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An ideal contact mechanical constraint supplying a normal reaction and no tangential friction. For a stationary constraining surface, the permitted velocity is a tangent vector, so the normal force has zero power and does no work. For a moving surface, normality alone does not imply zero power in the laboratory frame.

Purdue's Bicycle Wheel Gyroscope demonstration uses a bicycle wheel on a long axle, with a ball at one end seated in a freely rotating socket on a heavy stand. Spin the wheel by hand or with the supplied motor, seat the ball, and release the nearly horizontal axle. The readout is the axle sweeping around the vertical while the wheel continues spinning. Compare with the stopped wheel, which falls when one end is released; reverse the spin with the same support and the precession reverses. These are the page's documented apparatus, presentation, and reversal checks. ^[ref]

An alternative in Physics Demonstrations by Julien Clinton Sprott (section 1.3, Materials and Procedure) suspends one axle end by stainless steel wire and a fishing-line swivel. The swivel allows precession without winding up the suspension. Sprott describes the axle bobbing as nutation and precession becoming faster as spin decreases. A suspended support can move, whereas the calculation here assumes a fixed $O$; use the socket version for the closest realization. ^[ref]

A proposed quantitative extension is to record an overhead view of an axle marker against a fixed angle scale, together with a side view for $\theta (t)$. Time successive overhead passages to estimate $\Omega =2\pi /T$; count passages of a rim mark to estimate $|{\omega }_3|=2\pi f$. Record the spin sense too. Measure $M$ and the centre of mass offset $d$ of the whole moving assembly, and determine its axial moment of inertia $I_3$ independently. Wheel, axle, support, release angle, and observation interval should stay the same between spin trials. Measuring both $\Omega$ and $\theta (t)$ distinguishes azimuthal precession from nutation; an overhead movie alone can miss the latter. These camera readouts and parameter checks are proposed here, not reported measurements from the demonstration pages.

For slow precession, compare $\Omega I_3{\omega }_3$ with $Mgd$, and test whether reversing ${\omega }_3$ reverses $\Omega$ with nearly unchanged magnitude. A brief interval minimizes spin loss from friction; repeat at several starting spin rates rather than interpreting an entire slowing run as one steady state. A suspension restoring torque governed by its torsion constant, socket friction, support motion, a misplaced centre of mass, and a push at release can all supply extra torque or change the initial conditions. The swivel reduces the twist that activates the suspension's torsion constant; it does not demonstrate zero torque.

Steady precession keeps $\theta$ constant; nutation changes $\theta$ while the axle may also precess. Sudden release generally excites both. A small initial drop is consistent with conservation of energy and conserved vertical angular momentum: some gravitational potential energy supplies the motion around the vertical, and the axial contribution to vertical angular momentum adjusts. Bearing friction may damp the bobbing, but an ideal undamped heavy symmetric top need not settle into steady precession. Chapter 20: Rotation in space explicitly distinguishes uniform precession from release transients and nutation (section 20-3, paragraphs following equation 20.15).^[ref]

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

Oberlin College's documented demonstration uses a track, two collision carts, two force probes set to the $50\mathrm{N}$ range, added $500\mathrm{g}$ bars, and a laptop with a Vernier interface. Preparation consists of entering the measurement calibration, securing the force probes to the carts, and zeroing them. Rubber stoppers serve for elastic collisions, clay for inelastic collisions, and hooks with string for pulling. Comparing the two force records tests the mutual interaction; adding mass to one cart tests whether unequal masses change the equality. ^[ref]

Vernier's force probe uses a bending beam with strain gauges; changing electrical resistance produces a voltage readout calibrated as force. Its manual recommends zeroing in the working orientation. If recalibration is needed, use an unloaded reading and a known suspended gravitational weight; for the $50\mathrm{N}$ range its example is $1\mathrm{kg}$ giving $9.8\mathrm{N}$. Re-zero after changing orientation. The manual quotes calibrated accuracy $\pm 0.2\mathrm{N}$, distinct from the $0.05\mathrm{N}$ resolution of that range, and warns that the interface's data-collection rate can limit collision measurements. ^[ref]

The following sampling and control choices are proposed additions to that documented experiment. Level the track and check unloaded motion in both directions for slope and friction. Use a soft bumper and one interface for simultaneous records; start before contact and stop after separation. A trial sampling rate of $1000{\mathrm{s}}^{-1}$, if supported, would put about 100 samples across a $0.10\mathrm{s}$ interaction. Increase the sampling rate and repeat until the peak and integrated impulse comparison are stable; a large sample count alone does not establish adequate response. Measure each total moving mass, including the force probe and added bars.

Convert both channels to the same right-positive force convention. Oppositely facing force probes can have matching raw signs because their local positive directions differ; check each sign by a gentle push of known direction before the run. Overlay $F_{B\mathrm{on}A}(t)$ and $-F_{A\mathrm{on}B}(t)$, and also examine their signed sum. Integrate both over the same contact interval to compare impulses. Repeat with added mass on one cart, exchange the probes between carts, and check unloaded zeros before and after. These controls help distinguish a physical mass change from a channel offset or gain error.

Agreement within measurement error supports the pair-interaction model over the resolved interval; it does not establish exact equality at arbitrarily short time scales. Misalignment, a loose mounting, sensor-beam dynamics, saturation, unequal timing, and zero drift can spoil the comparison. Track friction and cable pulls are additional external forces; a probe measuring bumper contact need not measure the entire net force on its cart. If a massive connector carries appreciable changing momentum, compare the full cart–connector collection rather than assuming its endpoint forces are an instantaneous cart-to-cart pair. The constructed pulse diagram above illustrates what to compare, and supplies no measurements from this apparatus.

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

Research author of How transform fault shear influences where detachment faults form near mid-ocean ridges by Jana C. Schierjott and others.