# Question #d6539

Jul 24, 2015

When you say frictional force, I am assuming you just mean friction. I will also suppose we are using an example where we have a smooth wooden block sliding down a semi-rough inclined ramp. Thus I will be referring to static friction, which I will call ${F}_{s}$. Friction is in some sense a reaction force (although strictly speaking it is not, it is intuitive to call it that for a reason I will get to in a moment), because it decreases as the incline angle increases since the parallel component of the force due to gravity starts to vanish; at an angle perpendicular to the flat surface on which the ramp is laid, there is only the vertical component to the force due to gravity (${F}_{g}$).

The parallel component of the force due to gravity might be called ${F}_{g | |}$. When you draw the Free Body Diagram of this situation, you have ${F}_{g | |}$ pointing along the ramp downwards, ${F}_{g \bot}$ pointing inwards and perpendicular to the ramp surface, ${F}_{N}$ (the Normal Force) pointing opposite ${F}_{g \bot}$, and the static friction force (in this case), ${F}_{s}$, pointing along the ramp upwards. Since the direction of ${F}_{s}$ is opposite that of ${F}_{g | |}$, as ${F}_{g | |}$ vanishes, so does ${F}_{s}$.

You can tell that this occurs since the wooden block starts to slide down the ramp as the angle increases, indicating a lower coefficient of static friction (and thus a lower static friction) along the ramp and proving that its magnitude is proportional to the magnitude of its opposing force. That is why I said it was intuitive to call it a "reaction force", even though it is not reacting to an applied force.

A fun problem with friction is:

At what angle would a wooden block on a ramp not slide down, if ${\mu}_{s}$ is $0.2$?

${F}_{g | |} = m g \sin \theta = - {F}_{s}$
${F}_{g \bot} = m g \cos \theta = - {F}_{N}$
${F}_{N} = - m g \cos \theta$
${F}_{s} = {\mu}_{s} {F}_{N} = - {\mu}_{s} m g \cos \theta = - m g \sin \theta$

${\mu}_{s} = \tan \theta$
$\arctan \left({\mu}_{s}\right) = \theta$ for which the block is in static equilibrium.

$\approx {11.31}^{o}$

with positive $x$ being down the ramp and away from the ramp in the perpendicular direction ($y$).