# Changes

## Sliding motion along an inclined plane

, 03:22, 16 January 2012
Kinematics
{{perspectives}}
{{mechanics scenario}}
[[File:Blockonincline.png|thumb|500px|right|The brown triangle is the fixed inclined plane and the black block is placed on it with a dry surface of contact.]]
The scenario here is a dry block (with a stable surface of contact) on a dry ''fixed'' [[involves::inclined plane]], with $\theta$ being the angle of inclination with the horizontal axis.
The extremes are $\theta = 0$ (whence, the plane is horizontal) and $\theta = \pi/2$ (whence, the plane is vertical).

We assume that the block undergoes no rotational motion, i.e., it does not roll or topple. Although the diagram here shows a block with square cross section, this shape assumption is not necessary as long as we assume that the block undergoes no rotational motion.
==Similar scenarios==
! Scenario !! Key addition/complication !! Picture
|-
| [[sliding motion along a frictionless inclined plane]] || $\mu_k = \mu_s = 0$ || [[File:Blockonincline.png|100px]]|-| [[toppling motion along an inclined plane]] || The block can topple || [[File:Topplableblockonincline.png|100px]]|-| [[sliding motion for adjacent blocks along an inclined plane]] || Two blocks instead of one, with a normal force between them || [[File:Adjacentblocksonincline.png|100px]]
|-
| [[sliding-cum-rotational motion along an inclined plane]] || A sphere or cylinder on an inclined plane || [[File:Rotatoronincline.png|100px]]
|-
| [[pulley system on a double inclined plane]] || Two sliding blocks, connected by a string over a pulley, forcing a relationship between their motion || [[File:Pulleysystemondoubleinclinedplane.png|100px]]
|-
| [[sliding motion along a frictionless circular incline]] || Incline is frictionless, but circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]
|-
| [[sliding motion along a circular incline]] || Incline is circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]
|-
| [[block on free wedge on horizontal floor]] || A block on the incline of a wedge that is free to move on a horizontal floor. || [[File:Blockonwedge.png|100px]]
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]
===Taking components of the gravitational force===

{{gravitational key force concept|Gravity acts on the block with the same magnitude and direction regardless of whether the block is sliding up, sliding down, or remaining where it is. Thus, the analysis of the gravitational force and its components is completely universal.}}
The most important thing for the force diagram is understanding how the gravitational force, which acts vertically downward on the block, splits into components along and perpendicular to the incline. The component along the incline is $mg\sin \theta$ and the component perpendicular to the incline is $mg\cos \theta$. The process of taking components is illustrated in the adjacent figure
$\! N = mg \cos \theta$
where $N$ is the normal force between the block and the inclined plane, $m$ is the mass of the block, and $g$ is the [[acceleration due to gravity]]. $N$ acts inward on the inclined plane and outward on the block.  {{action reaction force concept|In this case, $N$ and $mg\cos \theta$ do ''not'' form an action-reaction pair. The reason that they are equal is different, it is because the net acceleration in the normal direction is zero.}} Some observations:
{| class="sortable" border="1"
===Component along (down) the inclined plane===
For the axis ''down'' the inclined plane, the gravitational force component is $mg\sin \theta$. 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 $f_s$ acts(subject to the constraint $f_s \le \mu_sN$), the case of ''sliding down'', where an upward kinetic friction $f_k = \mu_kN$ acts, and the case of ''sliding up'', where a downward kinetic friction $f_k = \mu_kN$ acts.
:
{| class="sortable" border="1"
! Case !! Convention on the sign of acceleration $a$ !! Equation using $f_s, f_k$ !! Constraint on $f_s, f_k$ !! Simplified, substituting $f_s,f_k$ in terms of $N,\mu_s,\mu_k$ !! Simplified, after combining with $\! N = mg \cos \theta$
|-
| Block stationary || no acceleration, so no sign convention || $\! mg \sin \theta = f_s$ || $f_s \le \mu_sN$ || $\! mg \sin \theta \le \mu_sN$ || $\! \tan \theta \le \mu_s$
|-
| Block sliding down the inclined plane || measured positive down the inclined plane, hence a positive number || $\! ma = mg \sin \theta - f_k$ || $\! f_k = \mu_kN$ || $\! ma = mg \sin \theta - \mu_kN$ || $\! a = g (\sin \theta - \mu_k\cos \theta) = g \cos \theta (\tan \theta - \mu_k)$
|-
| Block sliding up the inclined plane || measured positive down the inclined plane (note that acceleration opposes direction of motion, leading to retardation) || $\! ma = mg \sin \theta + f_k$ || $\! f_k = \mu_kN$ || $\! ma = mg \sin \theta + \mu_kN$ || $\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \mu_k)$
|}
==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==
$\! a = g\sqrt{1 + \mu_k^2}$

==Behavior for a block initially placed at rest==
| Time taken on return journey || $\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}$
|-
| Speed at instant of return || $\! u \sqrt{\frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}}$
|}