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Virtual displacement and generalised forces

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Gregg
#1
Apr26-12, 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?
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haruspex
#2
Apr27-12, 01:43 AM
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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.
Hassan2
#3
Apr27-12, 09:22 AM
P: 409
Quote Quote by Gregg View Post
##\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 Quote by Gregg View Post
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
#4
Apr27-12, 11:16 AM
P: 463
Virtual displacement and generalised forces

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

Quote Quote by Hassan2 View Post
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
#5
Apr27-12, 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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