Sliding (motion)

Free body diagram for a block subject to friction as it slides on an incline plane

A common problem presented in introductory physics classes is a block subject to friction as it slides up or down an inclined plane. This is shown in the free body diagram to the right.

The component of the force of gravity in the direction of the incline is given by:[4]

${\displaystyle F_{g}=mg\sin {\theta }}$

The normal force (perpendicular to the surface) is given by:

${\displaystyle N=mg\cos {\theta }}$

Therefore, since the force of friction opposes the motion of the block,

${\displaystyle F_{k}=\mu _{k}\cdot mg\cos {\theta }}$

To find the coefficient of kinetic friction on an inclined plane, one must find the moment where the force parallel to the plane is equal to the force perpendicular; this occurs when the block is moving at a constant velocity at some angle ${\displaystyle \theta }$

${\displaystyle \sum F=ma=0}$

${\displaystyle F_{k}=F_{g}}$ or ${\displaystyle \mu _{k}mg\cos {\theta }=mg\sin {\theta }}$

Here it is found that:

${\displaystyle \mu _{k}={\frac {mg\sin {\theta }}{mg\cos {\theta }}}=\tan {\theta }}$ where ${\displaystyle \theta }$ is the angle at which the block begins moving at a constant velocity[5]