What is D’Alembert’s Principle? / D’Alembert’s Principle Mathematical Representation / Derivation of D’Alembert Principle

D’Alembert’s Principle is a powerful concept in mechanics that helps us analyze the motion of bodies by converting a dynamic problem into an equivalent equilibrium problem.

It states that for a system of particles, the difference between the applied forces and the inertial forces (mass × acceleration) is zero when considered along any virtual displacement.

This principle is also called the Lagrange–D’Alembert Principle, named after the French physicist and mathematician Jean le Rond d’Alembert.

Connection with Newton’s Second Law

According to Newton’s Second Law of Motion:

$$F = ma$$

D’Alembert rearranged this equation as:

$$F - ma = 0$$

Here:

• F is the real (applied) force

• −ma is the inertial force, a fictitious force that acts opposite to the acceleration

By introducing the inertial force, a moving body can be treated as if it is in equilibrium, making analysis much easier.

Mathematical Representation of D’Alembert’s Principle

D’Alembert’s principle can be written as:

$$\sum (F_i - m_i a_i) \cdot \delta r_i = 0$$

Where:

• i represents the (i)th particle in the system

• Fᵢ is the total applied force on the (i)th particle

• mᵢ is the mass of the (i)th particle

• aᵢ is the acceleration of the (i)th particle

• mᵢaᵢ represents the inertial force

• δrᵢ is the virtual displacement of the (i)th particle

Derivation of D’Alembert’s Principle 

1. Consider all forces acting on each particle in the system.

2. Move the inertial force $m a$ to the left side of the equation.

3. The system now appears to be in quasi-static equilibrium.

4. Apply the principle of virtual work.

5. Separate applied forces and constraint forces.

6. The total virtual work becomes zero, which is D’Alembert’s principle.

Examples of D’Alembert’s Principle

1. One-Dimensional Motion

For a body moving vertically:

$$T - W = ma \quad \text{or} \quad T = W + ma$$

Where:

• T = tension in the string

• W = weight of the body

• ma = inertial force

2. Two-Dimensional Motion

For a rigid body moving in an x-y plane:

$$F_i = -m r_c$$

Where:

• Fᵢ is the applied force

• m is the mass of the body

• rₙ is the position vector of the center of mass

Applications of D’Alembert’s Principle

D’Alembert’s principle combines virtual work and inertial forces and is widely used in mechanics. Some important applications include:

• Motion of a mass falling under gravity

• Parallel axis theorem

• Motion of a bead on a frictionless vertical hoop

• Analysis of complex mechanical systems

Frequently Asked Questions (FAQs)

Q1. What is the principle of virtual work?

It states that when a system is in equilibrium, the total virtual work done by all forces acting on it is zero.

Q2. What is inertial force?

Inertial force is a fictitious force that acts opposite to acceleration and is equal to mass × acceleration.

Q3. Why is D’Alembert’s principle important?

It simplifies dynamic problems by converting them into equilibrium problems, making them easier to solve.

Q4. What is virtual displacement?

A virtual displacement is an imaginary, infinitesimally small change in the position of a system that is consistent with the constraints.

Q5. Which quantities are related in Newton’s second law?

Newton’s second law relates two vector quantities: force and acceleration.

Solved Numerical Problems on D’Alembert’s Principle

Problem 1: Mass Moving Vertically (1D Motion)

A mass of 5 kg is lifted vertically upward with an acceleration of 2 m/s².

Find the tension in the string.

(Take $g = 9.8 \, \text{m/s}^2$)

Solution:

Using D’Alembert’s principle:

$$T - W = ma$$

Where:

• $m = 5 \, \text{kg}$
  

• $a = 2 \, \text{m/s}^2$

• $W = mg = 5 \times 9.8 = 49 \, \text{N}$
  

Substitute values:

$$T - 49 = 5 \times 2$$
$$T - 49 = 10$$
$$T = 59 \, \text{N}$$

Answer:

The tension in the string is 59 N.

Problem 2: Lift Moving Downward

A lift of mass 800 kg moves downward with an acceleration of 1.5 m/s².
Find the tension in the supporting cable.
($g = 9.8 \, \text{m/s}^2$)

Solution:

For downward acceleration:

$$T - W = -ma$$

Calculate weight:

$$W = mg = 800 \times 9.8 = 7840 \, \text{N}$$

Substitute values:

$$T - 7840 = - (800 \times 1.5)$$
$$T - 7840 = -1200$$
$$T = 6640 \, \text{N}$$

Answer:

The tension in the cable is 6640 N.

Problem 3: Block on a Smooth Horizontal Surface

A block of mass 10 kg is pulled on a smooth horizontal surface with an acceleration of 3 m/s².
Find the applied force using D’Alembert’s principle.

Solution:

According to D’Alembert’s principle:

$$F - ma = 0$$
$$F = ma$$

Substitute values:

$$F = 10 \times 3$$
$$F = 30 \, \text{N}$$

Answer:

The applied force is 30 N.

Problem 4: Block on a Rough Horizontal Surface

A block of mass 4 kg moves on a rough horizontal surface with an acceleration of 1 m/s².
The coefficient of friction is 0.25.
Find the applied force.
($g = 9.8 \, \text{m/s}^2$)

Solution:

Step 1: Calculate frictional force

$$f = \mu mg = 0.25 \times 4 \times 9.8 = 9.8 \, \text{N}$$

Step 2: Apply D’Alembert’s principle

$$F - f - ma = 0$$
$$F = f + ma$$
$$F = 9.8 + (4 \times 1)$$
$$F = 13.8 \, \text{N}$$

Answer:

The applied force is 13.8 N.

Problem 5: Body Sliding Down an Inclined Plane

A block of mass 6 kg slides down a smooth inclined plane making an angle of 30° with the horizontal.
Find the acceleration using D’Alembert’s principle.
($g = 9.8 \, \text{m/s}^2$)

Solution:

Component of weight along the plane:

$$mg \sin \theta = ma$$

Cancel $m$:

$$a = g \sin \theta$$
$$a = 9.8 \times \sin 30^\circ$$
$$a = 9.8 \times 0.5$$
$$a = 4.9 \, \text{m/s}^2$$

Answer:

The acceleration of the block is 4.9 m/s².