with regards to a string, but in tension and in static equilibrium it allows a problem of this nature to be solved with techniques from a statics class. We did problems similar to the first example all the time in statics class.
A ball hanging from a string puts the string in tension. No one considered the deformation of the string in that class. It wasn’t relevant to solving for forces. You just needed to do a balance of forces and torques and solve that system of equations. That’s how we did it. It wasn’t until strength of materials and deformation of solid bodies that we worried about the much more challenging how things precisely deformed under load problems.
But back to the music instrument example. When you tune the string you increase or decrease the tension in the string. Which puts a force on the guitar. Which is why the body and neck have to have a certain strength. Nothing is moving because the system is in equilibrium. But there are definite forces/stress/strain present. It’s just the strain is ignored in a lot of these problems as being inconsequential.
Why 1% ? Because Ultra-high-molecular-weight polyethylenes, such as Spectra and Dyneema (strength per strand = 45.5 kg (100 lb), stretch = 1.0%), have been used since the 1990s. They are lighter, therefore faster, than Kevlar—and have a much longer life.
You can turn 1% stretch into an angle,theta… which is the deflection of the string when the string is 1% longer.
The tension is 1/2 times Load (a string on each side…) divided by sin (Theta).
See Finding the Tension of Two Strings with Different Angles — Mathwizurd
That example won’t help with the pre-tensioned constraint. Alright, let me get my calculator and come up with the exact solution for the initial problem.
Ok, if this were a homework problem you’d treat the given information as such. Therefore we are treating the string as practically inelastic and pretensioned at 900 lbf.
So the string is applying a force to each of the two anchor points of a magnitude 901.4 lbf at an angle 3.18º from vertical.
Now, it’s not exactly correct. But given the constraints should be a reasonable answer.
It would be interesting to test this with something like a guitar string under tension on a guitar, some weights, and a strain gauge. To see how well the simple non-elastic model works. I could be totally wrong. :eek:
Given that the string is inelastic, it will not deform with the 100 lb of pressure otherwise it would have to grow longer (Pythagorean theorem) viz. with lateral force the string acts like a rod. So wouldn’t the OP’s question reduce to “I have an unmovable object. I put 900 lb of force in one direction and 100 lb of force at 90 degrees to it. What is the total force?”
I’ll throw this out there.
That’s a Tyrolean Traverse. At least in rock climbing.
I used to be a member of the Colorado Ground Search and Rescue Team. An arm of CAP. Did a bit of rock rescue practice (mostly it was boring as shit trying to find crashed private planes).
Anyway, on one tyrolean practice, we did not have the rope tight enough. The guy went into the river. He was roped up from both sides and we pulled him out.
That little practice was the reason we created the Spinner Award. Ever seen a spinner fish lure on the end of a line? That’s pretty much what this was. But with a 170lb man.
Yes, when you’re dealing with mostly-rigid objects like rods. A steel rod is approximately rigid, so treating it as completely rigid is usually a reasonable simplification. But strings are not rigid at all. Taking something that’s not rigid at all and assuming that it’s perfectly rigid is an absurdly terrible simplification. Much better to approximate the string as a perfect string.
Hey HoneyBadgerDC - I always love reading about your work and adventures with bows. This question was really intriguing me and did the “math” and here are the quick answers :
1> If the string was made of NYLON, the force on the anchor points will be 907# (assuming # means pound force i.e. lbf)
2> For MILD STEEL, the force will be 1211# (i.e. it will break)
3> For GRAPHENE, the force will be 912#
Hope this helps; the details are in the subsequent post.