classical mechanics - What is a "Reversed Effective Force"?

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?

Answer

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: $$\left(\mathbf{F} - m\mathbf{a}- m\ddot{\mathbf x}'\right)\cdot\delta{\mathbf x'} = 0$$