Posted: February 19th, 2023

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SOLUTION

The differential equation for the problem of a cable under its own weight is:

y”(x) = -w(x)/T

where y(x) is the height of the cable at position x, w(x) is the weight density of the cable at position x, T is the tension in the cable, and the prime notation (‘) denotes differentiation with respect to x.

Assuming the cable is of uniform weight density, we can express w(x) as w(x) = w0, where w0 is a constant. Substituting this into the differential equation, we get:

y”(x) = -w0/T

Integrating both sides of the equation once with respect to x, we get:

y'(x) = -w0/T * x + C1

where C1 is a constant of integration. We can determine C1 by using the fact that the cable is hung at uniform height, which means that y'(0) = y'(L) = 0, where L is the length of the cable. Therefore, we have:

C1 = w0/T * L/2

Integrating both sides of the equation again with respect to x, we get:

y(x) = -w0/T * x^2/2 + C1 * x + C2

where C2 is another constant of integration. We can determine C2 by using the fact that the cable is hung at uniform height, which means that y(0) = y(L) = 0. Therefore, we have:

C2 = 0

Substituting the values of C1 and C2 into the equation, we get the function of the curve that will occur when the cable is hung by each end at uniform height:

y(x) = w0/T

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