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Section 9.1 Dry Friction

There are multiple types of friction, including:

  1. Dry friction: the force that opposes one solid surface sliding across another solid surface.

  2. Fluid friction: the friction between layers of a viscous fluid in motion.

  3. Internal friction: the force resisting internal deformation of a solid material.

Statics focuses on dry friction. Dry friction occurs between two surfaces in contact and always acts in the direction which opposes the relative motion of the contact surfaces. We can represent this distributed force as a concentrated friction force. Depending on the details of the situation, dry friction can hold the object in equilibrium or cause it to accelerate or decelerate. In Statics we are only concerned with objects in equilibrium; accelerating and decelerating objects will be covered in another course.

Consider the football training sled shown in Figure 9.1.1. Prior to use, the weight of the sled is supported by a normal force directly underneath the center of gravity, and the frictional between the ground and the sled is zero. When players push to move the sled to the right, dry friction will develop along the bottom surface of the sled in order to prevent the sled from sliding to the right. This frictional force is represented as a point normal force, offset from the center of gravity of the sled. This offset is required to maintain rotational equilibrium given the pushing force. If the pushing force is large enough, equilibrium will break and the sled will begin to slide to the right.

Figure 9.1.1. Dry friction appears between the bottom of a training sled and the field when motion is impending or occurring.

The friction forces acting on the motorcycle in Figure 9.1.2 are more complicated. Both wheels rotate clockwise, but the rear wheel is driven by the engine and chain, while the front wheel is rotated by the road friction. The friction force on the rear tire acts to the right and is what allows the bike to maintain speed or accelerate. The dry friction on the front tire acts to the left and retards the motion of the motorcycle.

Figure 9.1.2. Dry friction on a motorcycle occurs where the tires contact the road.

For computational simplicity we can decompose the force on an object being pushed across the ground into a vertical normal force and a horizontal friction force, as shown in the third FBD below. In reality, the force acting on an object approximates the non-vertical distributed force shown in the first FBD below. The second FBD shows the intermediate step.

Subsection 9.1.1 How can you tell from a problem statement which phase of friction to use?

The most commonly used model for dry friction is Coulomb friction, which includes phases (as shown in Figure 9.1.4 below): (1) static but not impending friction, (2) impending motion, and (3) kinetic friction. Coulomb friction relies on a friction coefficient \(\mu\) to serve as the proportionality constant between the normal and friction forces. The friction coefficient \mu is always greater than 0 and commonly less than 1. The friction coefficient can be greater than 1 for materials that exhibit positive adhesion to each other (like silicone rubber or glued surfaces).

Figure 9.1.4. A simple model of a book being pushed horizontally while sitting on a table. FBDs are provided for the system, book, and table. Additionally, a Coulomb friction graph is shown and the various phases are numbered.

Here is a more detailed discussion of what happens in each friction phase.

  1. At the origin of the graph in Figure 9.1.4, there is material interaction between the book and table, thus a friction force is available, but not engaged \(F_f=F_\text{push}=0\text{.}\)

  2. Assuming the pushing force \(F_\text{push}\) gradually increases from zero in this static-but-not-impending phase, the friction force increases proportionally to keep the forces balanced and the box static \(F_f=F_\text{push}\text{.}\)

  3. Next, the maximum friction possible is reached at impending motion and the pushing force matches the maximum friction force \(F_{f_\text{max}}=\mu_\text{s} N\text{.}\)

  4. Finally, if the pushing force is increased beyond impending motion, the book starts to move under the influence of kinetic friction \(F_{f_\text{max}}=\mu_k N\text{.}\)

Notice that impending motion friction is always greater than kinetic friction, this is because the coefficient \(\mu_\text{s} \gt \mu_k\) for most materials. Practically, this tells us that once a material starts to move it is easier to keep moving than it was to get it started from rest.

If you wonder why we include kinetic friction in a statics course, remember that a body in equilibrium might be moving with constant velocity.

The key to deciding what type of friction is appropriate for a given problem will depend on the specifics of the problem statement.

Table 9.1.5. Summary of the three phases of friction with the key to recognizing and examples of each.


Governing equation

Key to recognizing

Static-but-not- impending motion

None, \(F_f\) is independent of \(N\text{.}\)

Mention of sitting or static, but no extremal language


A moment of \(\Nm{20}\) is applied to a wheel held static by a brake arm. What is the friction force between the wheel and the brake arm? A box sits on a slope, find the resultant of the friction and normal forces on the box.

Impending motion

\(F_f = \mu_\text{s} N\)

The use of extremal language (related to maximum or minimum values) in the problem statements


What is the maximum force applied to the box before it will start to move? What is the minimum coefficient of static friction that will keep the box static? What is the lightest box which will not slip or tip on this slope? What is the most massive moment that the brake arm will hold static?

Kinetic friction

\(F_f = \mu_k N\)

Mention of the body moving at a constant velocity under the effect of friction. (You will learn to solve non-constant velocity friction problems in Dynamics).


A \(\kg{40}\) box is sliding down a 20° slope, what is the coefficient of friction to keep the velocity constant? A rope slips around a tree with a known friction coefficient and a constant velocity, what is the contact angle of the rope?

Note that when you solve static-but-not-impending motion problems (as opposed to impending or kinetic) that you will need one additional piece of data given in the problem to solve for both \(F_f \) and \(N\) as you will not be able to apply the equation \(F_f=\mu N\text{.}\)

Subsection 9.1.2 Friction angle and friction resultant

Friction and normal forces can be represented in two equivalent ways:

Friction and normal forces.

Using these perpendicular vectors is most convenient on block and wedge type problems and apply equally-well in all three phases of friction (static but not impending, impending motion, and kinetic friction)

Resultant force and friction angle.

Recall that resultants are the sum of two or more vectors. In this case, the friction resultant is \(\vec{R}=\vec{F}+\vec{N}\text{.}\) The friction angle \(\phi_s \) is the angle between the friction resultant and normal forces but only is applicable at impending motion. The friction resultant and friction angle are used for screw, flexible belt, and journal bearing type problems.

Figure 9.1.6. A block of weight \(W\) is pushed sideways by force \(P\text{.}\) The combined effect of the friction and normal forces can either be modeled as separate friction (\(F\)) and normal (\(N\)) forces (left) or a friction resultant force (\(R\)) and friction angle \((\phi_s)\text{.}\)

The friction angle \(\phi_s\) relates directly to the coefficient of static friction \((\mu_\text{s})\) as the friction angle \(\phi_s\) is the internal angle of the right triangle formed by the normal force \(N\text{,}\) the friction resultant \(R\text{,}\) and the friction force \(F\text{.}\) Hence:

\begin{equation} \tan \phi_s = \frac{F}{N} \qquad \phi_s = \tan^{-1}\left(\frac{F}{N} \right)\tag{9.1.1} \end{equation}

Subsection 9.1.3 Normal Forces

Commonly on two dimensional problems, normal forces are represented at the point of contact or in the middle of the contacting surface. However, normal forces are rarely point loads but instead are loads distributed across a surface. The point force vector we typically represent as a normal force on an FBD is the externally equivalent resultant of the distributed normal force (as demonstrated in Chapter 7, distributed loads). To demonstrate this further, take a look at Figure 9.1.7.

Figure 9.1.7. (A) A box at rest has a uniformly distributed normal force along the bottom of the box. Hence, the resultant normal force is located directly below the weight force, and the friction force (while available) is not engaged. (B) As a pushing force\(F_\text{push}\) is applied, the distributed normal force changes shape, and the resultant normal force \(N\) is shifted to the right to maintain equilibrium. (C) This trend continues as you push harder on the box as the normal force continues to shift right.