Sliding Friction: Definition, Coefficient, Formula (W/ Examples)

Sliding friction, more commonly referred to as kinetic friction, is a force that opposes the sliding motion of two surfaces moving past each other. In contrast, static friction is a type of friction force between two surfaces that are pushing against each other, but not sliding relative to each other. (Imagine pushing on a chair before it begins sliding across the floor. The force you apply before the sliding begins is opposed by static friction.)

Sliding friction typically involves less resistance than static friction, which is why you often have to push harder to get an object to start sliding than to keep it sliding. The magnitude of the force of friction is directly proportional to the magnitude of the normal force. Recall that the normal force is the force perpendicular to the surface that counteracts any other forces being applied in that direction.

The constant of proportionality is a unitless quantity called the coefficient of friction, and it varies depending on the surfaces in contact. (Values for this coefficient are typically looked up in tables.) The friction coefficient is usually represented by the Greek letter ​μ​ with a subscript ​k​ indicating kinetic friction. The frictional force formula is given by:

\(F_f=\mu_kF_N\)

Where ​_F_N_​ is the magnitude of the normal force, the units are in newtons (N) and the direction of this force is opposite the direction of motion.

Rolling resistance is sometimes referred to as rolling friction, though it is not exactly a friction force because it is not the result of two surfaces in contact trying to push against each other. It is a resistive force resulting from energy loss due to deformations of the rolling object and the surface.

Just as with friction forces, however, the magnitude of the rolling resistance force is directly proportional to the magnitude of the normal force, with a constant of proportionality that depends on the surfaces in contact. While ​_μ_r​ is sometimes used for the coefficient, it is more common to see ​C_rr_​, making the equation for the rolling resistance magnitude the following:

\(F_r=C_{rr}F_N\)

This force acts opposite the direction of motion.

Let's consider a friction example involving a dynamics cart found in a typical physics classroom and compare the acceleration with which it travels down a metal track inclined at 20 degrees for three different scenarios:

​Scenario 1:​ There are no friction or resistive forces acting on the cart as it rolls freely without slipping down the track.

First we draw the free-body diagram. The force of gravity pointing straight down, and the normal force pointing perpendicular to the surface are the only forces acting.

The net force equations are:

\(F_{netx}=F_g\sin{\theta}=ma\
F_{nety}=F_N-F_g\cos(\theta)=0\)

Straight away we can solve the first equation for acceleration and plug in values to get the answer:

\(F_g\sin{\theta}=ma\
\implies mg\sin(\theta)=ma\
\implies a=g\sin(\theta)=9.8\sin(20)=\boxed{3.35\text{ m/s}^2}\)

​Scenario 2:​ Rolling resistance acts on the cart as it rolls freely without slipping down the track.

Here we will assume a coefficient of rolling resistance of 0.0065, which is based on an example found in a paper from the U.S. Naval Academy.

Now our free-body diagram includes rolling resistance acting up the track. Our net force equations become:

\(F_{netx}=F_g\sin{\theta}-F_r=ma\
F_{nety}=F_N-F_g\cos(\theta)=0\)

From the second equation, we can solve for ​_F_N​, plug the result into the expression for friction in the first equation, and solve for ​a_​:

\(F_N-F_g\cos(\theta)=0\implies F_N=F_g\cos(\theta)\
F_g\sin(\theta)-C_{rr}F_N=F_g\sin(\theta)-C_{rr}F_g\cos(\theta)=ma\
\implies \cancel mg\sin(\theta)-C_{rr}\cancel mg\cos(\theta)=\cancel ma\
\implies a=g(\sin(\theta)-C_{rr}\cos(\theta))=9.8(\sin(20)-0.0065\cos(20))\
=\boxed{3.29 \text{ m/s}^2}\)

​Scenario 3:​ The cart's wheels are locked in place, and it slides down the track, impeded by kinetic friction.

Here we will use a coefficient of kinetic friction of 0.2, which is in the middle of the range of values typically listed for plastic on metal.

Our free-body diagram looks very similar to the rolling resistance case except that it is a sliding friction force acting up the ramp. Our net force equations become:

\(F_{netx}=F_g\sin{\theta}-F_k=ma\
F_{nety}=F_N-F_g\cos(\theta)=0\)

And again we solve for ​a​ in a similar fashion:

\(F_N-F_g\cos(\theta)=0\implies F_N=F_g\cos(\theta)\
F_g\sin(\theta)-\mu_kF_N=F_g\sin(\theta)-\mu_kF_g\cos(\theta)=ma\
\implies \cancel mg\sin(\theta)-\mu_k\cancel mg\cos(\theta)=\cancel ma\
\implies a=g(\sin(\theta)-\mu_k\cos(\theta))=9.8(\sin(20)-0.2\cos(20))\
=\boxed{1.51 \text{ m/s}^2}\)

Note that the acceleration with rolling resistance is very close to the frictionless case, while the sliding friction case is significantly different. This is why rolling resistance is neglected in most situations and why the wheel was a brilliant invention!