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Hanging Cables Consider a portion of cable At lowest point of cable, a horizontal force H acts to stop the cable moving to the right H W(x) At centre of mass of cable a weight force W acts At point (x,y) there is a tension force T tangential to the cable T (x,y)(x,y) Take origin at lowest point. x y

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T Since system is in horizontal and vertical equilibrium As the gradient at (x,y) is given by H W(x) (x,y)(x,y)

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But W varies as x changes, so differentiating Fundamental Equation H W(x) (x,y)(x,y)

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The catenary A flexible rope of constant density hangs between two points. Determine its shape. Let w be the mass per unit length of the rope where s is the length of rope The fundamental equation is The fundamental equation becomes

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Short section of cable ds dx dy Consider a short section of cable, so short that it is effectively a straight line

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Fundamental equation Taking the origin at the lowest point what are the initial conditions?

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Separate variables

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Solve Separate variables and integrate

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What is the length of the cable?

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If we measure arc length from the origin Separate variables and integrate

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Total length L D

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L D If we want a cable to span a distance L Knowing the two properties of the cable, weight density and the tension that it can withstand We can work out the length of cable required and the droop of the cable.

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Clicker Question 1 If x = e 2t + 1 and y = 2t 2 + t, then what is y as a function of x ? – A. y = (1/2)(ln 2 (x – 1) + ln(x – 1)) – B. y = ln 2 (x – 1)

Clicker Question 1 If x = e 2t + 1 and y = 2t 2 + t, then what is y as a function of x ? – A. y = (1/2)(ln 2 (x – 1) + ln(x – 1)) – B. y = ln 2 (x – 1)

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