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I have some confusion about the "Reversed effective force" as it appears in the derivation of D'Alembert's principle. In Goldstein d'Alembert's principle is given as:

$(F-\dot{p}) \cdot \delta r = 0$

First I have sources that seem to be contradictory.

http://books.google.com/books?id=o8RvD3X8ur8C&printsec=frontcover#v=onepage&q&f=false

This book, on page 8, indicates that "reversed effective forces" are not real forces. However, this book indicates otherwise.

http://books.google.com/books?id=4wkLl4NvmWAC&pg=PA345&lpg=PA345&dq=what+is+a+reversed+effective+force&source=bl&ots=obL2JFMwAw&sig=lpf5LvJJM_iNvyHDhXAV0gIaDO0&hl=en&sa=X&ei=FkMsVeG_C8ahNv3vgPAE&ved=0CCYQ6AEwAQ#v=onepage&q=what%20is%20a%20reversed%20effective%20force&f=false

On page 345, this book indicates "this is the force exerted by the moving body to resist the change in its state.

I think that the former source is probably correct, but I lack good intuition of D'Alembert's principle. For reference the reversed effective force is represented by $-ma$ or $-\dot{p}$. My best guess as to what the reversed effective force is that it it the force that is required to appear if you view the system from the perspective of the accelerating body. Is this correct? If it is then how is d'Alembert's principle applicable to inertial systems?

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  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Sep 21 '17 at 17:34
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It is possible to interpret D'Alembert's principle through the requirement that any particle is always in equilibrium in its own rest frame; it is, after all, at rest in this frame. However, as this frame is necessarily accelerating with respect to any inertial frame, there is an additional inertial force $-m\ddot{\mathbf x}$ on the particle. The requirement of static equilibrium in this frame at every instant now reads: $$\left(\mathbf{F} - m\ddot{\mathbf x}\right)\cdot\delta{\mathbf x}=0$$ where $\mathbf F$ is the resultant of the "other" i.e. non-inertial forces acting on it, as seen from any inertial frame. An important piece of interpretation here is that $\ddot{\mathbf x}$ is not the acceleration of the particle in the current frame of reference (in which it is zero), but its acceleration as seen from an inertial frame.

Now, observe that all systems will agree on what the rest frame of the particle is. Therefore, all frames of reference (given the above interpretation of $\ddot{\mathbf x}$) will agree on the above equation.

Now, as it happens, $\ddot{\mathbf x}$ is also the observed coordinate acceleration in all inertial frames, therefore one may simply use the equation directly in any inertial frame of reference. For accelerated frames (of acceleration $\mathbf a$), the coordinate acceleration $\ddot{\mathbf x}'$ is related to $\ddot{\mathbf x}$ by $$\ddot{\mathbf x}' = \ddot{\mathbf x} - \mathbf{a}$$ Therefore, D'Alembert's principle in terms of the acceleration observed in the frame is: $$\left(\mathbf{F} - m\mathbf{a}- m\ddot{\mathbf x}'\right)\cdot\delta{\mathbf x'} = 0$$

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  • $\begingroup$ By inertial force you mean a force like a Coriolis force meant to rectify having a reference frame in a rotating frame, except the inertial force in this case is meant to rectify the particle being at rest in its own form. Is this understanding correct? $\endgroup$ – Reid Erdwien May 3 '15 at 0:21
  • $\begingroup$ Yes, the inertial force in this case is required to correct for choosing an accelerating reference frame (the particle's rest frame) instead of an inertial frame. $\endgroup$ – AV23 May 3 '15 at 7:49
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It's the net resultant that a force would react with on application of a certain change in momentum in the body, on which the force is applied.

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    $\begingroup$ What is the force acting on then? What is causing the force? This answer is unclear. $\endgroup$ – Reid Erdwien Apr 15 '15 at 18:22
  • $\begingroup$ Ried Erdwien: I have tried to convey what reversed effective force is, the initial or initiating force (for it) can be anything. $\endgroup$ – Abhijeet Apr 16 '15 at 5:36
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$\dot{p}$ in d'Alembert's formula represents the derivative of the object's momentum with respect to time.

$\delta r$ represents what would have been the displacement of the object during the infinitesimal interval of $\dot{p}$.

So what d'Alembert is saying is that Force minus the effect of Force = zero.

If you analyze a system this way, you can treat it as though it is in equilibrium, because you have zeroed out its change.

Here's a good article on d'Alembert's original formulation: http://homes.chass.utoronto.ca/~cfraser/Dalembert.pdf

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