 Posers and Puzzles

1. 14 Mar '07 13:571 edit
A cable, 16 metres in length, hangs between two pillars that are both 15 metres high. The ends of the cable are attached to the tops of the pillars. At its lowest point, the cable hangs 7 metres above the ground.

How far are the two pillars apart?
2. 14 Mar '07 15:51
Originally posted by Mathurine
A cable, 16 metres in length, hangs between two pillars that are both 15 metres high. The ends of the cable are attached to the tops of the pillars. At its lowest point, the cable hangs 7 metres above the ground.

[b]How far are the two pillars apart?
[/b]
Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

7 = (a/2)*(e^0 + e^0)

7 = a

Now, subbing in (d/2, 15), we get:

15 = (7/2)*(e^(d/14) + e^(-d/14))

(30/7) = e^(d/14) + e^(-d/14)

e^(d/14)*(30/7) = e^(2d/14) + e^(0)

e^(2d/14) - (30/7)*e^(d/14) + 1 = 0

e^(d/14) = (30/7) +/- (SQRT((30/7)^2 - 4))/2 = k

d/14 = ln(k)

d = 14*ln(k) = +/- 19.541 (approx.)

"d" cannot be negative (although in this case it doesn't matter since the negative value gives an answer identical to the positive value), so the distance between the pillars must be +19.541 (approx.). However, since the cable is only 16 metres long, it is not possible to string it up between the two pillars. (I think the constants need some adjustment 😉)
3. 15 Mar '07 01:44
Originally posted by PBE6
Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

7 = (a/2)*(e^0 ...[text shortened]... ble to string it up between the two pillars. (I think the constants need some adjustment 😉)
Maybe it's a big rubber band cable🙂
4. 15 Mar '07 05:34
I've seen one like this before, and you don't need anything other than your basic add/subtract.

The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
5. 15 Mar '07 09:53
Originally posted by geepamoogle
I've seen one like this before, and you don't need anything other than your basic add/subtract.

The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
Haha, what?
6. 15 Mar '07 10:08
Originally posted by geepamoogle
I've seen one like this before, and you don't need anything other than your basic add/subtract.

The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
Nice one!
7. 15 Mar '07 14:01
Originally posted by geepamoogle
I've seen one like this before, and you don't need anything other than your basic add/subtract.

The 16 metre cable hangs down 8 metres, and hence there is only one easy-to-find solution (no special math functions needed).
Geez! Two goof-ups in a row. I was really off yesterday. 😳
8. 17 Mar '07 00:462 edits
Ok got the answer and yes it is really stupid. The poles are touching hence the wire hangs down 16/2 = 8m so 7m from the ground
9. 17 Mar '07 02:08
Originally posted by PBE6
Cable hanging under their own weight take the shape of a catenary, which is expressed mathematically as the hyperbolic cosine function:

f(x) = a*cosh(x/a) = (a/2)*(e^(x/a) + e^(-x/a))

We centre the low point of the catenary over the origin (x = 0), at a height of 7. The pillars lie at x = -(d/2) and x = (d/2). Subbing in (0,7), we get:

7 = (a/2)*(e^0 ...[text shortened]... ble to string it up between the two pillars. (I think the constants need some adjustment 😉)
10. 17 Mar '07 02:21
11. 22 Mar '07 15:092 edits