1. The problem statement, all variables and given/known data
when a cable with non-zero mass is connected to a pole at both ends, the shape it assumes is called a catenary.
it can be shown that for an electrical wire whose linear mass density is .9 kg/m strung between poles 30m apart(and making a 22 degree angle at each end) the mathematical equation is
y=(38.1m)[cosh(x/38.1m)-1]
a) an electircal wire of linear mass density .9 kg/m is strung, between poles 30m apart, from west to east on earth. initially the current flowing is zero and therefore the wires shape is that of a catenary. what is the weight of the wire?HINT: the length of the wire(which is not 30m) is found by integrating dl over the catenary. Use the fact the
dl=dx[sqrt(1+(dy/dx)^2) in order to have an integration over the x-axis.
It shows a picture with an equation for the wire. y=(38.1m)[cosh(x/38.1m)-1]
2. Relevant equations
equation for the wire ....y=(38.1m)[cosh(x/38.1m)-1]
length of the wire.... dl=dx[sqrt(1+(dy/dx)^2)
He also gave us all of the equations and proofs of hyperbolic functions.
3. The attempt at a solution
I was not sure what to do since i have never done a problem like this. I was neither shown in class how to do anything close to this.
I started with integrating the length of the wire function. Im having problems with the integration though. im not sure what to do after this or if its even right what im doing.
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Draw a straight line on the ground between the two telegraph poles. Mark off distances of DeltaX on it. You can then approximate the length of copper in the wire as a series of straight line segments of length DeltaL. Pythagoras:
DeltaL^2=DeltaX^2+DeltaY^2
DeltaL^2=DeltaX^2(1+(DeltaY/DeltaX)^2)
Obviously, the approximation improves as DeltaX gets smaller. In the limit of it being infintesimally small, each Delta becomes a d, and the equation above reduces to the one you gave in step 2. Because we are now working with infintesimals, the sum of all the elements becomes integration over x from one telegraph pole to the next.
You have an expression for y so you can get one for dy/dx. Sub that in and you have an integral in x that you should be able to do which gives you the length.
I was always taught dy/dx as a kind of arbitrary notation. Discovering that it was actually a division in some sense was something of a revelation.
By the way, any serious mathematician will wince at all of my explanation. Best not to tell them unless you want a lecture on calculus.
DeltaL^2=DeltaX^2+DeltaY^2
DeltaL^2=DeltaX^2(1+(DeltaY/DeltaX)^2)
Obviously, the approximation improves as DeltaX gets smaller. In the limit of it being infintesimally small, each Delta becomes a d, and the equation above reduces to the one you gave in step 2. Because we are now working with infintesimals, the sum of all the elements becomes integration over x from one telegraph pole to the next.
You have an expression for y so you can get one for dy/dx. Sub that in and you have an integral in x that you should be able to do which gives you the length.
I was always taught dy/dx as a kind of arbitrary notation. Discovering that it was actually a division in some sense was something of a revelation.
By the way, any serious mathematician will wince at all of my explanation. Best not to tell them unless you want a lecture on calculus.