---
title: "Dynamic equilibrium; moment of block B on rod"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:dynamic-equilibrium-block-b-rod-moment-about-o"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/dynamic-equilibrium-block-b-rod-moment-about-o
site: BCT Engineering
---

# Dynamic equilibrium; moment of block B on rod

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2069 Asar, 2+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/dynamic-equilibrium-block-b-rod-moment-about-o
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Discuss dynamic equilibrium with an example. In figure below, the motion of block B on a smooth horizontal table is controlled by the rod which rotates about a vertical axis at O. At the position shown, B has the given values of velocity and acceleration relative to the rod which is rotating with the given values of angular velocity ω and angular acceleration α. If B weighs 60 N, what moment does it exert about O? [Figure: rod from O at 30° to the horizontal, block B at 10 cm from O on the rod, relative velocity V = 22.5 cm/sec, relative acceleration a = 15 cm/sec², α = 1 rad/sec², ω = 3 rad/sec.] OR: see the next question.

## Short answer

In dynamic equilibrium, an imaginary force -m a is added so that ΣF - m a = 0. For the block, transverse acceleration a_θ = rα + 2ṙω = 1.45 m/s², transverse force F_θ = 8.87 N. Moment exerted about O is r F_θ = 0.89 N·m, which the block exerts on the rod.

## Answer

#### Dynamic equilibrium
When a body is accelerating, adding an imaginary inertia force (−m a) to the actual forces makes the body appear to be in equilibrium (D'Alembert's principle): ΣF − m a = 0. Example: a lift moving upward with acceleration a carries a person of mass m. The real forces are the normal reaction N and the weight mg. With the inertia force m a added downward, N − mg − m a = 0, so N = m(g + a); the person feels heavier. A car taking a curve is similar: the centrifugal (inertia) force m v²/r balances the friction force at the tyres.

#### Moment of block B about O
Block B (W = 60 N) slides on a smooth horizontal table along the rod. The vertical forces cancel, and the horizontal forces on B have radial and transverse components. Data in polar coordinates: r = 10 cm = 0.1 m, ṙ = 22.5 cm/s = 0.225 m/s, r̈ = 15 cm/s² = 0.15 m/s², ω = θ̇ = 3 rad/s, α = θ̈ = 1 rad/s². Mass m = 60/9.81 = 6.116 kg.

- a_r = r̈ − rω² = 0.15 − 0.1(9) = −0.75 m/s²
- a_θ = rα + 2ṙω = 0.1(1) + 2(0.225)(3) = 1.45 m/s²
- Transverse force (from the rod) F_θ = m a_θ = 6.116 × 1.45 = 8.87 N
- Radial force F_r = m a_r = −4.59 N (it passes through O, so it has no moment about O)

Moment about O:

M_O = r F_θ = 0.1 × 8.87 = **0.887 N·m**

Check using H_O = m r²θ̇: dH_O/dt = m(r²θ̈ + 2rṙθ̇) = 6.116(0.01 + 0.135) = 0.887 N·m ✓.

**Answer: M_O ≈ 0.89 N·m.** The rod exerts this moment on B in the sense of the rotation; B therefore exerts an equal moment of 0.89 N·m on the rod about O in the opposite sense (opposing the rotation).

## Check yourself

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

1. In the lift example for dynamic equilibrium, a lift accelerating upward with a gives the normal reaction N equal to what?
   A. N = m(g - a)
   B. N = mg
   C. N = m(g + a)
   D. N = m a
   Answer: C. With inertia force m a added downward, N - mg - m a = 0, so N = m(g + a) and the person feels heavier. (Key point: Lift example)

2. Which gives the transverse acceleration of B on the rotating rod?
   A. a_theta = r alpha + 2 r_dot omega = 1.45 m/s^2
   B. a_theta = r alpha - r omega^2
   C. a_theta = r_ddot - r omega^2
   D. a_theta = r omega^2
   Answer: A. a_theta = 0.1(1) + 2(0.225)(3) = 1.45 m/s^2, including the Coriolis term. (Key point: Transverse acceleration)

3. Which force component gives a moment about O?
   A. The radial force, because it is larger
   B. The weight of B
   C. The transverse force only, since the radial force passes through O
   D. Neither, since the table is smooth
   Answer: C. The radial force acts along the line through O so its moment arm is zero. M_O = r F_theta. (Key point: Which force gives moment about O)

4. What is the moment of B about O?
   A. 8.87 N.m
   B. 0.887 N.m, from r x m x a_theta = 0.1 x 6.116 x 1.45
   C. 0.459 N.m
   D. 60 x 0.1 = 6 N.m
   Answer: B. m = 60/9.81 = 6.116 kg, F_theta = 8.87 N, and M_O = 0.1 x 8.87 = 0.887 N.m. Weight is ignored as it is vertical. (Key point: Moment value)

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