Infinite Force??

I'm running into what I believe to be a physical paradox in what I believe to be a poorly stated physics problem.

https://brilliant.org/weekly-problems/2018-11-26/basic/?p=3

If you can't get to the problem from the link:

"Charlie hangs a string horizontally between two walls and uses an identical string to hang weights vertically (in the center between the walls).
As he adds weights, the tensions in both strings increase.
Which string will break first?

Firstly, it shouldn't ignore elastic theory, but that's not necessarily the issue I want to discuss ( unless it resolves the issue).

Once a weight is hung the tensile force in the rope strung between the walls will be given by:

F = m*g/(2*sin( δ ))

where;
F = Tensile Force
m = mass of hung weight
δ = the angle between horizontal and the Tensile Force
g = acceleration due to gravity

I guess I'm disturbed by the apparent conclusion that the lighter the hanging mass, the lesser the angle δ, and F goes to infinity.

Is there some hidden limit taking interpretation that tames the infinity I'm missing, Is it an artifact of the deflection assumption, or am I completely wrong, and everything is consistent and well?

You mean because it is saying divide by zero? Maybe if so, the formula is wrong. Besides, if you reduce the mass of the weights how would that ever break the strings which is the point of the experiment?

@joe-shmo

If the angle is small we have that:

T = mg/2δ

The angle δ is going to depend on the mass, and since for zero mass we expect the angle to be zero, to first order I'd expect a relationship of the form:

δ = a*m^b(1 + O(m))

It's easy to show that the tension in the string goes as m^(1 - b).

There are 3 cases, depending on whether the exponent b is less than, equal to, or greater than 1. If b is greater than 1 the tension diverges, if b is less than 1 the tension goes to zero in the limit that the mass goes to zero. If b is 1 the tension is given by:

T = g/2a

Since the tension being infinite is unphysical, we have b <= 1. The parameters a and b probably depend on whether the string is taut but tension free in the unloaded case.

@sonhouse

Yeah, the formula implies a few different things. If we hold the mass constant and decrease the deflection ( moving the string back toward horizontal) by tensioning the strings that the tensile force is going to grow rapidly to catastrophic levels and the string will rupture. I'm fine with this, as it will rupture long before it "never reaches infinity". I'm fine with that.

But I was having trouble was also reconciling the opposite in the formula, if the mass is small, the angle is going to approach zero. And since the angles dependency on mass it isn't explicitly derived in the formula it was steering me to an illogical proposition. Something like I could put a feather on the string, and the angle would be so small a virtual division by zero would occur and the string would explode from a feather! I intuitively knew this was not the case, so I was pretty conflicted about that interpretation of the model.

What I think Deep Thought ( I'll have to really try to unpackage it more tomorrow) is saying that the "hidden" dependency of the angle on the mass of the hung weight is such that when you let the mass approach zero the limit of "m/δ(m)" (whatever δ(m) - "the angle as a function of the mass" may be) in the formula approaches zero. Meaning the tensile force goes to zero, and the "apparent paradox" is resolved, and everything is wrong with world once again... meaning I again didn't stumble on to anything all that interesting per se!

@DeepThought


"δ = a*m^b(1 + O(m))"


I didn't recognize this at first, but I actually think I derived this earlier when I tried to take elasticity into consideration.

I started with :

2*F* sin( δ ) - mg = m* y" ...(1)

Where y is the coordinate measuring the vertical sag. I set y" = to zero to find the equilibrium position.

2*F* sin( δ ) - mg = 0 …(2)

Then I incorporated: stress-strain in the elastic range where hooks law applies.

E = Modulus of Elasticity
σ = normal stress
ε = strain
a = cross-sectional area

dσ = E* dε ----> dF = E*A*dL/L ...(3)

Then I formulated that

L + dL - L = L*dδ ----> dL/L = dδ …(4)

From which you can integrate (3) to find F( δ )

F( δ ) = E*A*δ …(5)

and substitute (5) into (2).

2*E*A*δ* sin( δ ) - mg = 0

And use the small angle approximation for sin ( δ ) = δ

2*E*A*δ^2 - mg = 0

δ = ( mg/( 2EA))^(1/2) which is of the form a*m^b with b<1

I didn't think to substitutes it back into the original formula and take the limit to see what happened.


How do you just come up with the proper form like you did on the fly?

@joe-shmo said
@DeepThought


"δ = a*m^b(1 + O(m))"


I didn't recognize this at first, but I actually think I derived this earlier when I tried to take elasticity into consideration.

I started with :

2*F* sin( δ ) - mg = m* y" ...(1)

Where y is the coordinate measuring the vertical sag. I set y" = to zero to find the equilibrium position.

2*F* sin( δ ) - mg = 0 …(2) ...[text shortened]... mit to see what happened.


How do you just come up with the proper form like you did on the fly?

I guessed. Basically, it has to be zero for zero mass so a Taylor series (y(0) + y'(0)*x + 1/2 * y''(0)*x^2 + ...) looks wrong as the constant term needs to be zero for zero mass, giving us the b=1 case from my earlier post, but that has a non-zero tension for no load so we need something else. The simplest function I could think of was a power law, which are ubiquitous in physics. I should have stuck my neck out and gone for b=1/2, as it seems the most likely value.

Your derivation looks correct.

@DeepThought

Thanks for that! I guess the difference is as a physicist your used to staring with nothing. I kind of need that less shaky theoretical ground to get my footing.

I just have a quick follow up. I don't know if you were able to reach the problem without creating an account or not, but the three possible answers were:

A) the Horizontal String
B) The Verticle String
C) Both will break at the same time

I think the proper answer is not a choice. I propose as stated, D) all of the above are possible, as it depends on the elasticity of the material. It looks like δ^2 approximates δ*sin( δ ) well until around 40 degrees. A string with a low MoE ( ductile material ) may be able to stretch without rupture such that δ is 30 degrees, at which point the Tensile force in all three strings becomes equal. So I believe all three cases are possible, and the problem is incorrect.

Do you see any flaws in that argument?

@joe-shmo

Not instantly, the vertical string will snap for some load m_b (mass breaking). The corresponding tension in the vertical string is just m_b*g. This tension is achieved in the horizontal string for some mass M with an angle w such that

m_b * g = M*g/(2*sin(w))

so:

M = 2*m_b * sin(w)

Using your formula for the angle:

2*E*A*δ* sin( δ ) - mg = 0

we can write:

sin(w) = Mg/2EAw

To obtain

M = 2m_bMg/2EAw

or
EAw=m_bg

w =m_b*g/EA

at which point we need some numbers. Specifically what the breaking strain is.

@DeepThought
If you have those specific sprain numbers, can that answer the original question as to which string will break? If carried to extremes, suppose the rope is elastic in the extreme and it just stretches to infinity, which at the point where there is a string of individual atoms held together by Van Der Waals and his buddies😉 so eventually all things being exactly equal all the atoms separate at the same time into a cloud......

Providing the horizontal string does not deviate from the
horizontal by more than 45 deg it will break first. This is
(or was) a typical GCSE 'A" level question in applied Maths. (UK)

I think you guys are over-complicating it.

More interesting would be a scenario with a
steadily increasing weight until the string broke.
Where the tensile force and angle are both functions
of the weight. (Do elastics obey Hookers Law?)

@wolfgang59 said
More interesting would be a scenario with a
steadily increasing weight until the string broke.
Where the tensile force and angle are both functions
of the weight. (Do elastics obey Hookers Law?)

Hookers law? The magnitude of the erection is proportional to the applied ...

Hooke's law applies only when the deformation is elastic.

I don't believe your 45° figure, it ought to be 30°, is that from memory?

@deepthought said
Hookers law? The magnitude of the erection is proportional to the applied ...

Hooke's law applies only when the deformation is elastic.

I don't believe your 45° figure, it ought to be 30°, is that from memory?

A glass of red too many.
Yes Hooke's Law .. I remember applies to springs ... elastics as well?

And yes 30deg. At 30deg the tension in all strings is equal.
(sin 30 = 0.5)

@wolfgang59 said
A glass of red too many.
Yes Hooke's Law .. I remember applies to springs ... elastics as well?

And yes 30deg. At 30deg the tension in all strings is equal.
(sin 30 = 0.5)

Comically elastic isn't elastic, at least not quite. The stress strain curve exhibits hysteresis. Some relevant pages on Wikipedia are:

https://en.m.wikipedia.org/wiki/Natural_rubber

See the section on properties which has links to three of the following pages:

https://en.m.wikipedia.org/wiki/Viscoelasticity
https://en.m.wikipedia.org/wiki/Hyperelastic_material
https://en.m.wikipedia.org/wiki/Payne_effect
https://en.m.wikipedia.org/wiki/Mullins_effect