---
title: "Sliding ladder by instantaneous centre"
subject: "Engineering Mechanics"
chapter: 5
key: "semester_3/engineering_mechanics:sliding-ladder-instantaneous-centre"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/sliding-ladder-instantaneous-centre
site: BCT Engineering
---

# Sliding ladder by instantaneous centre

Engineering Mechanics · Chapter 5: Kinematics of Rigid Bodies

- Asked in: year not given, 6 marks
- Video: [4- ICR Method: Introduction to concept of ICR by using example of Ladder leaning against wall.](https://www.youtube.com/watch?v=z4TCCGVNZMY) — vishal galande
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/sliding-ladder-instantaneous-centre
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-5

## Question

A ladder AB of length 4 m has its end A resting against a smooth vertical wall and end B on a smooth horizontal floor. End A slides down the wall at 2 m/s when the ladder makes 60° with the floor. Locate the instantaneous centre and find the angular velocity of the ladder, the velocity of end B and the velocity of the mid-point G of the ladder.

## Short answer

Instantaneous centre I found by intersecting horizontal from A and vertical from B; I is 2 m horizontally from A and 3.464 m vertically from B. Angular velocity = 1.00 rad/s, velocity of B = 3.46 m/s, velocity of midpoint G = 2.00 m/s.

## Answer

**Given:** $L = 4$ m, $\theta = 60^\circ$ with the floor, $v_A = 2$ m/s downward along the wall.

#### Locating the instantaneous centre
$v_A$ is vertical, so the perpendicular to it at A is the horizontal line through A. $v_B$ is horizontal, so the perpendicular at B is the vertical line through B. These meet at I.

```text
   wall
    |A
    | \         I = instantaneous centre
  --+--\-----I     (horizontal from A,
    |   \G   |      vertical from B)
    |    \   |
    +-----\--B--- floor
```

Coordinates (origin at the wall-floor corner): A $(0,\ 3.464)$, B $(2,\ 0)$, so I $= (2,\ 3.464)$ m.

- $IA = L\cos60^\circ = 2.00$ m
- $IB = L\sin60^\circ = 3.464$ m
- $IG = L/2 = 2$ m (I, A, B and the corner form a rectangle, and G is the midpoint of the diagonal AB)

#### Angular velocity

$$
\omega = \frac{v_A}{IA} = \frac{2}{2.00} = 1.00\ \text{rad/s (counter-clockwise)}
$$

#### Velocities of B and G

$$
v_B = \omega \times IB = 1.00 \times 3.464 = 3.464\ \text{m/s (horizontal, away from the wall)}
$$

$$
v_G = \omega \times IG = 1.00 \times 2 = 2.00\ \text{m/s}
$$

$v_G$ is perpendicular to IG.

**Answer: I is 2 m horizontally from A and 3.464 m vertically from B; ω = 1.00 rad/s; v_B = 3.46 m/s; v_G = 2.00 m/s.**

## Check yourself

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

1. How is I located for the sliding ladder?
   A. Midpoint of AB
   B. Intersection of the horizontal line through A and the vertical line through B
   C. At the corner
   D. At G
   Answer: B. v_A is vertical so its perpendicular is horizontal; v_B is horizontal so its perpendicular is vertical. (Key point: Locating I)

2. With L = 4 m and 60 deg to the floor, IA is:
   A. L sin60 = 3.464 m
   B. L cos60 = 2 m
   C. L = 4 m
   D. L/2 tan60
   Answer: B. I, A, B and the corner form a rectangle, so IA = L cos60 = 2 m and IB = L sin60 = 3.464 m. (Key point: Distances)

3. The angular velocity of the ladder is:
   A. v_A x IA = 4
   B. v_A / IA = 1 rad/s
   C. v_A / IB = 0.58 rad/s
   D. v_A / L = 0.5 rad/s
   Answer: B. omega = v/distance from the IC = 2/2 = 1 rad/s. (Key point: Angular velocity)

4. The velocities of B and of the midpoint G are:
   A. v_B = 3.46 m/s horizontal; v_G = 2 m/s perpendicular to IG
   B. v_B = 2 m/s; v_G = 3.46 m/s
   C. v_B = 2 m/s; v_G = 1 m/s
   D. v_B = 3.46 m/s; v_G = 0
   Answer: A. v_B = omega IB = 3.464 m/s; IG = L/2 = 2 so v_G = 2 m/s. (Key point: Velocity of B and G)

## Related in Engineering Mechanics

- [Ladder equilibrium by virtual work](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aladder-virtual-work-friction) — Engineering Mechanics, chapter 1
- [Velocity analysis of slider-crank mechanism](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acrank-slider-velocity-numerical) — Engineering Mechanics, chapter 5
- [Locating the instantaneous centre](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ainstantaneous-centre-locating-methods) — Engineering Mechanics, chapter 5
- [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6
- [Cylinder rolling down an incline](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Acylinder-rolling-down-incline) — Engineering Mechanics, chapter 6
- [Block sliding on incline compressing spring](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ablock-incline-spring-compression) — Engineering Mechanics, chapter 3

## Related in other subjects

- [Velocities in plane motion by instantaneous centre](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aladder-sliding-wheel-rolling-velocities) — Theory of Mechanism and Machine I, chapter 7
- [Slider crank velocity by instantaneous centre method](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Aslider-crank-velocity-instantaneous-centre) — Theory of Mechanism and Machine I, chapter 7
- [Rod and spring, angular velocity at vertical](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-explain-impulse) — Applied Mechanics (Dynamics), chapter 6
- [ICR; angular velocity of bar BD](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aicr-bar-ab-bd-inches) — Applied Mechanics (Dynamics), chapter 4