I know that there are many posts about virtual displacement, but I want to answer the question if of: is virtual displacement is always needed to get the same results? I am going through a PDF by Subhankar Ray & J. Shamanna on virtual work Here. So the main idea is how far we can get using allowed displacement opposed to virtual displacement:
If the constraints are not time dependent, then we have the similar property
\begin{equation}\tag{a} \sum^N_{k=1}(m_k\textbf{a}_k - \textbf{F}_k)\cdot d\textbf{r}_k = 0 \end{equation} 2.Using Generalized Coordinates
We then read on and we note that because of the constraints, not all the $d\textbf{r}_k$ are independent of eachother and so we then want to change it to independent parameters to remove this issue.
So to change the parameters we use
\begin{equation}
d\textbf{r}_k = \sum^N_{k=1}\frac{\partial \textbf{r}_k}{\partial q_j}dq_j +\frac{\partial\textbf{r}_k}{\partial t}dt
\end{equation}
opposed to
\begin{equation}\tag{23}
\delta\textbf{r}_k = \sum^N_{k=1}\frac{\partial \textbf{r}_k}{\partial q_j}\delta q_j
\end{equation}
thus we will get for $({a})$,
\begin{equation}\tag{b} \sum^N_{k=1}(m_k\textbf{a}_k - \textbf{F}_k)\cdot \left(\sum^N_{k=1}\frac{\partial \textbf{r}_k}{\partial q_j}dq_j +\frac{\partial\textbf{r}_k}{\partial t}dt\right) = 0 \end{equation}
My Question
From above, $(b)$, I am having a hard time picturing this. I feel like the "$\frac{\partial\textbf{r}_k}{\partial t}dt$" part may cause issues(just like when the constraint equation was time dependent and $d\textbf{r}_k$ was no longer tangential to the constraint equation)-or am I worrying about nothing. I would be grateful for any insight you can offer.
Edit Like I feel like that \begin{equation}\tag{b} \sum^N_{k=1}(m_k\textbf{a}_k - \textbf{F}_k)\cdot \left(\sum^N_{k=1}\frac{\partial \textbf{r}_k}{\partial q_j}dq_j \right) = 0 \end{equation}
should hold as we are moving the object with the idea that time is frozen.
Is there any sense to this line of reasoning?
