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Let $fdx$ be reversible work supplied to stretch a band. Its Helmholtz free energy obeys $dF=-SdT+fdx$. Equality of mixed derivatives gives the displayed relation. An entropy decrease on stretching therefore corresponds to increasing tensile force on heating at fixed extension. The sign follows the force/work convention; a restoring force exerted by the band has the opposite sign.

A pressure-containing shell whose thickness is small relative to its curvature radius, allowing a membrane-stress approximation. For a cylindrical closed-end tube, pressure force balance gives hoop stress $Pr/t$ and axial stress $Pr/(2t)$, with the same radius convention and neglected end disturbances.

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.

At an equilibrium ${\mathbf{r}}_{\ast }$, the total force vanishes. A small transverse displacement must produce a transverse restoring force; a small axial displacement must also restore the bead. Downstream of a tight focus, the backward optical gradient force can oppose the forward optical scattering force. Equality at one point is insufficient: the total axial force must decrease as the bead moves downstream through that point. The equilibrium can therefore lie downstream of the intensity maximum.

For uncoupled overdamped coordinates, ${\zeta }_i\dot{\delta r_i}=-k_i\delta r_i$ with ${\zeta }_i>0$ decays only when each $k_i>0$. More generally, linearizing gives $\dot{\delta \mathbf{r}}=MJ\delta \mathbf{r}$, where $M$ is the hydrodynamic mobility matrix and $J_{ij}=\partial F_i/\partial r_j$ at equilibrium. All eigenvalues of $MJ$ must have negative real parts for linear asymptotic stability. For a conservative force with symmetric positive-definite mobility, a positive-definite Hessian matrix of the potential energy supplies this condition. A general optical force need not possess a global potential energy.

Increasing numerical aperture admits more strongly converging light rays and a steeper axial variation of the field. But a larger nominal numerical aperture alone does not establish trapping: illumination of the objective pupil, optical aberration, particle properties and the actual in-liquid focus also matter. A harmonic optical trap describes local confinement after this stability question has been answered; it does not predict capture from arbitrary positions or an infinite holding force.

A propulsive force exerted on a body by an engine or another momentum-transfer mechanism. In a vehicle moving through a fluid, its direction need not coincide with the vehicle's actual velocity, and a transverse component can steer the motion. The component along the velocity contributes mechanical power $\mathbf{T}\cdot \mathbf{v}$; a purely transverse component changes direction without doing instantaneous work. Rocket thrust is one particular mechanism and includes momentum carried by expelled material.

The instantaneous acceleration contribution from rocket thrust $T$ acting on remaining rocket mass $M$ is $a_T=T/M$. Gravity and other external forces contribute separately. A constant rocket thrust therefore produces an increasing thrust acceleration as rocket propellant is consumed.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

A tide is a periodic change in ocean level driven mainly by the differential gravitational forces of the Moon and Sun, together with the Earth's rotation. Their tidal forces act across the extended planet; a uniform acceleration shared by the whole Earth does not by itself generate a tide. Ocean-basin resonance and coastal geometry modify the local response.

A vehicle with centre of mass midway between level rails separated by d, at height h, follows a turn of radius R. Vertical force balance and cross-sectional moment balance give inner rail reaction $mg/2-m{v}^{2}h/(Rd)$. Its first zero occurs at the displayed speed. This is a toppling criterion under maintained lateral contact, not a guarantee against sliding or suspension instability.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

A path along a circle can have variable areal velocity. As an illustrative controlled trajectory, keep a point mass on a circle of radius $R$ with angleIts acceleration in polar coordinates and the required net force arewhere ${\mathbf{e}}_r$ and ${\mathbf{e}}_{\theta }$ are the radial and transverse unit vectors. A circular guide can supply the radial force, while a tangential drive supplies $mR\alpha$. HenceFor $R=m={\omega }_0=\alpha =1$ in illustrative units, the three equal time intervals $[0,0.5]$, $[0.5,1]$ and $[1,1.5]$ sweep areas $0.3125$, $0.4375$ and $0.5625$. Their difference comes from the transverse force's positive torque, even though the radius and path shape stay fixed. The area under each areal velocity graph segment equals the corresponding swept area.

A torsion balance measures a weak torque by suspending a body from a thin fiber and reading its twist angle. The fiber supports the body's gravitational weight while permitting rotation about the suspension axis. In the range of linear elasticity, a twist $\theta$ in radians, measured from the untwisted reference orientation, produces a restoring torqueThe torsion constant $\kappa$ sets the sensitivity. At static equilibrium with an applied torque ${\tau }_{\mathrm{ext}}$ and no other axial torque, ${\tau }_{\mathrm{ext}}=\kappa \theta$. A perpendicular force $F$ at lever arm $a$ therefore gives $\theta =aF/\kappa$. The torsion balance measures a turning effect: a force whose line of action passes through the suspension axis gives no axial torque. The Cavendish torsion balance is the gravitational application; Coulomb's law supplies an electrical one. ^[ref]

The original schematic separates two ways of reading the same torque. In deflection operation the top attachment stays fixed and the body rotates. In null operation the experimenter rotates the top attachment by $\alpha$ while returning the body's laboratory angle $\beta$ to its original value. If the initially untwisted configuration has $\alpha =\beta =0$, the fiber twist is $\theta =\beta -\alpha$, not $\beta$ alone. ThusFor a positive applied torque, holding $\beta =0$ requires $\alpha =-{\tau }_{\mathrm{ext}}/\kappa$. Keeping the body at the same angle keeps the lever arm and source geometry fixed, making comparison of different applied forces easier. Either readout still requires a calibrated torsion constant and a stable reference orientation.

An illustrative calculation shows the amplification: with $a=0.050\mathrm{m}$, $F=2.0\times 10^{-9}\mathrm{N}$, and $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, the applied torque is $1.0\times 10^{-10}\mathrm{N}\mathrm{m}$ and the equilibrium twist is $0.010\mathrm{rad}\simeq 0.573^{\circ }$. These are stipulated values, not measured data. Reducing the torsion constant tenfold increases this twist tenfold, provided the fiber remains in its calibrated linear elasticity range. The cost is a slower response, explained by the oscillation period in torsion constant.

A concrete laboratory experiment is PASCO's Coulomb balance: level its counterbalanced rod, discharge both conducting spheres, and set the unloaded zero. Charge the spheres at maximum separation; switch off the charging supply. Move the second sphere to a known center separation $R$, then turn the suspension dial until the rod returns to its index mark. Record the dial change and $R$, recharge, and repeat at several separations. The manual recommends a draft-free room, clean insulating supports, and recharging before each reading: leakage, especially in humid conditions, otherwise changes the electric charges during the sequence. At close separations the conducting spheres redistribute charge, so the point charge approximation needs correction. ^[ref]

For constant electric charges and a fixed perpendicular lever arm, the Coulomb's law prediction is $|\theta |\propto R^{-2}$. Plotting the logarithm of twist magnitude against the logarithm of separation gives $\log |\theta |=\text{constant}-2\log R$, so the predicted slope is $-2$. Absolute force measurement additionally needs the torsion constant; comparing ratios can eliminate it if it remains stable. Repeating a reference separation between other settings distinguishes a separation effect from a changing signal. A declining reading at the reference separation would invalidate the assumption of unchanged electric charges, even if a distance plot looks smooth.

The historical electrical example is Coulomb's first memoir of 1785. His apparatus enclosed a suspended needle with a small ball and a second ball; an adjustable head twisted the suspension. After charging the balls by contact, he reported angular separations $36^{\circ }$, $18^{\circ }$, and $8.5^{\circ }$ with head settings $0^{\circ }$, $126^{\circ }$, and $567^{\circ }$. The total twists were approximately $36^{\circ }$, $144^{\circ }$, and $576^{\circ }$: fourfold restoring torque accompanied approximately halved separation. These are his reported observations, not data from our diagrams. ^[ref] A precise reconstruction must use center distances and the changing lever arm, rather than identify angular separation exactly with distance; Alberto A. Martínez's replication explicitly analyzes that geometry. ^[ref]

Sensitivity also exposes changes of the apparatus itself. Fleischer and colleagues' cryogenic torsion balance used a temperature-controlled enclosure to reduce vacuum-vessel top-plate temperature fluctuations to approximately $\pm 0.01\mathrm{K}$, limiting changes in suspension geometry from thermal expansion. They used a mild bake-out near $40^{\circ }\mathrm{C}$ to speed initial fiber relaxation and reduce slow zero motion. Their optical readout compared the pendulum mirror with an internal reference mirror to reject shared optical motion. Room-temperature thermal radiation entering through the optical opening heated the pendulum; a cold filter reduced that load. The authors could not directly measure the pendulum's temperature without disturbing it, and explicitly distinguished the cold shield's temperature from the body's. Cooling therefore requires checking the actual noise and thermal environment, rather than assuming that the fiber and body share the shield's temperature. ^[ref]

A useful control strategy is to record unloaded readings and temperature before and after each signal sequence, recalibrate the torsion constant under the same conditions, and compare forward and reverse settings. The algebra of torsion balance zero drift shows precisely which offsets such comparisons can cancel. A fine angular scale alone does not establish a small measurement error.

The torsion constant, or torsional spring constant, is the positive restoring torque per unit twist angle of a complete suspension in its linear elasticity range:Its SI unit is $\mathrm{N}m/rad$. The minus sign describes restoration toward the untwisted orientation. The derivative specifies a local stiffness if the response is not globally linear; using one constant over a finite range additionally requires a straight calibration curve. Return to the same zero after unloading tests recoverable elastic deformation; repeat loading in both directions to check whether the response depends on its previous loading. Staying below permanent deformation is necessary, but does not alone prove proportionality between torque and twist.

For an ideal uniform circular wire of length $\ell$, radius $r$, and shear modulus $G_s$, the filament torsional stiffness is $C=G_s{J}_p$, with polar second moment of area $J_p=\pi r^4/2$. Uniform twist gives $\theta /\ell$ as the twist per unit length, soHere $U$ is stored elastic energy. Doubling the wire's radius multiplies the ideal torsion constant by $16$; doubling its length halves it. This explains why a long thin fiber is useful for weak torque measurement. The formula assumes linear elasticity with uniform material and circular section; it does not calibrate arbitrary ribbons, clamps, or a compound suspension. Within this ideal model, the equipartition theorem predicts an equilibrium thermal mean-square angular displacement $⟨{\theta }^2⟩=k_B\mathcal{T}/\kappa$, where $k_B$ is the Boltzmann constant and $\mathcal{T}$ the uniform absolute temperature. A softer suspension magnifies both the signal and its thermal fluctuations.

Static calibration supplies known torques without assuming the unknown interaction being measured. Apply a known perpendicular force $F$ at lever arm $a$, record the change from the unloaded twist, and useSeveral positive and negative settings test proportionality; fit the slope of applied torque against twist and retain a possible intercept. A nonzero intercept must not be silently absorbed into the slope. Convert degrees to radians before reporting $\mathrm{N}m/rad$. PASCO's procedure turns the Coulomb balance on its side and adds small known masses to generate a gravitational weight at the sphere's lever arm, returning the rod to its index mark by turning the dial. Its tabulated coefficient is force per degree: multiply by the lever arm and $180/\pi$ to obtain our torsion constant. ^[ref]

The figure's ideal static calibration uses $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$: an applied torque of $2.0\times 10^{-10}\mathrm{N}\mathrm{m}$ gives $0.020\mathrm{rad}$. Both signs lie on the same straight line. The plotted points are calculated examples, not experimental observations; they cannot demonstrate a real fiber's linearity.

Dynamic calibration instead measures the oscillation period $T_0$ with known total suspended moment of inertia $I$. For negligible damping, small free oscillations about a fixed reference obey simple harmonic motion:Include the rod, mirrors, and any attachments in $I$ unless their contributions are demonstrably negligible. Keep the applied torque absent or independent of angle. If an external torque has derivative $g_{\tau }=d{\tau }_{\mathrm{ext}}/d\theta$ at equilibrium, linearization gives effective stiffness $\kappa -g_{\tau }$: timing alone then measures that combination, and stability requires $\kappa -g_{\tau }>0$. A constant applied torque merely shifts equilibrium. Non-negligible damping changes the measured period; damped torsional oscillation gives a correction under a specified damping model. Agreement between static and dynamic calibrations is a useful check on the assumed moment of inertia, reference stability, and restoring law, rather than a reason to ignore their assumptions.

Traction is the force per unit area transmitted across an oriented surface. For unit normal vector $\mathbf{n}$, the stress tensor gives $\mathbf{t}=\mathbf{\sigma }\mathbf{n}$. Its normal component is normal stress, and its tangential component is shear stress.

If the origin of a nonrotating reference frame has inertial position $\mathbf{a}(t)$ and a particle has relative position $\mathbf{R}$, then $\mathbf{r}=\mathbf{R}+\mathbf{a}$. Newton's second law becomes $m\ddot{\mathbf{R}}=\mathbf{F}-m\ddot{\mathbf{a}}$. The effective inertial force is therefore $-m\ddot{\mathbf{a}}$. No rotation-dependent terms arise because the coordinate axes remain parallel to the inertial axes.

For a fixed rigid-body shape and origin in an unbounded stationary Stokes flow, the Lorentz reciprocal theorem makes the off-diagonal blocks of the six-dimensional hydrodynamic resistance matrix transposes. Here $\mathbf{F},\mathbf{T}$ are the fluid's force and torque on the body. If every pure translation has zero torque about the chosen origin, then $B^T=0$, hence $B=0$: every pure rotation about that origin has zero resultant force. This conditional statement does not require such an origin to exist for every body.

Two fixed outer wires at $x=\pm d$ attract a middle wire carrying the same electric current $I$. If the middle wire is displaced in the perpendicular direction $y$, the net magnetic force per unit length is $-{\mu }_0{I}^{2}y/[\pi (d^2+y^2)]$. For $|y|\ll d$, the wire's linear mass density ${\mu }_m$ gives the displayed simple harmonic motion frequency. Motion towards either outer wire in the original plane is unstable. Gravity must be absent from or balanced in the oscillation direction.

The component of the internal resultant force perpendicular to an elastic beam's tangent. For a planar elastic beam, one consistent sign convention has bending moment $M=-B{w}^{\prime \prime }$ and $M^{\prime }=N$, so $N=-B{w}^{\prime \prime \prime }$. Reversing the section or bending moment convention reverses the corresponding signs.

A taut spring of equilibrium length $L$ and initial tension $T_0$ whose ends acquire a small relative transverse displacement $y$ tilts through $y/L$. Its leading transverse force is therefore $-T_{0}y/L$. The length change is quadratic in $y$, so using only incremental axial extension misses this first-order force. A Hooke spring with natural length $l_0$ has $T_0=k(L-l_0)$ and $k_{\perp }=k(1-l_0/L)$. The unstretched limit has no linear transverse stiffness under this geometry.