# Homework Help: Single cable

1. Jan 2, 2017

### wah31

Mod note: Moved from technical forum, so missing the HW template.
Hello everybody and happy new year

Here is a solution for the following case. Do you agree with ? Thank you in advance for your answer

The question is to give the values of T0 T1 and d in terms of g', L and G in these 2 cases:
1. g'=0
2. g‘ ≠ 0

See attached file..

Proposition:
Cable balance (without counterweight):

The y coordinate (height) of any point in terms of the horizontal force :
y = Th / g’ ∙ (cosh (g’ ∙ x / Th) - 1)

The cable length from origin (center):
s = Th / g’ ∙ sinh(g’ ∙ x / Th)

The total cable length:
S = (2Th / g’) ∙ sinh (g’ ∙ L / 2Th)

The cable span:
L = (2Th / g’) ∙ arcsinh (S ∙ g’/ 2Th)

The horizontal force :
Th = (g’ / 8d) ∙ (S2 – 4d2)
This tension is constant over the entire length of the cable.

The vertical force :
Tv = g’ ∙ S / 2
This tension increases with the height (from the center to the end of the cable)

The minimum tension is, of course, Th, at the center point where the cable doesn't support any of it's own weight.

Cable balance (with counterweight)

1. g’ = 0

Tension in the middle of the cable T0
Th = 0
Tv = G

T0 = (Th2 + Tv2)0.5 = G

Tension at the end of the cable T1
Th = 0
Tv = G

T1 = (Th2 + Tv2)0.5 = G

2. g‘ ≠ 0

Tension in the middle of the cable T0
Th = (g’ / 8d) ∙ (S2 – 4d2)
Tv = G

T0 = (Th2 + Tv2)0.5
T0 = [(g’ / 8d) ∙ (S2 – 4d2)]2 + G2]0.5

Tension at the end of the cable T1
Th = (g’ / 8d) ∙ (S2 – 4d2)
Tv = g’ ∙ S / 2 + G

T1 = (Th2 + Tv2)0.5
T1 = [(g’ / 8d) ∙ (S2 – 4d2)]2 + (g’ ∙ S / 2 + G)2]0.5

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2. Jan 2, 2017

### haruspex

If T1 is not equal to G, what will happen?

Is g' the weight per unit length of the rope? Please explain where your starting equations come from. Did you derive these or are you taking them as standard?

3. Jan 3, 2017

### wah31

That’s right, g’ is the cable weight per unit length

I think G is there to balance the cable weight i.e. the vertical component of tension Tv

These equations come from the catenary principle, have a look below:

4. Jan 3, 2017

### haruspex

When a rope passes over a stationary, frictionless pulley, what is the usual relationship between the tension on the one side and the tension on the other?

5. Jan 3, 2017

### wah31

It is the same ..
You mean that T1 is equal to G?

6. Jan 3, 2017

Yes.

7. Jan 3, 2017

### wah31

Now if I summarize we get:

For g’ = 0

Tension in the middle of the cable T0

Th = (g’ / 8d) ∙ (S2 – 4d2)=0

Tv = g’ ∙ S / 2 =0

T0 = 0

Tension at the end of the cable T1

Th = (g’ / 8d) ∙ (S2 – 4d2)=0

Tv = G

T1 = (Th2 + Tv2)0.5 = G

For g‘ ≠ 0

Tension in the middle of the cable T0

Th = (g’ / 8d) ∙ (S2 – 4d2)

Tv = 0

T0 = (g’ / 8d) ∙ (S2 – 4d2)

Tension at the end of the cable T1

Th = (g’ / 8d) ∙ (S2 – 4d2)

Tv = g’ ∙ S / 2

T1 = (Th2 + Tv2)0.5 =G

T1 = [(g’ / 8d) ∙ (S2 – 4d2)]2 + (g’ ∙ S / 2 )2]0.5 = G

Is that correct ?

8. Jan 3, 2017

### haruspex

You are getting nonsense answers for the easy case, g'=0. How can the tension in the middle be zero?
I think you are using equations that don't allow for a source of tension other than the weight of the cable.

In theother case, you have S in the answer. You need to eliminate that.

9. Jan 8, 2017

### wah31

I tried another approach that I believe is the right (see attached file):

For g’ = 0

Tension in the middle of the cable T0
H = (g’ / 8d) ∙ (S2 – 4d2)=0
V = G (the cable doesn't support any of it's own weight)
T0 = G

Tension at the end of the cable T1
H = (g’ / 8d) ∙ (S2 – 4d2)=0
V = G
T1 = G

It means that the tension is the same over the entire length of the cable (=G)

For g‘ ≠ 0

Tension in the middle of the cable T0
H = (g’ / 8d) ∙ (S2 – 4d2)
V = 0 (the cable doesn't support any of it's own weight)
T0 = (g’ / 8d) ∙ (S2 – 4d2)

Tension at the end of the cable T1
H = (g’ / 8d) ∙ (S2 – 4d2)
V = g’ ∙ S / 2
T1 = (H2 + V2)0.5 = G

T1 = [(g’ / 8d) ∙ (S2 – 4d2)]2 + (g’ ∙ S / 2 )2]0.5 = G

Now I need actually to eliminate S and present equations in terms of L ?

Do you have an idea ?

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