Distinguish static and sliding friction

A capacity is not the force currently acting

Push a resting block gently across a horizontal floor. It may remain still even though your hand applies a force. In the investigation, a +4 N push is balanced by −4 N of static friction. The net horizontal force is zero. The floor could supply a larger opposing force if needed, but it does not automatically apply its maximum.

The block has mass m = 2 kg. We use teaching gravitational acceleration g = 10 m/s². With no other vertical force and no vertical acceleration, the normal force is N = mg = 20 N. Here N names the normal-force magnitude; N beside a numerical value is the unit newton. Right is positive throughout.

The assigned static coefficient μₛ = 0.5 is dimensionless. The Greek letter μ is read “mu”; the subscript s denotes static contact. The maximum static-friction magnitude is μₛN = 0.5 × 20 = 10 N. While the surfaces stick, |fₛ| ≤ μₛN: the magnitude of the actual static friction can be anything from zero up to that limit. The bars around fₛ mean absolute value.

For a +4 N push, horizontal equilibrium requires 4 + fₛ = 0, so fₛ = −4 N. This is within the available capacity. Substituting −10 N instead would predict an unexplained acceleration to the left from a gentle rightward push. Check the regime and the needed force before substituting a maximum.

Approach the threshold from rest

Set current velocity to zero. Increase applied force from +4 to +10 N. Friction changes from −4 to −10 N, keeping net force and acceleration zero. At exactly the threshold, this ideal model chooses continued rest. Equality marks limiting equilibrium; the control does not represent an extra disturbance that starts motion.

At +12 N, the required balancing force would be −12 N. The static limit cannot supply it. The model switches to the response immediately after slipping starts. Its assigned kinetic coefficient is μₖ = 0.3, where k denotes sliding contact. Sliding-friction magnitude is μₖN = 6 N. The force is −6 N, leaving net force +6 N and acceleration 6/2 = +3 m/s².

The drop from the static limit to a smaller sliding force is an idealized contact rule. It is not a new force added to the old static force. Use one applicable friction regime. The chart compares applied, friction and net force on a fixed signed scale. Its net row is a sum of the first two rows, not an extra interaction in a free-body diagram.

Motion direction matters more than push direction

Keep the applied force at +4 N, but select current velocity +1 m/s. The block is already sliding right relative to the stationary floor. Friction points left at −6 N, so net force is −2 N and acceleration is −1 m/s². The block initially slows while still moving right.

Now select current velocity −1 m/s with the same push. Friction points right at +6 N because relative sliding is leftward. Applied force and friction add: net force is +10 N and acceleration is +5 m/s². This block also initially slows, but its velocity and acceleration have different signs from the previous case.

The rule is not “friction always opposes the applied force.” Sliding friction opposes relative slipping at the contact. A moving belt could exert forward friction on an object; our stationary-floor scenario does not model that belt. Nor does the dry-contact model describe rolling resistance or fluid drag.

Do not extend the sliding rule through a stop

The velocity control selects an independent instantaneous state. It does not advance a clock or preserve a previous trajectory. At +1 m/s with a +4 N push, acceleration is −1 m/s² while sliding continues. Under an unchanged push and these ideal coefficients, velocity reaches zero after one second. At that point, recompute the contact regime: a −4 N static force is sufficient to balance the push, so the block can remain at rest.

Extending −1 m/s² beyond that instant would predict reversal, but would keep using leftward kinetic friction after leftward motion began. That would violate the friction rule. A time simulation needs to detect the zero crossing and switch regimes. This widget deliberately reports the instantaneous response rather than displaying an incorrectly extrapolated journey.

The ledger also shows friction force multiplied by relative velocity, in watts. For this translating block and stationary floor, the product is nonpositive: sliding friction removes mechanical energy from the block. At rest it is zero even when static friction is nonzero. This calculation is not a temperature prediction and does not allocate thermal energy between the contacting bodies.

Use an empirical model with explicit limits

The OpenStax treatment of friction develops static and kinetic contact models. The coefficients here are assigned teaching values, not measured properties of named materials. Real contact depends on surface condition, contamination, speed and other factors. The simple constant-coefficient rule is useful within its assumptions, not an exact universal law.

An investigation of real contact should record repeated onset-force measurements separately from forces during steady sliding. A steady sliding force can estimate kinetic friction only when acceleration is negligible and other horizontal forces are accounted for. An onset threshold estimates a static limit; a small balancing force below onset does not determine that limit. A browser capture demonstrates a model consequence, not either physical measurement.

Transfer: A different 5 kg block has normal force 50 N, static coefficient 0.4 and kinetic coefficient 0.2. A horizontal +15 N push acts. Compare a block initially at rest with one already sliding left.

Check the friction transfer

The static capacity is 0.4 × 50 = 20 N. From rest, friction −15 N balances the push, so acceleration is zero. While sliding left, friction is +10 N, giving net force +25 N and acceleration +5 m/s². Once it stops, the static regime can balance the continuing +15 N push. The same applied force and mass do not determine the friction regime without contact-motion information.