= Wrench
{disambiguate=mechanics}
{title2=$\mathcal W=(\boldsymbol\tau,\mathbf F)$}
A <wrench (mechanics)> represents a system of loads on a <rigid body> by its resultant <force> and resultant <torque> about a stated reference point. In the moment-first convention, write
$$
\mathcal W_O=\begin{pmatrix}\boldsymbol\tau_O\\\mathbf F\end{pmatrix},\qquad
\mathbf F=\sum_i\mathbf F_i,\quad
\boldsymbol\tau_O=\sum_i\mathbf r_i\times\mathbf F_i+\sum_j\mathbf C_j.
$$
The $\mathbf C_j$ are applied <force couples>. The six entries package two different physical units: newtons for force and newton-metres for torque. The primary robotics text and its author video use this representation to transform measured forces and moments between reference frames.<Modern Robotics: Mechanics, Planning, and Control by Kevin M. Lynch and Frank C. Park>[§3.4, Wrenches; author video transcript, six-vector and frame transformation]{ref}
An original example has a force $(0,10,0)\,\mathrm N$ applied at $(0.30,0,0)\,\mathrm m$, plus a free couple $(0,0,2)\,\mathrm{N\,m}$. Its resultant torque about the origin is $(0,0,5)\,\mathrm{N\,m}$. Shift the reference point to $P=(0.10,0,0)\,\mathrm m$ without rotating the axes. Then
$$
\boldsymbol\tau_P=\boldsymbol\tau_O-\mathbf r_{OP}\times\mathbf F
=(0,0,4)\,\mathrm{N\,m}.
$$
The physical loading has not changed, but its torque coordinates depend on the reference point. A rotation of axes additionally rotates the vector components.
The dual velocity object is a <rigid-body twist>, ordered as $(\boldsymbol\omega,\mathbf v_O)$. Their pairing is mechanical <power>:
$$
\mathcal P=\boldsymbol\tau_O\cdot\boldsymbol\omega+\mathbf F\cdot\mathbf v_O.
$$
Since $\mathbf v_P=\mathbf v_O+\boldsymbol\omega\times\mathbf r_{OP}$, substituting the shifted torque cancels the extra scalar triple product and preserves power. This explains why changing the force-sensor frame must transform torque and force together.
A nonzero resultant force does not always replace the whole wrench by one force alone. Define <wrench pitch>
$$
h=\frac{\mathbf F\cdot\boldsymbol\tau_O}{\|\mathbf F\|^2},\qquad
\mathbf r_0=\frac{\mathbf F\times\boldsymbol\tau_O}{\|\mathbf F\|^2}.
$$
Then $\mathbf r_0\times\mathbf F=\boldsymbol\tau_O-h\mathbf F$. A force along the central line through $\mathbf r_0$ leaves the parallel residual couple $h\mathbf F$. For $\mathbf F=(0,0,10)\,\mathrm N$ and $\boldsymbol\tau_O=(2,0,3)\,\mathrm{N\,m}$, $\mathbf r_0=(0,0.20,0)\,\mathrm m$ and $h=0.30\,\mathrm m$. The remaining axial couple is $(0,0,3)\,\mathrm{N\,m}$. If the resultant force is zero, a nonzero pure couple must instead be retained directly.
A proposed check uses a calibrated force-torque sensor and known lever arms to compare the reported six components at two reference points. Agreement tests the transformation and loading model; it does not establish that two externally equivalent loads produce identical internal <stress> in a deformable part.
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