# 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==

The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.

{| class="sortable" border="1"
! 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]]
|-
| [[blocks on two inclines of a free wedge]] || Blocks on two inclines of a wedge that is free to move on a horizontal floor. || [[File:Blocksontwoinclinesoffreewedge.png|100px]]
|}
==Basic components of force diagram==
A good way of understanding the force diagram is using the coordinate axes as the axis along the inclined plane and normal to the inclined plane. For simplicity, we will assume a two-dimensional situation, with no forces acting along the horizontal axis that is part of the inclined plane (in our pictorial representation, this ''no action'' axis is the axis perpendicular to the plane).
===The four candidate forces===

Assuming no external forces are applied, there are four candidate forces on the block:

{| class="sortable" border="1"
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction
|-
| $mg$ || [[gravitational force]] || unconditional|| $mg$ where $m$ is the [[mass]] and $g$ is the [[acceleration due to gravity]] || vertically downward, hence an angle $\theta$ with the normal to the incline and an angle $(\pi/2) - \theta$ with the incline
|-
| $N$ || [[normal force]] || unconditional|| adjusts so that there is no net acceleration perpendicular to the plane of the incline || outward normal to the incline
|-
| $f_s$ || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || up the incline (note that this could change if external forces were applied to push the block up the incline).
|-
| $f_k$ || [[kinetic friction]] || sliding || $\mu_kN$ where $N$ is the normal force || opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down. Assuming the block was ''initially'' at rest, it can only slide down, hence ''up the incline'' is the only possibility.
|}

Here is the force diagram (without components) in the ''no sliding'' case:

[[File:Blockoninclinenoslidingforcediagram.png|400px]]

Here is the force diagram (without taking components) in the ''sliding down'' case:

[[File:Blockoninclineslidingdownforcediagram.png|400px]]

Here is the force diagram (without taking components) in the ''sliding up'' case:

[[File:Blockoninclineslidingupforcediagram.png|400px]]

[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]
===Taking components of the gravitational force===
[[File:mgcomponentsforincline{{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.png|thumb|400px|right|Component-taking for Thus, the analysis of the gravitational forceand 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
===Component perpendicular to the inclined plane===
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component $mg\cos \theta$ of gravitational force $mg$ should cancel out normal force.]]
In this case, assuming a stable surface of contact, and that the inclined plane does not break under the weight of the block, and no other external forces, we get the following equation from [[Newton's first law of motion]] applied to the axis perpendicular to the inclined plane:
$\! N = mg \cos \theta$
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 sliding down the inclined plane stationary || measured positive down the inclined planeno acceleration, hence a positive number so no sign convention || $\! ma = mg \sin \theta - = f_s$ || $f_s \mu_kN le \mu_sN$ || $\! a = g (mg \sin \theta - \mu_kle \cos mu_sN$ || $\theta) = g ! \cos tan \theta (\tan \theta - le \mu_k)mu_s$
|-
| Block sliding up down the inclined plane || measured positive down the inclined plane (note that acceleration opposes direction of motion, leading to retardation) 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 stationary sliding up the inclined plane || no measured positive down the inclined plane (note that accelerationopposes direction of motion, so no sign convention leading to retardation) || $\! ma = mg \sin \theta + f_k$ || $\le ! f_k = \mu_sNmu_kN$ || $\! ma = mg \sin \theta + \mu_kN$ || $\! a = g (\sin \theta + \mu_k \cos \theta) = g \cos \theta (\tan \theta + \le \mu_smu_k)$
|}
Here are some picturesA 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?
* If 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 known fixed to be 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== ===For a block sliding downward=== As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:  $\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)$ The quotient $a/g$ where $a$ is downward acceleration and $g$ is the acceleration due to gravity, as a function of the angle (in radians). Note that each line corresponds to a particular value of friction coefficient $\mu_k$. The acceleration is measured in the downward direction. Note that when the value is negative, the block undergoes retardation and hence it will not ''start'' moving if it is initially placed at rest.
[[File:Downaccelerationintermsofinclineangle.png|400px]]
* If We see that $a/g$ is an increasing function of $\theta$ for fixed $\mu_k$. The point where it crosses the axis is $\tan^{-1}(\mu_k)$. ===For a block sliding upward=== As we saw above, if the block is known to be sliding upwardabove, the acceleration (downward positive) is given by:  $\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)$ The quotient $a/g$ where $a$ is downward acceleration and $g$ is the acceleration due to gravity, as a function of the angle (in radians). Note that each line corresponds to a particular value of friction coefficient $\mu_k$. The acceleration is measured in the downward direction. Note that it is always negative, indicating that the block will undergo retardation.
[[File:Upaccelerationintermsofinclineangle.png|400px]]
The magnitude of acceleration (i.e., retardation) is maximum when $\theta = (\pi/2) - \tan^{-1}(\mu_k)$. In the extreme case that $\mu_k = 0$, retardation is maximum for $\theta = \pi/2$, and as $\mu_k$ increases, the angle for maximum retardation decreases. The magnitude of maximum possible retardation is:

$\! a = g\sqrt{1 + \mu_k^2}$
==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 || $\! u \sqrt{\frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}}$
|}