OlympIQ
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Static Friction Adjusts Up to μs N; Kinetic Friction Is a Fixed μk N

Friction questions turn on two distinctions: whether the object is at rest (friction adjusts, up to a limit) or sliding (friction takes a fixed value), and what the normal force really is, which is not always mg.

NSEJS · Laws of Motion and Friction · 11 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

Static versus kinetic friction

Static friction adjusts to whatever balances the applied force, up to the limit fₛ ≤ μs N; sliding friction is fₖ = μk N with μk smaller than μs.

A 20 kg body pushed with just enough force to start moving needs
F=0.6x200=120N;F = 0.6 x 200 = 120 N;

once moving, friction drops to 0.2 x 200 = 40 N, so with the same 120 N the net force is 80 N and a = 4 m/s². The coefficient of friction is a pure ratio of two forces, so it has no dimensions, and rolling friction is smaller than sliding friction.

Normal force changes the limit

A 2.5 kg block with a horizontal pull of 6 N and an upward force
P = 9 N has N = 25 - 9 = 16 N, so the limiting friction is 0.4 x 16 = 6.4 N

Because 6.4 N exceeds 6 N, the block stays at rest and friction equals the applied force, 6 N, not 6.4.

A block pressed against a vertical cart wall stays up only if μ N ≥ m g, and
N = m a, so a ≥ g/μ

Inclines

A block at rest on an incline has friction f = m g sinθ, not μ m g cosθ: a 10 N frictional force on a 30° incline means 5m = 10 and m = 2 kg, regardless of μ.

A block starts to slide when tan
θ=μs\theta = \mu s

If a block takes n times as long to slide down a rough 45° incline as down a smooth one, a_smooth = g sinθ and a_rough = g (sinθ - μ cosθ), and since t is proportional to 1/sqrt(a), n² = 1/(1 - μ), so μ = 1 - 1/n².

Hanging masses and stacked loads

A block on a table pulled by a hanging mass starts to slide when the pull equals μs M g. If a 6 kg mass hangs from a cord pulling a 4 kg block with μ = 0.4, an extra mass m on the block stops the acceleration when 0.4 (4 + m) g = 6 g, so m = 11 kg. A load placed on a 10 kg block that is pressed on a rough surface prevents slipping when the friction μ (10 + m) g equals 10 g, giving m = 10 kg for μ = 0.5. In movable-pulley set-ups, the hanging weight is shared by two string segments, so the tension on the block is half the hanging weight.

OlympIQ exam lens

  • ✓Static friction is f ≤ μs N and adjusts to the applied force; kinetic friction is μk N
  • ✓Check whether the applied force exceeds μs N before deciding the block moves
  • ✓Normal force N is not always mg: include vertical pushes, pulls and accelerations
  • ✓At rest on an incline: friction = m g sinθ; sliding starts at tanθ = μs
  • ✓Sliding-time ratio on a rough and a smooth 45° incline: μ = 1 - 1/n²
OlympIQ trap

Do not use μ N as the friction on a block that is not sliding. Friction on a stationary block equals the force it opposes, which is less than μ N unless the block is about to slip.

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