- #1
phantomvommand
- 242
- 38
- Homework Statement
- See picture below
- Relevant Equations
- F = kx
Resolving forces?
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).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.
Are you suggesting that the diameter of the semicircle is (D + 2L/pi)?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?
Yes, it is non obvious.phantomvommand said:understanding how a vertical tension in the string even arises
Nearly right… think that through again.phantomvommand said:the diameter of the semicircle is (D + 2L/pi)?
The thing is, as the picture shows, that bent spring will not take the shape of a semicircle.phantomvommand said:Homework Statement:: See picture below
Relevant Equations:: F = kx
Resolving forces?Is this even solvable? I don't know where to begin.
True, but the question explicitly states it is to be taken to be a semicircle.Lnewqban said:that bent spring will not take the shape of a semicircle.
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).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?
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:Are you suggesting that the tension is then kD*pi?
The next step would be to consider moments.phantomvommand said:how such a tension translates into a net vertical tension required at the ends with the string
A spring does not behave like a rubber band.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.
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:As I can see spaces between the coils still in the bent position
A moment had to be applied whether or not the coils are touching their neighbours on the inside of the curve.Lnewqban said:If that is true, at each end of the spring, a moment had to be applied
Lnewqban said:the spring can’t keep the semicircular shape
I don't understand those last two points. The sequence you described seems to beLnewqban said:the spring is in repose
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).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
That last step does not mean the spring is now relaxed.
- manually bending the spring into the arc
- attaching the string
- releasing the spring so that the string is in tension
Tension in a spring refers to the force that is exerted on the spring when it is stretched or compressed. It is a result of the spring's elasticity and is measured in units of Newtons (N).
When a spring is bent into a semicircle, the tension in the spring increases. This is because the spring is being stretched along a larger distance, causing the spring to experience a greater force and therefore, a higher tension.
The tension in a spring bent into a semicircle is affected by several factors, including the material of the spring, the diameter of the semicircle, and the angle at which the spring is bent. Additionally, the amount of weight or force applied to the spring can also impact the tension.
The tension in a spring bent into a semicircle can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension or compression. The formula for tension in a spring is T = kx, where T is the tension, k is the spring constant, and x is the displacement from the equilibrium position.
Yes, the tension in a spring bent into a semicircle can be changed by altering the factors that affect it, such as the material, diameter, and angle of the spring, as well as the amount of weight or force applied. Additionally, the tension can also be adjusted by changing the length of the spring or by adding or removing coils.