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SCIENCE · INVESTIGATE AND EXPLAIN

Decide whether static friction can maintain rest

Static friction prevents relative sliding between surfaces when the needed tangential force is within a limiting magnitude. The rule is |fₛ|≤μₛN, not fₛ=μₛN in every resting situation. A small push can be balanced by small static friction; no horizontal push may require no friction at all.

Learning outcome: At normal force 40 N, compare pushes 8,20 and 22 N. State the regime before calculating friction. Then hold the push at 22 N and increase normal force to 60 N. Explain why increasing the maximum allowed friction need not mean the actual friction reaches that maximum.

Starting recap

Review signed vectors, components, acceleration, graph slopes and system boundaries. Every force must name its agent and recipient.

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Try

With normal force 40 N, maximum static friction is 20 N. An 8 N push is balanced by 8 N static friction, so acceleration is zero. A 22 N push exceeds the bound: kinetic friction is 12 N and initial acceleration is(22−12)/4=2.5 m/s². The specified 4 kg mass stays fixed while the normal load is independently controlled.

Open the experiment

Predict

If the applied force is well below the static limit, does increasing the normal force change the actual static friction?

Experiment

At normal force 40 N, compare pushes 8,20 and 22 N. State the regime before calculating friction. Then hold the push at 22 N and increase normal force to 60 N. Explain why increasing the maximum allowed friction need not mean the actual friction reaches that maximum.

Determine whether static friction can hold

Worked teaching model · synthetic values · A 4 kg block initially at rest on a horizontal surface has μs=0.5 and μk=0.3. Assigned normal force is independently controlled by an extra vertical load/support and all vertical forces balance. At the limiting equality a rest solution still exists; above it sliding starts in the applied-force direction.

Reset keeps your prediction, captures and practice history.

Interaction arrows are grouped by recipient. For horizontal-only cart/block diagrams, balanced vertical interactions are omitted explicitly; this is a component diagram, not a claim that gravity vanished. For the spring case only the selected spring interaction is drawn; the opposite external holding force is tabulated but omitted from that diagram. For circular cases, the arrow is the required tension, not a claim it is available when the feasibility check fails. All arrows in one view share a scale; scale changes with controls. A zero force has no visible arrow. The force table supplies exact signs and units.

BlockF1: (8, 0) NF2: (-8, 0) N
Proposed-model interaction components — right and up are positive
RecipientInteractionFx (N)Fy (N)
BlockApplied horizontal force80
BlockSurface on block: static friction-80
Bounded synthetic force-model results
QuantityValueUnit
Maximum static-friction magnitude20N
Actual opposing-friction magnitude8N
Initial resulting acceleration0m/s²
Sliding-start flag00 no / 1 yes

Static friction adjusts to the applied force; the block can remain at rest.

f_s≤μ_sN is a bound, not an equality for every resting block. This test begins from rest each time; it is not a time-history simulation of a block already sliding.

Static friction adjusts to the applied force; the block can remain at rest.

f_s≤μ_sN is a bound, not an equality for every resting block. This test begins from rest each time; it is not a time-history simulation of a block already sliding.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: Decide whether static friction can maintain rest

Static friction prevents relative sliding between surfaces when the needed tangential force is within a limiting magnitude. The rule is |fₛ|≤μₛN, not fₛ=μₛN in every resting situation. A small push can be balanced by small static friction; no horizontal push may require no friction at all.

2. Explain and derive

For a block initially at rest on a horizontal surface with one positive horizontal applied force F, a rest solution requires friction−F. Check whether F≤μₛN. At the limiting equality, the ideal model still permits rest. Above the bound, rest cannot be maintained and positive sliding begins.

Once sliding begins, this teaching approximation uses kinetic-friction magnitude μₖN opposing the relative sliding direction. It is a different relation with a different coefficient. For the assigned μₛ=0.5 and μₖ=0.3, crossing the threshold changes the force model; the resulting acceleration need not rise smoothly from zero.

The experiment resets the block to rest for each setting. It cannot describe a block that is already moving and then experiences a smaller push; that case requires its velocity state as well. Friction opposes relative slipping or its tendency at the contact, not necessarily the object’s velocity in every frame. Static friction can accelerate a walking person forward.

3. Work an example

With normal force 40 N, maximum static friction is 20 N. An 8 N push is balanced by 8 N static friction, so acceleration is zero. A 22 N push exceeds the bound: kinetic friction is 12 N and initial acceleration is(22−12)/4=2.5 m/s². The specified 4 kg mass stays fixed while the normal load is independently controlled.

4. Predict and investigate

Predict first: If the applied force is well below the static limit, does increasing the normal force change the actual static friction?

At normal force 40 N, compare pushes 8,20 and 22 N. State the regime before calculating friction. Then hold the push at 22 N and increase normal force to 60 N. Explain why increasing the maximum allowed friction need not mean the actual friction reaches that maximum. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

A 4 kg block begins sliding right under a 26 N push. Its normal force is 40 N and μₖ=0.3. Find its initial acceleration.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

How would your analysis change if the block were already sliding left? Why is the initially-at-rest assumption necessary for the displayed rule?

Retain your first answer, force diagram, equations, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Supplied synthetic outputs do not certify actual laboratory skill.

Material pair and apparent contact area

The normal force points away from the surface and is perpendicular to it. Friction is tangential. In the supplied dry-friction approximation, coefficients depend on the contacting material pair, and the static coefficient is typically greater than the kinetic coefficient. These are empirical modeling statements, not universal guarantees for every lubricated, adhesive or deforming contact.

At the same normal load and material pair, changing the nominal contact area does not change μN in this approximation. Predict and check: with N=30 N,μs=0.6,μk=0.4, the static limit is 18 N and sliding magnitude is 12 N. Doubling nominal contact area leaves those values unchanged; a 7 N resting push still needs only 7 N static friction. Changing the material pair to μs=0.3,μk=0.2 instead gives a 9 N limit and 6 N sliding magnitude. Keep load, area and material interventions separate, and do not infer the same result when the constitutive assumptions cease to apply.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.7. Checked 30 October 2026. This is a content cross-reference, not AP authorization.

Apply

A resting block has μₛN=20 N and receives a 10 N horizontal push. What is the static-friction magnitude if rest is maintained?

Public practice checks a specific claim. It does not establish independent mastery.

How would your analysis change if the block were already sliding left? Why is the initially-at-rest assumption necessary for the displayed rule?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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