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A flexible chain of length $L$ is suspended between two poles of equal height separated by a distance $2 M$ (Figure I3). By Newton's laws, the chain describes ...

Question

A flexible chain of length $L$ is suspended between two poles of equal height separated by a distance $2 M$ (Figure I3). By Newton's laws, the chain describes a catenary $y=a \cosh \left(\frac{x}{a}\right)$, where a is the number such that $L=2 a \sinh \left(\frac{M}{a}\right) .$ The sag $s$ is the vertical distance from the highest to the lowest point on the chain. Suppose that $L=160$ and $M=50$. (a) Use Newton's Method to find a value of $a$ (to two decimal places) satisfying $L=2 a

A flexible chain of length $L$ is suspended between two poles of equal height separated by a distance $2 M$ (Figure I3). By Newton's laws, the chain describes a catenary $y=a \cosh \left(\frac{x}{a}\right)$, where a is the number such that $L=2 a \sinh \left(\frac{M}{a}\right) .$ The sag $s$ is the vertical distance from the highest to the lowest point on the chain. Suppose that $L=160$ and $M=50$. (a) Use Newton's Method to find a value of $a$ (to two decimal places) satisfying $L=2 a \sinh (M / a)$. (b) Use Eq. (3) and the Linear Approximation to estimate the increase in sag $\Delta s$ for changes in length $\Delta L=1$ and $\Delta L=5$. (c) $\quad$ Compute $s(161)-s(160)$ and $s(165)-s(160)$ directly and compare with your estimates in (b).



Answers

Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve $ y = f(x) $ that satisfies the differential equation
$ \frac {d^2 y}{dx^2} = \frac {pg}{T} \sqrt {1 + (\frac {dy}{dx})^2} $
where $ p $ is the linear density of the cable, $ g $ is the acceleration due to gravity, $ T $ is the tension in the cable at its lowest point, and the coordinates system is chosen appropriately. Verify that the function
$ y = f(x) = \frac {T}{pg} \cosh (\frac {pgx}{T}) $
is a solution of this differential equation.


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