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Static equilibrium is a state in which a mechanical system remains at rest in a specified reference frame. For a rigid body in an inertial frame, zero resultant external force and zero resultant external torque are the balance conditions:The body must also initially be at rest: zero acceleration alone does not exclude uniform translation or steady rotation.^[ref]

A free-body diagram identifies the body being analyzed and includes its external loads and support reactions. Internal action–reaction pairs cancel when both interacting parts belong to the chosen system. Choosing a different point $O$ for the torques changes their sum by a term proportional to the resultant force. Thus, when the resultant force is zero, a vanishing resultant torque about one point also vanishes about any other point.^[ref]

For an original example, take an ideal massless horizontal beam of length $4\mathrm{m}$, supported at its ends, with a downward $120\mathrm{N}$ load $1\mathrm{m}$ from the left support. Let the upward support reactions be $R_L$ and $R_R$. Vertical force balance and torque balance about the left end giveso $R_R=30\mathrm{N}$ and $R_L=90\mathrm{N}$. The nearer support carries more load because the farther reaction has the larger moment arm. Two equal and opposite forces alone would not guarantee equilibrium: a force couple can have zero resultant force and a nonzero torque.

A simple apparatus test suspends a beam from two spring balances and moves a known load along it. Account separately for the beam's own gravitational weight and its center of mass before comparing the measured reactions with the ideal calculation. The sum of the readings should track total downward load, while their difference changes with load position. Finally, static equilibrium does not imply dynamical stability: an exactly upright inverted pendulum balances at rest but departs from that state after a small disturbance.

Unilateral contact permits bodies to separate while preventing them from passing through each other. In an ideal rigid, nonadhesive model, the supporting normal contact force can push but cannot pull. It is therefore different from a permanently attached support: an upward pull may lift a body away instead of producing a downward support reaction.

Let $d$ be the signed normal gap, positive when the bodies are separated, and $N$ the outward compressive normal contact force. The nonpenetration constraint and absence of adhesion give the unilateral contact complementarity conditionsAn open gap requires $N=0$, while $N>0$ requires a closed gap. The pair $d=N=0$ is also allowed: touching does not by itself imply a loaded contact.^[ref] Tangential friction requires an additional relation such as Coulomb friction; it is not specified by the normal-gap conditions.

The equation of motion is still needed to determine the reaction. Consider a block of mass $m$ on a fixed horizontal floor, initially at rest at $d=0$, with an upward applied force $u$ through its center of mass. With upward positive, Newton's second law isIf the contact remains closed, $\ddot{d}=0$ would require $N=mg-u$. That is feasible for $u\le mg$. For $u>mg$, a negative normal reaction would violate unilateral contact, so the contact opens with $N=0$ and $\ddot{d}=(u-mg)/m>0$. At a closed contact with zero normal velocity, the appropriate acceleration-level contact complementarity iswhich yieldsThese acceleration conditions are being applied at a closed, initially stationary contact; they are not a rule forbidding downward acceleration during separated free flight.

For an original numerical example, let $m=2\mathrm{kg}$ and $g=9.81m/{\mathrm{s}}^2$. Pulling upward with $u=5\mathrm{N}$ leaves a normal reaction of $14.62\mathrm{N}$. At $u=25\mathrm{N}$, the reaction is zero and the initial upward acceleration is $2.69m/{\mathrm{s}}^2$. At $u=19.62\mathrm{N}$, both the reaction and the initial acceleration vanish, while touching remains possible.

A simple apparatus test places the block on a calibrated load cell and applies the upward pull through a spring balance, keeping the pull vertical through the center of mass to avoid an extra overturning torque. The support reading should decrease with the pull while contact persists, then vanish at lift-off. Recording the gap as well distinguishes an unloaded touch from a genuinely open contact. Real support elastic deformation and adhesion must be accounted for before treating that apparatus as the ideal rigid, nonadhesive model.

A penalty approximation of unilateral contact replaces perfect rigidity by a chosen stiff response, for exampleHere a small negative gap represents an allowed modeled overlap relative to undeformed reference surfaces. It is not an exact solution of $d\ge 0$. For the $14.62\mathrm{N}$ supported load above and chosen penalty stiffness $k=10^{5}N/m$, equilibrium requires $d=-0.1462\mathrm{mm}$. Increasing $k$ reduces that overlap at the same load. The penalty method and the ideal constraint are different approximations of the contact response.

Finally, impact requires a rule for the velocity change. The position-level inequalities alone do not decide whether an approaching body sticks or rebounds. A coefficient of restitution or another specified impact law determines that behavior, and the impulsive contact force is summarized by its impulse. Tedrake's particular time-stepping formulation identifies its block collisions as perfectly inelastic; that is a feature of the stated model, not a consequence of the three gap/reaction inequalities alone.^[ref]