Tension in a spring that's bent into a semicircle

In summary, you are suggesting that the tension in the string arises from the fact that the inner circumference of the spring is greater than the distance around its outer perimeter.
  • #1
phantomvommand
249
39
Homework Statement
See picture below
Relevant Equations
F = kx
Resolving forces?
Screenshot 2023-02-11 at 2.54.14 PM.png

Is this even solvable? I don't know where to begin.
 
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  • #2
phantomvommand said:
Homework Statement:: See picture below
Relevant Equations:: F = kx
Resolving forces?

View attachment 322080
Is this even solvable? I don't know where to begin.
Judging from the diagram, the spring starts compact, i.e. it cannot be compressed (much).
Consider the outermost parts of the spring in the curved position. What length is that now?
 
Last edited:
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  • #3
haruspex said:
Judging from the diagram, the spring starts compact, i.e. it cannot be compressed (much).
Consider the outermost parts of the spring in the curved position. What length is that now?
Are you suggesting that the diameter of the semicircle is (D + 2L/pi)?

I think I need help with understanding how a vertical tension in the string even arises. It is obvious intuitively, but when I break the spring down into an infinitesimal element I cannot figure out why.

Considering the small element of the spring where the string is attached to, it experiences a leftward spring / tension force, and a downward tension force from the string. Yet, it is in equilibrium. Presumably, the leftward tension is balanced by some rightward friction with the string? But how about the downward tension force? What gives rise to a necessary upward force which the downward string tension must balance?

If my method of analysis above is incorrect, I'd be happy to hear how else the tension in the string arises.
 
  • #4
phantomvommand said:
understanding how a vertical tension in the string even arises
Yes, it is non obvious.
When a helical spring is stretched, each small cylindrical element of the wire is twisted about its axis. So it involves shear moments, like a torsion wire. (Confusingly, a "torsion spring" works through bending moments.)
However, you are probably not expected to get into such detail.

phantomvommand said:
the diameter of the semicircle is (D + 2L/pi)?
Nearly right… think that through again.
But it is not the diameter that is of direct relevance; as I asked, what is the distance around the outer perimeter of the curved spring? Hence, how much has the spring been stretched on that side?
 
  • #5
phantomvommand said:
Homework Statement:: See picture below
Relevant Equations:: F = kx
Resolving forces?Is this even solvable? I don't know where to begin.
The thing is, as the picture shows, that bent spring will not take the shape of a semicircle.
The apex is under a greater amount of bending moment than the cross-sections closer to the ends.
What is the subject to which this problem is related?
 
  • #6
Lnewqban said:
that bent spring will not take the shape of a semicircle.
True, but the question explicitly states it is to be taken to be a semicircle.
 
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  • #7
haruspex said:
Yes, it is non obvious.
When a helical spring is stretched, each small cylindrical element of the wire is twisted about its axis. So it involves shear moments, like a torsion wire. (Confusingly, a "torsion spring" works through bending moments.)
However, you are probably not expected to get into such detail.Nearly right… think that through again.
But it is not the diameter that is of direct relevance; as I asked, what is the distance around the outer perimeter of the curved spring? Hence, how much has the spring been stretched on that side?
If I am understanding correctly, you are suggesting that assuming the spring is not stretched much, it's inner circumference is L, which gives an outer perimeter length of (L + D*pi).

Are you suggesting that the tension is then kD*pi?

I am a little confused as to how such a tension translates into a net vertical tension required at the ends with the string, and furthermore, why is this tension, which is calculated only for the outer perimeter of the spring, representative of the overall tension at all points in the spring?
 
  • #8
phantomvommand said:
Are you suggesting that the tension is then kD*pi?
Not quite. As you note, that is only for the most stretched arc through the spring. What would it be on average? Bear in mind the circular nature of each coil of the spring.
phantomvommand said:
how such a tension translates into a net vertical tension required at the ends with the string
The next step would be to consider moments.
Having found the effective tension in the spring, think of the spring as two rigid quadrants hinged by a single coil of the spring. Take the balance of moments about the innermost point of that coil.
 
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  • #9
phantomvommand said:
I think I need help with understanding how a vertical tension in the string even arises. It is obvious intuitively, but when I break the spring down into an infinitesimal element I cannot figure out why.

… What gives rise to a necessary upward force which the downward string tension must balance?

If my method of analysis above is incorrect, I'd be happy to hear how else the tension in the string arises.
A spring does not behave like a rubber band.
Besides accepting stretching, it can be compressed (if pitch of coils allows space for it), and can also be bent, or even twisted.

Our spring has been bent until reaching the shown shape.
As I can see spaces between the coils still in the bent position, I would assume that there is a neutral axis, a compressed concave side and a stretched convex side (just like it happens to a beam subjected to bending loads).

Please, see:
https://en.wikipedia.org/wiki/Bending

If that is true, at each end of the spring, a moment had to be applied, in order to achieve a semi-circular shape.
The magnitude of that moment should be equal to stretching force times D, or compressing force times D, or (stretching force + compressing force) times D/2.

If you now attach the string to both ends, and remove the bending moments applied to the ends, the spring can’t keep the semicircular shape, but a tension appears in the string.
There is now an internal moment that grows in magnitude from each end to the apex, where is value is maximum.

The tension of your string is now perfectly balancing all those internal moments, since the spring is in repose.
 
  • #10
Lnewqban said:
As I can see spaces between the coils still in the bent position
As I wrote in post #2, it doesn’t look that way to me (except near the ends, where the moment from the string is weak).
Lnewqban said:
If that is true, at each end of the spring, a moment had to be applied
A moment had to be applied whether or not the coils are touching their neighbours on the inside of the curve.
Lnewqban said:
the spring can’t keep the semicircular shape
Lnewqban said:
the spring is in repose
I don't understand those last two points. The sequence you described seems to be
  • manually bending the spring into the arc
  • attaching the string
  • releasing the spring so that the string is in tension
That last step does not mean the spring is now relaxed.
 
  • #11
haruspex said:
As I wrote in post #2, it doesn’t look that way to me (except near the ends, where the moment from the string is weak).

A moment had to be applied whether or not the coils are touching their neighbours on the inside of the curve.I don't understand those last two points. The sequence you described seems to be
  • manually bending the spring into the arc
  • attaching the string
  • releasing the spring so that the string is in tension
That last step does not mean the spring is now relaxed.
My post was directed to the OP, who stated he needs help understanding the reason for the string tension (please, see selected header quote in my last post).
As you may know by now, sometimes I induce confusion as I try to explain my ideas, as English is not my native language.
My apologies, @haruspex
 

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