$F_{s, \text{max}} = \mu_s N$ Where $\mu_s$ is the coefficient of static friction, and $N$ is the normal force.

$F_k = \mu_k N$ Where $\mu_k$ is the coefficient of kinetic friction. Typically, $\mu_k < \mu_s$.

Note: Note: If the block is NOT moving and NOT on the verge of moving, the actual friction force $F$ is determined strictly by the equations of equilibrium ($\Sigma F = 0$), and $F \le \mu_s N$.

A wedge is a simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force. They are used to lift heavy blocks, adjust elevations, or split materials.

The transmission of power through belts and pulleys, or the braking of a rotating drum using a band brake, relies entirely on friction over a curved surface. When a flexible belt is wrapped around a rough cylinder and motion is impending (the belt is about to slip), the tension on the pulling side ($T_2$) is greater than the tension on the yielding side ($T_1$) according to the equation: $T_2 = T_1 e^{\mu \beta}$

Where:

Unlike flat belts that sit on a cylindrical drum, a V-belt sits in a wedged groove with an angle $2\alpha$. The normal force is magnified by the wedge effect, which significantly increases the friction force without requiring more tension on the belt. The belt friction equation is modified to use an effective coefficient of friction $\mu'$: $T_2 = T_1 e^{\mu' \beta}$ Where: $\mu' = \frac{\mu}{\sin \alpha}$ Because $\sin \alpha < 1$, the effective friction $\mu'$ is much larger than the actual friction $\mu$.

A square-threaded screw is essentially a wedge wrapped around a cylinder. When a moment $M$ is applied to tighten the screw against an axial load $W$, impending motion is upward along the thread. The required moment is: $M = W r \tan(\theta + \phi_s)$ Where:

To loosen the screw (impending motion downward): $M' = W r \tan(\theta - \phi_s)$ If $\theta \le \phi_s$, the screw is self-locking, meaning it will not unscrew under the axial load $W$ without an applied moment.

When a cylinder or wheel rolls on a surface, deformation of both bodies occurs, shifting the normal reaction force slightly forward in the direction of motion by a distance $a$. This creates a resisting moment that must be overcome to maintain constant speed: $P \approx \frac{W a}{r}$ Where $P$ is the horizontal pulling force required, $W$ is the weight of the cylinder, $r$ is its radius, and $a$ is the coefficient of rolling resistance (which has units of length).

Consider a block of weight $W$, height $h$, and base width $b$. A horizontal force $P$ is applied at a height $y$ from the base.

Conclusion: The actual force $P$ required to cause motion is the smaller of $P_{\text{slip}}$ and $P_{\text{tip}}$. The smaller value determines the dominant failure mode.