Nairaland gurus...over to you

videomaid:
Nairaland gurus...over to you

zero meter

Answer is 0m.

Proof: if you divide the rope into two halves you have 40m in length for one half, which is10m short the full length of one of the poles the only way to have 10m distance from the ground is if both poles have 0m distance apart.

Thus, if you move both poles apart the rope stretches and the distance from the ground becomes greater than 10m.

THE ANSWER IS 0m THEY REALLY WANT TO KNOW IF YOU ARE A SMART EMPLOYEE

My response would be 70m apart

Analysis:

First of all you have an imaginary rectangle shape, with lenght 50m and the breadth is unknown.

The breadth = The space between the poles.

But the breadth is separated by a slacked cable. If the cable was at full stretch, the breadth could easily have been 80m. But it has an excess (slack region) which is directly proportional to the length between the base of the rectangle and the curve tip of the cable= 10m

So I would say the breadth or space between the pole= 80m - 10m = 70m.

My view though.

unclebuddy:
My response would be 70m apart

Analysis:

First of all you have an imaginary rectangle shape, with lenght 50m and the breadth is unknown.

The breadth = The space between the poles.

But the breadth is separated by a slacked cable. If the cable was at full stretch, the breadth could easily have been 80m. But it has an excess (slack region) which is directly proportional to the length between the base of the rectangle and the curve tip of the cable= 10m

So I would say the breadth or space between the pole= 80m - 10m = 70m.

My view though.

very wrong

Oh boy you wicked ooo! Did you have to add 'very' to the 'wrong'.

We are all here to learn, please guide us on the correct answer.

videomaid:
Nairaland gurus...over to you

0

Either 20m apart in my opinion. Can't go into the calculation. Could be wrong.

dejt4u:
zero meter

I used to see your posts and marvel at your calculations.

I am terrible at mathematics but when it comes to writing... Whewwwww. My narration is 100%

SirWhiteFish:
I used to see your posts and marvel at your calculations.

I am terrible at mathematics but when it comes to writing... Whewwwww. My narration is 100%

thank you so much for this feedback..

Modified: The answer is 40m
The farthest distance that can be between the poles is 80m (The cable will be uniformly horizontal and 50m above the ground). For every 10m decrease in distance between the poles (initially from 80m apart), the height of the cable(lowest/curved part of the cable) above the ground (initially 50m when the cable was uniformly horizontal) also decreases by 10m, so that when this height is 10m, the distance between the poles is 40m. Arithmetic Progression will be useful in the calculations.

40m.

If the 50m length poles are stretched without any dip, the distance between them is 80m.

Therefore, for every 10m reduction in distance between the poles there is a 10m sag.

If the length of the pole is 50m and the bottom of the sag to the ground is 10m, then the dip is 50m-10m=40m.

Since there is 40m dip in the cable, the distance between the poles is 80-40=40m.

P.s to those claiming to be zero, this is wrong. If the height of the poles are 50m and the poles are placed 0m to one another, the cable must surely make contact with the ground since the cable is 30m longer than the pole.

Herard:
40m.

If the 50m length poles are stretched without any dip, the distance between them is 80m.

Therefore, for every 10m reduction in distance between the poles there is a 10m sag.

If the length of the pole is 50m and the bottom of the sag to the ground is 10m, then the dip is 50m-10m=40m.
other
Since there is 40m dip in the cable, the distance between the poles is 80-40=40m.
other
P.s to those claiming to be zero, this is wrong. If the height of the poles are 50m and the poles are placed 0m to one another, the cable must surely make contact with the ground since the cable is 30m longer than the pole.

I beg to differ, sir.

For every 10m closure of the gap, there is a 5m sag of the rope. Reason is that the sagging occurs equally from each pole. hence, 80m closure= 80/2 = 40. And 80m closure means zero gap between the poles.

D answer na zelo

40m+40m=80m^2

ThrustReverser:
Modified: The answer is 40m
The farthest distance that can be between the poles is 80m (The cable will be uniformly horizontal and 50m above the ground). For every 10m decrease in distance between the poles (initially from 80m apart), the height of the cable(lowest/curved part of the cable) above the ground (initially 50m when the cable was uniformly horizontal) also decreases by 10m, so that when this height is 10m, the distance between the poles is 40m. Arithmetic Progression will be useful in the calculations.

It's zero. Check again. It's a simple but tricky question.

divide (fold) the rope into 2 equal part = 40m
the length of the pole = 50m

length of pole minus half length of rope = 10m

so the only way the distance of rope from the ground will be 10m is if the rope are folded into 2 equal part making the distance of pole to be 0m

or simply put joining the 2 poles together.

dejt4u:
zero meter

Aizenosa:
Answer is 0m.

Proof: if you divide the rope into two halves you have 40m in length for one half, which is10m short the full length of one of the poles the only way to have 10m distance from the ground is if both poles have 0m distance apart.

Thus, if you move both poles apart the rope stretches and the distance from the ground becomes greater than 10m.

You guys are smart and the answer was prompt. Highly commendable. Keep it up.

Happy Sunday!!

Aizenosa:
Answer is 0m.

Proof: if you divide the rope into two halves you have 40m in length for one half, which is10m short the full length of one of the poles the only way to have 10m distance from the ground is if both poles have 0m distance apart.

Thus, if you move both poles apart the rope stretches and the distance from the ground becomes greater than 10m.

Nice. Please, under what topic in mathematics is the question?

Assuming the cable is a uniform flexible cable, it's shape is the graph of a hyperbolic cosine function. From my calculations, the poles are about 43.94449154672m apart.

The exact value is 20ln(9). I can show my workings if you want.

Total depth =50
Cable length=80
one-side cable Length L =80/2
L=40m
Cable depth V= 50-10
V=40m

V=cosh (h)
h= acosh(V)

H=2h

or

L=sinh (h)
h= asinh (L)
H=2h

guiddoti:
Total depth =50
Cable length=80
one-side cable Length L =80/2
L=40m
Cable depth V= 50-10
V=40m

V=cosh (h)
h= acosh(V)

H=2h

or

L=sinh (h)
h= asinh (L)
H=2h

Martinez39s:

. Did I missed it? I have no calculator for inverse cosh, I would have calculated it. But the question is catenary question.

All those saying zero, hahaha.

I am still wondering how you guys got zero.
The poles are obviously not on the same radar, or, am I mixing things up here

Well, I think the answer is 45

guiddoti:
. Did I missed it? I have no calculator for inverse cosh, I would have calculated it. But the question is catenary question.

Indeed, it is a catenary problem. The point of my post was that I don't understand your workings. I just can't grasp the sound reasoning behind your workings.

All those saying zero, hahaha.

Lmao. The first two guys to comment, especially the second guy, had me laughing hard.

Martinez39s:
Indeed, it is a catenary problem. The point of my post was that I don't understand your workings. I just can't grasp the sound reasoning behind your workings.

Lmao. The first two guys to comment, especially the second guy, had me laughing hard.

Okay. That's the step solution to the problem, though. You have idea to it too, when you mentioned hyperbolic function or so.

guiddoti:
Okay. That's the step solution to the problem, though. You have idea to it too, when you mentioned hyperbolic function or so.

Well, a catenary is shape exactly like the graph of a hyperbolic cosine function. The catenary in the op is the graph of

y = 10Cosh(x/10)

Martinez39s:
Well, a catenary is shape exactly like the graph of a hyperbolic cosine function. The catenary in the op is the graph of

Not right.

Catenary

Y= n/p(cosh (xp/n)-1)

n= axial force
p= cable weight
x= horizontal
Y= cable depth

where n/p is neglected(assumed too small)
or

L=n/p(sinh(xp/n)

L=cable length

Because it sag at the middle, total length is divided into two. L= total cable length/2.

guiddoti:
Not right.

Catenary

Y= n/p(cosh (xp/n)-1)

n= axial force
p= cable weight
x= horizontal
Y= cable depth

where n/p is neglected(assumed too small)
or

L=n/p(sinh(xp/n)

L=cable length

Because it sag at the middle, total length is divided into two. L= total cable length/2.

Martinez39s:

Actually, I have used the catenary a lot to model lazy wave riser and presently using it to model S-shaped pipeline.