Virtual work and d’Alembert’s principle

One bead, one allowed direction

A 2 kg bead is confined to a fixed straight guide that rises four units for every three units to the right. Treat the bead as a point. The guide is bilateral: it prevents departure on either side. Assume no friction, constant mass and an inertial Cartesian frame, with x rightward and y upward. These are supplied Newtonian model conditions, not measured properties of a real guide.

A horizontal hand force H = (10, 0) N and weight W = (0, −20) N act on the bead. The guide supplies a physical reaction R. In this lesson F = H + W = (10, −20) N denotes the applied forces other than R. All three interactions are physical; this separation is bookkeeping for the projection method.

Write the position as r = q t̂, where q is signed distance along the guide in metres and t̂ = (3/5, 4/5) is its upward unit tangent. The symbol t̂ names a direction, not time. Its length is one because (3/5)² + (4/5)² = 1. A perpendicular unit direction is n̂ = (−4/5, 3/5). Thus x = 3q/5, y = 4q/5 and the restriction is −4x + 3y = 0.

Compare configurations at the same time

Hold the clock fixed and imagine changing q a little while keeping the bead on this guide. The permitted first-order change is δr = t̂ δq, so −4δx + 3δy = 0. The symbol δ distinguishes this virtual change from a change along the actual motion over elapsed time. It tests an allowed direction; forces do not have to drive the bead in that direction.

For arithmetic, use the scaled variation vector δq = 0.05 m and δr = (0.03, 0.04) m. These are coefficients of an arbitrarily small comparison: replace q by q + ε(0.05 m), with a dimensionless parameter ε tending to zero. The first-order position change is ε(0.03, 0.04) m. We quote the coefficient rather than pretending the bead jumps 0.05 m in zero time. The straight restriction makes this comparison exactly admissible for sufficiently small positive or negative ε.

For two fixed-axis vectors, the dot product A · B is A_xB_x + A_yB_y. A force’s virtual-work coefficient is δW = F · δr. It has units N m = J. In our scaled comparison, the first-order virtual work is εδW. It is not a claim that energy has actually been transferred.

Test equilibrium with every allowed variation

An ideal constraint is one whose reactions have zero total virtual work for every allowed virtual displacement. For this frictionless straight guide, R is normal to the guide while δr is tangent, so R · δr = 0. This is a stated property of the model; a restriction on position alone does not establish that every real contact is ideal.

If the bead is in static equilibrium, a = 0 and Newton’s equation gives F + R = 0. Projecting on δr then gives F · δr = 0 for every allowed variation. Conversely, in this regular bilateral guide model, zero applied projection for all tangent variations removes the tangent imbalance; the normal reaction supplies the remaining balance. A system at rest also needs zero initial velocity. Zero net force alone permits constant velocity.

Here the applied projection is F · t̂ = 10(3/5) − 20(4/5) = −10 N, so the supplied 10 N hand force does not produce static equilibrium. To hold the bead, change the horizontal force to a different value H₀: (3/5)H₀ − 16 N = 0 gives H₀ = 80/3 N. Requiring the balance for every allowed variation matters; one zero dot product in an unrelated direction would prove little.

Project the dynamic Newton equation

Acceleration a is the rate of change of velocity; its components have units m/s². For the same bead moving on the fixed straight guide, a is tangent. Newton’s equation is ma = F + R. Taking the tangent projection eliminates R: m a_q = F · t̂, where a_q is the signed acceleration along the guide. Thus a_q = (−10 N)/(2 kg) = −5 m/s² and a = (−3, −4) m/s².

The normal applied projection is F · n̂ = −20 N. Zero normal acceleration requires R = 20 N n̂ = (−16, 12) N. The physical resultant is therefore F + R = (−6, −8) N = ma. Rearrange the same equation as F − ma = −R and project: (F − ma) · δr = 0. This is the one-particle form of d’Alembert’s principle under the ideal-constraint assumption.

Here F − ma = (16, −12) N is perpendicular to the allowed direction. The subtraction is an effective balance, not a new physical interaction on the bead. The term −ma is sometimes called an inertia force in this construction. It is not an extra hand, gravity or guide force, and this calculation still uses inertial acceleration.

For a concrete actual motion consistent with these conditions, supply q(t) = 1 m + (0.2 m/s)t − (2.5 m/s²)t² and v_q(t) = 0.2 m/s − (5 m/s²)t only for 0 ≤ t ≤ 0.2 s. At t = 0.1 s, the bead moves downhill with v_q = −0.3 m/s. Our upward virtual direction remains allowed at that same instant. A virtual displacement is not chosen by the sign of the actual velocity.

Physical forces and a projected effective balance

The left diagram draws the three physical interactions on the bead. The right draws only the calculated vector F − ma, using a dashed line and open diamond head. Both panels use 7 pixels per newton. Dotted arrows with open triangular heads show the allowed upward virtual direction. Their lengths indicate direction only, not a measured displacement or a displacement scale.

The guide has the same 3:4 orientation in both panels. Small x and y axes show the coordinate directions; they are not position rulers. The effective vector is normal to the virtual direction, so its dot product is zero. It need not be the zero vector. The table gives exact text equivalents for the chosen scaled variation coefficient.

A moving ideal guide can transfer energy

Now change the model: translate the whole straight guide uniformly at 0.5 m/s along n̂ without rotating it. At any fixed time, the virtual direction is still t̂. During actual motion, however, the bead’s velocity can include the guide’s normal velocity. Supply an instantaneous tangent velocity of 0.2 m/s relative to the guide. The inertial velocity is v = (0.2 m/s)t̂ + (0.5 m/s)n̂ = (−0.28, 0.46) m/s.

Keep the same mass, hand force and weight. Because the translation is uniform and the straight guide has no normal acceleration, its reaction remains R = 20 N n̂. The reaction’s virtual-work coefficient remains zero. Its actual instantaneous power is instead R · v = (20 N)(0.5 m/s) = 10 W. Power means actual work per unit time and has units J/s = W.

The guide can therefore supply energy to the bead in this moving case. For the fixed guide, actual velocity is tangent and the reaction power is zero. The claim that an ideal reaction always does zero actual work would mix up these two situations. These are local model calculations, not laboratory power measurements.

State the restrictions before generalizing

For several particles with positive constant masses in an inertial frame, the corresponding statement is Σᵢ(Fᵢ − mᵢaᵢ) · δrᵢ = 0 for every allowed virtual displacement, when ΣᵢRᵢ · δrᵢ = 0. The symbol Σᵢ means add the contribution of each particle. Some ideal internal reactions can cancel in the sum even when an individual reaction’s virtual work is not zero.

To use independent configuration coordinates this way, take smooth equality constraints depending on position and possibly time, with independent regular restrictions in the region considered. Such restrictions are called holonomic. Regular means that the independent equations’ first-order coefficient rows remain linearly independent in that region, leaving a well-defined set of allowed tangent directions. A degree of freedom is one independent configuration coordinate. Our single equation −4x + 3y = 0 is regular because its coefficient vector (−4, 3) is nonzero. Copying that equation twice would not remove another degree of freedom.

This lesson does not extend the coordinate-elimination rule to every velocity-dependent restriction or a one-sided contact that can detach. If a contact has friction along an allowed displacement, retain that force in the projected balance rather than declaring its virtual work zero. Ideal constraints and admissible variations must be justified for the particular model.

Use the method without inventing forces

Choose the system and frame, list physical forces, and state the constraint. Construct the allowed virtual directions at fixed time. For an ideal constraint, project the static equation or the dynamic F − ma equation on every independent direction. Recover a reaction from the full vector equation when it is needed.

Units, standards, and physical constants and Vectors and vector spaces provide preparation for units and component dot products. The needed guide geometry and force equation have been introduced here. On a first reading, the bead, table and moving-guide comparison suffice; return to the several-particle statement when you need a larger system.

The next subject, The principle of stationary action, compares whole candidate histories. Here the virtual projection was taken at one fixed time. The two constructions should not be merged merely because both use a δ symbol.

Worked examples

Try it yourself

1. For the fixed guide, is the proposed vector (0.04, −0.03) m an allowed virtual variation coefficient? Check the restriction.

   Show the answer and reasoning

   No. −4(0.04 m) + 3(−0.03 m) = −0.25 m, which is not zero.

   The proposal points normal to the guide, not along its allowed tangent. A zero dot product in a disallowed direction would not establish the required projected balance.

2. Reverse the main variation coefficient to δr = (−0.03, −0.04) m without changing any forces or acceleration. Find F · δr and (F − ma) · δr.

   Show the answer and reasoning

   The applied coefficient becomes +0.50 J; the effective coefficient stays 0 J.

   A linear dot product changes sign when its test vector reverses. The zero d’Alembert balance holds for both directions, while the nonzero applied coefficient still excludes static equilibrium.

3. In another model, take mass 4 kg but explicitly keep H = (10, 0) N and W = (0, −20) N on the same fixed ideal guide. Find the tangent acceleration, Cartesian acceleration and reaction.

   Show the answer and reasoning

   a_q = −2.5 m/s², a = (−1.5, −2) m/s² and R = (−16, 12) N.

   The applied tangent projection remains −10 N, so divide by 4 kg. The normal projection remains −20 N, so the reaction is unchanged. This exercise stipulates the same downward force; it does not assume weight stays fixed when mass changes in an unchanged gravitational field.

4. In a modified 2 kg fixed-guide case, supply an additional friction force of magnitude 2 N along the upward tangent at an instant of downhill motion. Find a_q and that force’s coefficient for δq = +0.05 m.

   Show the answer and reasoning

   a_q = −4 m/s² and the friction coefficient is +0.10 J.

   Retain the supplied nonideal tangential force: (−10 + 2) N divided by 2 kg gives −4 m/s². Its dot product with the positive tangent variation is (2 N)(0.05 m). Positive virtual work here does not mean positive actual friction power during downhill motion.

5. For a separate guide translating uniformly at −0.5 m/s along n̂, keep the main forces and an instantaneous relative tangent speed +0.2 m/s. Find the guide reaction power and its virtual-work coefficient.

   Show the answer and reasoning

   Reaction power is −10 W; its virtual-work coefficient is still 0 J.

   The physical reaction remains 20 N n̂ because uniform translation adds no normal acceleration. Its power is (20 N)(−0.5 m/s), while every fixed-time virtual vector remains tangent. The guide now removes energy from the bead at this supplied instant.

6. A planar bead satisfies both −4x + 3y = 0 and −8x + 6y = 0. Do these leave zero or one independent configuration coordinate? Must the nonzero effective vector in the main model be an extra physical force?

   Show the answer and reasoning

   One independent coordinate remains. The second equation duplicates the first, and F − ma is an effective calculated vector rather than an extra physical interaction.

   The two coefficient rows are proportional, so there is only one independent regular restriction on two planar coordinates. The main effective vector is nonzero but perpendicular to all allowed tangent variations; its virtual projection vanishes.

What to carry forward

• A virtual variation tests allowed configurations at one fixed time; it is not an actual timed displacement.
• Virtual work is a force projection on that variation and has energy units without being an observed energy transfer.
• Ideal reactions have zero total virtual work for every admissible variation, under the declared constraint model.
• Static balance uses F; dynamic d’Alembert balance uses F − ma with inertial constant-mass acceleration.
• A moving ideal guide can do actual work even when its reaction’s virtual work is zero.
• Independent regular restrictions and the correct physical forces determine which projections are legitimate.