A uniform rigid rod of length $L$ lies at the edge of a frictionless table so length $x$ of the rod rests on the table and the rest is beyond its edge.
Intuition suggests that the rod will stay like this unless $x$ is smaller than $L/2$ which is when the centre of mass hangs off the table. However, if this were true, then I am left with a dilemma. While $x > L/2$, to ensure each infinitesimal section $\mu \delta x$ of the rod is in equilibrium ($\mu $ is mass density of rod), there must be a reaction force = $g\mu \delta x$ on it. But then the total force on the rod would be $-mg(L-x)$, so its centre of mass must fall.
What does this mean? Is it impossible for a uniform rod to partially hang off the edge of the table while being in equilibrium? Or has something gone wrong with my analysis?