---
title: "Moment of guy wire forces about tower base"
subject: "Applied Mechanics"
chapter: 3
key: "semester_1/applied_mechanics:guy-wires-tower-moment-about-base"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/guy-wires-tower-moment-about-base
site: BCT Engineering
---

# Moment of guy wire forces about tower base

Applied Mechanics · Chapter 3: Forces acting on particle and rigid body

- Asked in: 2074 Chaitra, 7 marks
- Video: [Statics | "A transmission tower is held by three guy wires attached to a pin at A and anchored..."](https://www.youtube.com/watch?v=bul_VYrPmv8) — Alex
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/guy-wires-tower-moment-about-base
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-3

## Question

Three guy wires are used in the support system for a television transmission tower that is 600 m tall. Wires A and B are tightened to a tension of 60 kN, whereas wire C has only 30 kN of tension. What is the moment of wire forces about the base O of the tower? The Y axis is collinear with AO. [Figure: tower of height 600 m (z axis) with base O; wires A, B, C anchored on the ground at 300 m from O; angles of 120° between the wire anchor lines, wire C at 30° from the x axis.]

## Short answer

Compute wire force components at the tower top using wire length 670.8 m. Sum the forces and take moment about O: M_O = 4025i − 6971j kN·m, magnitude 8050 kN·m, with no twisting moment about the tower axis.

## Answer

**Assumptions from the figure:** the tower is along z, height 600 m, top T(0, 0, 600). The three wires are anchored on the ground at 300 m from O, 120° apart in plan. The y axis is collinear with AO, so A is on the −y axis: A(0, −300, 0). Wire C is 30° above the x axis in plan, and B is symmetric with it: C(300 cos 30°, 300 sin 30°, 0) = (259.8, 150, 0), B(−259.8, 150, 0). Tensions: T_A = T_B = 60 kN, T_C = 30 kN.

#### Length of each wire
L = √(300² + 600²) = 670.8 m

#### Wire forces at the top (direction from top to anchor)
```text
F_A = 60 (0, −300, −600)/670.8      = (  0.00, −26.83, −53.67) kN
F_B = 60 (−259.8, 150, −600)/670.8  = (−23.24,  13.42, −53.67) kN
F_C = 30 (259.8, 150, −600)/670.8   = ( 11.62,   6.71, −26.83) kN
```
Resultant: ΣF = (−11.62, −6.71, −134.16) kN

#### Moment about the base O
All forces act at the top, r = (0, 0, 600) m, so
M_O = r × ΣF = (0, 0, 600) × (−11.62, −6.71, −134.16)
```text
Mx = 0×(−134.16) − 600×(−6.71)  = +4025 kN·m
My = 600×(−11.62) − 0×(−134.16) = −6971 kN·m
Mz = 0
```
|M_O| = √(4025² + 6971²) = 8050 kN·m

**Answer: M_O = 4025 i − 6971 j kN·m (magnitude 8050 kN·m). The moment vector lies in the horizontal xy plane, so there is no twisting moment about the tower axis.**

## Check yourself

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

1. How is each wire force written as a vector at the tower top?
   A. Tension × (vector from top to anchor)/(wire length 670.8 m)
   B. Tension × (vector from anchor to base)
   C. Tension × 600/300
   D. Tension divided by the wire length only
   Answer: A. Each force is its tension times the unit vector along the wire from the top to the anchor point. (Key point: Wire force vectors)

2. Why is each wire's length 670.8 m?
   A. sqrt(300^2 + 600^2): horizontal distance 300 m and height 600 m
   B. 300 + 600 divided by 2
   C. 600 / sin 30°
   D. Because the three wires are 120° apart
   Answer: A. The wire is the hypotenuse of a right triangle with legs 300 m and 600 m. (Key point: Wire length)

3. Why can M_O be found as (0,0,600) × ΣF?
   A. All three wire forces act at the tower top, so one position vector r = (0,0,600) m applies
   B. Because the wire forces cancel at the base
   C. Because the moment is only about the z axis
   D. Because ΣF is vertical
   Answer: A. The moment about O equals r × (sum of forces) when all forces share the same point of application. (Key point: Moment method)

4. Why is M_z = 0?
   A. r is along z, so r × F has no z component
   B. Because the three wires have equal tensions
   C. Because the resultant force is zero
   D. Because the base is fixed
   Answer: A. The cross product of a z-directed vector with any force is perpendicular to z, so the wires cause no twisting about the tower axis. (Key point: No twist)

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