Ronen Rappaport’s project: messing with strings

The impetus of my doing of this experiment was to see how material presented in my differential equations class is applicable to real life situations. Before embarking in the large project, I decided to pursue a problem that was within my scope of solving and reasoning, as well as something that I have come across before and was curious about. The perfect starting example that came across my mind was the Catenary curve; I have used this equation many times in my Calculus 142 and Physics 211 class, unaware of how it was derived and of its differential-related properties. This proof will reflect my understanding of how to derive the position equation for this catenary curve using the skills I have used in Math 242 and Physics 211 as my resources. I will use this example as starting point to develop a more complete picture on the behavior of strings under different forces.

Part I: The Catenary

The basic understanding of the catenary curve has allowed for many engineering and architectural feats to be accomplished. Whether it is a suspension bridge such as the commonly referenced Saarinen’s Gateway Arch in St. Louis or the suspension cables of the Golden Gate Bridge in San Francisco, catenary curves are applicable and widely used for statics in structural supports all over the world. The word “catenary” is derived from the Latin word “chain”; it is also called an alysoid or a chainette. It was recorded that Galileo claim of a chain adopting a parabolic shape was disproved by Jungius in 1669. Afterwards, the equation was obtained by Leibniz, Huygens, and Johann Bernoulli in 1691.

In this proof, I am assuming that this is an ideal chain. The chain will have a uniform linear density (measured, for example in kg/s²) and negligible transverse stiffness (which means tension in the $x$–direction will be constant)

Deriving the Catenary equation

• Let $T_x$ represent the tension in the horizontal direction. By the hypothesis above (the horizontal tension is constant along the entire string), we can express it as $T_x = T\cos\theta = H$ for some constant value $H$
• Linear weight density $W$ is uniform along the string. $W$ is in units of force of gravity over meters, since the only force over the distance of the string is the force of gravity in the $-y$ direction.
• $T_y$ is the tension in the $-y$ direction $T_y=-F_g$ where $F_g$ is the force of gravity in the downward direction. Since it is $-F_g,$ the force of $T_y$ negates the force of gravity, therefore it is facing up. $T_y=T\sin\theta.$ It can be expanded into $T_y=(\Delta x)(W)(-g),$ where $g$ is the acceleration due to gravity ($-9.8$ m/s²).
• $T_y$ increases towards the endpoints of the catenary since the sum of the masses are greatest there.
• The total tension is then $T_{net} =\sqrt{T_x^2+T_y^2}.$

We have already all the ingredients: If we assume now that $y= f(x)$ represents the shape of the string, we can then write

 $y'=\tan \theta,$ (1)

since $\tan\theta$ denotes the total tension $T_{net}$ which is tangential to the curve at every point.

Model for Solution

• $T_y=T\sin\theta$
• $T_x=T\cos\theta=H$ (constant)
• $W= \text{linear weight density}$ (mg/d) =(kg/s²)
• $s= \text{arclength}$
• $\frac{ds}{dx}= \sqrt{1+\big(\frac{dy}{dx}\big)^2}$
• $dx=$ velocity component of $T_x,$ always horizontal to the curve.
• $dy=$ velocity component of $T_y,$ always pointing downward from the curve, in the $-y$ direction.

Sketch of the proof

Assume that the points $A$ and $B$ are at the same height. To find the distance $a$ (the radius of the curve from the line between $A$ and $B$ to $C$), compute $H/W=a.$ This is the horizontal component of tension over the linear weight density of the string. By this logic, $T\sin\theta=Wx$ at any specific point between $[A,B].$ As a function of the curve, $T\sin\theta=Ws$ shows the vertical tension of the rope along the whole curve (the main component that influences the shape of the curve since it is the only fluctuating force along the curve).

Summarizing: we are searching for a function $y=f(x)$ subject to the constraints

 $T\sin\theta=Ws$ (2) $T\cos \theta=H$ (3)

Substituting with (3) in (2) gives $H\tan \theta = Ws$ and, by our definition of $a$ and equation (1), it must be then $ay' = s.$ The model is found after taking derivatives on this expression, which gives the differential equation $ay''=\frac{ds}{dx}=\sqrt{1+(y')^2}.$

This is a second-order differential equation without the variable $x,$ and we can solve it by performing the substitution $p=y',\frac{dp}{dx}=y'':$

$\begin{array}{c} \displaystyle{a \dfrac{dp}{dx} = \sqrt{1+p^2}} \\ \\ \displaystyle{a\int \frac{dp}{\sqrt{1+p^2}} = x+C } \\ \\ a \sinh^{-1} p = x+C \\ \\ y' = \sinh \big( \frac{x+C}{a} \big) \\ \\ y = \displaystyle{\int \sinh \big(\tfrac{x+C}{a}\big) \, dx} = a\cosh \big(\tfrac{x+C}{a}\big) + D \end{array}$

So the catenary is nothing but a hyperbolic cosine. Note the vertical and horizontal shifts given by the constants of integration $D$ and $C$ respectively, as well as the different compression factors (vertical and horizontal) given by the quantity $a.$