This question was asked years ago  by another user but I cannot find satisfying answer so I ask again. The problem is from Arnold's &quot;Mathematical Methods of Classical Mechanics&quot;, p. 97 in Chapter 21 on the D'Alembert's principle: A rod of weight P, tilted at an angle of $60^{\circ}$ to the plane of a table, begins to fall with initial velocity zero. Find the constraint force of the table at the initial moment, considering the table as (a) absolutely smooth (b) absolutely rough (In the first case, the holonomic constraint holds the end of the rod on the plane of the table, and in the second case, at a given point.)  My solution for question (b) is,  Since the end is fixed, the virtual variation is orthogonal to the rod, hence the total force of the constraint force and the gravity must be orthogonal to the rod. On the other hand, according to the D'Alembert's principle, the constraint force must be orthogonal to the virtual variation, so the constraint force must be along the rod. Combine 1 and 2 and by elementary geometry, we can find that the constraint force is $\frac{\sqrt{3}}{2}P$.    My questions are:    Is my solution to (b) right? How to solve (a)?
I really hope someone will take a quick look at the following, I would just love to better understand it...  This exercise is from Arnold's "Mathematical Methods of Classical Mechanics", p. 97 in the chapter on d'Alemberts principle:  A rod of weight P, tilted at an angle of 60° to the plane of a table, begins to fall with initial velocity zero. Find the constraint force of the table at the initial moment, considering the table as  (a) absolutely smooth  (b) absolutely rough  (In the first case, the holonomic constraint holds the end of the rod on the plane of the table, and in the second case, at a given point.)"      I must admit, I am pretty unsure on how to do calculations using this "fancy" mathematical kind of physics.  First off, I'm lost with (a). But I'll have a go at (b): As far as I understood, d'Alemberts principle states, that if $M$ is a constraining manifold, $x(q)$ is a curve in $M$ and $\xi$ is a vector perpendicular to $T_xM$, then $x$ satisfies Lagrange's equations  $\frac{d}{dt} \frac{\partial L}{\partial \dot q} = \frac{\partial L}{\partial q} \qquad L = \frac{{\dot x}^2}{2} - U(x)$  iff for the following inner product, we have  $\left(m \ddot x + \frac{\partial U}{\partial x}, \xi \right) = 0$  I guess that's about right so far? The constraint would in this case essentially be $\mathbb S^1$, since the rod moves on a circe around the point in contact with the table. Can we assume that all the rod's mass is at the center of mass?  Would we then have $U(x) = -gx_2$ in this case (where $x_2$ is the vertical component of $x$)?  If yes, then $\partial U / \partial x = -ge_2$. Where $e_1, e_2$ are the horizontal and vertical unit vectors, respectively.  At the inital moment, we have $x = \cos(60°) e_1 + \sin(60°) e_2$, so by d'Alembert's principle we must have  $\left(m \ddot x + \frac{\partial U}{\partial x}, \cos(60°) e_1 + \sin(60°) e_2 \right) = 0$  or written differently  $m \ddot {x_1} \cos(60°) + m \ddot{x_2} \sin(60°) - \sin(60°)g = 0$    So: Is this correct so far? Or am I way off? Is (a) handled any differently up to this point?  Thanks for reading (and hopefully thanks for your helpful reply)!  If anyone could recommend a good problem book (with solutions), in which this kind of mathematical approach is used (I don't know if this is how physicists would actually compute stuff??), I would greatly appreciate it.  Kind regards,  Sam
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