# Friction Continued

The following example illustrates friction.

### Example: Skiing Exercise

A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.

Strategy

The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force $$N$$ as $${f}_{\text{k}}={\mu }_{\text{k}}N$$; thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier’s weight perpendicular to the slope. (See the skier and free-body diagram in the figure below.) The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). $$\mathbf{\text{N}}$$ (the normal force) is perpendicular to the slope, and $$\mathbf{\text{f}}$$ (the friction) is parallel to the slope, but $$\mathbf{\text{w}}$$ (the skier’s weight) has components along both axes, namely $${\mathbf{\text{w}}}_{\text{⊥}}$$ and $${\mathbf{\text{W}}}_{//}$$. $$\mathbf{N}$$ is equal in magnitude to $${\mathbf{w}}_{\perp }$$, so there is no motion perpendicular to the slope. However, $$\mathbf{f}$$ is less than $${\mathbf{W}}_{\text{//}}$$ in magnitude, so there is acceleration down the slope (along the x-axis).

That is,

$$N={w}_{\perp }=w\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º}=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º}.$$

Substituting this into our expression for kinetic friction, we get

$${f}_{\text{k}}={\mu }_{\text{k}}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º},$$

which can now be solved for the coefficient of kinetic friction $${\mu }_{\text{k}}$$.

Solution

Solving for $${\mu }_{k}$$ gives

$${\mu }_{\text{k}}=\cfrac{{f}_{\text{k}}}{N}=\cfrac{{f}_{\text{k}}}{w\phantom{\rule{0.15em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º}}=\cfrac{{f}_{\text{k}}}{\text{mg}\phantom{\rule{0.15em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º}.}$$

Substituting known values on the right-hand side of the equation,

$${\mu }_{\text{k}}=\cfrac{\text{45.0 N}}{\left(\text{62 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^{2}\right)\left(0\text{.}\text{906}\right)}=0\text{.}\text{082}.$$

Discussion

This result is a little smaller than the coefficient listed in the table in the previous lesson for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly.

In situations like this, where an object of mass $$m$$ slides down a slope that makes an angle $$\theta$$ with the horizontal, friction is given by $${f}_{\text{k}}={\mu }_{\text{k}}\text{mg}\phantom{\rule{0.15em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\theta$$.

All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this tutorial’s Problems and Exercises.

### Optional Take-Home Experiment

An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in the example above, the kinetic friction on a slope $${f}_{\text{k}}={\mu }_{\text{k}}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\theta$$.

The component of the weight down the slope is equal to $$\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.15em}{0ex}}\theta$$ (see the free-body diagram in the figure above). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out:

$${f}_{\text{k}}={\text{Fg}}_{\text{x}}$$

$${\mu }_{\text{k}}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$$

Solving for $${\mu }_{\text{k}}$$, we find that

$${\mu }_{\text{k}}=\cfrac{\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }{\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta }=\text{tan}\phantom{\rule{0.25em}{0ex}}\theta .$$

Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find $${\mu }_{\text{k}}$$.

Note that the coin will not start to slide at all until an angle greater than $$\theta$$ is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Think about how this may affect the value for $${\mu }_{\text{k}}$$ and its uncertainty.

We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force.

### Making Connections: Submicroscopic Explanations of Friction

The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics.

These characteristics not only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat.

The figure below illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion.

When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area. Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied force, the area of actual contact increases as does friction.

But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices.

When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces.

The figure below shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this tutorial.

The variation in shear stress is remarkable (more than a factor of $${\text{10}}^{\text{12}}$$ ) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction. The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic nature of friction.

### PhET Explorations: Friction

Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces).

Forces and Motion

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