Question regarding d'Alembert's principle I am new to the subject of Classical Mechanics, I started with Principle of Least Action and now I am learning d'Alembert's Principle. Forgive my ignorance ,but I find it counterintuitive, according to Britannica:
https://www.britannica.com/science/dAlemberts-principle
The second law states that the force $F$ acting on a body is equal to the product of the mass m and acceleration a of the body, or $F = ma$; in d’Alembert’s form, the force F plus the negative of the mass m times acceleration a of the body is equal to zero: $F - ma = 0$. In other words, the body is in equilibrium under the action of the real force $F$ and the fictitious force $-ma$. The fictitious force is also called an inertial force and a reversed effective force.
But what is happening here? why isn't there any mention of non inertial frame,if the force is fictitious then how can we consider it regardless of the frame?
What is happening here, is there a word missing here (Non-Inertial Frame), in mechanics up to the level of AP ,we are told to ignore Fictitious Forces when we consider Inertial Frames.
Is there an intuitive explanation for d'Alembert Principle?
 A: In the way that the link says, the principle doesn't add much. But the wikipedia article about D'Alembert Principle expains better the idea. If a vector is zero, its dot product with another vector is also zero.
$(\mathbf F - m\mathbf a)\boldsymbol {.\delta r} = 0$
That expression transforms a vectorial equation in a scalar one, by projecting the vectors over some convenient direction. For example, for a simple pendulum, we can choose the tangential direction of the mass, because it is the only possible displacement direction:

$(\mathbf F - m\mathbf a)\boldsymbol {.\delta r} = (m\mathbf g + \mathbf T - m\mathbf a)\boldsymbol {.\delta r} = 0$
As $\mathbf T$ is orthogonal to $\boldsymbol \delta r$, its dot product is zero:
$m\mathbf g\boldsymbol {.\delta r} - m\mathbf a\boldsymbol {.\delta r} = 0$
$$m\mathbf g\boldsymbol {.\delta r} = mgsin(\theta)L\delta \theta = mgLsin(\theta)\omega \delta t$$
$$m\mathbf a\boldsymbol {.\delta r} = m \frac{\mathbf {dv}}{dt}\boldsymbol {.\delta r} = m \frac{\mathbf {dv}}{dt}\frac{\boldsymbol {.\partial r}}{\partial t}\delta t = m \frac{\mathbf {dv}}{dt}\boldsymbol {.v}\delta t = \frac{1}{2}m\frac{\boldsymbol {\partial(v.v)}}{\partial t}\delta t = \frac{1}{2}m \frac{\partial (v^2)}{\partial t}\delta t$$
As $v = \omega L$
$$m\mathbf a\boldsymbol {.\delta r} = \frac{1}{2}mL^2 \frac{\partial (\omega^2)}{\partial t}\delta t = mL^2 \omega \frac{\partial \omega}{\partial t}\delta t$$
So, the complete equation:
$$mgLsin(\theta)\omega \delta t - mL^2 \omega \frac{\partial \omega}{\partial t}\delta t = 0 \implies \frac{\partial \omega}{\partial t} = \frac{g}{L}sin(\theta)$$
