When two dry surfaces touch, the contact force can be separated into a normal component $N$ and a tangential friction force $f$. A common introductory approximation is the Coulomb friction model.

Friction opposes relative sliding or the tendency to slide at the contact, not necessarily the motion of the object's center of mass.

Static friction adjusts to the situation

If the surfaces are not sliding relative to one another, the friction is static. Its magnitude is not generally equal to $\mu_sN$. Instead,

$$\boxed{|f_s|\le \mu_sN},$$

where $\mu_s$ is the coefficient of static friction.

The actual value of $f_s$ is whatever the equations of motion require to maintain no slip, provided the required value does not exceed the limit $\mu_sN$.

For example, suppose a box rests on a horizontal floor and the maximum available static friction is $30,\mathrm N$. If you push horizontally with $10,\mathrm N$ and the box remains at rest, then

$$f_s=10,\mathrm N,$$

not $30,\mathrm N$. If the required friction reaches the limiting value,

$$|f_s|=\mu_sN,$$

the contact is at impending slip. A larger tangential demand cannot be sustained by the static-friction model, so sliding begins.

Kinetic friction during sliding

Once the surfaces slide relative to each other, a common approximation is

$$\boxed{|f_k|=\mu_kN},$$

where $\mu_k$ is the coefficient of kinetic friction. Its direction is opposite the relative sliding velocity at the contact.

For many ordinary material pairs under similar conditions, $\mu_k$ is smaller than the maximum static-friction coefficient $\mu_s$, but this is an empirical tendency rather than a universal law of nature.

The normal force must be found first

Both friction formulas depend on $N$, and the normal force is not automatically equal to $mg$.

If a block of mass $m$ rests on an incline of angle $\alpha$, perpendicular force balance gives

$$N=mg\cos\alpha.$$

The component of gravity tending to pull the block down the slope is

$$mg\sin\alpha.$$

For the block to remain at rest, static friction must supply

$$f_s=mg\sin\alpha.$$

This is possible only while

$$mg\sin\alpha\le\mu_smg\cos\alpha.$$

At the threshold of sliding,

$$\tan\alpha=\mu_s.$$

Thus the angle at which a block just begins to slide provides an experimental way to estimate the static friction coefficient.

Example

A $4.0,\mathrm{kg}$ block rests on a $20^\circ$ incline with $\mu_s=0.50$ and $\mu_k=0.30$.

The normal force is

$$N=(4.0)(9.81)\cos20^\circ\approx36.9,\mathrm N.$$

The downhill component of gravity is

$$mg\sin20^\circ\approx13.4,\mathrm N.$$

The maximum static friction is

$$\mu_sN\approx(0.50)(36.9)=18.5,\mathrm N.$$

Because only $13.4,\mathrm N$ is required, the block remains at rest and

$$f_s=13.4,\mathrm N.$$

If the angle is increased until sliding begins, the friction changes regime. During downward sliding,

$$f_k=\mu_kN$$

acts uphill.

Limits of the model

The coefficients $\mu_s$ and $\mu_k$ are dimensionless empirical parameters. Real friction can depend on surface condition, lubrication, temperature, sliding speed, contact history, deformation, and other effects. The Coulomb model is therefore an approximation whose usefulness depends on the physical situation.

The essential solving rule is: determine the required static friction from the mechanics first, compare it with the maximum available value, and use the kinetic model only after relative sliding occurs.