Line of action

For a nonzero force applied at position ${\mathbf{r}}_0$, its line of action is the straight line through that point parallel to the force:The force direction and line position together determine its torque about a chosen origin. They contain more mechanical information than the magnitude alone.

The moment about the origin is $\mathbf{\tau }={\mathbf{r}}_0\times \mathbf{F}$. Moving the application point by a displacement parallel to $\mathbf{F}$ leaves the moment unchanged because $\Delta \mathbf{r}\times \mathbf{F}=0$. The principle of transmissibility uses this equivalence for the external motion or equilibrium of an ideal rigid body. The lever-arm formulation expresses the same geometry: the torque magnitude is $Fd$, where $d$ is perpendicular distance from the origin to the line.^[ref]

For a constructed two-dimensional example, a force $(0,10)\mathrm{N}$ acts at $(0.30,0.20)\mathrm{m}$. Its line is $x=0.30\mathrm{m}$ and its moment is ${\tau }_z=0.30(10)=3.0\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80)\mathrm{m}$ changes neither force nor moment. Moving it instead to $x=0.50\mathrm{m}$ gives $5.0\mathrm{N}\mathrm{m}$, so the unchanged force magnitude does not imply the same rigid-body effect.

An equal and opposite pair on different parallel lines forms a couple: the resultant force is zero but a moment remains. Consequently, not every loading system can be replaced by one force with one line of action. A free-body diagram should retain the relevant forces and couples.

Transmissibility is an external rigid-body statement. Applying a load at a different location on a deformable object can change internal stress, deformation and local contact even if the resultant and moment are preserved. A static load-cell comparison and a deformation measurement can therefore answer different questions about apparently equivalent loading arrangements.

The principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.