Hanging cables Imagine a cable, like a telephone line or

Chapter 7, Problem 83E

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Hanging cables     Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is a constant w and the horizontal tension at its lowest point is a vector of length \(H\). If we choose a coordinate system for the plane of the cable in which the \(x-axis\) is horizontal, the force of gravity is straight down, the positive \(y-axis\) points straight up, and the lowest point of the cable lies at the point \(y = H/w\) on the \(y-axis\) (see accompanying figure), then it can be shown that the cable lies along the graph of the hyperbolic cosine

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning “chain.”

a. Let \(P(x, y)\) denote an arbitrary point on the cable. The next accompanying figure displays the tension at \(P\) as a vector of length (magnitude) \(T\), as well as the tension \(H\) at the lowest point \(A\). Show that the cable’s slope at \(P\) is

b. Using the result from part (a) and the fact that the horizontal tension at \(P\) must equal \(H\) (the cable is not moving), show that \(T = wy\). Hence, the magnitude of the tension at \(P(x, y)\) is exactly equal to the weight of \(y\) units of cable.

Equation Transcription:

Text Transcription:

H

x-axis

y-axis

y = H/w

P(x,y)

P

T

H

A

T = wy

y