# I The hanging chain/string

#### aliens123

Let's say I am trying to derive the equation for the hanging cable. The solution is a catenary curve. My question is:

In the setup for this problem, one would begin by considering some small segment of the string and analyzing the forces on it. Because it is in static equilibrium, the forces must all sum to zero. Because of the symmetry of the problem, we will consider the origin (0,0) to be the midway point where the chain is the lowest. Then we want to find y=ƒ(x); in other words, some equation describing the height of the chain as a function of x. Let us call the mass per unit length μ, the tension force T, the force due to gravitation g, and Δx the length of the string segment.

The gravitational force would act directly downward, and the tension force would act in two different directions at different angles. Here would be the equations describing the force

x dimension:
-Tcos(Θ) +Tcos(Θ+ΔΘ)=0
y dimension:
-μ(Δx)g - Tsin(Θ) +Tsin(Θ+ΔΘ)=0

However, at this point I realize that I must be in error. If we assume that the tension force is the same (and that T≠0) throughout the string, then in the 'x' dimension it will be impossible for the forces to sum to zero. Because in general cos(Θ)≠cos(Θ+ΔΘ), we have a contradiction:
-Tcos(Θ) +Tcos(Θ+ΔΘ)=0
T(-cos(Θ) +cos(Θ+ΔΘ))=0
T=0
or
-cos(Θ) +cos(Θ+ΔΘ)=0
cos(Θ)=cos(Θ+ΔΘ)

So what assumption is wrong here? Thanks for any help.

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#### drvrm

The gravitational force would act directly downward, and the tension force would act in two different directions at different angles. Here would be the equations describing the force
Is it correct to assume that tension T will be same through out the string/chain?
as i find that at the lowest point the tension is horizontal and is not supporting the vertically downward force-weight and it must be pulled by a different T' (vertical component)....and as you move to say top point the tension is closer to vertical

#### vanhees71

Science Advisor
Gold Member
I've no clue how to solve this problem using forces. It's much simpler to use the variational principle that a stationary state of the chain is given by the curve minimizing the potential energy. Assume that the chain has a constant mass distribution $\mu$ along it and let the curve be described by $y=y(x)$ with and the gravitational field $\vec{g}=-g \vec{e}_y=\text{const}$.
$$U=\int_{0}^{x_0} \mathrm{d} x \mu g y \sqrt{1+y^{\prime 2}}.$$
Since the endpoints and the total length are fixed we have a usual variational principle with the Lagrangian, using a Lagrange parameter $\mu g \lambda$ for the constraint
$$\int_0^{x_0} \mathrm{d} x \sqrt{1+y^{\prime 2}}=l,$$
leading to the Lagrangian
$$L=\mu g (y-\lambda) \sqrt{1+y^{\prime 2}}.$$
Since this doesn't depend explicitly on $x$, the "Hamiltonian"
$$H=p \dot{y}-L=\text{const}, \quad p=\frac{\partial L}{\partial y'}.$$
Now we have
$$p=\mu g \frac{(y-\lambda) y'}{\sqrt{1+y^{\prime 2}}}$$
and thus
$$H=\mu g (y-\lambda) \left (\frac{y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}-{\sqrt{1+y^{\prime 2}}} \right)=-\frac{\mu g (y-\lambda)}{\sqrt{1+y^{\prime 2}}}=\mu g a=\text{const}.$$
This implies the ODE
$$a^2(1+y^{\prime 2})=(y-\lambda)^2.$$
Separation of variables implies
$$\frac{\mathrm{d}y}{\sqrt{(y-\lambda)^2/a^2-1}}=\mathrm{d} x,$$
and integrating on both sides gives
$$a\, \mathrm{arcosh}\left (\frac{y-\lambda}{a} \right)=x+c \; \Rightarrow y=\lambda+a \cosh \left (\frac{x+c}{a} \right).$$
The constants $\lambda$, $c$, and $a$ can be used to fix the entpoints and total length of the chain.

#### aliens123

Is it correct to assume that tension T will be same through out the string/chain?
as i find that at the lowest point the tension is horizontal and is not supporting the vertically downward force-weight and it must be pulled by a different T' (vertical component)....and as you move to say top point the tension is closer to vertical
I am not quite sure, I too do not see how the tension can be the same everywhere. Also, as you pointed out, how is it that at the midway point the center is suspended at all if theoretically the slope of the line at that point is zero (and so there can be no vertical component of T)?

#### aliens123

I've no clue how to solve this problem using forces. It's much simpler to use the variational principle that a stationary state of the chain is given by the curve minimizing the potential energy. Assume that the chain has a constant mass distribution $\mu$ along it and let the curve be described by $y=y(x)$ with and the gravitational field $\vec{g}=-g \vec{e}_y=\text{const}$.
$$U=\int_{0}^{x_0} \mathrm{d} x \mu g y \sqrt{1+y^{\prime 2}}.$$
Since the endpoints and the total length are fixed we have a usual variational principle with the Lagrangian, using a Lagrange parameter $\mu g \lambda$ for the constraint
$$\int_0^{x_0} \mathrm{d} x \sqrt{1+y^{\prime 2}}=l,$$
leading to the Lagrangian
$$L=\mu g (y-\lambda) \sqrt{1+y^{\prime 2}}.$$
Since this doesn't depend explicitly on $x$, the "Hamiltonian"
$$H=p \dot{y}-L=\text{const}, \quad p=\frac{\partial L}{\partial y'}.$$
Now we have
$$p=\mu g \frac{(y-\lambda) y'}{\sqrt{1+y^{\prime 2}}}$$
and thus
$$H=\mu g (y-\lambda) \left (\frac{y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}-{\sqrt{1+y^{\prime 2}}} \right)=-\frac{\mu g (y-\lambda)}{\sqrt{1+y^{\prime 2}}}=\mu g a=\text{const}.$$
This implies the ODE
$$a^2(1+y^{\prime 2})=(y-\lambda)^2.$$
Separation of variables implies
$$\frac{\mathrm{d}y}{\sqrt{(y-\lambda)^2/a^2-1}}=\mathrm{d} x,$$
and integrating on both sides gives
$$a\, \mathrm{arcosh}\left (\frac{y-\lambda}{a} \right)=x+c \; \Rightarrow y=\lambda+a \cosh \left (\frac{x+c}{a} \right).$$
The constants $\lambda$, $c$, and $a$ can be used to fix the entpoints and total length of the chain.
Thank you for your reply. Unfortunately, I have not taken a course in classical mechanics yet, and so I do not know what a Langrangian/Hamiltonian is. Here is an example of a force analysis of the problem, but I don't quite understand some of the steps they take: http://homepage.divms.uiowa.edu/~stroyan/CTLC3rdEd/ProjectsOldCD/estroyan/cd/13/index.htm

P.S.
What commands did you use for LaTex on this forum?

#### Chestermiller

Mentor
I am not quite sure, I too do not see how the tension can be the same everywhere. Also, as you pointed out, how is it that at the midway point the center is suspended at all if theoretically the slope of the line at that point is zero (and so there can be no vertical component of T)?
What would your equation in the x direction look like if the tension were not constant?

#### aliens123

What would your equation in the x direction look like if the tension were not constant?
Something like:
-T(x)cos(Θ) +T(x+Δx)cos(Θ+ΔΘ)=0

Where T is a function of x. I am not really committed to an intuition either way as to whether or not the tension force has to be constant throughout the string.

I am not really sure how the midpoint is suspended at all if there is no component of vertical tension.
If we consider some length of string segment Δx about the origin, and let Δx→0, then for every finite Δx there is a component of vertical tension, because there is a slight curvature.

But I guess the question of what happens to the exact infinitesimally small Δx is just that it has no mass and therefore no gravitational acceleration. So maybe I just answered my own question?

#### houlahound

Wouldn't the tension at a link depend on the mass of all the links below it?

#### Chestermiller

Mentor
Something like:
-T(x)cos(Θ) +T(x+Δx)cos(Θ+ΔΘ)=0

Where T is a function of x. I am not really committed to an intuition either way as to whether or not the tension force has to be constant throughout the string.

I am not really sure how the midpoint is suspended at all if there is no component of vertical tension.
If we consider some length of string segment Δx about the origin, and let Δx→0, then for every finite Δx there is a component of vertical tension, because there is a slight curvature.

But I guess the question of what happens to the exact infinitesimally small Δx is just that it has no mass and therefore no gravitational acceleration. So maybe I just answered my own question?
This works the same way the curvature of the path works in centripetal acceleration. So, you are correct about the effect of curvature.

From your equation, $$T\cos{\theta}=T_0$$where T is the tension at the minimum point.

#### peregoudov

Sure the problem can be solved with elementary force approach without any variational principles involved. And much simplier as well though some integration is anyway needed.

Unfortunately the equations in starting post are all erroneous. The correct equations are
$$-T\cos\theta+(T+\Delta T)\cos(\theta+\Delta\theta)=0,$$
$$-\mu g\,\Delta s-T\sin\theta+(T+\Delta T)\sin(\theta+\Delta\theta)=0,$$
or, up to linear terms
$$-T\,\Delta\theta\,\sin\theta+\Delta T\,\cos\theta=0,$$
$$-\mu g\,\Delta s+T\,\Delta\theta\,\cos\theta+\Delta T\,\sin\theta=0.$$
These equations differ from equations by aliens123 in two aspects: we assume tension $T$ varies along the string and gravitational force is proportional to the length of the string not to it's x-projection. The first equation, as it was mentioned above, can be solved to $T\cos\theta=T_0$. Substituting this in the second one we can find the dependence of $\theta$ on $s$.

But there exists much more simple way. Let's consider finite part of the string of length $s$ starting from its bottommost point. Then equlibrium equations are
$$T_x=T_0,\quad T_y=\mu gs.$$
Since $T$ is directed along the string
$$\tan\theta=\frac{T_y}{T_x}=\frac{\mu gs}{T_0}.$$
The same result of course follows from equations for small part of the string written above.

It is convenient to measure all distances in $T_0/\mu g$ units, then $T_0/\mu g$ disappears from formulae. Now we have to calculate string shape. We write
$$\Delta x=\cos\theta\,\Delta s=\frac{\Delta s}{\sqrt{1+\tan^2\theta}}=\frac{\Delta s}{\sqrt{1+s^2}},$$
$$\Delta y=\sin\theta\,\Delta s=\frac{s\,\Delta s}{\sqrt{1+s^2}}.$$
An elementary integration is needed here. We obtain
$$x=\mathop{\rm arcsinh}s,\quad y=\sqrt{1+s^2}.$$
Little algebra leads to $y=\cosh x$.

#### aliens123

These equations differ from equations by aliens123 in two aspects: we assume tension $T$ varies along the string and gravitational force is proportional to the length of the string not to it's x-projection. The first equation, as it was mentioned above, can be solved to $T\cos\theta=T_0$. Substituting this in the second one we can find the dependence of $\theta$ on $s$.
Haha yes, I realized as much today, but you beat me to it! I am going to try and not look at the rest of your post because I want to derive it on my own, but it looks like you know what you are doing. Also, what commands do you use on this forum for Latex support?
This works the same way the curvature of the path works in centripetal acceleration.
Can you explain what you mean by this?
Wouldn't the tension at a link depend on the mass of all the links below it?
Yes, I think so.

#### David Neves

I would definitely recommend that you study calculus of variations, because that is the easiest most straight forward way to solve this problem, even if it can be solved in other ways. The calculus of variations will not only give you the mathematical underpinning needed to understand physics, but an intuitive understanding of why these things are true, including things as basic as why a circle is the closed plane curve with maximum area for a given perimeter.

#### AlecS0602

I am researching this topic in my differential equations class. It is almost always referred to as "the hanging chain" and the position of the chain is expressed as a Bessel function. Is this unique to only chains or does it apply to cords of any type? Would a hanging rope also be modeled by a bessel function? Thanks

#### Chestermiller

Mentor
I am researching this topic in my differential equations class. It is almost always referred to as "the hanging chain" and the position of the chain is expressed as a Bessel function. Is this unique to only chains or does it apply to cords of any type? Would a hanging rope also be modeled by a bessel function? Thanks
Tell us more about the Bessel function solution, and how it is derived.

#### Chestermiller

Mentor
This transient problem is not the same as the steady state catenary problem that has been discussed throughout this thread. It has no relevance to the discussion.

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