CivilForge

Coulomb Friction

F \leq μ_{s} N (equality only at impending slip)

Statics p. 97 prints F ≤ μs N and calls the largest value the limiting friction. Equality is impending slip, not the force at rest. Dynamics p. 110 adds F < μs N (no slip), F = μs N (impending), and F = μk N (sliding). Belt friction F1 = F2 e^{μθ} (θ in radians) and the square-thread screw are on Statics p. 98; they are not the dry-friction inequality tested as the Civil statics bullet. On a slope, impending motion is the derived condition tan α = μs.

μ_{s}

Static friction coefficient

μ_{k}

Kinetic friction coefficient

N

(N)

Normal force

Step-by-step solved example

A 200 kg crate rests on a horizontal floor, μ_s = 0.35. What horizontal force starts motion? g = 9.81 m/s².

Horizontal 200 kg crate: N = mg and impending slip at P = μs N.

1. Normal force
N = mg = 200×9.81 = 1962 N.
2. Impending friction
Impending motion: F = μ_s N = 0.35×1962 = 686.7 N → 687 N. At rest with a smaller push, F would be less than μs N.
$F = 687 N$

Answer: 687 N

10 practice questions

0/10 correct

1.If μ_s = 0.4 and N = 500 N, max static friction is

2.A block on a rough horizontal surface with no applied load has friction

3.Angle of repose is

4.Typically μ_k compared with μ_s is

5.Wedge of angle θ, μ = 0, self-locking never occurs because

6.A 50 lb block, μ_s = 0.25, just moves under P. P is

7.On a 20° slope, μ_s = 0.5, a block

Block on a 20° incline. Weight acts straight down. Normal is perpendicular to the surface. Friction acts up the plane. μs = 0.5.

8.Belt friction (no slip impending) uses

9.Normal force on a horizontal floor under a 10 kN crate is

10.If applied force is 30 N and μ_s N = 50 N, friction is