Mech - User contributions [en]

MediaWiki:Sitenotice

2024-10-06T22:11:12Z

Vipul:

Want site search autocompletion? See [[Project:Enabling site search autocompletion|here]]<br/>
Encountering 429 Too Many Requests errors when browsing the site? See [[Project:429 Too Many Requests error|here]]

User:Vipul/Sandbox

2024-10-06T22:10:57Z

Vipul:

Testing

* <math>\sqrt{7 + 2}!! + 0 = 720</math>
* <math>5^{1 + 2} = 125</math>
* <math>6^{2 + 1} = 216</math>
* <math>(3 + 4)^3 = 343</math>
* <math>3! * 6 = 36</math>
* <math>9^{\sqrt{7 + 2}} = 729</math>

MediaWiki:Sitenotice

2024-09-30T01:24:12Z

Vipul:

'''This site is in the process of being migrated to a new server. Edits made until this notice has been removed may be lost.'''<br/>
Want site search autocompletion? See [[Project:Enabling site search autocompletion|here]]<br/>
Encountering 429 Too Many Requests errors when browsing the site? See [[Project:429 Too Many Requests error|here]]

MediaWiki:Sitenotice

2024-09-06T01:03:27Z

Vipul:

Mech:Enabling site search autocompletion

2024-09-06T01:02:47Z

Vipul: Created page with "Content copied from Ref:Ref:Enabling site search autocompletion. Images used are specific to this site (Mech). Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on. ==What's wrong with site search autocompletion and how to fix it== ===What's wrong=== When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the..."

Content copied from [[Ref:Ref:Enabling site search autocompletion]]. Images used are specific to this site (Mech).

Site search autocompletion is currently broken by default on this site. This page includes details on how to get it to work, and what's going on.

==What's wrong with site search autocompletion and how to fix it==

===What's wrong===

When you start typing something in the site search bar, you'll see it stuck at "Loading search suggestions" as shown in the screenshot below:

[[File:Site search autocompletion broken.png]]

Note that the actual search is still working -- you just have to hit Enter after typing the search query and it'll go to the search results page. It's the autocompletion before you hit Enter that is broken.

===How to fix it===

To fix it, you need to follow these steps:

* Write to vipulnaik1@gmail.com asking for a login to the site. Please include the following with your request: preferred username, preferred initial password (you can change it after logging in), real name (if you want it entered), email address to use (if you want an actual email address by which you can be contacted), and whether you want edit access as well. You don't need edit access for enabling site search autocompletion.
* Log in to the site. Then go to [[Special:Preferences]]. Go to the Appearance section and switch the Skin from "Vector (2022)" to "Vector legacy (2010)".
* Make sure to hit "Save" at the bottom.
* Now you can reload the page or load a new page.

Site search autocompletion should now work. Here's an example:

[[File:Site search autocompletion working.png]]

==More background==

We've recently upgraded the MediaWiki version of this wiki from 1.35.13 to 1.41.2 (see [[Special:Version]]). The upgrade allows us to migrate the wiki to a more modern operating system version running PHP 8. With the current setup for MediaWiki 1.41.2, we're in this situation:

* The "Vector legacy (2010)" skin has site search autocompletion working, but it doesn't render well on small screens. Specifically, even on small mobile screens, it still shows the left menu, and doesn't properly use the MobileFrontend extension settings.
* The "Vector (2022)" skin doesn't have site search autocompletion working (see screenshots in preceding section) but it does render fine on mobile devices.

It is possible to set only one default skin (that is applicable to all non-logged-in users and is the default for logged-in users who have not configured a skin for themselves). So, the selection of default skin comes down to whether it's more important for casual users to have the mobile experience working or to have site search autocompletion working. Based on a general understanding of user behavior, we believe that having a usable mobile experience is more important for casual users than having site search autocompletion.

However, for power users who are using the site extensively, site search autocompletion may be important. That's why we've written this page giving guidance on how to set up site search autocompletion.

File:Site search autocompletion working.png

2024-09-06T01:02:43Z

Vipul:

File:Site search autocompletion broken.png

2024-09-06T01:02:21Z

Vipul:

Mech:429 Too Many Requests error

2024-09-06T00:59:38Z

Vipul: Created page with "This content is copied from Ref:Ref:429 Too Many Requests error. If you get a 429 Too Many Requests error when browsing this site, read on. You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work. If you are an actual h..."

This content is copied from [[Ref:Ref:429 Too Many Requests error]].

If you get a 429 Too Many Requests error when browsing this site, read on.

You're probably seeing this error because a large number of requests have been made from your IP address over a short period of time. That's probably a lot of requests from you or others who share your IP address (such as your home wi-fi network). Waiting a minute and then retrying should generally work.

If you are an actual human being with a legitimate reason to be browsing the site heavily, first, thank you and sorry about this! We set rate limits to prevent bots, spiders, spammers, and malicious actors from consuming too much of our server's resources so that our server's resources can be devoted to real humans like you. Consider writing to vipulnaik1@gmail.com with your IP address to have the IP address whitelisted. You can get your IP address by [https://www.google.com/search?q=my+ip+address Googling "my IP address"] (scroll down a little bit to where Google includes the IP address in a box). NOTE: If you have both an IPv4 address and an IPv6 address, you should send both; the server supports both IPv4 and IPv6, so either may end up getting used. To check if you have an IPv6 address, try visiting [https://ipv6.google.com/ ipv6.google.com].

If your IP address changes, or you are away from your home network, then you'll get rate-limited again. So if you find yourself getting rate-limited after already having been whitelisted, check if you are on a different IP address than the one for which you requested whitelisting.

MediaWiki:Sitenotice

2024-09-06T00:57:59Z

Vipul: Blanked the page

User:Vipul/Sandbox

2024-09-06T00:55:04Z

Vipul:

Testing

* <math>\sqrt{7 + 2}!! + 0 = 720</math>
* <math>5^{1 + 2} = 125</math>
* <math>6^{2 + 1} = 216</math>
* <math>(3 + 4)^3 = 343</math>
* <math>3! * 6 = 36</math>

User:Vipul/Sandbox

2024-09-06T00:50:15Z

2024-07-06T01:26:56Z

Vipul:

Testing

<math>\frac{e^{2\pi \sqrt{3}}}{2t^3 + 1 + 4}</math>

<math>\sqrt{7 + 2}!! + 0 = 720</math>

<math>5^{1 + 2} = 125</math>

User:Vipul/Sandbox

2024-04-28T05:25:47Z

Vipul:

Testing

<math>\frac{e^{2\pi \sqrt{3}}}{2t^3 + 1 + 4}</math>

<math>\sqrt{7 + 2}!! + 0 = 720</math>

Quiz:Sliding motion along an inclined plane

2024-04-27T23:39:58Z

Vipul: /* Force diagram and acceleration analysis */

See [[sliding motion along an inclined plane]] for background information.

==Force diagram and acceleration analysis==

<quiz display=simple>

{Consider a situation where <math>\mu_s = mu_k = \mu</math> is the coefficient of static as well as kinetic friction between the block and the incline. What can we say about the normal force between the block (of mass <math>m</math>) and the incline as a function of the angle of incline <math>\theta</math> (this is the angle that the incline makes with the horizontal)?
|type="()"}
- The magnitude of normal force equals <math>mg</math> and is independent of <math>\theta</math>
+ The magnitude of normal force decreases from <math>mg</math> to 0 as <math>\theta</math> increases from 0 to <math>\pi/2</math>
- The magnitude of normal force increases from 0 to <math>mg</math> as <math>\theta</math> increases from 0 to <math>\pi/2</math>
- The magnitude of normal force increases for <math>\theta</math> increasing from 0 to <math>\tan^{-1}(\mu)</math> and decreases for <math>\theta</math> increasing from <math>\tan^{-1}(\mu)</math> to <math>\pi/2</math>
- The magnitude of normal force decreases for <math>\theta</math> increasing from 0 to <math>\tan^{-1}(\mu)</math> and increases for <math>\theta</math> increasing from <math>\tan^{-1}(\mu)</math> to <math>\pi/2</math>

{Consider a situation where <math>\mu_s = \mu_k = \mu</math> is the coefficient of static as well as kinetic friction between the block and the incline. Once the angle of the incline (denoted <math>\theta</math>) exceeds <math>\tan^{-1}(\mu_s)</math>, the magnitude of downward acceleration <math>a</math> is an increasing function of <math>\theta</math>. What is its derivative with respect to <math>\theta</math>?
|type="()"}
- <math>g(\sin \theta + \mu \cos \theta)</math>
- <math>g(\sin \theta - \mu \cos \theta)</math>
- <math>g(\cos \theta - \mu \sin \theta)</math>
+ <math>g(\cos \theta + \mu \sin \theta)</math>

</quiz>

Sliding motion along a frictionless inclined plane

2021-09-21T03:02:13Z

Vipul: /* The two candidate forces */

{{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 frictionless surface of contact.]]

The scenario here is a dry block (with a stable surface of contact) on a dry ''fixed'' frictionless inclined plane, with <math>\theta</math> being the angle of inclination with the horizontal axis.

The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (whence, the plane is vertical).

Though the picture depicts a square cross section, this shape assumption is not necessary.

A more general scenario that includes the case of an inclined plane with friction is [[sliding motion along an inclined plane]].

==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 two candidate forces===

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

{| class="sortable" border="1"
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction
|-
| <math>mg</math> || [[gravitational force]] || unconditional|| <math>mg</math> where <math>m</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\theta</math> with the normal to the incline and an angle <math>(\pi/2) - \theta</math> with the incline
|-
| <math>N</math> || [[normal force]] || unconditional|| adjusts so that there is no net acceleration perpendicular to the plane of the incline || outward normal to the incline
|}

[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]

===Taking components of the gravitational force===

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 <math>mg\sin \theta</math> and the component perpendicular to the incline is <math>mg\cos \theta</math>. 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 <math>mg\cos \theta</math> of gravitational force <math>mg</math> 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:

<math>\! N = mg \cos \theta</math>

where <math>N</math> is the normal force between the block and the inclined plane, <math>m</math> is the mass of the block, and <math>g</math> is the [[acceleration due to gravity]]. <math>N</math> acts inward on the inclined plane and outward on the block. Some observations:

{| class="sortable" border="1"
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments
|-
| <math>\theta = 0</math> (horizontal plane)|| <math>N = mg</math> || The normal force exerted on a horizontal surface equals the mass times the acceleration due to gravity, which we customarily call the [[weight]].
|-
| <math>\theta = \pi/2</math> (vertical plane) || <math>N = 0</math> || The block and the inclined plane are barely in contact and hardly pressed together.
|-
| <math>\theta</math> increases from <math>0</math> to <math>\pi/2</math> || <math>N</math> reduces from <math>mg</math> to <math>0</math>. The derivative <math>\frac{dN}{d\theta}</math> is <math>-mg\sin \theta</math> || The force pressing the block and the inclined plane reduces as the slope of the incline increases.
|}
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.

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

For the axis ''down'' the inclined plane, the gravitational force component is <math>mg\sin \theta</math>. There are no other forces in this direction, so, if <math>a</math> denotes the acceleration measured positive in the downward direction, we get:

<math>ma = mg \sin \theta</math>

After cancellation of <math>m</math>, we get:

<math>a = g \sin \theta</math>

Note that if the block is sliding upward (for instance, if given an initial upward velocity) this acceleration functions as retardation, whereas if the block is sliding downward (which may happen if the block is placed at rest, or given an initial downward velocity, or of it turns back after sliding upward) then the acceleration increases the speed.

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

===Kinematics===

The kinematic evolution in the second case is given as follows, if we set <math>t = 0</math> as the time when the block is placed, we have the following (here, the row variable is written in terms of the column variable):

{| class"sortable" border="1"
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)
|-
| <math>\! v</math> || <math>\! v</math> || <math>\! gt\sin \theta</math> || <math>\! \sqrt{2sg\sin \theta}</math> || <math>\! \sqrt{2hg}</math> || <math>\! \sqrt{2xg\tan \theta}</math>
|-
| <math>\! s</math> || <math>\! v^2/(2g\sin \theta)</math> || <math>\! (1/2)g (\sin \theta) t^2</math>|| <math>\! s</math> || <math>\! h/\sin \theta</math> || <math>\! x/\cos \theta</math>
|-
| <math>\! h</math> || <math>\! v^2/(2g)</math> || <math>\! (1/2) g \sin^2\theta</math> || <math>\! s \sin \theta</math> || <math>\! h</math> || <math>\! x \tan \theta</math>
|-
| <math>\! x</math> || <math>\! v^2/(2g\tan \theta)</math> || <math>\! (1/2)g \cos \theta \sin \theta</math> || <math>\! s \cos \theta</math> || <math>\! h \cot \theta</math> || <math>\! x</math>
|}

Sandbox

2017-11-24T14:04:12Z

Vipul: Created page with "<math>e^{\pi\sqrt{2}}</math>"

User:Vipul/sandbox

2016-09-05T19:41:18Z

Vipul:

2011-12-20T00:49:34Z

Vipul:

Mech:Error log

2011-12-20T00:49:09Z

Vipul: Created page with "Did you find something that looks like an error (or possible error) to you? Report the error quickly [http://www.surveymonkey.com/s/subwiki-error-report here] (anonymous, does ''..."

2011-12-05T18:31:52Z

Vipul:

Video:Sliding motion along a frictionless inclined plane

2011-12-05T18:25:45Z

Vipul: Created page with "==Instructional video links== ===Basic coverage: Khan Academy videos=== {| class="sortable" border="1" ! Video embedded in this page (click SHOW MORE to view) !! Video link (on..."

==Instructional video links==

===Basic coverage: Khan Academy videos===

{| class="sortable" border="1"
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information
|-
| <toggledisplay>{{#widget:YouTube|id=TC23wD34C7k}}</toggledisplay> || [http://www.khanacademy.org/video/inclined-plane-force-components?playlist=Physics Khan Academy video: inclined plane force components] (click through to view comments, more) || full video (12:42) || concentrates on the set up of the gravitational force and the normal reaction force
|-
| <toggledisplay>{{#widget:YouTube|id=Mz2nDXElcoM}}</toggledisplay> || [http://www.khanacademy.org/video/ice-accelerating-down-an-incline?playlist=Physics Khan Academy video: ice accelerating down incline] || full video (10:37) || considers the kinematics, without friction.
|}

Template:Perspectives

2011-12-05T18:05:20Z

Vipul:

{{quotation|'''ALSO CHECK OUT''': {{#ifexist:Questions:{{PAGENAME}}|[[Questions:{{PAGENAME}}|Questions page (list of common doubts/questions/curiosities)]] <nowiki>|</nowiki>}}{{#ifexist:Quiz:{{PAGENAME}}|[[Quiz:{{PAGENAME}}|Quiz (multiple choice questions to test your understanding)]] <nowiki>|</nowiki>}}{{#ifexist:Pedagogy:{{PAGENAME}}|Pedagogy page (discussion of how this topic is or could be taught)]]<nowiki>|</nowiki>}}{{#ifexist:Video:{{PAGENAME}}|[[Video:{{PAGENAME}}|Page with videos on the topic, both embedded and linked to]]}}}}

Template:Perspectives

2011-12-05T18:04:07Z

Vipul:

{{quotation|'''ALSO CHECK OUT''': {{#ifexist:Questions:{{PAGENAME}}|[[Questions:{{PAGENAME}}|Questions page (list of common doubts/questions/curiosities)]] <nowiki>|</nowiki>}}{{#ifexist:Quiz:{{PAGENAME}}|[[Quiz:{{PAGENAME}}|Quiz (multiple choice questions to test your understanding)]] <nowiki>|</nowiki>}}{{#ifexist:Pedagogy:{{PAGENAME}}|Pedagogy page (discussion of how this topic is or could be taught)]]<nowiki>|</nowiki>}}{{#ifexist:Video:{{PAGENAME}}|[[Page with videos on the topic, both embedded and linked to]]}}}}

Sliding motion along an inclined plane

2011-12-05T18:03:12Z

Vipul:

{{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 <math>\theta</math> being the angle of inclination with the horizontal axis.

The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (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]] || <math>\mu_k = \mu_s = 0</math> || [[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
|-
| <math>mg</math> || [[gravitational force]] || unconditional|| <math>mg</math> where <math>m</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\theta</math> with the normal to the incline and an angle <math>(\pi/2) - \theta</math> with the incline
|-
| <math>N</math> || [[normal force]] || unconditional|| adjusts so that there is no net acceleration perpendicular to the plane of the incline || outward normal to the incline
|-
| <math>f_s</math> || [[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).
|-
| <math>f_k</math> || [[kinetic friction]] || sliding || <math>\mu_kN</math> where <math>N</math> 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===

{{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 <math>mg\sin \theta</math> and the component perpendicular to the incline is <math>mg\cos \theta</math>. 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 <math>mg\cos \theta</math> of gravitational force <math>mg</math> 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:

<math>\! N = mg \cos \theta</math>

where <math>N</math> is the normal force between the block and the inclined plane, <math>m</math> is the mass of the block, and <math>g</math> is the [[acceleration due to gravity]]. <math>N</math> acts inward on the inclined plane and outward on the block.

{{action reaction force concept|In this case, <math>N</math> and <math>mg\cos \theta</math> 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"
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments
|-
| <math>\theta = 0</math> (horizontal plane)|| <math>N = mg</math> || The normal force exerted on a horizontal surface equals the mass times the acceleration due to gravity, which we customarily call the [[weight]].
|-
| <math>\theta = \pi/2</math> (vertical plane) || <math>N = 0</math> || The block and the inclined plane are barely in contact and hardly pressed together.
|-
| <math>\theta</math> increases from <math>0</math> to <math>\pi/2</math> || <math>N</math> reduces from <math>mg</math> to <math>0</math>. The derivative <math>\frac{dN}{d\theta}</math> is <math>-mg\sin \theta</math> || The force pressing the block and the inclined plane reduces as the slope of the incline increases.
|}
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.

===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==

===For a block sliding downward===

As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:

<math>\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)</math>

The quotient <math>a/g</math> where <math>a</math> is downward acceleration and <math>g</math> 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 <math>\mu_k</math>. 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]]

We see that <math>a/g</math> is an increasing function of <math>\theta</math> for fixed <math>\mu_k</math>. The point where it crosses the axis is <math>\tan^{-1}(\mu_k)</math>.

===For a block sliding upward===

As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:

<math>\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)</math>

The quotient <math>a/g</math> where <math>a</math> is downward acceleration and <math>g</math> 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 <math>\mu_k</math>. 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 <math>\theta = (\pi/2) - \tan^{-1}(\mu_k)</math>. In the extreme case that <math>\mu_k = 0</math>, retardation is maximum for <math>\theta = \pi/2</math>, and as <math>\mu_k</math> increases, the angle for maximum retardation decreases. The magnitude of maximum possible retardation is:

<math>\! a = g\sqrt{1 + \mu_k^2}</math>
==Behavior for a block initially placed at rest==

===Statics and dynamics===

If the block is initially placed gently at rest, we have the following cases:

{| class="sortable" border="1"
! Case !! What happens
|-
| <math>\tan \theta < \mu_s</math>, or <math>\theta < \tan^{-1}(\mu_s)</math> || The block does not start sliding down, and instead stays stable where it is. The magnitude of static friction is <math>mg \sin \theta</math>, and its direction is upward along the inclined plane, precisely balancing and hence canceling the component of gravitational force along the inclined plane.
|-
| <math>\tan \theta > \mu_s</math> or <math>\theta > \tan^{-1}(\mu_s)</math> || The block starts sliding down, and the downward acceleration is given by <math>a = g(\sin \theta - \mu_k \cos \theta) = g \cos \theta (\tan \theta - \mu_k)</math>.
|}

The angle <math>\tan^{-1}(\mu_s)</math> is termed the [[angle of repose]], since this is the largest angle at which the block does not start sliding down.

===Kinematics===

The kinematic evolution in the second case is given as follows, if we set <math>t = 0</math> as the time when the block is placed, we have the following (here, the row variable is written in terms of the column variable):

{| class"sortable" border="1"
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)
|-
| <math>\! v</math> || <math>\! v</math> || <math>\! gt\cos \theta(\tan \theta - \mu_k)</math> || <math>\! \sqrt{2sg\cos \theta (\tan \theta - \mu_k)}</math> || <math>\! \sqrt{2hg(1 - \mu_k \cot \theta)}</math> || <math>\! \sqrt{2xg(\tan \theta - \mu_k)}</math>
|-
| <math>\! s</math> || <math>\! v^2/(2g\cos \theta (\tan \theta - \mu_k))</math> || <math>\! (1/2)g \cos \theta (\tan \theta - \mu_k)t^2</math>|| <math>\! s</math> || <math>\! h/\sin \theta</math> || <math>\! x/\cos \theta</math>
|-
| <math>\! h</math> || <math>\! v^2 \tan \theta/(2g(\tan \theta - \mu_k))</math> || <math>\! (1/2) g \cos \theta \sin \theta (\tan \theta - \mu_k)</math> || <math>\! s \sin \theta</math> || <math>\! h</math> || <math>\! x \tan \theta</math>
|-
| <math>\! x</math> || <math>\! v^2/(2g (\tan \theta - \mu_k))</math> || <math>\! (1/2)g \cos^2 \theta (\tan \theta - \mu_k)</math> || <math>\! s \cos \theta</math> || <math>\! h \cot \theta</math> || <math>\! x</math>
|}

===Energy changes===

==Given an initial speed downward==

===Statics and dynamics===
If the block is given an initial downward speed, the acceleration is <math>g \cos \theta (\tan \theta - \mu_k)</math> downward, and its long-term behavior is determined by whether <math>\tan \theta < \mu_k</math> or <math>\tan \theta > \mu_k</math>:

* If <math>\tan \theta < \mu_k</math>, then the block's speed reduces with time. The magnitude of retardation is <math>g \cos \theta (\mu_k - \tan \theta)</math>. If the incline is long enough, this retardation continues until the block reaches a speed of zero, at which point it comes to rest and thence stays at rest.
* If <math>\tan \theta > \mu_k</math>, then the block's speed increases with time. The magnitude of acceleration is <math>g \cos \theta (\tan \theta - \mu_k)</math>. The block thus does not come to a stop and keeps going faster as it goes down the incline.

===Kinematics===

{{fillin}}

===Energy changes===

{{fillin}}

==Given an initial speed upward==

===Statics and dynamics===

If the block is given an initial upward speed, the acceleration is <math>g \cos \theta (\tan \theta + \mu_k)</math> downward, i.e., the retardation is <math>g \cos \theta(\tan \theta + \mu_k)</math>. We again consider two cases:

{| class="sortable" border="1"
! Case !! What happens
|-
| <math>\! \tan \theta < \mu_k</math> or <math>\! \theta < \tan^{-1}(\mu_k)</math> || The block's speed reduces with time as it rises. If the incline is long enough, the block eventually comes to a halt and stays still after that point.
|-
| <math>\! \tan \theta</math> is between <math>\! \mu_k</math> and <math>\! \mu_s</math> || The behavior is somewhat indeterminate, in the sense that whether the block starts sliding down after stopping its upward slide is unclear.
|-
| <math>\! \tan \theta > \mu_s</math> || The block's speed reduces with time as it rises, until it comes to a halt, after which it starts sliding downward, just as in the case of a block initially placed at rest.
|}

===Kinematics===

Suppose the initial upward speed is <math>u</math>. Because of our ''downward positive'' convention, we will measure the velocity as <math>-u</math>. Then, we have:

{| class="sortable" border="1"
! Quantity !! Value
|-
| Time taken to reach highest point || <math>\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}</math>
|-
| Displacement to highest point achieved (along incline) || <math>\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}</math>
|-
| Vertical height to highest point achieved || <math>\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}</math>
|-
| Horizontal displacement to highest point achieved || <math>\frac{u^2}{2g(\tan \theta + \mu_k)}</math>
|}

In case that <math>\tan \theta > \mu_s</math>, the block is expected to slide back down and return to the original position. The kinematics of the downward motion are the same as those for a block initially placed at rest. Two important values are given below:

{| class="sortable" border="1"
! Quantity !! Value
|-
| 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 \frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}</math>
|}

===Energy changes===

{{fillin}}

==External links==

===Instructional video links===

See [[Video:Sliding motion along an inclined plane]].