core30 min studyLesson 113 in Physics

Friction is a branch decision

Use the static-friction inequality before a sliding law, determine the direction from relative slip, and recognize where Coulomb friction stops being predictive.

On this page

What you’ll understand

  • Test a no-slip hypothesis against the static-friction bound.
  • Choose kinetic-friction direction from contact slip velocity.
  • Identify normal-load changes and limitations of constant-coefficient friction.

A box can remain still under several different horizontal pushes. Its friction force therefore cannot always equal one fixed coefficient times its weight. A useful dry-contact model has separate sticking and sliding branches. First ask whether a no-slip motion is dynamically possible; only then choose the force law. This procedure turns friction problems from a memorized arrow rule into a precise test of an assumed contact state.

Sticking determines a required force

In the elementary Coulomb model, static friction adjusts within a bounded range to prevent relative tangential slipping at the contact. Its direction opposes the slipping that would occur without it, not necessarily the body's laboratory velocity. A box at rest on an accelerating conveyor may need forward static friction to share the belt's acceleration. A rolling driven wheel can also receive forward static friction from the road. Determine the force needed for the proposed no-slip motion from Newton's law, then compare its magnitude with the maximum permitted by the contact model.

The normal reaction is a separate unknown. On a horizontal floor with no other vertical force and no vertical acceleration, it equals weight. An upward angled pull reduces the normal reaction and therefore reduces the static-friction capacity. On an incline, the perpendicular force balance supplies the normal reaction. Do this calculation before applying the inequality. If the required static force is smaller than the maximum, use the required value itself. Replacing it by the maximum in every sticking situation invents an acceleration that can violate the very no-slip assumption being tested.

∣fs∣≤μsN,sticking requires ∣frequired∣≤μsN|f_s|\leq\mu_s N,\qquad \text{sticking requires } |f_{\mathrm{required}}|\leq\mu_sN
The coefficient μs is dimensionless and N is a nonnegative normal load in newtons. Equality describes impending slip in this idealized model; most static contacts operate strictly below the bound.↔ Scroll wide equations horizontally.

Sliding changes the constitutive rule

Once the surfaces slide relative to one another, a common approximation gives kinetic friction a constant magnitude proportional to normal load. Its direction is opposite the relative tangential sliding velocity at that contact. For a block sliding right on a stationary table, friction points left. If a belt moves right faster than a right-moving block, the block slips left relative to the belt and friction on the block points right. Naming both contacting surfaces and their relative motion removes the ambiguity in the phrase friction opposes motion.

When the relative slip speed reaches zero, the sliding rule alone no longer decides what happens next. Retest the static condition. If the sticking force required at that instant is allowed, the surfaces can stick; otherwise slipping resumes in the direction consistent with the net force and constraint. Numerical solvers need event handling or a carefully chosen regularized contact model here. Simply continuing a constant kinetic force through zero speed can create artificial reversals and chatter. These are failures of branch handling, not evidence that dry friction really supplies an arbitrary oscillating force at rest.

fk=−μkN vslip∣vslip∣(∣vslip∣>0)\mathbf f_k=-\mu_kN\,\frac{\mathbf v_{\mathrm{slip}}}{|\mathbf v_{\mathrm{slip}}|}\qquad (|\mathbf v_{\mathrm{slip}}|>0)
This isotropic kinetic-friction approximation applies only to nonzero relative tangential slip. It does not define a direction at zero slip, where the sticking feasibility test must be reconsidered.↔ Scroll wide equations horizontally.

Separate an effective law from microscopic explanation

Constant friction coefficients summarize a restricted range of loads, surfaces, speeds, and environmental conditions. They do not follow directly from Newton's laws. Real contacts deform, form microscopic junctions, heat up, wear, and can depend on dwell time or humidity. Apparent contact area can be unimportant in one dry-contact regime and important in another. State which measurements calibrated the coefficient and avoid extending the model automatically to lubricated bearings, rubber tires over every speed, or nanoscale contacts. A simple model can be useful without being a universal law of matter.

Energy transferred out of mechanical motion by dry friction does not disappear. For a sliding block on a stationary rough plane with constant friction magnitude, friction's work is minus that magnitude times the sliding distance; total kinetic-energy change also includes work by gravity and any other forces. A downhill block can therefore speed up while friction dissipates energy. On a moving belt, the motor can also supply energy, and friction's laboratory work on an individual body can be positive. Friction opposes relative slip and produces nonnegative contact dissipation in this ideal model; individual-body work depends on its displacement in the chosen frame. Boundary selection remains as important for energy accounting as for force diagrams.

Predict the transition before calculating acceleration

A 5 kg5\,\mathrm{kg} box rests on a horizontal floor. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}, μs=0.4\mu_s=0.4, and μk=0.3\mu_k=0.3. Compare steady horizontal pushes of 15 N15\,\mathrm N and 25 N25\,\mathrm N applied from rest.

  1. Vertical balance gives N=50 NN=50\,\mathrm N. The largest static-friction magnitude is 0.4(50)=20 N0.4(50)=20\,\mathrm N; this is a capacity, not the actual force under every push.

  2. For a 15 N15\,\mathrm N push, sticking requires 15 N15\,\mathrm N of opposing friction, which is within the bound. The box remains at rest and its acceleration is zero.

  3. For a 25 N25\,\mathrm N push, sticking would require more than 20 N20\,\mathrm N, so it fails. During rightward slip, kinetic friction is 15 N15\,\mathrm N leftward. The acceleration is (25−15)/5=2 m s−2(25-15)/5=2\,\mathrm{m\,s^{-2}}.

The result

The same contact supports 15 N of static friction in the first case and 15 N of kinetic friction in the second, but the equal numerical values arise from different rules and produce different motions.

Explore the model

Interactive lab

Interactive lab · vectors

Resolve a slanted applied pull

Set a normalized pull to magnitude 2 and angle 30°, giving upward component 1; change to 90°, giving 2. Predict how that larger upward component reduces the normal load of a hypothetical supported box. The lab displays only vector projections; it does not implement a friction threshold or stick-slip switching.

Resolve a slanted applied pullPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCompute the normal load before using a friction bound.

Check your understanding

A box remains at rest while a horizontal push increases from 2 N to 6 N, both below its static-friction limit. What happens to the friction magnitude?

Try it yourself

A 2 kg2\,\mathrm{kg} box initially rests on a rough 30∘30^\circ incline with μs=0.7\mu_s=0.7, μk=0.4\mu_k=0.4, and g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Can it remain at rest? If it is instead already sliding downward, find its acceleration.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compare the required uphill static force with μsN; use the kinetic law only for the separately specified sliding state.

Compare with the solution

The downhill weight component is mgsin⁡30∘=10 Nmg\sin30^\circ=10\,\mathrm N and N=mgcos⁡30∘=103 NN=mg\cos30^\circ=10\sqrt3\,\mathrm N. Static capacity is 73≈12.1 N7\sqrt3\approx12.1\,\mathrm N, so rest is possible with 10 N uphill friction. During downward sliding, friction is 43≈6.93 N4\sqrt3\approx6.93\,\mathrm N uphill and acceleration is (10−43)/2≈1.54 m s−2(10-4\sqrt3)/2\approx1.54\,\mathrm{m\,s^{-2}} downhill.

Take it with you

  • Static friction satisfies an inequality and adjusts to the required sticking force.
  • Kinetic friction opposes relative contact slip, which can differ from laboratory motion.
  • Constant friction coefficients are calibrated effective laws with limited domains.

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