OlympIQ
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Find the Largest Acceleration Friction Can Give, Then Check the Whole System

When one block rides on another, friction is the only horizontal force on the top block, so it limits how fast the top block can accelerate. That limit gives the greatest force you can apply without the blocks slipping.

NSEJS · Laws of Motion and Friction · 4 linked questions · 100% coverage
Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction
Study deeper: NCERT Science · Laws of Motion and Friction

Maximum common acceleration

For a block of mass m on top of a block M with friction coefficient μ between them, the top block can accelerate at most at μ g, because its only horizontal force is friction up to μ m g. The largest force on the pair is therefore (M + m) μ g.

For a 2 kg block A on a 5 kg block B with
μ = 0.60 on a smooth plane, F_max = 7 x 0.6 x 10 = 42 N

For a 4 kg block on a 10 kg block with μ = 0.3, the limiting acceleration is 3 m/s² and F_max = (10 + 4) x 3 = 42 N, which equals the equation (F - 12)/10 = 12/4.

Does the lower block even move?

When both blocks rest on a rough surface, first compare the friction the ground can provide, μ_ground (M_total) g, with the largest friction the top block can transmit or the applied force. A 5 kg block A on a 10 kg block B with μ_AB = 0.8, μ_ground = 0.4 and a 50 N pull on A: the ground can resist 0.4 x 15 x 10 = 60 N, which is more than 50 N, so B stays at rest. The friction on A is then 0.8 x 5 x 10 = 40 N, and the acceleration of A is (50 - 40)/5 = 2 m/s².

A block held on an accelerating wedge

A block on a smooth wedge of inclination θ stays at rest relative to the wedge if the wedge accelerates horizontally with a = g tanθ, because then the horizontal component of the normal force supplies exactly the force m a needed.

Equivalently, a = g / cotθ

OlympIQ exam lens

  • ✓Top block's maximum acceleration from friction = μ g
  • ✓Largest force before slipping on a stack: F = (total mass) x μ g (when the force is applied to the bottom block)
  • ✓Check whether the ground can hold the lower block before assuming both move
  • ✓Block at rest on a smooth accelerating wedge: a = g tanθ
OlympIQ trap

Do not assume both blocks accelerate together. First test whether the friction needed to give the top block the same acceleration exceeds μ m g; if it does, they slip and each block has its own acceleration.

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