Kinetic and static friction

Friction acts along the surface between two objects in contact. It comes in two forms with different rules, and telling them apart is the whole job. Kinetic friction acts when the surfaces slide. Its size is fixed: $f_k = \mu_k N$, where $N$ is the normal force pressing them together and $\mu_k$ is the coefficient of kinetic friction. It does not depend on how fast they slide, and it points opposite the sliding.

Static friction acts when the surfaces are not sliding. It has no single value. It gives back exactly enough force to stop slipping, up to a limit: $f_s \le \mu_s N$, with the maximum $\mu_s N$ reached only at the verge of slipping. Usually $\mu_s > \mu_k$, so it takes more force to start something sliding than to keep it going.

Both forms use the normal force $N$. On flat ground with no vertical push, $N = mg$. But $N$ is whatever the surface must push with to keep the object from sinking in, and that is not always $mg$ (§5). Since friction is $\mu N$ either way, a wrong $N$ makes the friction wrong too.

Two questions run through the topic: how big is the friction, and which way does it point? Static friction makes both tricky. Its size is as much as needed, up to $\mu_s N$ (§2), and its direction opposes the tendency to slip, which is often the way the object actually moves (§4).

The most common friction mistake is to compute $\mu_s N$ and report it as the friction, no matter the situation. But $\mu_s N$ is the maximum static friction, not its actual value. While the object stays put, static friction equals only what it takes to balance the other forces.

Push a heavy crate that will not move. Push gently and it pushes back gently; push harder and it pushes back harder. The friction rises to match you. Since the crate stays still, static friction equals whatever you apply: push with $5$ N and friction is $5$ N; push with $20$ N and it is $20$ N, as long as the crate has not broken free.

It can only do this up to its ceiling $\mu_s N$. So finding static friction is a two-step slip check. First find the maximum, $\mu_s N$. Then compare it with the force trying to cause sliding. If that force is smaller, the object stays put and $f_s$ equals it. If it would exceed $\mu_s N$, the object breaks free and kinetic friction $\mu_k N$ takes over.

So $\mu_s N$ answers one question: how hard can you push before it slips? It is not the friction for every gentler push. Reporting $\mu_s N$ for a crate sitting still is the trap; the real friction there is whatever balances the push.

Setup. A $10$ kg box sits on level ground with $\mu_s = 0.5$ and $\mu_k = 0.3$. Use $g = 10$ N/kg. (a) You push horizontally with $30$ N. Does it move, and what is the friction? (b) You push with $60$ N instead.

Find the ceiling first. On flat ground $N = mg = 10 \cdot 10 = 100$ N, so the maximum static friction is $\mu_s N = 0.5 \cdot 100 = 50$ N. Compare both pushes against this.

(a) Push $= 30$ N. Since $30 < 50$, the box does not slip. Static friction matches the push: $f_s = 30$ N, and the box stays put. It is not $50$ N.

(b) Push $= 60$ N. Since $60 > 50$, the box breaks free and slides. Kinetic friction acts: $f_k = \mu_k N = 0.3 \cdot 100 = 30$ N. The net force is $60 - 30 = 30$ N, so $a = 30/10 = 3$ m/s$^2$. The friction dropped from its $50$ N peak to $30$ N the instant the box moved.

The trap to avoid. In part (a) the answer is $30$ N, not the maximum $50$ N. The product $\mu_s N$ is only the friction at the verge of slipping; below it, static friction gives back exactly what is applied.

Kinetic friction is easy: it points opposite the sliding. A block skidding right feels kinetic friction left. Static friction is trickier. It points opposite the tendency to slip, and that tendency often runs opposite the way the object moves, so the friction ends up pointing forward.

Walking is the clearest case. Your foot pushes backward on the floor; by Newton’s third law the floor pushes your foot forward, and that forward static friction moves you ahead. A car is the same: the tire pushes backward on the road, and the road’s static friction pushes the tire forward to speed the car up. A box in an accelerating truck bed rides forward on static friction too, since that is the only horizontal force on it.

The pattern: whenever something is pushed forward by contact without slipping, static friction points along the motion, not against it. Rolling without slipping does not always behave this way, though. The static friction on a rolling object can point forward, backward, or be zero depending on the other forces and torques, so fall back on the slip test below. Only when surfaces actually slide does friction oppose the motion, and even then it opposes the sliding of one surface past the other.

The trap is the blanket rule that friction always opposes motion. That holds for kinetic friction on a sliding object, but not for the static friction that walks you forward or drives a car. To find the direction, ask which way the surfaces would slip if friction vanished, then point friction the opposite way.

Both friction laws use $N$, and the most expensive error is to set $N = mg$ out of habit. The normal force is whatever the surface pushes with, straight out from itself, to keep the object from sinking in. On flat ground with no vertical push that equals $mg$. Tilt the surface or push at an angle and it does not.

On an incline. On a ramp tilted at angle $\theta$, gravity still points straight down, but only the part pressing into the surface counts. That part is $mg\cos\theta$, so $N = mg\cos\theta$, always less than $mg$. The friction available is then $f = \mu N = \mu m g\cos\theta$, not $\mu m g$. The steeper the ramp, the smaller $N$, and the less friction the surface can give.

With an angled push. Push down on a box at an angle and you press it harder into the floor: a downward part $F\sin\alpha$ gives $N = mg + F\sin\alpha$. Pull up at an angle and you lighten it: $N = mg - F\sin\alpha$. Either way the friction $\mu N$ changes with the push, even though the weight has not.

The trap is writing $f = \mu m g$ everywhere. That is only the flat-ground, no-vertical-push case. The fix: always find $N$ first from the forces pressing into the surface, then use $f = \mu N$. On an incline that means $N = mg\cos\theta$; under an angled push, add or subtract its vertical part.

One limit holds for all of these: a surface can push but never pull, so $N$ can never be negative. If the perpendicular balance would require $N < 0$, the object has lifted off the surface and $N = 0$ instead.

Setup. A $5$ kg block rests on a ramp tilted at $\theta = 37^\circ$, with $\mu_s = 0.5$ and $\mu_k = 0.4$. Use $g = 10$ N/kg, $\cos 37^\circ = 0.8$, $\sin 37^\circ = 0.6$. (a) Does the block slide? (b) If it does, find its acceleration down the ramp.

Compare the two forces along the ramp. Gravity’s pull down the slope is $mg\sin\theta = 5 \cdot 10 \cdot 0.6 = 30$ N. The most static friction can hold back is $\mu_s N = \mu_s m g\cos\theta = 0.5 \cdot 5 \cdot 10 \cdot 0.8 = 20$ N. Here $N = mg\cos\theta = 40$ N, not $mg = 50$ N.

(a) Slip check. The driving force down the slope is $30$ N; the maximum static friction is only $20$ N. Since $30 > 20$, static friction cannot hold the block and it slides.

(b) Acceleration while sliding. Kinetic friction now acts up the slope: $f_k = \mu_k m g\cos\theta = 0.4 \cdot 40 = 16$ N. Along the ramp, $ma = mg\sin\theta - f_k = 30 - 16 = 14$ N, so $a = 14/5 = 2.8$ m/s$^2$ down the slope. In symbols, $a = g(\sin\theta - \mu_k\cos\theta)$.

The trap to avoid. Using $N = mg = 50$ N would give $\mu m g$ instead of $\mu m g\cos\theta$, overstating the friction by a quarter and maybe hiding that the block slides. On an incline the normal force is $mg\cos\theta$: the perpendicular direction sets $N$, the along-ramp direction sets the motion.