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Virtual displacement and generalised forces 
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#1
Apr2612, 04:08 PM

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? 


#2
Apr2712, 01:43 AM

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P: 9,645

F[itex]_{k}[/itex]=δW/δq[itex]_{k}[/itex] looks wrong to me. Probably a typo of δ for ∂ in both places. ∂W/∂q[itex]_{k}[/itex] would be the limit of δW/δq[itex]_{k}[/itex] as δq[itex]_{k}[/itex] tends to zero.



#3
Apr2712, 09:22 AM

P: 409




#4
Apr2712, 11:16 AM

P: 463

Virtual displacement and generalised forces



#5
Apr2712, 11:54 AM

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:
[itex]\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}}[/itex] 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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