$\newcommand{\Dev}{\operatorname{Dev}}$ $\newcommand{\dev}{\operatorname{dev}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\Grad}{\operatorname{Grad}}$ $\newcommand{\divg}{\operatorname{div}}$ $\newcommand{\Ei}{\operatorname{Ei}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\Divg}{\operatorname{Div}}$ $\newcommand{\cay}{\operatorname{cay}}$ $\newcommand{\rot}{\operatorname{rot}}$

When a joint has a translational degree of freedom along $\mathbf{n}^{\left(1\right)}$ , there may be a need to prescribe the force $f$ along that direction. This means that the reaction force $\mathbf{f}^{\left(1\right)}$ may be supplemented with the force $f\,\mathbf{n}^{\left(1\right)}$ , $$\begin{equation} \mathbf{f}^{\left(1\right)}=f\mathbf{n}^{\left(1\right)}\,,\label{eq:pfj-1} \end{equation}$$ and the linearization produces $$\begin{equation} D\mathbf{f}^{\left(1\right)}=-f\,\hat{\mathbf{n}}^{\left(1\right)}\cdot\Delta\boldsymbol{\theta}^{\left(1\right)}\,.\label{eq:pfj-2} \end{equation}$$ Thus, $$\begin{equation} \boxed{\mathbf{K}_{f\theta}^{\left(1\right)}=-f\,\hat{\mathbf{n}}^{\left(1\right)}}\,,\label{eq:pfj-3} \end{equation}$$ and $$\begin{equation} \boxed{\mathbf{K}_{fr}^{\left(1\right)}=\mathbf{K}_{fr}^{\left(2\right)}=\mathbf{K}_{f\theta}^{\left(2\right)}=\mathbf{0}}\,.\label{eq:pfj-4} \end{equation}$$