---
title: "Dynamic equilibrium; equations of motion"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:dynamic-equilibrium-equation-of-motion"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/dynamic-equilibrium-equation-of-motion
site: BCT Engineering
---

# Dynamic equilibrium; equations of motion

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

- Asked in: 2068 Shrawan
- Video: [Enginering Mechanics Dynamics D'Alembert Principle 1](https://www.youtube.com/watch?v=UJb-Mdd4imk) — Mechanical Engineering by Ashish Purohit
- 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-equation-of-motion
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Define dynamic equilibrium. Also state equation of motion for rectilinear and curvilinear motion of particle.

## Short answer

Dynamic equilibrium means adding -m a to real forces gives ΣF - m a = 0. Rectilinear: ΣF = m a. Curvilinear: tangential ΣF_t = m dv/dt, normal ΣF_n = m v²/ρ; polar: ΣF_r = m(r̈ - rθ̇²), ΣF_θ = m(rθ̈ + 2ṙθ̇).

## Answer

#### Dynamic equilibrium
Newton's second law for a particle is ΣF = m a, which can be written as ΣF − m a = 0. If an imaginary **inertia force** −m a (equal in magnitude to mass × acceleration and opposite in direction to the acceleration) is added to the real forces, the body can be treated as being in equilibrium. This state is called **dynamic equilibrium** (D'Alembert's principle). The equations of statics can then be used to solve kinetics problems. Example: for a block pulled along a rough floor with acceleration a, P − F − m a = 0.

#### Equations of motion
**Rectilinear motion (along x):**
- ΣF = m a
- ΣFₓ = m ẍ = m dv/dt = m v dv/dx
(The sum of the forces perpendicular to the path is zero.)

**Curvilinear motion:**
- Rectangular components: ΣFₓ = m aₓ = m ẍ, ΣF_y = m a_y = m ÿ, ΣF_z = m a_z = m z̈
- Tangential and normal components: ΣF_t = m a_t = m dv/dt; ΣF_n = m a_n = m v²/ρ
- Radial and transverse components (polar): ΣF_r = m(r̈ − rθ̇²); ΣF_θ = m(rθ̈ + 2ṙθ̇)

where ρ is the radius of curvature of the path.

## Check yourself

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

1. What is the key idea of dynamic equilibrium?
   A. A body at rest has zero force
   B. Adding an inertia force -m a makes sum F - m a = 0 so statics can be used
   C. Forces always cancel in motion
   D. Velocity is constant
   Answer: B. The inertia force is equal to m a and opposite to acceleration (D'Alembert's principle). (Key point: Dynamic equilibrium)

2. Which pair gives the equations of motion in tangential and normal components?
   A. F_t = m v^2/rho, F_n = m dv/dt
   B. F_t = m dv/dt, F_n = m v^2/rho
   C. F_t = m v^2/rho, F_n = 0
   D. F_t = m r theta_ddot, F_n = m r theta_dot^2
   Answer: B. a_t = dv/dt along the path and a_n = v^2/rho toward the centre of curvature. (Key point: Tangential and normal equations)

3. Which is the radial equation of motion?
   A. F_r = m(r_ddot + r theta_dot^2)
   B. F_r = m(r theta_ddot + 2 r_dot theta_dot)
   C. F_r = m(r_ddot - r theta_dot^2)
   D. F_r = m r_dot
   Answer: C. F_r = m(r_ddot - r theta_dot^2) and F_theta = m(r theta_ddot + 2 r_dot theta_dot). (Key point: Polar equations)

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