Match a friction threshold to a moving-stage integral

Start with the interaction balance

The horizontal object has weight downward and a normal interaction upward. With no vertical acceleration or other vertical forces, N=mg. An applied force rises as F=rt. Before motion, static friction adjusts to oppose that force, with |fs|≤mu_s N. It is not automatically equal to mu_s N at every time. Rest persists while a compatible static force can balance the applied interaction.

Locate the transition

For m=2 and g=10, the static bound is 20mu_s newtons. The threshold time is tr=20mu_s/r. At equality, static equilibrium is marginally possible; the subsequent increasing force triggers the moving branch in this ideal model. No impulsive force is introduced, so position and velocity remain continuous at release.

Integrate the kinetic stage using elapsed time

The supplied kinetic coefficient is.15, so kinetic friction has magnitude 3 N while motion remains rightward. Let z=t−tr. Applied force is 20mu_s+rz, not rz alone. Therefore acceleration is(20mu_s−3)/2+(r/2)z. With release velocity and displacement zero, integrate to v=[(20mu_s−3)/2]z+(r/4)z² and Δx=[(20mu_s−3)/4]z²+(r/12)z³.

Check the original packet

For mu_s=.4 and r=2, threshold 8 N occurs at 4 s. The acceleration immediately after release is(8−3)/2=2.5 m/s² under the ideal friction drop. After one additional second, v=2.5+.5=3 m/s and displacement 1.25+1/6=17/12 m. The global clock reads 5 s, but the kinetic initial-value problem has elapsed time 1 s.

Transfer and regime limits

For mass 3,mu_s=.5,g10 and ramp 5, the static threshold 15 N occurs at 3 s. The later moving trajectory would also require a kinetic coefficient; a static coefficient alone cannot determine it. The model’s legal range keeps rightward motion and rising applied force, so no velocity reversal requires a new friction sign.

Interpret real contact evidence

The jump from limiting static to kinetic friction is idealized. Real surfaces can depend on speed, history, deformation and lubrication. A threshold fit does not certify a material parameter for every condition. Do not use a virtual coefficient or stopping result as practical machinery guidance; retain the normal-force assumptions, force direction and tested regime.

Test the ideal contact assumptions

The normal force points outward from the contact surface, perpendicular to it; friction acts tangent to that surface. In the ideal dry-friction model used here, the coefficients characterize the pair of materials and their surface condition, not a universal property of the moving body. Typically mu_s exceeds mu_k, but the selected values are stipulated model inputs rather than a law for every real contact.

Predict the change in friction when the same body is turned onto a face of twice the nominal contact area while N remains20 N and the surface pair remains unchanged. With mu_s=.4 and mu_k=.3, the maximum static magnitude remains8 N and sliding magnitude remains6 N: nominal area does not enter this ideal model. A5 N tangential demand is balanced by5 N static friction, not automatically8 N. Changing the pair to mu_s=.25 lowers the static limit to5 N; changing the actual normal would be a different intervention. Real adhesion, deformation, temperature or wear can violate these ideal assumptions, so this calculation is not a material rating or an instruction to test an object physically.

Public numerical transfer practice

Try a new practice case

A horizontal mass 3 kg has mu_s=.5,g10 and applied force ramp 5 N/s. Find threshold time.

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Sources and evidence boundary

Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.