THE HANGING CABLE PROBLEM FOR PRACTICAL …

1 Atlantic of , Number1, HANGING CABLE PROBLEM FOR PRACTICALAPPLICATIONSNeil chatterjee and Bogdan G. NitaDepartment of Mathematical SciencesMontclair State UniversityMontclair, NJ investigate the ` HANGING CABLE ' PROBLEM for PRACTICAL applica-tions. We focus on determining the minimum distance between two verticalpoles which will prevent a CABLE , HANGING from the top of these poles, to touchthe ground. We consider two set-ups, starting with the case of equal poles thengeneralizing to unequal poles. In both cases we assume that the only knownquantities are the heights of the poles and the length of the many PRACTICAL applications it is necessary to determine therelationship between the length of a CABLE HANGING from two vertical poles, theheight of the poles and the lowest distance between the CABLE and the ground.

2 Suchapplications include electrical or telephone cables which, when HANGING too low,could cause property damage or injuries and even deaths. In this paper we studythis relationship, write the equations describing it and solve them start by looking at a specific set-up: given that a CABLE of length 120mishanging from two poles of height 50m, what is the minimum distance betweenthe two poles which will prevent the CABLE from touching the ground? Obviouslythis PROBLEM can be stated in a different form involving taller poles and the lowestdistance between the CABLE and the ground: given that a CABLE of length 120mis HANGING from two poles of height 60m, what is the minimum distance betweenthe two poles which will prevent the CABLE from getting closer than 10mfrom theground?

3 After solving the PROBLEM in the first form, we give the general formulasfor calculating the minimum distance between the poles for any CABLE length andpoles height. Finally we generalize this PROBLEM to the case of unequal poles andfind the general formula for calculating the minimum distance between the polesfor any CABLE length and any two vertical perfectly flexible and inextensible CABLE of uniform density and cross sectionhanging freely from two poles assumes the shape of a catenary. The equation (upto translations and rotations) of a catenary in Cartesian coordinates has the form(see [3] section page 464)y=acosh(xa)(1)where cosh is the hyperbolic cosine function.

4 The scaling factorais usually inter-preted as the ratio between the horizontal component of the tension on the cableand the weight of the CABLE per unit length. For a given value of the parametera2000 Mathematics Subject Classi : 00A69; Secondary: 00A71, words and , HANGING shape of the catenary is known. However, in the PROBLEM we are investigating,we are regardingaas an unknown which depends on the distance between the polesthat the CABLE is HANGING from .In this paper we deal with the limiting case in which the catenary is tangent tothex-axis (the ground) and therefore we subtractato arrive at the equation wewill use in the problemy=acosh(xa) a:(2) history of the word catenary comes from the latinword catena which means chain.

5 The equation describing the curve has beenknown and well understood for a long his last book, published in 1638,Discourses and Mathematical DemonstrationsRelating to Two New Sciences(see [1]), Galileo made reference to the curve givenby a HANGING chain, and stated that a HANGING chain resembles a parabola, correctlyobserving that this approximation improves as the curvature gets smaller and isalmost exact when the elevation is less than 45 . A few years later the Germanmathematician Joachim Jungius proved that the shape of the HANGING CABLE is nota parabola. However he was unable to derive the equation of the curve.

6 His resultsand conclusions were published 12 years after his death, in 1691, three heavyweights of classical mathematical physics, Gottfried Wil-helm Leibniz, Christiaan Huygens, and Johann Bernoulli, derived the equation inresponse to a challenge by Jakob Bernoulli. Huygens first used the term catenaria in a letter to Leibniz in 1690, and David Gregory, a Scottish mathematician andastronomer, wrote a treatise on the catenary in 1690. However Thomas Jefferson isusually credited with the English word catenary .The HANGING CABLE derivation arises from analyzing it in the sense of a physicalproblem.

7 The only forces acting on a HANGING CABLE at a certain point are its weightand the tension in the CABLE . The resultant of these forces must equal to zeroconsidering the CABLE is at rest. By knowing their sum a differential equation ariseswith the unique solution of cosine hyperbolic. For a more detailed history and thederivation of the equation describing the shape of the catenary curve see reference[2]. case: equal that the length of the CABLE is 120mandthe two poles have equal height of 50m. We also assume that each pole is locatedat distancexfrom the midpoint which is assumed to be they-axis (see Figure 1).

8 With the given data, we can model the PROBLEM using two we consider the equation which calculates half of the length of the cableusing the arclength formula for the catenary described by the function in Equation(2): x0 1 + (dydt)2dt= 60:(3)Second, we consider the equation which describes the height of the poles or theheight of the CABLE at distancexfrom the midpointy(x) = 50:(4)72 NEIL chatterjee AND BOGDAN G. NITAx6050 Figure HANGING CABLE PROBLEM for equal poles: an exampleAfter integrating Equation (3) and substituting the expression foryfrom Equa-tion (2) into Equation (4) our two main equations becomeasinh(xa)= 60(5)andacosh(xa)= 50 +a:(6)We arrive at this system of equations with an objective: first it is necessary to findtheavalue which models the HANGING CABLE or catenary, and only then we can solveforx.

9 To solve the system, we divide both equations bya, then square both of themand subtract them. Using the hyperbolic identitycosh2(t) sinh2(t) = 1(7)we find(50 +aa)2 (60a)2= 1;(8)which givesa= 11:(9)Pluggingainto Equation (5) we can solve forxand findx= 11 ln(11) 26:375:(10)The distance between the two poles when the CABLE is tangent to the ground istherefore2x 52:75:(11)We can generalize this example and find a general method for calculating thedistance between two equal poles given the poles heightzand the CABLE length 2ywhen the CABLE barely touches the ground (See Figure 2). As before, we can writea two equations modelling our PROBLEM :asinh(xa) =y(12)andacosh(xa) a=z:(13)CATENARY73yxzFigure HANGING CABLE PROBLEM for equal poles: general caseIsolating each trigonometric hyperbolic term, squaring and subtracting, accordingto the identity in Equation (7), we find(z+aa)2 (ya)2= 1(14)Solving fora, we arrive ata=y2 z22z.

10 (15)We can now solve for the general form of 2x, the distance between poles, by pluggingtheavalue from Equation (15) into Equation (12) to find2x= 2aln(ya+ (ya)2+ 1)(16)or, after using the formula forafrom Equation (15),2x=(y2 z2) ln(2yzy2 z2+ 4y2z2y4 2y2z2+z4+ 1)z(17)As a final comment we notice that ify= 60 andz= 50 formulas (15) and (17) giveusa= 11 and 2x 52:75:just as case: unequal this section we investigate a similar problembut this time with poles of different heights. Assume that the length of the CABLE is140mand the two poles have heights of 50mand 70m(see Figure 3). Again, ourgoal is to determine the minimum distance between the two poles that will preventthe CABLE from touching the ground.