Static friction — Science definition

The friction force between two surfaces that are not yet sliding; it adjusts up to a maximum to prevent motion.

Where it is taught

In use

In a simple model, sliding friction has magnitude approximately μkN, while static friction adjusts up to a limit μsN.

Friction & Common Forces

Questions using static friction

Problem 1

A box requires 12 N static friction, while μ_sN=20 N. What is actual static friction? Show the reasoning, not just the conclusion.
Hints

Static friction instead adjusts up to its limiting value.

Answer and explanation

12 N, enough to balance the required tangential force without exceeding the maximum.

Problem 2

A 5 kg block has μs=0.40 and μk=0.30 on a level surface. For a 15 N push, find static friction; for a 25 N push during sliding, find acceleration.
Answer and explanation

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction. During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction.

During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 3

Explain the following claim and supply a valid example or counterexample: Static friction adjusts up toμsN; it is not always equal to that maximum.
Answer and explanation

Static friction is 2N opposite the push, so net force is zero. It need not reach 5N. Static friction adjusts up toμsN; it is not always equal to that maximum.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 4

Independent chapter check: A 5 kg block has μs=0.40 and μk=0.30 on a level surface. For a 15 N push, find static friction; for a 25 N push during sliding, find acceleration.
Answer and explanation

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction. During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction.

During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Taught alongside

Elsewhere

Browse all courses