Solving Cable Balance for T0 and T1

[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 = T_h / g’ ∙ (cosh (g’ ∙ x / T_h) - 1)

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

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

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

The horizontal force :
T_h = (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, T_h, 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
T_h = 0
T_v = G

T₀ = (T_h² + T_v²)^0.5 = G

Tension at the end of the cable T1
T_h = 0
T_v = G

T₁ = (T_h² + T_v²)^0.5 = G


2. g‘ ≠ 0

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

T₀ = (T_h² + T_v²)^0.5
T₀ = [(g’ / 8d) ∙ (S² – 4d²)]² + G²]^0.5

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

T₁ = (T_h² + T_v²)^0.5
T₁ = [(g’ / 8d) ∙ (S² – 4d²)]² + (g’ ∙ S / 2 + G)²]^0.5

 

[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?

 

[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 T_v

These equations come from the catenary principle, have a look below:
http://home.earthlink.net/~w6rmk/math/catenary.htm

 

[haruspex]

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

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?

 

[wah31]

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

 

[haruspex]

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

Yes.

 

[wah31]

Now if I summarize we get:

For g’ = 0

Tension in the middle of the cable T₀

T_h = (g’ / 8d) ∙ (S² – 4d²)=0

T_v = g’ ∙ S / 2 =0

T₀ = 0

Tension at the end of the cable T₁

T_h = (g’ / 8d) ∙ (S² – 4d²)=0

T_v = G

T₁ = (T_h² + T_v²)^0.5 = G

For g‘ ≠ 0

Tension in the middle of the cable T₀

T_h = (g’ / 8d) ∙ (S² – 4d²)

T_v = 0

T0 = (g’ / 8d) ∙ (S² – 4d²)

Tension at the end of the cable T₁

T_h = (g’ / 8d) ∙ (S² – 4d²)

T_v = g’ ∙ S / 2

T₁ = (T_h2 + T_v²)^0.5 =G

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

Is that correct ?

 

[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.

 

[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 ?