6.1 Laws of dry friction and applications

Types of Friction

Friction as an Opposing Force

Friction is a force that opposes the relative motion between two surfaces in contact. It acts parallel to the contact surfaces and points opposite to the direction of motion or attempted motion. Every friction problem in this course starts by identifying this direction correctly on your free body diagram.

Static, Kinetic, and Rolling Friction

Static friction acts between two surfaces that are at rest relative to each other. It prevents motion from starting, and its magnitude adjusts to match whatever force is trying to cause sliding. This is a critical detail: static friction is not a fixed value. It ranges from zero up to a maximum given by:

$f_{s,\max} = \mu_s \cdot N$

where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Until the applied force exceeds $f_{s,\max}$, the object stays put, and the friction force simply equals whatever it needs to be to maintain equilibrium.

Kinetic friction acts between two surfaces that are already sliding relative to each other. Unlike static friction, kinetic friction has a single, constant magnitude for a given pair of surfaces:

$f_k = \mu_k \cdot N$

where $\mu_k$ is the coefficient of kinetic friction. For nearly all material pairs, $\mu_k < \mu_s$, which is why it takes more force to start sliding something than to keep it sliding.

Rolling friction acts when an object rolls along a surface. It's caused by slight deformation at the contact zone and is typically much smaller than sliding friction. You won't encounter it as often in a statics course, but it explains why wheels are so much more efficient than dragging something.

Coefficient of Friction

The coefficient of friction ($\mu$) is a dimensionless number that characterizes how "grippy" a pair of surfaces is. Some reference values to build intuition:

• Rubber on dry concrete: $\mu_s \approx 0.6 - 0.8$
• Steel on steel (dry): $\mu_s \approx 0.6$
• Steel on ice: $\mu_s \approx 0.03 - 0.05$

These values are determined experimentally. The coefficient depends on the material pair, not on the contact area. A wider block on the same surface has the same $\mu$ as a narrow one.

Laws of Dry Friction

Direction and Magnitude of the Friction Force

Coulomb's model of dry friction (the one used throughout this course) rests on a few key principles:

• The friction force always acts tangent to the contact surface, opposing the direction of motion or the tendency toward motion.
• The maximum static friction force is proportional to the normal force: $f_{s,\max} = \mu_s \cdot N$.
• The kinetic friction force is proportional to the normal force: $f_k = \mu_k \cdot N$.
• The friction force is independent of the apparent contact area.

The angle of friction $\phi_s$ is sometimes useful. It's defined as $\tan(\phi_s) = \mu_s$. Geometrically, $\phi_s$ is the angle between the resultant contact force (friction + normal) and the normal direction at the point of impending motion. If an applied force's line of action falls within the "cone of friction" defined by $\phi_s$, the object won't slide no matter how large the force is.

Determining Sliding Conditions

To figure out whether an object will slide, follow these steps:

1. Draw a complete free body diagram, including weight, normal force, applied forces, and the friction force (direction opposing the tendency of motion).
2. Assume the object is in equilibrium and solve for the required friction force using $\Sigma F_x = 0$ and $\Sigma F_y = 0$.
3. Compare the required friction force to $f_{s,\max} = \mu_s \cdot N$.
4. If the required friction force $\leq f_{s,\max}$, the object stays in equilibrium. The actual friction force is whatever you solved for in step 2.
5. If the required friction force $> f_{s,\max}$, the object slides, and the friction force becomes $f_k = \mu_k \cdot N$.

Friction's Effect on Motion

Friction's Influence on Velocity and Acceleration

When friction is the only horizontal force on a sliding object, it produces a deceleration. Using Newton's second law:

$a = \frac{f_k}{m} = \frac{\mu_k \cdot N}{m}$

For an object on a flat surface where $N = mg$, this simplifies to $a = \mu_k \cdot g$. The deceleration depends only on $\mu_k$ and gravity, not on the object's mass.

When a driving force is also present (like a car's engine), friction reduces the net force available for acceleration. The net acceleration becomes:

$a = \frac{F_{applied} - f_k}{m}$

Friction and Direction of Motion

If a force is applied at an angle to a surface, friction still acts along the surface opposing the component of motion (or tendency of motion) parallel to that surface. For a block pushed at an angle $\theta$ below the horizontal on a flat surface, the normal force increases because the vertical component of the push adds to the weight:

$N = mg + F\sin\theta$

This means the friction force also increases, which is why pushing something at a steep downward angle can make it harder to slide.

Work Done by Friction and Stopping Distance

Friction always does negative work on a sliding object because the force and displacement point in opposite directions. The energy removed by friction is converted to heat.

For an object sliding to a stop on a flat surface from initial speed $v_0$, you can find the stopping distance $d$ using the work-energy theorem:

$\frac{1}{2}mv_0^2 = \mu_k \cdot mg \cdot d$

Solving for $d$:

$d = \frac{v_0^2}{2\mu_k g}$

Notice that mass cancels out. A heavy box and a light box with the same initial speed on the same surface will slide the same distance.

Friction Force in Equilibrium

Equilibrium Conditions

In statics, you're mostly dealing with objects that aren't moving. For a rigid body in equilibrium, three conditions must hold:

• $\Sigma F_x = 0$
• $\Sigma F_y = 0$
• $\Sigma M_O = 0$ (moments about any point)

Friction is one of the unknown forces you solve for using these equations.

Determining the Friction Force in Equilibrium

Here's the process for a typical equilibrium-with-friction problem:

1. Draw the free body diagram. Include the weight at the center of gravity, normal forces at each contact, friction forces opposing the tendency of motion at each contact, and any applied loads.
2. Choose a coordinate system and a convenient moment point (often at a contact point to eliminate unknowns).
3. Write the three equilibrium equations.
4. Solve for the unknowns, including the friction force(s).

Checking the Validity of the Equilibrium Solution

After solving, you must verify that your answer is physically possible:

1. Check that each friction force satisfies $f \leq \mu_s \cdot N$ at its respective contact.
2. Check that each normal force is compressive ($N \geq 0$). A negative normal force means the surface would need to pull on the object, which a simple contact can't do.
3. If either check fails, the assumed equilibrium configuration is impossible, and the object will move.

Classic example: a ladder against a wall. A uniform ladder of length $L$ and weight $W$ leans against a smooth (frictionless) wall at angle $\theta$ with the floor. The floor has friction with coefficient $\mu_s$. Taking moments about the base eliminates the floor reactions and lets you solve for the wall's normal force. Then the horizontal equilibrium equation gives you the required friction at the floor. The ladder stays in place only if that required friction doesn't exceed $\mu_s$ times the floor's normal force. If you decrease $\theta$ (make the ladder more horizontal), the required friction increases, which is why shallow-angle ladders are more likely to slip.