Kinetic and static friction

Friction is a contact force between two surfaces. It comes in two modes. Static friction acts when the surfaces are not slipping past each other. Kinetic friction acts when they are. The two are governed by different rules and they are tested in different traps.

Both depend on the normal force pressing the surfaces together. Both are tangential to the contact surface (perpendicular to the normal). Neither depends on the contact area between the surfaces. That last point is counterintuitive: a brick on its broad face and the same brick on its narrow side feel the same friction force from the same surface. Same brick, same weight, same coefficient, same normal force. The microscopic reason: more area means more contact points, but less pressure on each one, and the two effects cancel.

Both also share something: friction acts along the contact surface, opposing relative motion (or its tendency). For a block sliding east on a stationary floor, that boils down to "friction on block points west." For a foot pushing off the ground while walking, it does not. § 5 takes that case apart.

The two coefficients, $\mu_s$ and $\mu_k$, are properties of the surface pair. They are dimensionless. Typical values run from about $0.05$ (steel on ice, kinetic) to about $1.0$ (rubber on dry asphalt, static). For most pairs $\mu_s$ is somewhat larger than $\mu_k$, a fact that powers the slip-and-drop you'll see in § 3.

Push lightly on a heavy box on a rough floor. Nothing happens. Push harder. Still nothing. Push harder still: at some threshold, the box gives way and starts to slide. During the whole time before that threshold, static friction was doing exactly the work needed to keep the box at rest. It grew with your push and matched it.

The rule is

$$f_s \leq \mu_s F_N.$$

This is an inequality. The product $\mu_s F_N$ is a maximum, not a value. Static friction takes whatever size the other forces require to keep the object at rest, up to that ceiling.

Two analogies make the inequality easier to hold onto. A wall pushing back on you: the wall provides exactly enough force to keep you from passing through it. Push harder, the wall pushes harder. The wall does not exert its maximum strength all the time. Static friction works the same way. Or: a withdrawal limit on a bank account. The limit is a ceiling, not a balance. You can withdraw any amount up to it.

The classic AP trap: a problem gives you a $5$ N push, a $30$ N normal force, and $\mu_s = 0.4$. A student writes $f_s = \mu_s F_N = 12$ N reflexively, concludes the net force on the block is $5 - 12 = -7$ N, and predicts the block accelerates backward at $1.4\,\text{m/s}^2$. The block does not accelerate backward. Static friction is $5$ N exactly, balancing the push, and the block stays at rest.

How to use the rule cleanly: write the equilibrium equations as if the block is at rest, solve for $f_s$, then check whether your $|f_s|$ is at most $\mu_s F_N$. If yes, the at-rest assumption was correct. If no, the block has already slipped and you should switch to kinetic friction.

Once the surfaces are sliding past each other, the friction force takes its kinetic value:

$$f_k = \mu_k F_N.$$

This is an equation, not an inequality. The force has a definite size, set by the normal force and the kinetic-friction coefficient for the pair. It does not depend on the speed of sliding, and it does not depend on the contact area.

The direction is set by relative motion: kinetic friction on a given surface points opposite to that surface's motion relative to the other one. For a block sliding east on a stationary floor: friction on the block points west. The third-law partner (the friction the block exerts on the floor) points east, and acts on the floor.

For most surface pairs, $\mu_k$ is somewhat smaller than $\mu_s$. The drop is visible in the figure as the discontinuity at the slip threshold. Practically, this is why pushing a heavy box at the threshold often produces an unexpected jolt: you were pushing at $\mu_s F_N$ to overcome the static cap, and the moment the block slips, friction drops to $\mu_k F_N$, leaving you with an unbalanced forward force that the box feels as a sudden acceleration.

The kinetic-friction force does not pick a direction based on what the block "wants" to do. It is set by what the surfaces are doing relative to each other. § 5 takes that idea further.

A block of mass $m$ sits on an incline whose angle $\theta$ is slowly increased from zero. The surface pair has static-friction coefficient $\mu_s$. At what angle does the block first slip?

Set up axes along and perpendicular to the slope. Resolve gravity:

• Along the slope (down-slope component): $mg \sin\theta$
• Perpendicular to the slope (into the surface): $mg \cos\theta$

Perpendicular balance (no acceleration through the surface):

$$F_N = mg \cos\theta.$$

While the block is still at rest, along-slope balance gives

$$f_s = mg \sin\theta.$$

Static friction is doing exactly what it needs to. As $\theta$ grows, the down-slope component grows, and $f_s$ grows with it. Meanwhile, the maximum allowable static friction $\mu_s F_N = \mu_s mg \cos\theta$ shrinks (slowly) because $\cos\theta$ shrinks. Eventually $f_s$ catches up to its ceiling.

The critical condition: $f_s = \mu_s F_N$.

$$mg \sin\theta_c = \mu_s\, mg \cos\theta_c$$

The mass cancels:

$$\tan\theta_c = \mu_s.$$

The slip angle depends only on the surface pair, not on what is sitting on the incline. A heavy crate and a light book tip at the same angle on the same slope. That cancellation is the same one that makes a feather and a hammer fall at the same rate (§ 4 of Topic 2.6), and it has the same source: gravitational and inertial mass are the same.

Past the slip angle, $f_s$ would have to exceed $\mu_s F_N$ to keep the block at rest, so the block instead starts to slide. Now use $f_k = \mu_k F_N$ for the friction; the down-slope acceleration becomes $a = g(\sin\theta - \mu_k \cos\theta)$, which is positive (the block speeds up) for any $\theta > \arctan\mu_k$.

The slogan "friction opposes motion" is half right. The full rule:

Friction on a given surface opposes the motion of that surface relative to the surface in contact with it.

For most everyday cases the two readings agree. A block sliding east on a stationary floor moves east relative to the floor; friction on the block points west. Same answer either way.

For walking, the two readings diverge.

When you take a step, your planted foot presses backward against the ground. The contact patch of the foot is not actually slipping (this is static friction, not kinetic), but it is trying to slip backward relative to the ground. Static friction on the foot from the ground points opposite to that tendency: forward. That forward static friction is what accelerates you while walking.

If you tried to walk on perfectly frictionless ice, you could not push off at all. You could swing your legs and arms, but no external horizontal force would act on you, so your center of mass would not move. The same logic applies to a car. The drive wheels push backward against the road; static friction on the tires from the road points forward, and that forward friction is the external force that accelerates the car. The engine drives the wheels but is internal to the car. Without ground friction, no acceleration.

Why static, not kinetic? Because the contact patch of a properly walking foot (or a rolling tire on dry pavement) is not slipping. It presses into the ground without sliding. Kinetic friction would be the case if your foot dragged backward across the floor as you tried to push off, the way a foot might on a wet bathtub.

The check that always works: ask which surface is trying to move which way relative to the other, then point friction on each surface opposite to its own tendency. For the sliding block, the block surface tries to move east relative to the floor; friction on the block points west. For the walking foot, the foot surface tries to move backward relative to the ground; friction on the foot points forward. Same rule, two different answers, different setups.

A box of mass $m$ rests on a level floor. The kinetic-friction coefficient between box and floor is $\mu_k$. A rope attached to the box pulls with force $F$ at an angle $\theta$ above the horizontal. The box slides at constant velocity. Find $F$ in terms of $m$, $\theta$, $g$, and $\mu_k$.

The free-body forces on the box are gravity, the normal force from the floor, the rope's pull (which has both horizontal and vertical components), and kinetic friction. The vertical and horizontal axes can be analyzed separately.

Vertical balance. The box is not accelerating vertically (it stays on the floor). Forces in the $y$-direction are gravity $-mg$, the upward component of the pull $+F \sin\theta$, and the normal force $+F_N$. Their sum is zero:

$$F_N + F \sin\theta - mg = 0,$$

so

$$F_N = mg - F \sin\theta.$$

This is the punchline. The normal force is not $mg$. The rope's upward component partially holds the box up, leaving the floor with less work to do, so $F_N$ is reduced. The trap that the AP exam loves to spring here is the automatic move $F_N = mg$ (Pitfall 03), which gives a wrong friction.

Horizontal balance. Constant velocity means no horizontal acceleration. Forces in the $x$-direction are the horizontal component of the pull $+F \cos\theta$ and kinetic friction $-\mu_k F_N$. Their sum is zero:

$$F \cos\theta - \mu_k F_N = 0.$$

Substitute the expression for $F_N$:

$$F \cos\theta = \mu_k\,(mg - F \sin\theta).$$

Solve for $F$:

$$F = \dfrac{\mu_k\, mg}{\cos\theta + \mu_k \sin\theta}.$$

Two quick checks. At $\theta = 0$, the rope pulls horizontally, $\sin\theta = 0$ and $\cos\theta = 1$, so $F = \mu_k mg$. That is just $\mu_k F_N$ with $F_N = mg$, which is the standard level-ground answer. Good.

At small but nonzero $\theta$, the denominator grows: $\mu_k \sin\theta$ becomes positive and adds to $\cos\theta$, which itself shrinks only slightly at small angles. So $F$ shrinks. Pulling at an angle is more efficient than pulling horizontally: the upward component reduces the normal force, which reduces the friction the rope has to overcome. There is an optimal angle (eventually $\cos\theta$ shrinks fast enough that $F$ starts growing again), but the point: an upward-angled pull lifts some of the weight off the surface, which lightens the friction.

If you tried this with $F_N = mg$ pre-baked in, you would get $F = \mu_k mg / \cos\theta$, which only grows with $\theta$ and predicts the opposite trend. The error is the automatic $F_N = mg$. A cosine factor in the denominator alone misses that the normal force is changing too.

Three failure modes show up over and over on friction problems. Each one has a family of distractors built around it on the AP exam.