---
title: "D'Alembert's principle; gears with couple"
subject: "Applied Mechanics (Dynamics)"
chapter: 5
key: "semester_3/applied_mechanics_dynamics:dalembert-gear-a-b-couple-8nm"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/dalembert-gear-a-b-couple-8nm
site: BCT Engineering
---

# D'Alembert's principle; gears with couple

Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations

- Asked in: 2079 Baishakh, 4+8 marks
- Video: [D'Alemberts Principle for Plane motion of rigid body | Applied Mechanics Dynamics | Prashant YT |](https://www.youtube.com/watch?v=9tHx43nPIxg) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/dalembert-gear-a-b-couple-8nm
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5

## Question

Explain D' Alemberts principle with necessary equations. Gear A has a mass of 10 kg and a radius of gyration of 80 mm. The system is at rest when a couple M of magnitude 8 Nm is applied to gear B. Neglecting friction. Take $r_A = 250$ mm and $r_B = 100$ mm. Determine: a) The time required for the angular velocity of gear C [sic, gear B] to reach 600 rpm. b) The tangential force which gear B exerts on gear A. [Figure: large gear A (radius 250 mm) meshing with small gear B (radius 100 mm); couple M on B.]

## Short answer

D’Alembert’s principle adds reversed inertia force –m a at G and inertia couple –¯ I α to real forces, giving dynamic equilibrium: ∑F – m a = 0, ∑M_G – ¯ I α = 0. For the gears, the tangential force that gear B exerts on A is F = 80.00 N; the time for gear B to reach 600 rpm is t = 0.0804 s.

## Answer

**D'Alembert's principle** converts a dynamics problem into a statics problem. A body in motion is in dynamic equilibrium if, in addition to the real forces and moments, we add *inertia effects* equal and opposite to the effective forces:

- Inertia force $-m\bar{\vec a}$ acting at the mass centre $G$.
- Inertia couple $-\bar I\vec\alpha$ (for plane motion).

For a rigid body in plane motion the equations of dynamic equilibrium are

$$
\sum F_x - m\bar a_x = 0, \qquad \sum F_y - m\bar a_y = 0, \qquad \sum M_G - \bar I\alpha = 0
$$

which are the same as $\sum \vec F = m\bar{\vec a}$ and $\sum M_G = \bar I\alpha$, but written as "sum of forces including inertia terms is zero". The moment equation can then be taken about any point, as in statics. Special cases: pure translation ($\alpha=0$): only the inertia force; rotation about a fixed axis through $G$: only the inertia couple $-I\alpha$.

```text
   real forces + (-m a) at G + (-I alpha) couple  =>  equilibrium
```

#### Numerical part
Data: $m_A = 10$ kg, $\bar k_A = 0.080$ m, $r_A = 0.250$ m, $r_B = 0.100$ m, $M = 8$ N m on gear $B$. The data give no mass for gear $B$, so its inertia is neglected (if it is included, add $I_B$ in the equations below). The tangential force between the teeth is $F$.

$$
\bar I_A = m_A\bar k_A^2 = 10(0.080)^2 = 0.064\ \text{kg m}^2
$$

```text
        M (on B)
         ( B  )---F--->(tooth)
                      ( A )
```

#### (b) Tangential force on A
Gear $B$ (massless): $M - F r_B = 0$

$$
F = \frac{M}{r_B} = \frac{8}{0.100} = 80.00\ \text{N}
$$

#### (a) Time to reach 600 rpm
Gear $A$: $F r_A = \bar I_A\alpha_A$

$$
\alpha_A = \frac{F r_A}{\bar I_A} = \frac{80.00\times0.250}{0.064} = 312.50\ \text{rad/s}^2
$$

The teeth have the same tangential acceleration: $\alpha_B r_B = \alpha_A r_A$

$$
\alpha_B = \alpha_A\frac{r_A}{r_B} = 312.50\times\frac{0.250}{0.100} = 781.25\ \text{rad/s}^2
$$

$$
\omega_B = 600\ \text{rpm} = \frac{600\times2\pi}{60} = 62.83\ \text{rad/s}, \qquad t = \frac{\omega_B}{\alpha_B} = \frac{62.83}{781.25} = 0.0804\ \text{s}
$$

**Answer: $t = 0.0804$ s; tangential force $F = 80.00$ N.**

## Check yourself

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

1. D'Alembert's principle turns a dynamics problem into a statics problem by adding:
   A. Extra external loads
   B. Inertia force -m a at G and inertia couple -I alpha
   C. Friction forces
   D. Only the weight
   Answer: B. Inertia effects equal and opposite to the effective forces give dynamic equilibrium. (Key point: D'Alembert principle)

2. Gear B is treated as massless, so M - F r_B = 0. What is F?
   A. 80 N
   B. 32 N
   C. 800 N
   D. 8 N
   Answer: A. F = M / r_B = 8 / 0.100 = 80 N. (Key point: Tangential force)

3. With I_A = 0.064 kg m^2, F = 80 N, r_A = 0.25 m, alpha_A is:
   A. 312.5 rad/s^2
   B. 125 rad/s^2
   C. 781.25 rad/s^2
   D. 1250 rad/s^2
   Answer: A. alpha_A = F r_A / I_A = 20/0.064 = 312.5 rad/s^2. (Key point: Alpha of gear A)

4. How is alpha_B related to alpha_A (meshed teeth)?
   A. alpha_B = alpha_A
   B. alpha_B = alpha_A r_A / r_B
   C. alpha_B = alpha_A r_B / r_A
   D. alpha_B = alpha_A (r_A/r_B)^2
   Answer: B. Teeth have equal tangential acceleration: alpha_B r_B = alpha_A r_A, so alpha_B = 781.25 rad/s^2, giving t = 62.83/781.25 = 0.0804 s. (Key point: Gear ratio)

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