## Poll

5 votes (26.31%) | |||

2 votes (10.52%) | |||

1 vote (5.26%) | |||

3 votes (15.78%) | |||

7 votes (36.84%) | |||

4 votes (21.05%) | |||

2 votes (10.52%) | |||

5 votes (26.31%) | |||

2 votes (10.52%) | |||

4 votes (21.05%) |

**19 members have voted**

Quote:unJonWhile I too have heard that girth matters, it appears that length trumps all, when it comes to the catenary position.

I have heard girth is just as important, if not more.

If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:

The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.

f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)

f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh

^{-1}5

Excel says this is 22.9243, so the distance between the poles would be 45.8486

However, I am not sure if that value is in degrees or radians.

That’s the answer I got as well.Quote:ThatDonGuyI tried to figure this one out, but something just doesn't make sense.

If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:

The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.

f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)

f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh^{-1}5

Excel says this is 22.9243, so the distance between the poles would be 45.8486

However, I am not sure if that value is in degrees or radians.

Quote:ThatDonGuyI tried to figure this one out, but something just doesn't make sense.

If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:

The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.

f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)

f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh^{-1}5

Excel says this is 22.9243, so the distance between the poles would be 45.8486

However, I am not sure if that value is in degrees or radians.

Here's the Wizard original question:

Quote:WizardA 100 meter rope is suspended from the top of two 50-meter poles. The lowest point of the rope is 10 meters from the ground. How far apart are the poles?...

Here's a hint on how to check your solution: If you change the 50-meter poles to 40-meter poles, the lowest point of the rope would then be tangent to the ground. But the distance between the poles would remain the same. When we solve this problem, the answer should be the same as for the 50-meter poles.

Quote:ChesterDogHere's a hint on how to check your solution: If you change the 50-meter poles to 40-meter poles, the lowest point of the rope would then be tangent to the ground. But the distance between the poles would remain the same. When we solve this problem, the answer should be the same as for the 50-meter poles.

That's another reason I'm pretty sure I'm going about it the wrong way; if I use f(x) = a cosh(x/a), but f(0) = 0, then a = 0 - but a can't be zero because then x/a is undefined.

Quote:ThatDonGuyThat's another reason I'm pretty sure I'm going about it the wrong way; if I use f(x) = a cosh(x/a), but f(0) = 0, then a = 0 - but a can't be zero because then x/a is undefined.

The Wizard gave us the general equation for a catenary as y = a cosh(x/a) - a. You can see that this is the equation for the catenary when it is tangent to the ground since when x = 0, then y = 0.

I worked on the problem many hours before giving up and Googling. The solution is so obvious to me in hindsight.

f(x) = a cosh(x/a) - a

f'(x) = sinh(x/a)

Let 2d be the distance between the poles; the rope goes through (0,0) and (d,40)

f(0) = 0

a (cosh (d/a) - 1) = 40

a cosh (d/a) = 40 + a

a

^{2}cosh

^{2}(d/a) = a

^{2}+ 80a + 1600

a

^{2}(cosh

^{2}(d/a) - 1) = 80a + 1600

a

^{2}sinh

^{2}(d/a) = 80a + 1600

50 = INTEGRAL(0,d) {sqrt(1 + sinh

^{2}(x/a)) dx}

50 = a INTEGRAL(0,d) {cosh(x/a)) d(x/a)}

50 = a INTEGRAL(0,d/a) {cosh t dt}

50 = a sinh (d/a)

2500 = a

^{2}sinh

^{2}(d/a)

Thus 80a + 1600 = 2500 -> a = 45/4

45/4 cosh (4d/45) = 205/4

cosh (4/45 d) = 205/45 = 41/9

d = 45/4 cosh

^{-1}41/9

Distance between poles = 2d = 49.4376

Quote:ThatDonGuyIn that case...

f(x) = a cosh(x/a) - a

f'(x) = sinh(x/a)

Let 2d be the distance between the poles; the rope goes through (0,0) and (d,40)

f(0) = 0

a (cosh (d/a) - 1) = 40

a cosh (d/a) = 40 + a

a^{2}cosh^{2}(d/a) = a^{2}+ 80a + 1600

a^{2}(cosh^{2}(d/a) - 1) = 80a + 1600

a^{2}sinh^{2}(d/a) = 80a + 1600

50 = INTEGRAL(0,d) {sqrt(1 + sinh^{2}(x/a)) dx}

50 = a INTEGRAL(0,d) {cosh(x/a)) d(x/a)}

50 = a INTEGRAL(0,d/a) {cosh t dt}

50 = a sinh (d/a)

2500 = a^{2}sinh^{2}(d/a)

Thus 80a + 1600 = 2500 -> a = 45/4

45/4 cosh (4d/45) = 205/4

cosh (4/45 d) = 205/45 = 41/9

d = 45/4 cosh^{-1}41/9

Distance between poles = 2d = 49.4376

Nice! That agrees with CrystalMath's answer.