---
title: "Cable 80 m: lowest point, length, tensions"
subject: "Theory of Structures I"
chapter: 7
key: "semester_4/theory_of_structures_i:cable-80m-different-levels-udl-1-5"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-80m-different-levels-udl-1-5
site: BCT Engineering
---

# Cable 80 m: lowest point, length, tensions

Theory of Structures I · Chapter 7: Suspension Cable Systems

- Asked in: 2070 Magh, 10 marks
- Video: [Cables - Problem No 2 (supported on different levels & subjected to Uniformly Distributed Load - UDL](https://www.youtube.com/watch?v=Wq6LFPrWAJs) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-80m-different-levels-udl-1-5
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7

## Question

A cable is hanging from two points A and B, 80 m apart horizontally, left end A being lower than the right end by 10 m. It supports a uniform load of 1.5 kN/m along the horizontal span. Determine: (i) the position of the lowest point if it sags 7.5 m, (ii) length of the cable, (iii) horizontal tension and tension at the two ends.

## Short answer

Lowest point is 31.65 m horizontally from the lower end A. Cable length is 85.09 m. Horizontal tension H = 100.18 kN; tension at lower end A is 110.86 kN and at higher end B is 123.68 kN.

## Answer

**Set-up.** A is the lower end and B is 10 m higher. The lowest point O sags 7.5 m below A, so B is $7.5 + 10 = 17.5$ m above O. Load $w$ = 1.5 kN/m on the horizontal span; $y = \dfrac{wx^2}{2H}$.

#### (i) Position of the lowest point
$$
\frac{x_A}{x_B} = \sqrt{\frac{7.5}{17.5}} = 0.6547,\qquad x_A + x_B = 80
$$
$$
x_B = \frac{80}{1+0.6547} = 48.35\ \text{m},\qquad x_A = 31.65\ \text{m}
$$
The lowest point is 31.65 m horizontally from the lower end A.

#### (iii) Horizontal tension and end tensions
$$
H = \frac{w x_A^2}{2y_A} = \frac{1.5(31.65)^2}{2(7.5)} = 100.18\ \text{kN}
$$
(Check from B: $\dfrac{1.5(48.35)^2}{2(17.5)} = 100.18$ kN.)

| End | Vertical $V = wx$ (kN) | $T = \sqrt{H^2+V^2}$ (kN) |
|---|---|---|
| A (lower) | 47.48 | 110.86 |
| B (higher) | 72.52 | 123.68 |

#### (ii) Length of the cable
With $s = 2y/x$ at each end and $l = \dfrac{x}{2}\left[\sqrt{1+s^2}+\dfrac{\sinh^{-1}s}{s}\right]$:

| Side | x (m) | y (m) | s | Length (m) |
|---|---|---|---|---|
| O to A | 31.65 | 7.5 | 0.4739 | 32.799 |
| O to B | 48.35 | 17.5 | 0.7239 | 52.289 |

Total length = 85.09 m.

**Answer:** (i) lowest point 31.65 m from A; (ii) cable length = 85.09 m; (iii) $H$ = 100.18 kN, $T_A$ = 110.86 kN, $T_B$ = 123.68 kN.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Lower end A is 10 m below B, and the lowest point sags 7.5 m below A. How high is B above the lowest point?
   A. 10 m
   B. 7.5 m
   C. 17.5 m
   D. 2.5 m
   Answer: C. B is 10 m above A, and A is 7.5 m above O, so y_B = 7.5 + 10 = 17.5 m. (Key point: Height of B above lowest point)

2. How are x_A and x_B found?
   A. x_A/x_B = 7.5/17.5, with x_A + x_B = 80
   B. x_A/x_B = sqrt(7.5/17.5) = 0.6547, with x_A + x_B = 80
   C. x_A = x_B = 40 m
   D. x_A/x_B = sqrt(17.5/7.5), with x_A + x_B = 80
   Answer: B. Because y is proportional to x^2, x_A/x_B = sqrt(y_A/y_B). This gives x_A = 31.65 m and x_B = 48.35 m, so the lowest point is nearer the lower end. (Key point: Position of lowest point)

3. Which calculation gives H, and what is the check?
   A. H = w x_A^2/(2 y_A) = 100.18 kN, and from B gives the same value
   B. H = w x_A/y_A = 6.3 kN
   C. H = w x_A^2/(2 y_A) from A only; B gives a different value because the ends are at different levels
   D. H = w (80)^2/(8 x 7.5) = 160 kN
   Answer: A. H is the same along the whole cable, so w x^2/(2y) from either side gives 100.18 kN. This is a useful check on x_A and x_B. (Key point: Horizontal tension)

4. Why is the tension at B (123.68 kN) greater than at A (110.86 kN)?
   A. B is higher, so its horizontal component is larger
   B. B is farther from the lowest point, so its vertical component V = wx (72.52 kN) is larger while H is the same
   C. B carries more load per metre
   D. The cable is longer on the A side
   Answer: B. H is constant (100.18 kN). T = sqrt(H^2 + V^2) with V = w x, and x_B > x_A, so V_B > V_A and T_B > T_A. (Key point: End tensions)

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