Problem — Coefficient of Static Friction

Problem

A crate of mass $m = 20\;\text{kg}$ is on the verge of slipping when a horizontal force $F = 92\;\text{N}$ is applied to it. Using $g = 9{,}8\;\text{m/s}^2$, determine the coefficient of static friction between crate and floor.

Solution

Step 1 — Normal reaction. The crate is on a horizontal plane and does not accelerate vertically: the normal reaction balances the weight.

$R = m\,g = 20\;\text{kg} \times 9{,}8\;\text{m/s}^2 = 196\;\text{N}$

Step 2 — Condition for slipping. “On the verge of slipping” means that static friction has reached its maximum value and still exactly balances the applied force:

$F = f_{s,\max} = \mu_s R$

Step 3 — Coefficient of static friction.

$\mu_s = \frac{F}{R} = \frac{F}{m\,g} = \frac{92\;\text{N}}{196\;\text{N}} \approx 0{,}47$

$\ev{\mu_s \approx 0{,}47}$

The coefficient is dimensionless, as it should be, being the ratio of two forces.

Links

Topics: Friction Concepts: Static friction · Normal force