Constraints and Constraint Forces

Every physics engine you have ever seen — game engines, robot simulators, animation software — is at heart a constraint-force solver. At each frame, the engine collects thousands of contacts, joints, and no-penetration conditions, writes them as $\mathbf{F}^{(c)} = \lambda\,\nabla f$ with unknown multipliers, and solves a large complementarity problem for the $\lambda$'s: exactly our unilateral logic ($N \ge 0$, contact or no contact) industrialized. Simulators like MuJoCo — the standard training ground for 2026's humanoid-robot policies — spend most of their compute deciding what the constraint forces are, because the applied forces (gravity, motors) are the easy part. A walking robot is a constraint negotiation in real time: each planted foot is a unilateral, friction-limited contact, and the controller must keep every required friction force inside its $\mu_s N$ budget — our rolling-cylinder check, run a thousand times a second on a machine that falls over when the answer is no. Molecular dynamics plays the same game at the ångström scale. The fastest vibrations in a biomolecule (hydrogen bond stretches) would force absurdly small time steps, so simulation packages constrain those bond lengths rigidly and use SHAKE/RATTLE-type algorithms to compute the Lagrange multipliers — the interatomic constraint forces — that hold them fixed. A protein-folding simulation is, in this precise sense, a pendulum problem with a few hundred thousand strings.

Take stock of four systems you have met — a pendulum, a bead on a curved wire, a block on an incline, a wheel rolling down a slope — and ask of each: what force keeps the geometry honest? The string's tension keeps the bob at distance $L$. The wire's sideways push keeps the bead on the curve. The incline's normal force keeps the block out of the ground. Static friction at the contact point keeps the wheel from slipping. Four different names, one job description: each force does whatever it takes, no more and no less, to enforce a geometric rule.

Now zoom out to the view that organizes everything. Describe the whole system by a single point in configuration space — one axis per coordinate of every particle. A constraint carves out a surface in that space, and the system's point is forbidden to leave it. The constraint force is the wall that confines the point, and like any ideal wall it pushes only perpendicular to itself, never along it. That one geometric sentence contains the two facts you will use a hundred times: the force's direction is fixed by the geometry (normal to the surface, along the string, at the contact point), while its magnitude is whatever the motion demands. This is why constraint forces feel slippery on first meeting: you cannot look them up. Gravity is prescribed; the normal force is negotiated. Press down on a table and it pushes back harder; jump, and for an instant it pushes not at all. The table is not consulting a formula — it is deforming microscopically, by exactly enough to stop your hand, and the "constraint" is our idealization of that stiff spring in the limit of zero give. Whenever you see an unknown $N$, $T$, or $f_s$ in a problem, read it as a question the geometry will answer. The last piece of the picture is the no-work property. A gigantic normal force on a sliding block does zero work, because it is perpendicular to every displacement the block can actually make. A pendulum string's tension does zero work, because the bob always moves perpendicular to the string. The contact-point friction of pure rolling does zero work, because the point it acts on is instantaneously at rest. Ideal constraint forces are enforcers, not investors: they redirect motion without ever paying into or withdrawing from the kinetic-energy account. That is the property d'Alembert isolated, and it is the trapdoor through which constraint forces will later vanish from mechanics entirely.

Before computing anything we need the grammar: what kinds of restriction exist, and how much freedom each one costs. The cleanest kind of constraint is a plain equation among coordinates. The pendulum's string enforces $x^2 + y^2 - L^2 = 0$; a bead on a parabola enforces $y - ax^2 = 0$; a particle glued to a table enforces $z = 0$. All of these fit one template:

where: $\mathbf{r}_i$ — the position of particle $i$; $f$ — a smooth function whose zero set is the allowed surface; $t$ — time, present only if the constraint itself moves (a pendulum whose pivot is being cranked, an elevator floor). Constraints of this form are called holonomic — expressible as coordinate equations. If $f$ has no explicit $t$ the constraint is scleronomic (rigid, time-frozen); if $t$ appears it is rheonomic (the wall itself moves, and — watch this — a moving wall can do real work on you). Each independent holonomic equation flattens the accessible configuration space by one dimension. Count what survives and you have the single most useful planning number in mechanics — how many coordinates you will actually need:

where: $n$ — degrees of freedom; $N$ — number of particles; $k$ — number of independent holonomic constraints. A free particle has $n=3$. A pendulum bob in a plane has $3 - 2 = 1$: the angle $\theta$. A rigid body has $n = 6$ no matter how many atoms it contains, because the enormous web of fixed-distance constraints is massively redundant — three numbers place the center of mass, three orient the body. When a problem hands you a tangle of pulleys and rods, computing $n$ first tells you exactly how many equations of motion you are ultimately solving for. Not every restriction is an equation among coordinates. A wheel rolling on a straight line must match its rim speed to its travel:

where: $x$ — position of the wheel's center; $R$ — its radius; $\theta$ — its rotation angle. On a straight line this integrates at once to $x = R\theta + \mathrm{const}$, so it is secretly holonomic. But let the wheel steer — a coin free to roll and turn on a tabletop, a car in a parking lot — and the same no-slip rule becomes a velocity condition that no coordinate equation can capture. The proof is your own experience: a car cannot slide sideways (the constraint forbids it at every instant), yet by rolling forward, turning, and backing up, you can move the car sideways into a parallel-parking spot. Where you can get depends on the path, not just on the rules — the fingerprint of a non-holonomic constraint. Such constraints still remove freedom from the velocities, but not from the configurations, and they demand more careful machinery (Lagrange multipliers, or momentum methods) later on. Finally, the property that gives constraint forces their special status. Freeze time, and imagine any small displacement $\delta\mathbf{r}_i$ of the particles that respects the constraints — a virtual displacement, a test wiggle along the allowed surface. Ideal constraint forces are exactly those that do no work under any such wiggle:

where: $\mathbf{F}^{(c)}_i$ — the constraint force on particle $i$; $\delta\mathbf{r}_i$ — any virtual displacement consistent with the constraints at fixed time. Frictionless surfaces, inextensible strings, rigid rods, and the static-friction contact of pure rolling all pass this test; sliding (kinetic) friction fails it, which is precisely why kinetic friction must be carried as an ordinary applied force. For a single particle on a surface $f(\mathbf{r})=0$, equation (4) pins down the direction of the enforcer completely:

where: $\nabla f$ — the gradient of the constraint function, which points normal to the surface $f=0$; $\lambda$ — an undetermined multiplier carrying the magnitude and sign, fixed only when you solve the equations of motion. Equation (5) is the mathematical form of "walls push perpendicular": the tangent directions are exactly the virtual displacements, and a vector orthogonal to all of them must lie along the normal. The number $\lambda$ is the ancestor of the Lagrange multiplier you will meet in analytical mechanics — same object, grander machinery.

Which tool answers which question

1 · Two blocks and a pulley (confidence builder)

A block of mass $m_1 = 3.0\ \mathrm{kg}$ sits on a frictionless table, connected by a light string over an ideal pulley at the table's edge to a hanging block of mass $m_2 = 2.0\ \mathrm{kg}$. Find the acceleration and the string tension.

Step 1. Count the unknowns. The motion unknowns are two accelerations and one tension — but the string's fixed length (Derivation A, fixed-pulley case) forces both blocks to share one acceleration magnitude $a$. Two unknowns remain: $a$ and $T$. Two Newton equations will close the problem.

Step 2. Newton for each body along its direction of motion. Table block, horizontal: $T = m_1 a$. Hanging block, vertical (down positive): $m_2 g - T = m_2 a$.

Step 3. Add the equations — the constraint force $T$ cancels, exactly as an internal enforcer should:

Step 4. Recover the constraint force from either equation: $T = m_1 a = 3.0 \times 3.92 = \mathbf{11.8\ N}$. Cross-check with the other body: $T = m_2(g - a) = 2.0 \times (9.81 - 3.92) = 11.8\ \mathrm{N}$. Consistent.

Step 5. Read the result like a physicist. $T = 11.8\ \mathrm{N}$ sits strictly between $0$ and $m_2 g = 19.6\ \mathrm{N}$: the string pulls hard enough to drag $m_1$ forward but not hard enough to hold $m_2$ static — the constraint force found the one value that lets both bodies obey the same geometry. If someone had "looked up" $T = m_2 g$, block 2 would never fall and block 1 would accelerate at $6.5\ \mathrm{m/s^2}$ — two bodies, two different stories, and a snapped constraint.

2 · Sliding off a hemisphere — when the constraint lets go (exam level)

A child's marble starts from rest at the very top of a frictionless hemisphere of radius $R = 2.0\ \mathrm{m}$ (a playground dome) and slides down the outside. At what angle from the vertical does it leave the surface, how fast is it moving at that moment, and how far below the top is it?

Step 1. Set up the constraint while it lasts. On the dome the marble moves on a circle of radius $R$, so at angle $\theta$ from the vertical its radial (center-pointing) equation of motion reads $mg\cos\theta - N = \dfrac{m v^2}{R}$, where $N$ is the outward normal force — the constraint force keeping the marble out of the dome.

Step 2. Get $v(\theta)$ from energy conservation — legitimate because the normal force is perpendicular to the motion and does no work: $\tfrac12 m v^2 = m g R\,(1 - \cos\theta)$, so $v^2 = 2gR(1-\cos\theta)$.

Step 3. Solve for the constraint force as a function of angle:

The enforcer weakens as the marble speeds up: gravity's radial component must both hold the circle and feed the growing centripetal demand, and $N$ absorbs the slack.

Step 4. Apply the unilateral test. A dome can push but never pull, so contact survives only while $N \ge 0$. Departure happens at $N = 0$:

Remarkably, the angle is universal — no mass, no radius, no planet enters.

Step 5. The numbers for $R = 2.0\ \mathrm{m}$. Speed at departure: $v^2 = 2gR(1 - \tfrac23) = \tfrac{2gR}{3} = \tfrac{2 \times 9.81 \times 2.0}{3} = 13.08\ \mathrm{m^2/s^2}$, so $v = \mathbf{3.6\ m/s}$. Height fallen: $R(1-\cos\theta^*) = 2.0 \times \tfrac13 = \mathbf{0.67\ m}$. From that point on the marble is a projectile — the constraint equation is deleted from the problem, and $N$ with it.

Step 6. Notice the method, because it recurs everywhere: solve assuming the constraint holds, watch the constraint force, and the moment it would change sign, switch problems. The same logic times a roller coaster cresting a hill ($N = 0$ at $v^2 = gR$), a car topping a rise, and cargo floating off a truck bed on a bump.

Texts. Morin, Introduction to Classical Mechanics, ch. 3 (F = ma with constraint-force examples) and ch. 8 (rolling) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.3–1.4 (constraints, d'Alembert's principle) · Taylor, Classical Mechanics, ch. 7 · Landau & Lifshitz, Mechanics, §1–§5 · Kleppner & Kolenkow, An Introduction to Mechanics, ch. 2–3. The key results collected here are: (1) movable-pulley kinematics $\ddot y_e = -2\ddot y_p$, from differentiating the rope-length invariant $2y_p + y_e = \mathrm{const}$; (2) free-wedge formulas $A = mg\sin\theta\cos\theta/(M+m\sin^2\theta)$, $a_{\mathrm{rel}} = (M+m)g\sin\theta/(M+m\sin^2\theta)$, $N = Mmg\cos\theta/(M+m\sin^2\theta)$, with the numeric case $M=4.0$ kg, $m=1.0$ kg, $\theta=30^\circ$ giving $A = 1.00\ \mathrm{m/s^2}$, $a_{\mathrm{rel}} = 5.77\ \mathrm{m/s^2}$, $N = 8.00\ \mathrm{N}$; (3) rolling solid cylinder $a = \tfrac23 g\sin\theta$, $f_s = \tfrac13 mg\sin\theta$, slip threshold $\tan\theta \le 3\mu_s$ using $I = \tfrac12 mR^2$; (4) hemisphere departure condition $\cos\theta^* = \tfrac23$ ($\theta^* = 48.2^\circ$) with $v = 3.6\ \mathrm{m/s}$ and drop $0.67\ \mathrm{m}$ for $R = 2.0$ m; (5) two-block pulley $a = 3.92\ \mathrm{m/s^2}$, $T = 11.8\ \mathrm{N}$ for $m_1 = 3.0$ kg, $m_2 = 2.0$ kg. All numerical values use $g = 9.81\ \mathrm{m/s^2}$.

Builds on: Newton's Second Law · Newton's Third Law · Vectors, Coordinate Systems and Kinematics · Normal Force · Circular Motion and Centripetal Acceleration. Leads to: Generalized Coordinates and Configuration Space · D'Alembert's Principle · Lagrangian Mechanics · Rigid Body Dynamics.