Virtual displacement and generalised forces

[Gregg]

I am unsure about the virtual displacement and work definition even after looking through the definition and seeming to understand it. If we have

$\delta W = \displaystyle \sum_{i} \vec{F}_i \cdot \delta \vec{r}_i$,

I can use,

$\delta \vec{r}_i = \sum_{i} \frac{\partial \vec{r}_i}{\partial q_k} \delta q_k$,

and get to

$\delta W = \displaystyle \sum_{k}\sum_{i} \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_k} \delta q_k$.

So,

$\delta W = \sum_{k} \mathcal{F}_k \delta q_k$.

Then in the derivation it says that this imples that

$\sum_{i} \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_k}= \mathcal{F}_k = \frac{ \delta W}{\delta q_k}$.


I thought that $\delta W = \sum_{k} \frac{\partial W}{\partial q_k} \delta q_k$ and $\mathcal{F}_k = \frac{\partial W}{\partial q_k}$. This seems to imply that:

$\delta W = \sum_{i} \frac{\delta W}{\delta q_k} \delta q_k$,

so where is the distinction, because I can't work out when to use the deltas or ds?

 

[haruspex]

F_{k}=δW/δq_{k} looks wrong to me. Probably a typo of δ for ∂ in both places. ∂W/∂q_{k} would be the limit of δW/δq_{k} as δq_{k} tends to zero.

 

[Hassan2]

$\sum_{i} \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_k}= \mathcal{F}_k = \frac{ \delta W}{\delta q_k}$.

Here $\delta W$ is the work done by all forces for virtual displacement $\delta q_{k}$ while no change in other generalized coordinates.

I thought that $\delta W = \sum_{k} \frac{\partial W}{\partial q_k} \delta q_k$

Don't forget that W depends on forces too, and the forces may depend on $q_{k}$. Partial derivative is not correct here then.

 

[Gregg]

F_{k}=δW/δq_{k} looks wrong to me. Probably a typo of δ for ∂ in both places. ∂W/∂q_{k} would be the limit of δW/δq_{k} as δq_{k} tends to zero.

The deltas are for virtual work / displacement not small change in $q_k$ etc.

Here $\delta W$ is the work done by all forces for virtual displacement $\delta q_{k}$ while no change in other generalized coordinates.

Don't forget that W depends on forces too, and the forces may depend on $q_{k}$. Partial derivative is not correct here then.

Right, so what is actually implied by $\frac{\delta W}{\delta q_{k}}$? If partial derivative is not correct, is there a correct way in terms of the partial derivatices to express this? and is this the formal way to do so with the $\delta$s?

 

[Hassan2]

I'm not sure but in my opinion, since δqk is arbitrary, we can set infinitesimal value to it and write the ratio in terms of partial derivatives:

\frac{\delta W}{\delta q_{k}} \rightarrow \frac{\partial W}{\partial q_{k}}+\sum_{i}\frac{\partial W}{\partial F_{i}}\frac{\partial F_{i}}{\partial q_{k}}

However I have never seem such formula perhaps because it's not useful. We often don't have W as a function readily.