# Comment on previous posted problem

#### Jomo

##### Elite Member

Some agreed that the distance between the two poles is 0. I disagree with that as I think that the distance between the two poles is twice the thickness of the cable. Am I again being too picky?

#### topsquark

##### Full Member
I dont' know anything about the thickness of the cable, but any hanging cord (wire, chain, etc.) is a section of a catenary curve. It can also be closely approximated by a parabola, which is much easier to integrate to get the arc length.

-Dan

#### Jomo

##### Elite Member
I dont' know anything about the thickness of the cable, but any hanging cord (wire, chain, etc.) is a section of a catenary curve. It can also be closely approximated by a parabola, which is much easier to integrate to get the arc length.

-Dan
Dan, why approximate the arc length as we are giving that length. It is 80 m long.
I am wondering why you say the hanging cable is a section of a catenary curve?
Steve

#### Dr.Peterson

##### Elite Member
If we take the cable's thickness (T) into account, then when the poles are two thicknesses apart (assuming that is measured between the near sides), we also have to take into account the semicircle formed by the lowest part of the cable (assuming it is the centerline of the cable that is exactly 80 meters long, and it is sufficiently flexible to bend that sharply). I figure the cable will then take up a length of 40 - pi/2 T + T, so the lowest point on the cable will be a little higher than 10 meters. That is, the bend takes up more cable than it contributes to the vertical fall.

So there may in fact be no solution (unless you borrow a physicist's zero-thickness cable).

On the other hand, do we know how the cable is attached at the top?

#### MarkFL

##### Super Moderator
Staff member
(unless you borrow a physicist's zero-thickness cable).
That's what I did.

#### Dr.Peterson

##### Elite Member
If we take the cable's thickness (T) into account, then when the poles are two thicknesses apart (assuming that is measured between the near sides), we also have to take into account the semicircle formed by the lowest part of the cable (assuming it is the centerline of the cable that is exactly 80 meters long, and it is sufficiently flexible to bend that sharply). I figure the cable will then take up a length of 40 - pi/2 T + T, so the lowest point on the cable will be a little higher than 10 meters. That is, the bend takes up more cable than it contributes to the vertical fall.
...
So there may in fact be no solution (unless you borrow a physicist's zero-thickness cable).

On the other hand, do we know how the cable is attached at the top?
I used the word "may" intentionally -- something felt wrong, because my conclusion was the opposite of my original thought. It turns out that I made a mistake in the calculation; it should have been 40 - pi/4 T + T, which is more than 40, not less than 40.

So the real answer is that at two thicknesses apart, the cable hangs a little too low, and you actually would need to move the poles further apart, which requires working with the arc length of the catenary.

#### yma16

##### Junior Member
Why do people spend time on this? The answer is 0.

#### Dr.Peterson

##### Elite Member
It's a joke ... and, yes, sometimes people waste time on jokes.

On the other hand, it's not a bad idea to take some time occasionally reminding ourselves that mathematical solutions are different from real-world solutions, and the latter can be a lot harder (typically requiring numerical methods). In this case, the lesson is, the more real you try to make it, the more difficulties you find.

#### yma16

##### Junior Member
If they change 10 to 9, it is really a joke.

#### Otis

##### Senior Member
… distance between the two poles is 0. I disagree … I think [it's] twice the thickness of the cable …
$$\;$$The cable might be attached to the "front" of each pole (similarly to having rectangular-solid poles). If such poles were brought together, the cable would hang in front of them, instead of inbetween.
$\;$

#### Jomo

##### Elite Member
$$\;$$The cable might be attached to the "front" of each pole (similarly to having rectangular-solid poles). If such poles were brought together, the cable would hang in front of them, instead of inbetween.
$\;$
Clever thought! Good for you. However the picture does not show that.

#### Dr.Peterson

##### Elite Member
Actually, that's what the picture looks like to me! The cables seem to be attached at the center of the top of each pole, and if they don't fall down along the front (or back) of each, then there must be a slot down the middle.

But this means that if the cable has zero thickness, then there is no solution, because the top ends of the cable can't coincide, and the lowest part can't be at 10 meters. We have to ignore the width of the poles, as well as the thickness of the cable, for the conventional "trick" solution. (I think we all knew that we were doing so anyway.)