Chain Sliding Off Table

A uniform chain with linear density λ = M/L lies on a horizontal table with a smooth vertical edge. The hanging segment of length s pulls with weight λgs; the segment on the table (length L−s) contributes normal force and friction μλg(L−s). Treating the whole chain as one system, the net horizontal acceleration along the slide is a = (g/L)(s − μ(L−s)) once static friction is overcome (s > μ_s(L−s)).

Who it's for: Intro mechanics; capstan/chain puzzles and friction demos.

Key terms

  • static friction
  • kinetic friction
  • linear mass density
  • critical overhang
  • sliding

How it works

A uniform chain lies on a horizontal table with a smooth edge. The hanging part pulls with weight proportional to its length; the part on the table presses down and can exert friction. The model treats the whole chain as one mass: net force is λg(s − μ(L−s)) along the slide, giving a = (g/L)(s − μ(L−s)) while moving. Static equilibrium holds until s exceeds μ_s times the length still on the table.

Key equations

Slip when: s > μ_s (L − s) ⇒ s_crit = μ_s L / (1 + μ_s)

a = (g/L)(s − μ_k(L − s)), ṡ = v, s̈ = a

Frequently asked questions

Why does acceleration increase as more chain hangs?
The driving term grows with s while friction on the table shrinks as (L−s). Both effects make the net force larger until the last chain leaves the table.
Is this model exact for a real chain?
The edge is assumed smooth; the chain bends as a rigid L-shape. Real chains have edge curvature and internal friction, but the 1D balance captures the main competition between weight and table friction.