Find centroid of catenary (y=cosh(x))

[oddjobmj]

Homework Statement

A uniform chain hangs in the shape of the catenary y=cosh(x) between x=−1 and x=1

Find [itex]\bar{y}[/itex]

Homework Equations

∫[itex]\bar{y}[/itex][itex]\rho[/itex]ds=∫y[itex]\rho[/itex]ds

The Attempt at a Solution

I can define ds (some small segment of the arc) by [itex]\sqrt{1+sinh(x)^2}[/itex]dx.

Also, y is given as y=cosh(x)

If I take the integral above from -1 to 1 as:

∫cosh(x)*[itex]\sqrt{1+sinh(x)^2}[/itex]dx

I get ~2.8

The problem I have with this answer is that the maximum y of this catenary is y=cosh(1) which is below y=2. The minimum value of the catenary is y=cosh(0)=1 so the y centroid should be somewhere between those two points.

What am I doing wrong? Thanks!

 

[BvU]

What does your ##\int ds## yield ?

 

[oddjobmj]

What does your ##\int ds## yield ?

Arc length. This turns out to be about 2.4 between x=-1 and x=1.

edit: from my book: (The example they use is y=x^2 between 0 and 1.)

Am I supposed to go about calculating the integral and then setting it equal to the form with [itex]\bar{y}[/itex] in it to solve for [itex]\bar{y}[/itex]?

 

[Ray Vickson]

Arc length. This turns out to be about 2.4 between x=-1 and x=1.

edit: from my book: (The example they use is y=x^2 between 0 and 1.)

Am I supposed to go about calculating the integral and then setting it equal to the form with [itex]\bar{y}[/itex] in it to solve for [itex]\bar{y}[/itex]?

Isn't that exactly what the last equation says?

 

[dirk_mec1]

You have to solve for y_COG from the equation as written in your notes.

 

[LCKurtz]

If you factor the constant ##\bar y## out of the second equation of 3.6 and solve for ##\bar y## you will have the formula you need. You have just calculated part of it.

 

[BvU]

Arc length. This turns out to be about 2.4 between x=-1 and x=1.

Correct. And the 2.8 you had before divided by the 2.4 arc length gives you ##\bar y## as per your own relevant equation in post #1. Done !