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Force balance equates the rate of change of momentum to the sum of applied forces. A static continuum segment with internal resultant $\mathbf{F}(s)$ and force per unit length $\mathbf{f}(s)$ obeys ${\partial }_s\mathbf{F}+\mathbf{f}=0$.

In a planar potential flow, superpose a positive-$x$ uniform flow of speed $U$ and a line source of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular cylinder with zero circulation. The Milne-Thomson circle theorem gives the complex potential of a two-dimensional potential flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ be the complex velocity. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite residues of $q^2$ is $2Um$. At the external pole $b$, write $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.

A force per unit length describes a distributed force on a filament and has units $\mathrm{N}{\mathrm{m}}^{-1}$. For a charged filament with linear charge density $\lambda$ and longitudinal current $I$, the external Lorentz force is $\mathbf{f}=\lambda \mathbf{E}+I\hat{\mathbf{t}}\times \mathbf{B}$. Length, density and force must all refer to the same measurement frame. An infinite uniform filament generally has a finite force density but no finite total force.

In the upper half-plane, a line source of outward volume flux $Q$ at height $a$ above an impermeable wall has an equal image source below it. On the wall $u=Qx/[\pi (x^2+a^2)]$ and the normal velocity vanishes. The steady Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{x}^2/[2{\pi }^2(x^2+a^2{)}^2]$. Sincethe ambient-relative force per unit length pulls the wall toward the source with the displayed magnitude. The torque about the source's projection onto the wall is zero in the symmetric Cauchy principal value sense. Individual one-sided moments diverge logarithmically. This planar result differs from the three-dimensional hydrodynamic attraction of a plane wall to a point source.

Orient $C$ counterclockwise around the body. For an arc element $ds$, the body-outward unit normal vector obeys $\mathbf{n}ds=(dy,-dx)$. The pressure acts normally into the body, so its force per unit length in the suppressed third direction isIntegrating givesTaking positive torque to mean counterclockwise about the origin, the infinitesimal moment is $dM=xd{F}_y-yd{F}_x=p(xdx+ydy)$. HenceAll three expressions are per unit span. They are the normal-pressure contributions assumed by the inviscid model; a viscous traction would require extra terms. A constant pressure contributes zero net force and torque on a closed body, since $\oint dx=\oint dy=\oint d(x^2+y^2)=0$.

Since the velocity tends to zero at infinity, the steady Bernoulli equation gives $p+\rho |\mathbf{u}{|}^2/2=p_{\infty }$. On the wall, therefore,Compute the finite source-induced force relative to uniform ambient pressure, or equivalently use the gauge pressure $p-p_{\infty }$. The fluid lies above the wall, so its normal force on the wall points in $-{\mathbf{e}}_y$ for positive pressure. The negative excess pressure found here consequently produces an upward net force, toward the source.

For the symmetric section $-R\le x\le R$, there is no horizontal force, and the vertical excess force isHere an antiderivative of $x^2/(x^2+a^2{)}^2$ is $(1/(2a))\arctan (x/a)-x/[2(x^2+a^2)]$. Taking $R\to \infty$ gives the hydrodynamic attraction of a wall to a line source:Each component is a force per unit length of the wall in the out-of-plane direction. The result is an attraction even though the source sends fluid away from itself: faster tangential motion reduces the wall pressure.

The excess torque about the origin isfor every $R$, because the integrand is odd. Thus the symmetric infinite-wall limit isThe symmetric prescription matters: the separate one-sided moment integrals grow logarithmically, whereas the force integral converges absolutely. This is the appropriate Cauchy principal value for the infinite wall. If one instead asks for the absolute fluid force with no balancing ambient pressure on the other side, its vertical component on a finite section is $-2R{p}_{\infty }+F_y(R)$ and has no finite infinite-wall limit. The finite answer above is the flow-induced load.

Let $F_x,F_y$ denote the force per unit length exerted by the fluid on the cylinder, and let $\rho$ be the constant mass density. For a counterclockwise contour $C$ around the unit circle, the Blasius force theorem givesAn efficient use of the residue theorem finds the sum of the enclosed residues by subtracting the external pole at $b$ from the sum of all finite residues. At large $z$,The sum of all finite residues of $q^2$ is therefore $2U$. At $b$, write $q=(z-b{)}^{-1}+H(z)$, with $H$ a holomorphic function near $b$. ThenThusUsing $a=1/b$ and $1/(b-a)-1/b=1/[b(b^2-1)]$ givesThe vanishing vertical force also follows from reflection symmetry about the real axis. This force on a unit cylinder due to a uniform flow and an external source uses the counterclockwise orientation in the Blasius force theorem to fix the sign of the horizontal force.

The infinitely long filaments have no finite total force; the meaningful result is force per unit length in the laboratory. Evaluate only A's field at B, excluding B's singular self-field. B is at $y=0,z=-d/2$, so defineIn both cases A moves at $+v\hat{\mathbf{x}}$, hence at BFor a filament with linear charge density ${\lambda }_B$ and convective electric current $I_B$, the Lorentz force per laboratory length isIn Case 1, ${\lambda }_B=\gamma \eta$ and $I_B=\gamma \eta v$. Electric repulsion and magnetic attraction givesoThe cancellation uses ${\gamma }^2(1-{\beta }^2)=1$. In Case 2, B is stationary, ${\lambda }_B=\eta$ and $I_B=0$, givingBoth results have units $\mathrm{N}{\mathrm{m}}^{-1}$ and are repulsive, away from A, irrespective of the common sign of $\eta$. In Case 1 the velocity independence of force per laboratory length does not violate the transverse force transformation: a material segment of proper length ${\ell }_0$ occupies laboratory length ${\ell }_0/\gamma$, so its force is the rest-frame force divided by $\gamma$. The calculation interprets the stated $\eta$ as each filament's final proper density; holding a laboratory density fixed during acceleration would be a different preparation.

For one wire, Ampère's circuital law on a concentric circle of radius $r$ givesThe other wire experiences the Lorentz force on its electric current, $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Its direction is perpendicular to this magnetic field, so the force per unit length isThe right-hand rule gives attraction for parallel electric currents in the same direction and repulsion for opposite directions.

For the upper wire, the upward magnetic force per unit length balances its weight per unit length, $\lambda g$, where $g$ is the gravitational acceleration. ThereforeThis assumes the lower wire is held fixed and both wires remain straight. Within the vertical-motion model, increasing the separation weakens the upward magnetic force, so this equilibrium is vertically restoring.

For a particle instantaneously at rest in a primed frame moving at $v\hat{\mathbf{x}}$ relative to the laboratory, the ordinary three-force transforms as $F_{\parallel }=F_{\parallel }^{\prime }$ and $F_{\perp }=F_{\perp }^{\prime }/\gamma$. The spatial components of the four-force are ${\gamma }_u\mathbf{F}$, whereas its time component is ${\gamma }_u\mathbf{F}\cdot \mathbf{u}/c$. For a co-moving material segment, length contraction also changes its laboratory length, so transforming total force is not the same as transforming force per unit length.