4
\begin{eqnarray}
F_{\theta}=[(2m+M)R^2+2J_w+2n^2J_m]\ddot{\theta}+(MLR \cos ( \psi - \zeta) -2n^2J_m)  \ddot{\psi}  -MLR \dot{\psi}^2 sin ( \psi - \zeta) + (M+m)g \sin \zeta  \label{eqn13}\\
F_{\psi} = (MLR cos \psi-2J_m)\ddot{\theta}+(ML^2+J_{ \psi}+2J_m)\ddot{ \psi} -MgL sin \psi-ML^2\dot{ \phi}^2 sin \psi cos \psi\label{eqn14} 
\end{eqnarray}
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  • 1
    Could you better explain where the problem is? In any case, don't use eqnarray, see eqnarray vs align
    – egreg
    Jun 9, 2016 at 23:35

1 Answer 1

7

Is this what you are looking for?

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}



\begin{document}

Some text before the equations.
\begin{align}
    F_{\theta} &=
        \begin{aligned}[t]
            &\bigl[(2m+M)R^2+2J_w+2n^2J_m\bigr]\ddot{\theta} \\
            &\quad +\bigl(MLR \cos ( \psi - \zeta) -2n^2J_m\bigr) \ddot{\psi} \\
            &\quad -MLR \dot{\psi}^2 \sin ( \psi - \zeta)
                    + (M+m)g \sin \zeta
        \end{aligned}
        \label{eqn13} \\[\jot]
    F_{\psi} &=
        \begin{aligned}[t]
            &(MLR \cos \psi-2J_m)\ddot{\theta}
                    +(ML^2+J_{ \psi}+2J_m) \ddot{\psi} \\
            &\quad -MgL \sin \psi-ML^2\dot{ \phi}^2 \sin \psi \cos \psi
        \end{aligned}
        \label{eqn14} 
\end{align}
Some text after the equations.

\end{document}

Output:

Output of the code

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  • 2
    Don't forget to use \sin \cos.
    – Sigur
    Jun 10, 2016 at 0:03

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