---
title: "Ladder equilibrium by virtual work"
subject: "Engineering Mechanics"
chapter: 1
key: "semester_3/engineering_mechanics:ladder-virtual-work-friction"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ladder-virtual-work-friction
site: BCT Engineering
---

# Ladder equilibrium by virtual work

Engineering Mechanics · Chapter 1: Virtual Work

- Asked in: year not given, 5 marks
- Video: [Problem 8 on virtual work method,  ladder, Engineering mechanics](https://www.youtube.com/watch?v=O005SFVGCTs) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/ladder-virtual-work-friction
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-1

## Question

A uniform ladder AB, 5 m long and weighing 400 N, rests with end A on a rough horizontal floor and end B against a smooth vertical wall. The ladder makes an angle of 60° with the floor. Using the principle of virtual work, find the horizontal frictional force at A required to keep the ladder in equilibrium.

## Short answer

Give the ladder a virtual rotation δθ. Virtual work of friction and weight sum to zero: F L sinθ δθ – W (L/2) cosθ δθ = 0, giving F = W/(2 tanθ). With W=400 N, θ=60°, F = 115.5 N directed toward the wall.

## Answer

**Method:** the floor friction force $F$ at A does work only if A is allowed to move, so give the ladder a virtual rotation $\delta\theta$ and equate the total virtual work to zero. The normal reaction at A and the wall reaction at B (smooth) do no work because they are perpendicular to the displacements of A and B (B moves only along the wall, A only along the floor).

```text
        wall
         |\ B
         | \
         |  \  W at G
         |   \
         |  60\
   ------+------A------> floor
            F <-- (friction toward wall)
```

Let the ladder make angle $\theta$ with the floor and length $L = 5$ m. Take the horizontal distance of A from the wall as $x_A = L\cos\theta$ and the height of the centre of gravity G as $y_G = \tfrac{L}{2}\sin\theta$.

$$
\delta x_A = -L\sin\theta\,\delta\theta \qquad \delta y_G = \tfrac{L}{2}\cos\theta\,\delta\theta
$$

Friction at A acts toward the wall (opposite to the sense in which A tends to slip), i.e. in the $-x$ direction, so its virtual work is $-F\,\delta x_A$. The weight acts downward, so its virtual work is $-W\,\delta y_G$.

$$
\begin{aligned}
\delta U &= -F\,\delta x_A - W\,\delta y_G = 0 \\
F L \sin\theta\,\delta\theta &- W\tfrac{L}{2}\cos\theta\,\delta\theta = 0 \\
F &= \frac{W}{2\tan\theta}
\end{aligned}
$$

Substituting $W = 400$ N and $\theta = 60^\circ$:

$$
F = \frac{400}{2\tan 60^\circ} = \frac{400}{3.464} = 115.5\ \text{N}
$$

**Check by statics:** wall reaction $N_B = F$ (horizontal equilibrium). Moments about A: $N_B L\sin\theta = W\tfrac{L}{2}\cos\theta$, giving $N_B = 400/(2\tan 60^\circ) = 115.5$ N. This agrees.

**Answer: Friction force at A = 115.5 N, directed toward the wall.**

## Check yourself

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

1. In the virtual work solution of the ladder, why are the normal reaction at A and the wall reaction at B left out?
   A. They are equal in magnitude
   B. They are perpendicular to the displacements of A and B and so do no work
   C. They are smaller than the friction
   D. They act through the centre of gravity
   Answer: B. A moves only along the floor and B only along the smooth wall, so both reactions are perpendicular to their points' motion. (Key point: Which forces do work)

2. Which coordinates are differentiated to get the virtual displacements?
   A. x_A = L cos(theta) and y_G = (L/2) sin(theta)
   B. x_A = L sin(theta) and y_G = L cos(theta)
   C. x_A = L/2 cos(theta) and y_G = L sin(theta)
   D. x_A = L tan(theta) and y_G = L/2
   Answer: A. A is at horizontal distance L cos(theta) from the wall and G is at height (L/2) sin(theta). (Key point: Coordinates)

3. Setting the virtual work to zero gives which friction formula?
   A. F = W tan(theta) / 2
   B. F = W / (2 tan(theta))
   C. F = W sin(theta)
   D. F = 2W tan(theta)
   Answer: B. F L sin(theta) = W (L/2) cos(theta) gives F = W / (2 tan(theta)). (Key point: Result of the equation)

4. For W = 400 N and theta = 60 deg, what is the friction at A, and how is it checked?
   A. 200 N; by resolving along the ladder
   B. 115.5 N; by statics with N_B = F and moments about A
   C. 230.9 N; by taking moments about B only
   D. 346.4 N; by vertical equilibrium
   Answer: B. 400/(2 tan 60) = 115.5 N. Statics gives N_B = F from horizontal equilibrium and the same value from moments about A. (Key point: Numerical value and check)

## Related in Engineering Mechanics

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- [Virtual work and its principle](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Avirtual-work-principle-definition) — Engineering Mechanics, chapter 1
- [Work-energy principle, power and efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Awork-energy-principle-power-efficiency) — Engineering Mechanics, chapter 3
- [Oblique impact of ball on a wall](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aball-oblique-impact-wall) — Engineering Mechanics, chapter 4
- [Inertia force and lift problem](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ainertia-force-lift-problem) — Engineering Mechanics, chapter 2
- [Conservative force and pendulum energy](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Apendulum-energy-conservation-tension) — Engineering Mechanics, chapter 3

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