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Loading noise is the acoustic dipole radiation of the force exerted by a body on its surrounding fluid. For a compact, slowly moving source, the radiating acoustic density perturbation is $n\cdot \dot{\mathcal{F}}(t-R/c_0)/(4\pi c_0^{3}R)$, where $\mathcal{F}$ is the total force. If this total is constant, finite source-delay corrections can still create higher acoustic multipole sources.

For a hollow slender viscous thread, $A=R_e^2-R_i^2$ and both surfaces contribute axial surface tension. The liquid cut force relative to exterior pressure, including these surface terms, is $\pi [3\mu A{w}_z+(p_i-p_e)R_i^2+\gamma (R_e+R_i)]$. The axial traction on the sloping inner gas surface contributes $-\pi (p_i-p_e)(R_i^2{)}_z$ to force balance. For constant gas pressures the extra derivative cancels, giving $3\mu (A{w}_z{)}_z+\rho gA+\gamma (R_e+R_i{)}_z=0$. The displayed $T$ is this effective tension after accounting for gas-pressure work, not the unadjusted liquid cut force.

For an axisymmetric slender viscous thread, the Young–Laplace equation and radial extensional viscous stress give $p=p_e+\gamma /a-\mu w_z$. The axial liquid stress relative to ambient pressure is $3\mu w_z-\gamma /a$. Its cross-sectional integral plus the circumferential surface tension force $2\pi \gamma a$ gives the displayed tensile force. Thus the net capillary contribution is $\pi \gamma a$, and axial force balance with downward gravity is $T_z+\pi \rho g{a}^2=0$.

For uniform circular motion of radius $R$ and angular speed $\omega$, the acceleration is directed toward the centre and has magnitude ${\omega }^{2}R=v^2/R$. Writing $\mathbf{r}$ for the position relative to the centre gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. The required resultant force is $m\ddot{\mathbf{r}}$; it may be supplied by a normal force, gravity, or another physical interaction.

The static friction force at impending sliding has magnitude ${\mu }_s$ times the normal contact force. In a simple grain model this coefficient represents resistance to downstream motion; the model excludes lift and detailed contact torques.

Two diametrically opposite rotating blades with equal tangential loads have zero total rotating force. Their leading compact acoustic dipoles cancel, but their retarded positions and moving-surface retarded Jacobians leave an acoustic quadrupole. For line load $2Fr/a^2$ and angular speed $\Omega$, the first radiating density is $2F{\Omega }^{2}a{\sin }^2\theta \sin (2\Omega {\tau }_0)/(3\pi c_0^{4}R)$. The twofold geometry gives frequency $2\Omega$, while projection into the rotor plane gives the amplitude directivity ${\sin }^2\theta$.

A control volume is a chosen region over which fluid mass conservation and momentum balance are integrated. It may be fixed or move with a prescribed velocity. Boundary fluxes must use velocity relative to the moving boundary. In Stokes flow, negligible inertia reduces the momentum balance to the sum of body forces and boundary tractions, making a carefully chosen control volume useful for finding drag without integrating directly over a curved body.

An asymmetric periodic planar filament can have nonzero cross-tangent average $\delta$ despite periodic vertical position. Its translational resistance per arclength period is $c_{\perp }\Lambda [I+(\rho -1)M]$, with $M_{xx}=\beta$, $M_{xy}=\delta$ and $M_{yy}=1-\beta$. Full force balance can therefore yield transverse translation, so an axial-only propulsion formula needs a symmetry or constraint assumption.

The in-plane tensile force per unit transverse length in a thin elastic sheet produces the linear transverse restoring force $T{\eta }_{xx}$ per unit area. A spatially uniform displacement has zero curvature and hence no elastic-sheet tension force. This explains why infinite elastic-sheet tension blocks fixed nonzero-wavenumber acoustic membrane waves but does not make the uniform mode rigid. The sheet's inertia, by contrast, acts also on uniform oscillations.

The electric field is the force per unit charge on a stationary test particle.

For the clamped-free beam under uniform load and endpoint force, the additional minimized internal energy isIts derivative is the endpoint displacement caused by the endpoint force:

For uniform angular velocity $\omega$, potential $\varphi$, and additional forces doing no work in the rotating frame,The Coriolis force does no work; the centrifugal potential accounts for the remaining apparent force. This energy is conserved precisely when the potential has no explicit time dependence in the rotating coordinates. An inertial time-independent conservative force need not meet that condition unless its potential is also invariant under the frame's rotation. The balance prevents applying a rotating-frame conservation law to a nonaxisymmetric stationary inertial potential.