# 0.7 3.8 friction  (Page 4/10)

 Page 4 / 10
${f}_{\text{k}}={\mu }_{\text{k}}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.15em}{0ex}}\text{25º},$

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

Solution

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

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

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

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

Discussion

This result is a little smaller than the coefficient listed in [link] 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 chapter’s Problems and Exercises.

## 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 [link] , 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 [link] ). 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

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

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. Discuss 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.

[link] 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.

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