A 5kg block on wood has μ_s = 0.5. What is the maximum static friction force?

F_s = μ_s × N = 0.5 × 5 × 9.8.

Why does adding weight to the block change the friction force but not the coefficient?

F_friction = μN; more weight increases N, which increases F_friction; μ stays constant.

Experimentally, why is μ_s > μ_k for most material pairs?

At rest, surface asperities interlock more deeply; motion reduces contact depth.

Pro 🔒~20 min

Friction Lab

Measure static and kinetic friction coefficients

How it works

Static friction prevents an object from moving when a force is applied up to a maximum threshold F_s,max = μ_s N. Once motion begins, kinetic friction F_k = μ_k N acts, typically with μ_k < μ_s. Both friction forces are proportional to the normal force N but independent of contact area. The coefficients μ_s and μ_k depend only on the materials in contact.

Upgrade to Pro to access this experiment

Step-by-step

Increase the applied force slowly.
The block remains stationary while friction matches it (static region).
When the applied force exceeds F_s,max the block starts sliding and friction drops to the kinetic value.
Record the threshold force to calculate μ_s, and the sliding force to find μ_k.

Key formulas

$F_{s,max} = \mu_s N$ Maximum static friction
$F_k = \mu_k N$ Kinetic friction force
$N = mg + W_{extra}$ Normal force

Frequently asked questions

A 5kg block on wood has μ_s = 0.5. What is the maximum static friction force?: F_s = μ_s × N = 0.5 × 5 × 9.8.
Why does adding weight to the block change the friction force but not the coefficient?: F_friction = μN; more weight increases N, which increases F_friction; μ stays constant.
Experimentally, why is μ_s > μ_k for most material pairs?: At rest, surface asperities interlock more deeply; motion reduces contact depth.