Static friction vs kinetic friction — which is larger and why

Question

Compare static friction and kinetic friction. Which one is larger? Explain the molecular reason for this difference and state the relationship between the two coefficients.

Solution — Step by Step

Static friction ( $f_s$ ) is the friction force that acts on an object at rest, preventing it from sliding. It adjusts itself to match the applied force — from zero all the way up to a maximum value $f_{s,\text{max}}$ .

Kinetic friction ( $f_k$ ) is the friction force that acts on an object that is already sliding. Unlike static friction, kinetic friction is approximately constant for a given pair of surfaces and doesn’t depend on speed (for most everyday situations).

Static friction is always greater than or equal to kinetic friction:

f_{s,\text{max}} \geq f_k

Or equivalently, the coefficients satisfy:

\mu_s \geq \mu_k

For most material pairs, $\mu_s$ is roughly 20–40% larger than $\mu_k$ . For example, rubber on concrete: $\mu_s \approx 0.8$ , $\mu_k \approx 0.6$ .

At the microscopic level, even smooth surfaces have asperities (tiny peaks and valleys). When two surfaces are in static contact:

The microscopic projections interlock and form cold-weld bonds — tiny junctions where surface atoms are very close together, almost forming temporary bonds.
These interlocking points can be numerous and deep because the surfaces have had time to “settle in.”
The maximum static friction represents the force needed to break all these junctions simultaneously and initiate sliding.

Once sliding begins (kinetic friction regime):

The junctions are continuously forming and breaking as the surfaces move past each other.
At any instant, fewer junctions exist in a fully locked state — the surface atoms don’t have time to settle and form deep interlocks before they’re torn apart.
This results in a lower average resistive force → $f_k < f_{s,\text{max}}$ .

This difference explains why it takes more effort to start an object sliding than to keep it sliding. Once you break static friction and the object moves, kinetic friction takes over and is smaller — that’s why objects sometimes “jerk” suddenly when you push hard enough. This is also why anti-lock braking systems (ABS) in cars work: they prevent wheels from locking (going from kinetic to static) to maintain maximum braking force.

Why This Works

The transition from static to kinetic friction involves a discontinuity — you can think of it as crossing an energy barrier. The maximum static friction is the peak of that barrier; kinetic friction is the steady-state lower value once you’re on the “other side.”

Formally, the laws of friction:

$f_s \leq \mu_s N$ (static friction is self-adjusting, up to the limit)
$f_k = \mu_k N$ (kinetic friction is constant at this value)

Both are proportional to the normal force $N$ , but $\mu_s > \mu_k$ .

Alternative Method

This can also be understood through the concept of contact area at the microscopic level. When surfaces are at rest, the actual contact area (sum of asperity contacts) is larger than when surfaces are sliding — because sliding breaks some contacts before new ones form. Larger contact area = larger adhesive force = larger friction. Static contact maximises the real contact area; kinetic contact reduces it.

JEE Main has asked both the inequality $\mu_s > \mu_k$ as an MCQ fact and problems where you must compare the forces needed to start vs. maintain sliding. CBSE Class 11 expects you to know the definition difference and the coefficient inequality with a brief explanation.

Common Mistake

A common error: treating $f_s$ as always equal to $\mu_s N$ . That’s only the maximum static friction. If you apply a force smaller than $\mu_s N$ and the object doesn’t move, the static friction force equals the applied force (Newton’s third law balance) — not $\mu_s N$ . The formula $f_{s,\text{max}} = \mu_s N$ gives only the upper limit, not the actual value when the object is at rest.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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