Maximum static friction

Problem

Maximum static friction. Four blocks resting on different surfaces, same mass $m = 5\;\text{kg}$:

• (A) rubber on dry asphalt, $\mu_s = 1{,}0$;
• (B) rubber on wet asphalt, $\mu_s = 0{,}5$;
• (C) steel on steel, $\mu_s = 0{,}7$;
• (D) wood on wood, $\mu_s = 0{,}4$.

Rank the maximum static friction forces from smallest to largest.

Solution

On a horizontal plane the normal reaction balances the weight, $N = mg$, so the maximum static friction force is $F_{s,\max} = \mu_s N = \mu_s\, m g$. For equal mass (and hence equal $N$), the force depends only on the coefficient $\mu_s$: it is enough to rank the $\mu_s$ values. Using $g = 9{,}8\;\text{m/s}^2$, the common factor is $mg = 5 \cdot 9{,}8 = 49\;\text{N}$:

$F_A = 1{,}0 \cdot 49 = 49\;\text{N} \qquad F_B = 0{,}5 \cdot 49 = 24{,}5\;\text{N}$

$F_C = 0{,}7 \cdot 49 = 34{,}3\;\text{N} \qquad F_D = 0{,}4 \cdot 49 = 19{,}6\;\text{N}$

Ranking the coefficients $0{,}4 < 0{,}5 < 0{,}7 < 1{,}0$, i.e. the forces from smallest to largest:

$\ev{F_D < F_B < F_C < F_A \quad (19{,}6;\; 24{,}5;\; 34{,}3;\; 49\;\text{N})}$

Physical note: this is consistent with experience that rubber on dry asphalt (A) offers the greatest grip, while wet asphalt (B) drastically reduces it — which is why braking on a wet road is more dangerous.

Links

Topics: Friction Concepts: Static friction Skills: Ranking reasoning