D’Alembert’s principle

1. Bring acceleration into virtual work

The virtual-work principle found the condition for an Atwood machine to remain in equilibrium: equal masses. When the masses are unequal, the system accelerates. The geometry and tension cancellation have not changed. What changes is the balance equation from which we start.

D’Alembert's principle uses Newton's equation and the same virtual projection to obtain dynamics without explicitly solving for ideal reactions. This is the final step of Unit 1 and the direct conceptual bridge to Lagrangian mechanics.

Prerequisites: Newton's second law, Chapters 4–5, and the pendulum tangent direction. All calculations here use an inertial frame and constant particle masses.

In equilibrium, applied forces have no resultant tendency along allowed directions. In motion, that tendency is not zero: it produces mass times acceleration. D’Alembert's principle projects their difference onto the allowed directions.

2. Derivation of D’Alembert’s principle

Consider constant-mass particles in an inertial frame. Write each total force as the sum of an applied force $\mathbf F_a$ and constraint reaction $\mathbf R_a$. Assume the reactions have zero total virtual work for all admissible virtual displacements.

For particle $a$,

Move the acceleration term to the force side:

This has the algebraic appearance of force balance. The term $-m_a\mathbf a_a$ is sometimes called D’Alembert's inertial force. In this inertial-frame derivation it is not a new interaction from another body, nor a Newton's-third-law partner. It is the acceleration term rewritten.

Take the dot product with $\delta\mathbf r_a$ and sum over particles:

The last sum is zero by ideality. The result is

This equation must hold for every admissible virtual displacement at the instant considered. With constant mass, $m_a\mathbf a_a=\dot{\mathbf p}_a$, so the momentum form replaces $m_a\mathbf a_a$ by $\dot{\mathbf p}_a$.

Only the components of applied force minus momentum change along permitted directions survive. Normal reactions disappear because their virtual work vanishes, not because the motion has no normal acceleration.

At static equilibrium, every acceleration is zero and the formula reduces to the principle of virtual work. For an unconstrained particle, $\delta\mathbf r$ may point in any direction. Therefore the vector coefficient must vanish, recovering $\mathbf F=m\mathbf a$.

3. Generalized-coordinate form

Assume independent generalized coordinates for ideal holonomic constraints. Substitute $\delta\mathbf r_a=\sum_i(\partial\mathbf r_a/\partial q_i)\delta q_i$ into D’Alembert's principle and collect variations:

Since the $\delta q_i$ are independent, each coefficient must vanish:

This is a usable equation of motion. The expression on the right is an inertial contribution appropriate to the chosen coordinate. It is not always simply “mass times $\ddot q_i$.” Angles introduce length factors, curved coordinates introduce geometrical terms, and coupled coordinates can involve several accelerations.

For dependent or non-holonomic variations, one cannot set every displayed coefficient to zero independently. That shortcut is restricted to independent generalized coordinates with the assumed admissible variations.

4. Applications

Example 1 · Horizontal mass–spring motion

A mass $m$ moves on a smooth horizontal line and is attached to a spring of stiffness $k$. Let $x$ denote its displacement from equilibrium. The spring supplies the only force component along the permitted motion.

Use $\delta\mathbf r=(\delta x,0)$; vertical weight and support reaction contribute no work along it. The horizontal force is $-kx$. D’Alembert's principle gives $(-kx-m\ddot x)\delta x=0$ for every $\delta x$. Hence

For $m=2\,\mathrm{kg}$ and $k=18\,\mathrm{N\,m^{-1}}$, $\ddot x+9x=0$ and angular frequency $\omega=3\,\mathrm{s^{-1}}$. The acceleration opposes extension, so the sign is restoring. The general solution $x=A\cos\omega t+B\sin\omega t$ follows by differentiating twice and substituting; initial position and velocity fix $A$ and $B$.

The corresponding energies are $T=\tfrac12m\dot x^2$, $V=\tfrac12kx^2$. These expressions will provide an alternative route to the equation of motion in Unit 2.

Example 2 · Motion on an inclined plane

A block of mass $m$ slides on a smooth incline at angle $\alpha$. Choose $q$ increasing down the slope, so that positive acceleration corresponds to downward motion.

Let $\hat{\mathbf t}$ be the constant unit vector down the plane. Then $\delta\mathbf r=\hat{\mathbf t}\delta q$ and $\mathbf a=\hat{\mathbf t}\ddot q$. The component of gravity along the slope is $mg\sin\alpha$; the normal reaction is perpendicular. D’Alembert gives $(mg\sin\alpha-m\ddot q)\delta q=0$, so

For $\alpha=30^\circ$, acceleration is $4.9\,\mathrm{m\,s^{-2}}$ with $g=9.8\,\mathrm{m\,s^{-2}}$. The mass cancels. A horizontal plane gives zero downhill acceleration.

If kinetic friction of magnitude $\mu_fmg\cos\alpha$ acts uphill while the block slides down, include it explicitly: $\ddot q=g(\sin\alpha-\mu_f\cos\alpha)$. This expression applies only during that assumed sliding direction. Static friction requires its own inequality and cannot be assigned the kinetic value automatically.

Example 3 · Dynamics of the Atwood machine

Two masses $m_1,m_2$ are connected by a taut, massless, inextensible string over an ideal fixed pulley. Take $q=x_1$ downward for mass 1. The string constraint then gives $x_2=b-q$, allowing the motion to be described by one coordinate.

Then $\delta x_1=\delta q$, $\delta x_2=-\delta q$, $\ddot x_1=\ddot q$, and $\ddot x_2=-\ddot q$. The string tensions cancel in total virtual work. Keeping gravity as applied force gives

Expand the second term before collecting: it is $-m_2g\delta q-m_2\ddot q\delta q$. Therefore

and the acceleration is

For $m_1=3\,\mathrm{kg}$ and $m_2=2\,\mathrm{kg}$, $\ddot q=1.96\,\mathrm{m\,s^{-2}}$ downward for mass 1. To recover tension, use Newton's equation for that mass: $m_1g-\mathcal T=m_1\ddot q$. Hence $\mathcal T=m_1(g-\ddot q)=23.52\,\mathrm N$. In general, $\mathcal T=2m_1m_2g/(m_1+m_2)$.

Equal masses give zero acceleration. For positive masses, $|\ddot q|<g$. Reversing the mass labels reverses the acceleration sign. The corresponding energies are $T=\tfrac12(m_1+m_2)\dot q^2$ and, up to a constant, $V=-(m_1-m_2)gq$.

Example 4 · Exact equation of pendulum motion

A bob of mass $m$ is attached to a fixed pivot by a rod of length $a$ and moves in a vertical plane under gravity. With $\theta$ measured from the downward vertical, the equation of motion follows from projection onto the tangent direction.

Use $\mathbf r=(a\sin\theta,-a\cos\theta)$ and $\delta\mathbf r=(a\cos\theta,a\sin\theta)\delta\theta$. Differentiate the velocity components from Chapter 4:

Project acceleration onto $\partial\mathbf r/\partial\theta$:

The two velocity-squared terms cancel; the squared-trigonometric terms sum to one. The projection is therefore $a^2\ddot\theta$. Gravity gives $Q_\theta=(0,-mg)\cdot(a\cos\theta,a\sin\theta)=-mga\sin\theta$. The generalized D’Alembert equation becomes $-mga\sin\theta=ma^2\ddot\theta$, yielding

This equation is exact within the ideal model. For $|\theta|\ll1$ in radians, $\sin\theta\approx\theta$, giving simple harmonic motion with $\omega=\sqrt{g/a}$. This approximation must not be used for large angles without checking its accuracy.

Only tangential acceleration survives the projection. Radial centripetal acceleration is present physically but is removed from this equation by the chosen tangent variation. If tension is needed afterward, radial Newtonian balance gives $\mathcal T=ma\dot\theta^2+mg\cos\theta$ for a string, and the taut-string condition must be checked.

The corresponding energies are $T=\tfrac12ma^2\dot\theta^2$ and $V=mga(1-\cos\theta)$ when the lowest point is the potential reference.

5. Why this leads naturally to Lagrangian mechanics

The pendulum and Atwood calculations eliminated the reactions successfully, but still required explicit accelerations in Cartesian form. That becomes lengthy for several coordinates. We would like a way to assemble the inertial terms from one scalar quantity, kinetic energy, and conservative applied forces from another, potential energy.

The velocity identities derived in Chapter 4 provide that bridge. Unit 2 will use $T$, $V$, and $L=T-V$ to obtain equations in the chosen coordinates directly. The physical system has not changed; the organization of the calculation has improved.

6. Scope of the dynamical projection

D’Alembert's principle does not turn a moving system into an actual static one. It writes a dynamical identity in a force-balance form. Do not say the acceleration is zero after introducing the inertial term.

Do not erase friction with the normal reaction. Do not set coefficients of dependent variations to zero separately. Do not assume $\mathbf a_a=\ddot q_i\hat{\mathbf e}_i$ in curved coordinates. The pendulum example shows the additional velocity-squared components explicitly.

If an accelerating reference frame is used, its appropriate fictitious forces must first be included consistently. All main derivations here use an inertial frame, so that separate complication is unnecessary.

7. Practice problems

See Chapter 6 hints and solutions.

1. 6.1 Is $-m\mathbf a$ a new physical interaction in the inertial-frame derivation?

2. 6.2 What does D’Alembert's principle become in static equilibrium?

3. 6.3 Can an ideal reaction enforce nonzero normal acceleration?

4. 6.4 When may coefficients of $\delta q_i$ be set to zero separately?

5. 6.5 Explain why the principle is a reformulation of Newton's law rather than a replacement physical law.

6. 6.6 Why does the Atwood acceleration involve the sum of the masses in the denominator?

7. 6.7 How can a reaction eliminated from the motion equation be calculated afterward?

8. 6.8 Derive D’Alembert's principle and its generalized-coordinate form, stating the role of ideality and independence.

9. 6.9 Derive the exact pendulum equation by projecting the Cartesian acceleration onto the virtual displacement.

10. 6.10 An Atwood machine has masses $5\,\mathrm{kg}$ and $3\,\mathrm{kg}$. Find acceleration and tension for $g=9.8\,\mathrm{m\,s^{-2}}$.

11. 6.11 A smooth incline makes $20^\circ$ with the horizontal. Find the downhill acceleration for $g=9.8\,\mathrm{m\,s^{-2}}$.

12. 6.12 A $0.25\,\mathrm{kg}$ mass is attached to a spring with $k=9\,\mathrm{N\,m^{-1}}$. Find the motion if $x(0)=0.04\,\mathrm m$ and $\dot x(0)=0$.

13. 6.13 State, derive, and interpret D’Alembert's principle. Apply it to an Atwood machine, including checks on your result.

14. 6.14 Explain the connected route from constraints to D’Alembert's principle and why a Lagrangian description is useful next.

15. 6.15 Add a constant horizontal force $P$ to the pendulum. Obtain the exact angular equation and show that its equilibrium agrees with Chapter 5.

16. 6.16 A particle slides on the fixed parabola $y=cx^2$ under gravity. With $q=x$, derive its equation of motion using generalized D’Alembert, including the velocity-squared term.

8. From force balance to energy methods

D’Alembert's principle retains Newton's dynamics while projecting the equations onto admissible displacement directions. It removes ideal reactions, but does not remove the physical normal acceleration or turn motion into static equilibrium. Reactions can be recovered afterward from the components of Newton's equations that were not retained in the projection.

The calculations for the Atwood machine and pendulum show both the usefulness and the remaining labour of this approach. The geometry is already encoded in the coordinates, yet Cartesian accelerations still have to be differentiated and projected. Expressing these inertial terms through kinetic energy leads to the Lagrangian formulation, developed in Unit 2. The Unit 1 problems collect applications of the present method.