Virtual displacement and generalised forces

 P: 463 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?
 Homework Sci Advisor HW Helper Thanks P: 9,849 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.
P: 409
 Quote by Gregg ##\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.

 Quote by Gregg 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.

P: 463
Virtual displacement and generalised forces

 Quote by 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.
The deltas are for virtual work / displacement not small change in ##q_k## etc.

 Quote by Hassan2 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?
 P: 409 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.

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