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

1. Sep 1, 2014

### oddjobmj

1. The problem statement, all variables and given/known data
A uniform chain hangs in the shape of the catenary y=cosh(x) between x=−1 and x=1

Find $\bar{y}$

2. Relevant equations
∫$\bar{y}$$\rho$ds=∫y$\rho$ds

3. The attempt at a solution
I can define ds (some small segment of the arc) by $\sqrt{1+sinh(x)^2}$dx.

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

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

∫cosh(x)*$\sqrt{1+sinh(x)^2}$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!

Last edited: Sep 1, 2014
2. Sep 1, 2014

### BvU

What does your $\int ds$ yield ?

3. Sep 1, 2014

### oddjobmj

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 $\bar{y}$ in it to solve for $\bar{y}$?

Last edited: Sep 1, 2014
4. Sep 1, 2014

### Ray Vickson

Isn't that exactly what the last equation says?

5. Sep 1, 2014

### dirk_mec1

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

6. Sep 1, 2014

### 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.

7. Sep 1, 2014

### BvU

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 !