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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?

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What you are not accounting for is the force binding each portion to the others next to it. This is because you imagine it rigid, in which case these forces are infinite and the rod behaves ideally: all mass is concentrated in the center and it does not fall unless this center of mass is out. In fact, torque analysis is more appropriate here: the whole torque on the suspended part equals the torque on the resting one due to gravity and table reaction together.

Now, allow for the rod to be flexible, like a rubber, and you would see the end out of the table bend. Its shape will be determined by how strong the pieces are bound to each other. And interestingly enough, now the conditions for falling are different. It would make a nice problem to solve!

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Both mathematically, and in actuality, from a torque perspective, all of the mass of the rod acts like it is concentrated at the center of mass of the rod. Thus, as long as the center of mass of the rod is over some portion of the table, it will not fall.

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If the rod is in equilibrium then (i) the total of reaction forces from the table must equal the weight of the rod (there are no horizontal forces), and (ii) the total of moments about any point must also be zero.

The reaction force on any element of the rod does not have to equal the weight of that element. The reaction force can be more or less than this. Forces can be transmitted through the rod to redistribute the weight of the section which is hanging off the table.

The total weight of the rod (and the reaction forces which oppose it) will not be distributed evenly across the part still on the table. The highest reaction forces will occur towards the edge of the table.

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