Position of Object Suspended on a String (Need Another Answer)

Question

I'm going to try to make as few errors in typing this as possible, so please bear with me and ask me to clarify/correct whatever needed.

Q: If an object is suspended on a string hung between two upright poles of varying heights, what is the position of the object in terms of the following variables:

$i$ - distance between poles

$\Delta j$ - difference in pole height

$l$ - length of string

A visual before I clarify a few minor details:

And some details:

The string is attached at the top of each pole.

We are assuming that the string has negligible mass, meaning our two segments are linear and not catenaries on their own.

The image depicts one string with an object hung on it, not two separate strings

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Now, I spent about a month generating and tweaking one method for giving the position as a pair of coordinates in terms of $i, \Delta j,$ and $l$ with respect to the midpoint of the pole tips. I will post this solution shortly.

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Where you come in

I want to see what other ways the object's position may be calculated. For particularly good solutions, I may award bounty. Here are a few things I ask of your answer:

$1$. As long as you use $i$, $\Delta j$, and $l$, you may use whatever other variables needed to reach your final answer as long as it ultimately is in terms of those 3 variables.

$2$. Your coordinate system may be with respect to any point in the system, just specify. For example, mine is with respect to the midpoint of the two pole tips.

$3$. Usage of visuals is strongly recommended wherever necessary.

$4$. Please post what branches of math were critical parts of your method. There are actually a number of very different ways to approach this problem (I am working on a second one using very different methods than what I'll post). I ask this so myself and others who may come across something we're unfamiliar with can research it.

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Thank you so much for taking the time to read this, and I anticipate seeing great solutions!

Answer 1

This can be solved by applying Newton’s laws and a bit of geometry.

If the poles are the same height, symmetry dictates that equilibrium is attained halfway between them. We are treating the string segments as two-force members, so in particular the force transmitted by them is independent of their length. From this we can deduce that in all configurations the angles $\alpha$ and $\alpha'$ in the diagram below are equal. Also, the mass must be lower than the tops of both poles, otherwise there will be a net sideways force that will move it toward the lower post. This allows us to “straighten” the string, after which the solution follows easily. Assume that the lower pole is on the left and let $x$ and $y$ be the horizontal and vertical distances, respectively, from the top of the left-hand (lower) pole. The Pythagorean theorem gives us $$i^2+(2y+\Delta j)^2=l^2$$ or $$y=\frac12\left(\sqrt{l^2-i^2}-\Delta j\right).\tag{1}$$ Similar triangles give us $$\frac xy = \frac i{2y+\Delta j}$$ so $$x=i\frac y{2y+\Delta j}=\frac i2\left({\sqrt{l^2-i^2}-\Delta j\over\sqrt{l^2-i^2}}\right).\tag{2}$$ I’ll leave it to you to verify that equations (1) and (2) also hold when $\Delta j\le0$, i.e., that we can drop the assumption that the left-hand pole is lower.

A few simple tests support the plausibility of this solution. First, when $\Delta j=0$, we have $x=\frac i2$ and $y=\frac12\sqrt{l^2-i^2}$ as expected. When $\Delta j\gt0$, as $l$ approaches the distance between pole tops, both $x$ and $y$ approach zero, while when $\Delta j\lt0$, $x$ approaches $i$ and $y$ approaches $\left\vert\Delta j\right\vert$, i.e., the equilibrium point gets closer and closer to the top of the lower pole as the string gets shorter. For $\Delta j\gt0$, $x\lt\frac i2$ and for $\Delta j\lt0$, $x\gt\frac i2$, i.e., the equilibrium point is always closer to the lower pole. Finally, as $l$ increases, $x$ approaches $\frac i2$ from either side.

Addition: Noticing that the locus of possible positions of the mass is an ellipse as in this answer leads to another proof that the angles in the diagram are equal. The mass will be at rest at a point at which the gravitational potential is at a minimum. This is a constrained optimization problem to which the Lagrange multiplier method can be applied: a minimum will be attained at a point on the ellipse where its normal is parallel to the gradient of the potential, which is in the direction of increasing height. However, every normal to an ellipse bisects the angle to the foci, making $\alpha$ and $\alpha'$ equal as well.

Answer 2

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This solution relies heavily on analytic geometry and differential calculus

Let's look at that OP image again:

One should note that the sum of the length of the two string segments (one leading from pole 1 to object, the other from object to pole 2) is a constant. This means given taut string segments, the graph of possible string segments is an ellipse with pole endpoints being the foci. We can use differential calculus to find its minimum and thus, the coordinates of the object. Consider this visual:

Now let's establish some variables for our work.

Our three known variables will be: $$\color{red}{i=\text{distance between poles/endpoints}}$$ $$\color{#F80}{\Delta j=\text{difference in height of poles/endpoints}}$$ $$\color{pink}{l=\text{catenary length}}$$

Other variables we'll later get in terms of the three knowns:

$$\color{purple}{a=\text{length of semimajor axis of ellipse}}$$ $$\color{blue}{b=\text{length of semiminor axis of ellipse}}$$ $$\color{yellow}{c=\text{distance from center to foci/endpoints}}$$ $$r=\text{distance between foci} \text{(Not shown)}$$ $$\color{green}{\theta=\text{angle of elevation/depression between endpoints}}$$

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Using analytic geometry and trigonometry, the following relations arise:

$$\sin \theta=\frac{\Delta j}{r}$$

$$\cos \theta=\frac{i}{r}$$

$$a=\frac{l}{2}$$

$$c=\frac{\sqrt{i^2+\Delta j^2}}{2}$$

$$b=\sqrt{\frac{l^2-i^2-\Delta j^2}{4}}$$

Now for the math part. The equation for a rotated ellipse is as follows:

$$\frac{(x\cos\theta + y\sin\theta)^2}{a^2} + \frac{(x\sin\theta - y\cos\theta)^2}{b^2} = 1 $$

Finding a common denominator and plugging in those basic trig identities:

$$b^2\left(x\left(\frac{i}{r}\right)+y\left(\frac{\Delta j}{r}\right) \right)+a^2\left(x\left(\frac{\Delta j}{r}\right)-y \left(\frac{i}{r}\right)\right)=a^2b^2$$

Now let's implicitly differentiate and isolate $\frac{dy}{dx}$.

$$\frac{dy}{dx}=\frac{-b^2i^2x-b^2i\Delta jy-a^2\Delta j^2x+a^2i\Delta jy}{b^2i\Delta jx+b^2\Delta j^2y+a^2i\Delta jx-a^2i^2y}$$

Setting slope equal to $0$ and isolating $y$:

$$y=\frac{b^2i^2x+a^2\Delta j^2x}{a^2i\Delta j-b^2i\Delta j}$$

Plugging into original ellipse equation and isolating x:

$$x=\frac{i\Delta j(a^2-b^2)}{\sqrt{a^2i^4\Delta j^2+2a^2i^2\Delta j^4+a^2\Delta j^6+b^2i^2\Delta j^4+2b^2i^4\Delta j^4+b^2i^6 }}$$

Plugging in identities for $a$ and $b$ and simplifying:

$$x=\frac{i\Delta j(i^2+\Delta j^2)^{3/2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }}$$

A little work shows that $y$ is this:

$$y=\frac{(-i^2\Delta j^2+i^2l^2+\Delta j^2l^2-i^4)\sqrt{i^2+\Delta j^2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }}$$

ANSWER

Thus, our position in terms of $i,\Delta j$, and $l$ $\;\;\forall\;\theta\in[0,\frac{\pi}{2})$ with respect to the midpoint of pole endpoints:

$$x=\frac{-i\Delta j(i^2+\Delta j^2)^{3/2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }}$$

$$y=\frac{(-i^2\Delta j^2+i^2l^2+\Delta j^2l^2-i^4)\sqrt{i^2+\Delta j^2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }} $$

$\;\;\forall\;\theta\in[\frac{\pi}{2},\pi)$ with respect to the midpoint of pole endpoints:

$$x=\frac{i\Delta j(i^2+\Delta j^2)^{3/2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }}$$

$$y=\frac{(-i^2\Delta j^2+i^2l^2+\Delta j^2l^2-i^4)\sqrt{i^2+\Delta j^2}}{2\sqrt{3i^4\Delta j^2l^2+3i^2\Delta j^4l^2- 3i^4\Delta j^4-3i^6\Delta j^2-i^2\Delta j^6+i^6l^2+\Delta j^6l^2-i^8 }} $$

I ignored showing steps since they are very extensive and would require hours to format properly (at some points I had to use size 5 font lines were so long). I have them all in Word though, so if anyone would like to see them I will be happy to provide them.

Answer 3

A diagram is shown below.

equations are listed below.

$$\begin{cases} \tan \alpha = \frac {\bigtriangleup j}b \\ \sin \alpha = \frac b{l_1} \\ \cos \alpha = \frac a{l_2} \\ 2l_2+l_1 = l \\ 2a +b = i \end{cases}$$

Here we have five unknowns: $\alpha, a, b, l_1, l_2$ and five equations.So there is possibility to write mass position a in terms $\bigtriangleup j, i$ and $l$.

Thus, $\alpha$, a and b are implicitly defined by, $$i-\frac{\bigtriangleup j}{\tan \alpha}=l \cos \alpha - \frac {\bigtriangleup j}{\tan ^2 \alpha}$$

and the solution is,

$$x = \frac b2 =\frac12(\frac{\bigtriangleup j}{\tan \alpha})$$

$$y = \frac{\bigtriangleup j}{2 \tan \alpha}-\frac{l}{2}\sin \alpha$$

In this method, we need to balance the weight. The only condition is that the angles should be the same.