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Home/Classical Mechanics/Block Stack & Friction

Block Stack & Friction

Identical blocks are stacked; the bottom block is pulled horizontally. Neighboring blocks are coupled by stiff damped springs, but the shear transmitted in each joint is limited by Coulomb friction: |T| ≤ μN, where N is the normal force at that interface (weight of all blocks above). The ground uses its own μ. When the pull is large, lower layers can slide faster than upper ones — compare with the textbook “rigid stack” acceleration (F − μ_gr Nmg)/(Nm) and the requirement a ≤ μ_block g for no inter-layer slip.

Who it's for: Intro mechanics; friction demos and “pull the bottom book” puzzles.

Key terms

  • static friction
  • kinetic friction
  • normal force
  • stack
  • shear
4
2.5 kg
45 N
0.35
0.28
0.4
0.32
2800
120

Rigid-stack estimate (no inter-slip): a ≈ (F − μ_k,gr Nmg)/(Nm); all layers move together only if a ≤ μ_s,blk g.

a_rigid (estimate)1.36 m/s²
μ_s,blk · g (cap)3.43 m/s²
Shear T₀₁ (link)0.0 N
v₁0.000 m/s
v_N0.000 m/s

Shortcuts

  • •Space — start
  • •R — reset

How it works

A stack of identical blocks rests on a rough table. You pull the bottom block horizontally. Neighboring blocks are coupled by stiff springs with damping, but the shear force in each joint is capped by static or kinetic friction: |T| ≤ μN, with N equal to the weight of the blocks above that joint. If the bottom is pulled hard enough, joints slip: lower blocks can move faster than upper ones. Compare the rough “rigid stack” acceleration estimate with the cap μ_s g on inter-layer slip.

Key equations

N_i = (N−i)·mg at interface i; |T_i| ≤ μ N_i; ground: |F| ≤ μ_g N_tot

Together without slip: a ≤ μ_s,blk g, F > μ_s,gr N_tot m g to start sliding on table

Frequently asked questions

Why not pure rigid-body dynamics?
A full analytic piecewise solution for N>2 is messy. The spring–damper links with friction caps give stable motion and visible slip while matching the same friction limits as the textbook model.