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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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.
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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.
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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.
| Recipient | Interaction | Fx (N) | Fy (N) |
|---|---|---|---|
| Block | Applied horizontal force | 8 | 0 |
| Block | Surface on block: static friction | -8 | 0 |
| Quantity | Value | Unit |
|---|---|---|
| Maximum static-friction magnitude | 20 | N |
| Actual opposing-friction magnitude | 8 | N |
| Initial resulting acceleration | 0 | m/s² |
| Sliding-start flag | 0 | 0 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.
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How would your analysis change if the block were already sliding left? Why is the initially-at-rest assumption necessary for the displayed rule?
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