Editing Inclined plane (section)

==Inclined plane with friction==
Where there is [[friction]] between the plane and the load, as for example with a heavy box being slid up a ramp, some of the work applied by the input force is dissipated as heat by friction, <math>W_\text{fric}</math>, so less work is done on the load.  Due to [[conservation of energy]], the sum of the output work and the frictional energy losses is equal to the input work 
:<math>W_\text{in} = W_\text{fric} + W_\text{out} \,</math>
Therefore, more input force is required, and the mechanical advantage is lower, than if friction were not present.
With friction, the load will only move if the net force parallel to the surface is greater than the frictional force <math>F_\text{f}</math> opposing it.<ref name="Ambekar" /><ref name="Bansal">{{cite book   
  | last = Bansal
  | first = R.K 
  | title = Engineering Mechanics and Strength of Materials
  | publisher = Laxmi Publications
  | year = 2005
  | pages = 165–167
  | url = https://books.google.com/books?id=PajxWzB6jusC&q=inclined+plane+friction&pg=PA165
  | isbn = 978-81-7008-094-7}}</ref><ref name="Gujral">This derives slightly more general equations which cover force applied at any angle: {{cite book   
  | last = Gujral
  | first = I.S. 
  | title = Engineering Mechanics
  | publisher = Firewall Media
  | year = 2008
  | pages = 275–277
  | url = https://books.google.com/books?id=JM0OG-XUyu0C&q=%22inclined+plane%22+%22angle+of+repose%22+coefficient+of+friction&pg=PA275
  | isbn = 978-81-318-0295-3}}</ref> The maximum friction force is given by 
:<math>F_f = \mu F_n \,</math>
where <math>F_\text{n}</math> is the [[normal force]] between the load and the plane, directed normal to the surface, and <math>\mu</math> is the [[coefficient of static friction]] between the two surfaces, which varies with the material.  When no input force is applied, if the inclination angle  <math>\theta</math> of the plane is less than some maximum value <math>\phi</math> the component of gravitational force parallel to the plane will be too small to overcome friction, and the load will remain motionless.  This angle is called the [[angle of repose]] and depends on the composition of the surfaces, but is independent of the load weight. It is shown below that the [[tangent (trigonometry)|tangent]] of the angle of repose  <math>\phi</math> is equal to <math>\mu</math>  
:<math>\phi = \tan^{-1} \mu \,</math>
With friction, there is always some range of input force <math>F_\text{i}</math> for which the load is stationary, neither sliding up or down the plane, whereas with a frictionless inclined plane there is only one particular value of input force for which the load is stationary.

===Analysis===
[[File:Free body1.3.svg|right|thumb|Key: ''F''<sub>n</sub> = ''N'' = [[Normal force]] that is perpendicular to the plane, ''F''<sub>i</sub> = ''f'' = input force,  ''F''<sub>w</sub> = ''mg'' = weight of the load, where m = [[mass]], g = [[gravity]] ]]

A load resting on an inclined plane, when considered as a [[Free body diagram|free body]] has three forces acting on it:<ref name="Ambekar" /><ref name="Bansal" /><ref name="Gujral" />
*The applied force, <math>F_\text{i}</math> exerted on the load to move it, which acts parallel to the inclined plane.
*The weight of the load, <math>F_\text{w}</math>, which acts vertically downwards
*The force of the plane on the load.  This can be resolved into two components:
**The normal force <math>F_\text{n}</math>  of the inclined plane on the load, supporting it.  This is directed perpendicular ([[surface normal|normal]]) to the surface.
**The frictional force, <math>F_\text{f}</math> of the plane on the load acts parallel to the surface, and is always in a direction opposite to the motion of the object.  It is equal to the normal force multiplied by the [[coefficient of static friction]] μ between the two surfaces.
Using [[Newton's second law of motion]] the load will be stationary or in steady motion if the sum of the forces on it is zero.  Since the direction of the frictional force is opposite for the case of uphill and downhill motion, these two cases must be considered separately: 
*'''Uphill motion:''' The total force on the load is toward the uphill side, so the frictional force is directed down the plane, opposing the input force.
{| class="wikitable" width=100%;
|-
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{{show
|Derivation of mechanical advantage for uphill motion
|
The equilibrium equations for forces parallel and perpendicular to the plane are
:<math>\sum F_{\|} =  F_i - F_f - F_w \sin \theta = 0  \,</math>
:<math>\sum F_{\perp}  =  F_n - F_w \cos \theta = 0  \,</math>

:Substituting  <math>F_f = \mu F_n \,</math> into first equation

:<math>F_i - \mu F_n - F_w \sin \theta = 0 \,</math>

:Solving second equation to get <math>F_n = F_w \cos \theta \,</math> and substituting into the above equation

:<math>F_i - \mu F_w \cos \theta - F_w \sin \theta = 0 \, </math>

:<math>\frac {F_w}{F_i} = \frac {1} {\sin \theta + \mu \cos \theta} \,</math>

:Defining <math>\mu = \tan \phi \,</math>
:<math>\frac {F_w}{F_i} = \frac {1} {\sin \theta + \tan \phi \cos \theta} = \dfrac{1}{\sin \theta + \dfrac {\sin \phi}{\cos \phi} \cos \theta} = \frac {\cos \phi} {\sin \theta \cos \phi + \cos \theta \sin \phi } \,</math>
:Using a sum-of-angles [[trigonometric identity]] on the denominator,
}}
|}
:The mechanical advantage is
{{Equation box 1 |indent =: |cellpadding = 0 |border = 1 |border colour = black |background colour = transparent
|equation = &nbsp;<math>\mathrm{MA} =  \frac {F_w}{F_i} = \frac {\cos \phi} { \sin (\theta + \phi ) } \, </math>
}}
:where <math>\phi = \tan^{-1} \mu \,</math>.  This is the condition for ''impending motion'' up the inclined plane.  If the applied force '''''F<sub>i</sub>''''' is greater than given by this equation, the load will move up the plane.

*'''Downhill motion:''' The total force on the load is toward the downhill side, so the frictional force is directed up the plane.
{| class="wikitable" width=100%;
|-
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{{show
|Derivation of mechanical advantage for downhill motion
|
The equilibrium equations are
:<math>\sum F_{\|} =  F_i + F_f - F_w \sin \theta = 0  \,</math>
:<math>\sum F_{\perp}  =  F_n - F_w \cos \theta = 0  \,</math>

:Substituting  <math>F_f = \mu F_n \,</math> into first equation

:<math>F_i + \mu F_n - F_w \sin \theta = 0 \,</math>

:Solving second equation to get <math>F_n = F_w \cos \theta \,</math> and substituting into the above equation

:<math>F_i + \mu F_w \cos \theta - F_w \sin \theta = 0 \, </math>

:<math>\frac {F_w}{F_i} = \frac {1} {\sin \theta - \mu \cos \theta} \,</math>

:Substituting in <math>\mu = \tan \phi \,</math> and simplifying as above
:<math>\frac {F_w}{F_i} = \frac {\cos \phi} {\sin \theta \cos \phi - \cos \theta \sin \phi } \,</math>
:Using another [[trigonometric identity]] on the denominator,
}}
|}
:The mechanical advantage is
{{Equation box 1 |indent =: |cellpadding = 0 |border = 1 |border colour = black |background colour = transparent
|equation = &nbsp;<math>\mathrm{MA} =  \frac {F_w}{F_i} = \frac {\cos \phi} { \sin (\theta - \phi ) } \,</math>
}}
:This is the condition for impending motion down the plane; if the applied force '''''F<sub>i</sub>''''' is less than given in this equation, the load will slide down the plane.  There are three cases:
:#<math>\theta < \phi\,</math>: The mechanical advantage is negative.  In the absence of applied force the load will remain motionless, and requires some negative (downhill) applied force to slide down.
:#<math>\theta = \phi\,</math>: The '[[angle of repose]]'.  The mechanical advantage is infinite. With no applied force, load will not slide, but the slightest negative (downhill) force will cause it to slide.
:#<math>\theta > \phi\,</math>: The mechanical advantage is positive.  In the absence of applied force the load will slide down the plane, and requires some positive (uphill) force to hold it motionless