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A triangular localized potential energy has forces $+F$ behind and $-F$ ahead of the trap centre. With $v_c=F/\zeta =L{f}_c$, $\alpha =2X_0/L$ and $\beta =f_T/f_c$, its circular response is $f_p/f_c=\alpha \beta /({\beta }^2-1+\alpha )$ for $\beta >1$. A trapped particle has $f_p/f_c=\beta$ for $0\le \beta \le 1$, interpreting the discontinuous force at the cusp by a sticking or rounded-well limit.

Turgor pressure is the internal hydrostatic pressure above the external pressure in a biological cell. When water enters a walled biological cell by osmosis, expansion loads the cell wall. Rising pressure can oppose further net entry while the cell membrane retains its selective permeability. Mechanical support therefore does not require the cell wall to block water.

As an original mechanical check, idealize a small flat patch of cell wall of surface area $A$ under uniform turgor pressure $\Delta P=P_{\mathrm{in}}-P_{\mathrm{out}}$. Its outward force is $F=\Delta PA$. Doubling $A$ at fixed pressure doubles the load; equal internal and external pressures give zero net pressure load. This is a force balance, not a measured wall strength or a complete model of a curved, growing biological cell. It says nothing about which solutes cross the cell membrane.

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.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

Inside a homogeneous spherical mass, gravity is proportional to position. Projection onto a fixed straight chord gives $\ddot{y}=-{\omega }_0^{2}y$, independently of its perpendicular offset from the sphere centre. Release from one surface endpoint at rest reaches the opposite endpoint after $\pi /{\omega }_0$. The transverse force is supplied by the tunnel constraint. This idealization neglects removed tunnel mass, rotation, density variation and dissipative forces.

A flat circular contact of radius $a$ carrying total normal force $N$ has uniform pressure $p=N/(\pi a^2)$. In steady relative rotation, a ring of radius $s$ and thickness $ds$ experiences tangential kinetic friction $dF=\mu p2\pi sds$. Its resisting torque is $d\tau =sdF$. Integration from zero to $a$ gives ${\tau }_f=2\mu Na/3$. A different pressure distribution changes the result; for an axisymmetric distribution use ${\tau }_f=2\pi \mu {\int }_0^{a}p(s)s^{2}ds$.

A radial co-rotating cable of constant linear mass density $\lambda$ extends from planetary radius $R$ to $b={\eta }^{\prime }R$, with zero applied endpoint tension. If the geostationary orbit has radius $\eta R$, integration of the effective force gives ${\eta }^{\prime }({\eta }^{\prime }+1)=2{\eta }^3$ for a nonzero cable. At position $r$ its tension is $\lambda [GM(1/R-1/r)-{\omega }^2(r^2-R^2)/2]$. Differentiation shows the maximum at the geostationary orbit. An anchor load or counterweight alters the integrated condition.

When a vacuum vessel's window shatters, a nearly atmospheric initial pressure difference gives inward pressure force $\Delta pA$ on frontal area $A$. Dividing by the glass mass $M$ estimates its initial collective acceleration. An ideal atmospheric pressure-work budget $p_{0}V$ for vessel volume $V$ would give a collective kinetic energy estimate $M{v}_{\mathrm{rms}}^2/2\lesssim p_{0}V$ if assigned entirely to the fragments. This is a model-dependent energy budget, not a universal bound on every fragment's speed: air motion, fragmentation, pressure equalization and wall impacts alter the transfer, and unequal sharing can accelerate a small fragment faster than the collective estimate.

A body exchanging mass with material of inertial velocity $\mathbf{u}$ satisfies the displayed momentum conservation law for one stream, with signed $\dot{m}$ positive for entry and negative for exit. The force $\mathbf{F}$ excludes the incoming or outgoing momentum flux. A short time interval gives $d(m\mathbf{v})/dt=\mathbf{F}+\mathbf{u}\dot{m}$, proving the formula. For stationary incoming matter, $d(m\mathbf{v})/dt=\mathbf{F}$; for matter leaving at the body's own velocity, $m\dot{\mathbf{v}}=\mathbf{F}$. The rocket equation and accreting-drop motion are applications.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

For a particle of mass $m$ with linear drag $-\zeta v$, the velocity relaxation time is the exponential relaxation time ${\tau }_v=m/\zeta$. Under a constant applied force $F$ and zero-mean random forcing, the averaged Langevin equation is $md⟨v⟩/dt=-\zeta ⟨v⟩+F$. Solving givesThus the initial mean-velocity deviation decays by a factor $e^{-1}$ after time ${\tau }_v$. Since the force-per-velocity coefficient $\zeta$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $m/\zeta$ has units of time. If the damping parameter is instead defined as a rate $\gamma =\zeta /m$, the same time is $1/\gamma$. For a sphere obeying Stokes's law, ${\tau }_v=m/(6\pi \eta R)$. The fluctuation-dissipation relation for a Langevin particle gives the same decay time for its equilibrium velocity covariance.

A velocity selector uses perpendicular electric field and magnetic field to transmit a narrow range of speeds along a chosen beam direction. Assume uniform static electric field $\mathbf{E}=E\hat{\mathbf{y}}$ and magnetic field ${\mathbf{B}}_s=B_s\hat{\mathbf{z}}$, with $E,B_s>0$, and an incident velocity $\mathbf{v}=v\hat{\mathbf{x}}$. The cross product gives $\mathbf{v}\times {\mathbf{B}}_s=-v{B}_s\hat{\mathbf{y}}$, so the transverse Lorentz force isFor an electron, the electric force points downward and the magnetic force upward. A faster electron initially bends upward; a slower one initially bends downward. At the selected speed, both forces cancel along the whole straight trajectory. The cancellation is independent of mass and nonzero electric charge; it selects velocity, not a particular charge-to-mass ratio. A component of velocity parallel to the magnetic field is not constrained by the cancellation, so the incident beam must also have its direction restricted.

A real exit slit accepts a finite range of speeds and incident directions. A zero displacement at one screen alone does not establish zero force throughout the path: a curved trajectory could return to that screen position. The ideal velocity selector interpretation requires a straight aligned beam, known overlapping electric field and magnetic field, and sufficiently narrow entrance and exit slits. Its useful role in the Thomson charge-to-mass experiment teaching model is to supply speed independently of the subsequent magnetic bending.

If $\delta D<0$, the leading interaction of two sedimenting spheres has a formal vertical equilibrium at $r_{\ast }=-3D/(2\delta )$. Choose labels with $\delta >0,D<0$ and put sphere one above sphere two. On that invariant vertical line, $\dot{r}=-\delta +\delta r_{\ast }/r$, so both sides approach $r_{\ast }$ and the radial eigenvalue is $-\delta /r_{\ast }$. A small tilt instead grows at rate $\delta /(2r_{\ast })$: the full configuration-space equilibrium is a saddle, not an attracting node. A physically valid far-field bound pair requires $r_{\ast }\gg a_1+a_2$, or $|D|\gg (a_1+a_2)|\delta |$. Reversing both driving forces reverses the vector field, interchanging radial attraction and repulsion, consistent with kinematic reversibility of Stokes flow.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]

Viscous damping means a force proportional to, and opposite to, velocity: $F_{\mathrm{d}}=-b\dot{x}$ with $b\ge 0$. Its coefficient has units $\mathrm{kg}{\mathrm{s}}^{-1}$, equivalently $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The name describes a force law; eddy-current damping can approximate the same law without a viscous fluid. Constant $m$, $b$ and $k$ produce a linear ordinary differential equation, allowing superposition of responses. Kinetic friction with approximately constant magnitude and quadratic drag obey different laws, so their decay need not have the exponential decay or critical threshold derived here.

The proportionality between drag force and relative velocity in a linear low-speed regime. Its translational units are $\mathrm{kg}{\mathrm{s}}^{-1}$; thermal equilibrium gives the Einstein diffusion relation $D=k_{B}T/\gamma$.

If resisting force per unit moving length is $\beta v$, the total resistance is $\beta l\dot{y}$ and $\ddot{y}+\Gamma \dot{y}+{\omega }_0^{2}y=0$, where $\Gamma =\beta /(\rho A)$. For $\Gamma <2{\omega }_0$, an initial velocity $v_0$ from zero displacement gives $y=(v_0/{\omega }_d)e^{-\Gamma t/2}\sin {\omega }_{d}t$, with ${\omega }_d^2={\omega }_0^2-{\Gamma }^2/4$. Its quality parameter is $Q={\omega }_0/\Gamma$, and the amplitude envelope decays at half the energy-decay exponent. The effective damping model is an assumption, not a derivation of detailed viscous flow near the wall.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.

Compare planar extension $\mathbf{u}=(ax,-ay,0)$ with solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$:Extension changes lengths and produces viscous normal stresses $\pm 2\mu a$. Rotation preserves distances and gives zero viscous stress, despite its nonzero velocity gradient and vorticity $2\Omega {\mathbf{e}}_z$. For $a=\Omega =1{\mathrm{s}}^{-1}$ and $\mu =0.10\mathrm{Pa}\mathrm{s}$, extension gives ${\sigma }_{\mathrm{visc},xx}=0.20\mathrm{Pa}$ and ${\sigma }_{\mathrm{visc},yy}=-0.20\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a pressure gradient supplying its centripetal acceleration.

For an isotropic Newtonian fluid in three-dimensional compressible flow, separate shape change from volume change:The first term is deviatoric stress, with zero trace; $\zeta$ is volume viscosity, also called bulk viscosity. For uniform expansion $\mathbf{u}=b\mathbf{x}$, $D=bI$ and $\nabla \cdot \mathbf{u}=3b$, so the shear contribution vanishes and the viscous stress is $3\zeta bI$. Thus dynamic viscosity and volume viscosity describe distinct responses, and incompressible flow removes only the volume-change term. ^[ref]

Weightlessness is the absence of supporting or constraining forces that would give a body an apparent weight. It does not require zero gravitational acceleration. A freely falling cabin and a small freely falling object inside share the same acceleration in a uniform field, so neither needs to support the other.

The circular orbit as continuous free fall gives a sustained example. Real spacecraft have small residual relative accelerations from tidal forces, atmospheric drag force and onboard disturbances; the ideal common-fall explanation does not assert that every point in an extended spacecraft experiences exactly the same Newtonian gravitational field. ^[ref]