Friction

Friction is the contact force that opposes relative motion (or its tendency) between two surfaces. It is the reason a brake stops a car, a foot grips the ground when you walk, and a book on a tilted table doesn't always slide.

Concept

When two surfaces are in contact, the contact force has two components: the normal force $N$ perpendicular to the surface, and the friction force $f$ parallel to it.

Three regimes of friction:

• Static friction $f_s$: opposes any tendency of relative motion. It adjusts itself to match the applied force up to a maximum, $f_s \leq \mu_s N$, where $\mu_s$ is the coefficient of static friction.
• Kinetic (sliding) friction $f_k$: acts when surfaces slide. Its magnitude is approximately constant: $f_k = \mu_k N$. Generally $\mu_k < \mu_s$.
• Rolling friction $f_r$: acts when a body rolls without slipping. It is much smaller than sliding friction: $\mu_r \ll \mu_k$.

Angle of friction $\lambda$: the angle the total contact force makes with the normal when slipping is just about to begin.

$\tan \lambda = \mu_s$

Angle of repose $\alpha$: the inclination of an inclined plane at which a body just starts to slide down on its own.

$\tan \alpha = \mu_s$

So angle of repose equals angle of friction.

Derivation

Angle of repose. Place a block of mass $m$ on a plane inclined at angle $\alpha$. The forces along and perpendicular to the plane are:

• weight component along the incline (down the slope): $mg \sin\alpha$
• weight component perpendicular (into the slope): $mg \cos\alpha$
• normal force: $N = mg \cos\alpha$
• friction (up the slope, opposing impending motion): $f_s$

For the block to just begin sliding, $f_s = \mu_s N$:

$mg \sin\alpha = \mu_s \cdot mg \cos\alpha$

$\boxed{\tan \alpha = \mu_s}$

The angle does not depend on the mass.

Worked Example

A block of mass 10 kg rests on a horizontal floor. $\mu_s = 0.4$, $\mu_k = 0.3$. (a) What minimum horizontal force is needed to start moving it? (b) Once moving, what horizontal force keeps it moving at constant velocity? Take $g = 10 \, \text{m/s}^2$.

Solution:

Normal force: $N = mg = 100 \, \text{N}$.

(a) To just start moving, applied force must overcome maximum static friction:

$F_{\min} = \mu_s N = (0.4)(100) = 40 \, \text{N}$

(b) Once sliding, kinetic friction acts:

$f_k = \mu_k N = (0.3)(100) = 30 \, \text{N}$

For constant velocity, applied force equals kinetic friction: $F = 30 \, \text{N}$.

The transition force drops from 40 N to 30 N — that's why a heavy box "jerks" forward once you finally get it moving.

Common Confusions

• "Friction always opposes motion." Friction opposes relative motion (or its tendency) between surfaces. When you walk, friction on your foot pushes you forward (the foot tries to slip backward).
• Static friction is not always at its maximum. $f_s$ adjusts itself to whatever is needed, up to $\mu_s N$. A book on a tilted table feels static friction equal to $mg \sin\theta$, not $\mu_s mg \cos\theta$, unless on the verge of slipping.
• Friction depends on normal force, not on area. $f = \mu N$ has no area in it (to first approximation).
• Rolling friction. A perfectly rigid wheel on a perfectly rigid surface would have zero rolling friction. Real surfaces deform slightly, so $\mu_r$ is small but nonzero.

Key Takeaways

• $f_s \leq \mu_s N$ (static), $f_k = \mu_k N$ (kinetic), with $\mu_k < \mu_s$.
• Angle of friction $\lambda$ and angle of repose $\alpha$ both satisfy $\tan = \mu_s$.
• Friction is independent of contact area (to first order) and depends on normal force.
• Friction enables walking, braking, and rolling; it opposes relative motion at the contact.
• Rolling friction is much smaller than sliding friction — hence wheels.