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→Kinematics

{{perspectives}}

{{mechanics scenario}}

===Component along (down) the inclined plane===

For the axis ''down'' the inclined plane, the gravitational force component is <math>mg\sin \theta</math>. The magnitude and direction of the friction force are not clear. We have three cases, the case of ''no sliding'', where an upward static friction <math>f_s</math> acts(subject to the constraint <math>f_s \le \mu_sN</math>), the case of ''sliding down'', where an upward kinetic friction <math>f_k = \mu_kN</math> acts, and the case of ''sliding up'', where a downward kinetic friction <math>f_k = \mu_kN</math> acts.

:

{| class="sortable" border="1"

! Case !! Convention on the sign of acceleration <math>a</math> !! Equation using <math>f_s, f_k</math> !! Constraint on <math>f_s, f_k</math> !! Simplified, substituting <math>f_s,f_k</math> in terms of <math>N,\mu_s,\mu_k</math> !! Simplified, after combining with <math>\! N = mg \cos \theta</math>

|-

| Block stationary || no acceleration, so no sign convention || <math>\! mg \sin \theta = f_s</math> || <math>f_s \le \mu_sN</math> || <math>\! mg \sin \theta \le \mu_sN</math> || <math>\! \tan \theta \le \mu_s</math>

|-

| Block sliding down the inclined plane || measured positive down the inclined plane, hence a positive number || <math>\! ma = mg \sin \theta - f_k</math> || <math>\! f_k = \mu_kN</math> || <math>\! ma = mg \sin \theta - \mu_kN </math> || <math>\! a = g (\sin \theta - \mu_k\cos \theta) = g \cos \theta (\tan \theta - \mu_k)</math>

|-

| Block sliding up the inclined plane || measured positive down the inclined plane (note that acceleration opposes direction of motion, leading to retardation) || <math>\! ma = mg \sin \theta + f_k</math> || <math>\! f_k = \mu_kN</math> || <math>\! ma = mg \sin \theta + \mu_kN</math> || <math>\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \mu_k)</math>

|}

==What about the inclined plane?==

A natural question that might occur at this stage is: what is the [[force diagram]] of the inclined plane? And, what is the effect on the inclined plane of the normal force and friction force exerted by the block on it?

The key thing to note is that by assumption, the inclined plane is ''fixed''. This means that ''whatever mechanism is being used to fix the inclined plane'' is generating the necessary forces to balance the forces exerted by the block, so that by [[Newton's first law of motion]], the inclined plane does not move. For instance, if the inclined plane is fixed to the floor with screws, then the [[contact force]] exerted from the floor ([[normal force]] or friction) exactly balances all the other forces on the inclined plane (including its own [[gravitational force]] and the [[contact force]] from the block).

==Value of the acceleration function==

<math>\! a = g\sqrt{1 + \mu_k^2}</math>

==Behavior for a block initially placed at rest==

| Time taken on return journey || <math>\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}</math>

|-

| Speed at instant of return || <math>\! u \sqrt{\frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}}</math>

|}

===Instructional video links===

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