https://mech.subwiki.org/w/api.php?action=feedcontributions&user=Vipul&feedformat=atomMech - User contributions [en]2019-08-22T08:56:05ZUser contributionsMediaWiki 1.29.2https://mech.subwiki.org/w/index.php?title=Sandbox&diff=560Sandbox2017-11-24T14:04:12Z<p>Vipul: Created page with "<math>e^{\pi\sqrt{2}}</math>"</p>
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<div><math>e^{\pi\sqrt{2}}</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=User:Vipul/sandbox&diff=552User:Vipul/sandbox2016-09-05T19:41:18Z<p>Vipul: </p>
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<div><math>e^{\sqrt{\pi}/312}</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=User:Vipul/sandbox&diff=551User:Vipul/sandbox2016-09-05T19:41:09Z<p>Vipul: Created page with "<math>e^{\pi/312}</math>"</p>
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<div><math>e^{\pi/312}</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=User:Vipul/Sandbox&diff=550User:Vipul/Sandbox2016-09-05T16:14:37Z<p>Vipul: </p>
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<div>Testing<br />
<br />
<math>\frac{e^{2\pi \sqrt{3}}}{2t^3 + 1 + 4}</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=User:Vipul/Sandbox&diff=549User:Vipul/Sandbox2016-09-05T16:11:17Z<p>Vipul: </p>
<hr />
<div>Testing<br />
<br />
<math>\frac{e^{2\pi \sqrt{3}}}{2t^3 + 1}</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=User:Vipul/Sandbox&diff=548User:Vipul/Sandbox2016-09-05T16:09:41Z<p>Vipul: Created page with "Testing"</p>
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<div>Testing</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Sliding_motion_along_a_frictionless_circular_incline&diff=547Sliding motion along a frictionless circular incline2014-09-26T19:03:12Z<p>Vipul: </p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article describes the sliding motion of a small block placed on a frictionless circular incline. This is similar to the situation of [[sliding motion along an inclined plane]], except that the inclined plane has constant slope whereas in the case of the circular incline, the slope is constantly changing.<br />
<br />
The more general version with friction is computationally much harder in terms of its kinematic evolution, and is discussed under [[sliding motion along a circular incline]].<br />
<br />
If <math>\theta</math> denotes the angle that the radial line to the circular incline makes with the vertical, then <math>\theta</math> is also the angle made by the tangent line with the horizontal. Locally, the analysis when the block is at such a point on the circular incline looks very similar to the analysis of [[sliding motion along a frictionless inclined plane]] where the plane makes an angle of <math>\theta</math> with the horizontal. The key difference is that, since the motion is circular, the normal component of acceleration is not zero; rather, it is given by <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius.<br />
<br />
[[File:Blockoncircularincline.png|thumb|500px|right]]<br />
<br />
==Basic components of force diagram==<br />
<br />
A good way of understanding the force diagram is using the coordinate axes as the axis tangent to the circular incline and normal to the circular incline (which in this case is radial). For simplicity, we will assume a two-dimensional situation, with no forces acting along the horizontal axis that is part of the circular incline (in our pictorial representation, this ''no action'' axis is the axis perpendicular to the plane).<br />
<br />
===The two candidate forces===<br />
<br />
Assuming no external forces are applied, there are two candidate forces on the block:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|}<br />
<br />
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]<br />
===Taking components of the gravitational force===<br />
<br />
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<br />
<br />
===Component perpendicular to the circular incline===<br />
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component <math>mg\cos \theta</math> of gravitational force <math>mg</math> should cancel out normal force.]]<br />
<br />
In this case, assuming a stable surface of contact, and that the circular incline 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 circular incline:<br />
<br />
<math>\! N - mg \cos \theta = mv^2/r</math><br />
<br />
where <math>N</math> is the normal force between the block and the circular incline, <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 circular incline and outward on the block. <br />
<br />
===Component along (down) the circular incline===<br />
<br />
For the axis ''down'' the circular incline, 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 tangential acceleration (which is also the change in ''speed'') measured positive in the downward direction, we get:<br />
<br />
<math>ma = mg \sin \theta</math><br />
<br />
After cancellation of <math>m</math>, we get:<br />
<br />
<math>a = g \sin \theta</math><br />
<br />
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.<br />
<br />
The overall system of equation is thus:<br />
<br />
<math>\! mg \cos \theta - N = \frac{mv^2}{r}</math><br />
<br />
and:<br />
<br />
<math>\! a = g\sin \theta</math><br />
<br />
where we have:<br />
<br />
<math>\! a = \frac{dv}{dt}</math><br />
<br />
==For a block initially placed at rest with radial making angle <math>\alpha</math> with vertical==<br />
<br />
If the block is placed right on top (<math>\alpha = 0</math>), it is in equilibrium, but the equilibrium is an unstable equilibrium, so even a very minor perturbation will cause the block to slide down. We assume that it starts sliding down with a zero initial velocity and deduce the kinematics.<br />
<br />
===Speed-position relationship===<br />
<br />
We have that:<br />
<br />
<math>v = r \frac{d\theta}{dt}, \qquad a = \frac{dv}{dt}, \qquad a = g \sin \theta</math><br />
<br />
Eliminating <math>t</math>, we get:<br />
<br />
<math>v \frac{dv}{d\theta} = gr \sin \theta</math><br />
<br />
Integrating, we get that:<br />
<br />
<math>v^2 = 2gr(\cos \alpha - \cos \theta)</math><br />
<br />
or:<br />
<br />
<math>v = \sqrt{2gr(\cos \alpha - \cos \theta)}</math><br />
<br />
This can also be deduced from energy considerations: the height difference between angles <math>\alpha</math> and <math>\theta</math> is <math>r(\cos \alpha - \cos \theta)</math>, so the loss in potential energy is <math>mgr(\cos \alpha - \cos \theta)</math>. This becomes kinetic energy, since there is no loss due to friction, and we get:<br />
<br />
<math>\frac{1}{2}mv^2 = mgr(\cos \alpha - \cos \theta)</math><br />
<br />
Solve to get <math>v</math> above.<br />
<br />
In the case <math>\alpha = 0</math>, this becomes:<br />
<br />
<math>v = \sqrt{2gr(1 - \cos \theta)}</math><br />
<br />
===Flying-off point===<br />
<br />
When the speed is sufficiently high, the block flies off the circular incline. To determine this, we need to set <math>N = 0</math>. We get:<br />
<br />
<math>mg \cos \theta = \frac{mv^2}{r}</math><br />
<br />
This simplifies to:<br />
<br />
<math>\! g \cos \theta = 2gr(\cos \alpha - \cos \theta)/r</math><br />
<br />
Cancel the <math>g</math> and <math>r</math> and we get:<br />
<br />
<math>\! \cos \theta = 2(\cos \alpha - \cos \theta)</math><br />
<br />
We get:<br />
<br />
<math>\! \cos \theta = \frac{2}{3}\cos \alpha</math><br />
<br />
So we get:<br />
<br />
<math>\! \theta = \arccos[\frac{2}{3} \cos \alpha]</math><br />
<br />
At this point, the speed is:<br />
<br />
<math>\! v = \sqrt{2gr\cos \alpha/3}</math><br />
<br />
In particular, when <math>\alpha = 0</math>, we get <math>\theta = \arccos(2/3)</math> and <math>v = \sqrt{2gr/3}</math>.<br />
<br />
===Reintroducing time: kinematics===<br />
<br />
The time taken to reach from <math>\alpha</math> to <math>\beta</math> can be computed using the following integral:<br />
<br />
<math>\! \int_\alpha^\beta \frac{d\theta}{\sqrt{2gr(\cos \alpha - \cos \theta)}} = t</math><br />
<br />
This integral need not be computable by elementary means, though it is so for some values of <math>\alpha</math>. The case <math>\alpha = 0</math> is discussed below:<br />
<br />
{{fillin}}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Newton%27s_third_law_of_motion&diff=546Newton's third law of motion2012-08-18T00:20:47Z<p>Vipul: /* Misconceptions */</p>
<hr />
<div>{{quick phrase|action and reaction are equal and opposite}}<br />
==Definition==<br />
<br />
'''Newton's third law''' or '''Newton's third law of motion''' gives a relationship between the forces that two bodies exert on each other. It says that if body <math>A</math> exerts a [[force]] on body <math>B</math>, then body <math>B</math> exerts a force on body <math>A</math>, and further, that the forces are equal in magnitude and opposite in direction.<br />
<br />
An important corollary of this formulation is that every force on a body is exerted ''by'' something. In other words, it isn't possible for an object to feel a force that isn't being exerted by anything.<br />
<br />
==Particular cases==<br />
<br />
===Force types===<br />
<br />
{| class="sortable" border="1"<br />
! Type of force !! Pair of bodies !! Directions of the action-reaction pair !! Magnitude of the action-reaction pair<br />
|-<br />
| [[Gravitational force]] || Two masses that exert a gravitational force on each other. In most earthly situations, one of the bodies is the earth. || On each body, in the direction toward the center of mass of the other body. Thus the force acts inward on both bodies along the line joining the [[center of mass|centers of mass]] of the two bodies || <math>Gm_1m_2/r^2</math> where <math>G</math> is the gravitational constant, <math>m_1, m_2</math> are the masses and <math>r</math> is the distance between the centers of mass<br />
|-<br />
| [[Normal force]] || Bodies that share a surface of contact. || On each body, the force acts in the outward direction along the surface of contact. || The magnitude is determined to be such that the net relative acceleration along the direction perpendicular to the surface of contact is zero. It thus adjusts in response to the other forces being exerted on the bodies.<br />
|-<br />
| [[Static friction]] || Bodies that share a surface of contact. || On each body, the force acts in a direction parallel to the surface of contact, opposing the tendency of slippage of surfaces. || The magnitude is determined so that the net relative acceleration of surfaces along the surface of contact is zero. It thus adjusts in response to the other forces being exerted on the bodies. There is, however, an upper limit: the [[limiting coefficient of static friction]] times the [[normal force]] magnitude.<br />
|-<br />
| [[Kinetic friction]] || Bodies that share a surface of contact. || On each body, the force acts in a direction parallel to the surface of contact, opposing the actual direction of slippage of surfaces. || The magnitude is a fixed multiple of the [[normal force]], where the constant of proportionality is the [[coefficient of kinetic friction]] and is determined by the nature of the surfaces in contact.<br />
|-<br />
| [[Tension]] || An inextensible string and a body attached to one end of the string || The string pulls the body inward; the body pulls the string outward. || Determined from other constraints.<br />
|}<br />
<br />
===Concrete examples===<br />
<br />
{| class="sortable" border="1"<br />
! Situation !! Pair of bodies !! Action-reaction pair !! What does Newton's third law tell us?<br />
|-<br />
| Any object close to the surface of the earth || the object and the earth || The gravitational force exerted by the earth on the object, and the gravitational force exerted by the object on the earth || The two forces are equal in magnitude and opposite in direction. In other words, the "downward" force experienced by the object due to gravity has a corresponding "upward" force experienced by the earth. However, due to the huge mass of the earth, the resultant acceleration of the earth is too negligible to be noticed.<br />
|-<br />
| A block resting on a fixed horizontal floor || the block and the floor || the upward normal force exerted by the table on the block and the downward normal force exerted by the block on the floor || The two forces are equal in magnitude and opposite in direction. The normal force adjusts in magnitude to counteract other forces, in this case gravitational forces, so the block experiences no net acceleration and remains stable (by [[Newton's first law of motion]]). The floor does not accelerate downward either, presumably because whatever mechanism is fixing it is also generating forces that counteract the downward force exerted by the block.<br />
|-<br />
| The earth and the moon || the earth and the moon || The gravitational force exerted by the earth on the moon, and the gravitational force exerted by the moon on the earth. || The two forces are equal in magnitude and opposite in direction. The effect on the moon -- the moon orbiting the earth -- is more visible because the moon has less mass. The effect on the earth -- including tides -- is less salient because of the larger mass of the earth.<br />
|}<br />
<br />
==Misconceptions==<br />
<br />
<center>{{#widget:YouTube|id=8bTdMmNZm2M}}</center><br />
<br />
===Misleading action-reaction formulation===<br />
<br />
The law is often stated as ''action and reaction are equal and opposite'' where "action" refers to one of the forces and "reaction" refers to the other force. However, this formulation is misleading because it suggests that one of the forces happens ''first'' and the other force happens in ''response'' to it. This is incorrect. The correct formulation is that both forces occur together and are a ''pair'', called an "action-reaction pair." Any physical phenomenon that causes a force also causes the corresponding reaction force.<br />
<br />
For instance, if I push a wall, the wall pushes me back. In terms of human intention, I might say that I was the "cause" of the pair of forces, so the force I exert is the "action" and the force exerted by the wall is the "reaction" force. However, as far as physics is concerned, the role of the two forces is completely symmetric.<br />
<br />
===Larger objects and larger forces===<br />
<br />
One of the common misconceptions surrounding Newton's third law is that the larger object must exert the larger force. There are two possible sources of this misconception:<br />
<br />
# The force exerted by the larger force has more of an ''effect'' on the smaller object than the force exerted by the smaller object. This is due to [[Newton's second law of motion]], which says that the magnitude of acceleration experienced due to a given force is inversely related to the mass.<br />
# It ''is'' true that if a smaller object were replaced by a larger object in a given setting, the larger object would exert more force than the smaller object. For instance, the forces exerted in a head-on collision of a light car and a heavy truck are greater than the forces exerted in a head-on collision of two cars. Newton's third law, in contrast, is comparing the forces between two objects ''within'' a given situation, rather than comparing across situations.<br />
<br />
===Normal "reaction" forces===<br />
<br />
Another common misconception is that forces that arise to cancel the effects of other forces are examples of Newton's third law. For instance, if a block is placed on a horizontal table, the table exerts an upward [[normal force]] on the block to counteract the downward gravitational force on the block.<br />
<br />
The normal force and gravitational force do ''not'' form an action-reaction pair and do ''not'' illustrate Newton's third law. The simplest way of seeing this is that both forces act ''on the same object''. Rather, the fact that they balance each other is due to [[Newton's first law of motion]], which causes the normal force to adjust in magnitude to cancel the downward force exerted due to gravity.<br />
<br />
Similar comments apply to [[static friction]] forces that arise to counteract external forces that would create a tendency for slipping.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Newton%27s_third_law&diff=545Newton's third law2012-08-18T00:18:10Z<p>Vipul: Redirected page to Newton's third law of motion</p>
<hr />
<div>#redirect [[Newton's third law of motion]]</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Newton%27s_third_law_of_motion&diff=544Newton's third law of motion2012-08-18T00:17:50Z<p>Vipul: </p>
<hr />
<div>{{quick phrase|action and reaction are equal and opposite}}<br />
==Definition==<br />
<br />
'''Newton's third law''' or '''Newton's third law of motion''' gives a relationship between the forces that two bodies exert on each other. It says that if body <math>A</math> exerts a [[force]] on body <math>B</math>, then body <math>B</math> exerts a force on body <math>A</math>, and further, that the forces are equal in magnitude and opposite in direction.<br />
<br />
An important corollary of this formulation is that every force on a body is exerted ''by'' something. In other words, it isn't possible for an object to feel a force that isn't being exerted by anything.<br />
<br />
==Particular cases==<br />
<br />
===Force types===<br />
<br />
{| class="sortable" border="1"<br />
! Type of force !! Pair of bodies !! Directions of the action-reaction pair !! Magnitude of the action-reaction pair<br />
|-<br />
| [[Gravitational force]] || Two masses that exert a gravitational force on each other. In most earthly situations, one of the bodies is the earth. || On each body, in the direction toward the center of mass of the other body. Thus the force acts inward on both bodies along the line joining the [[center of mass|centers of mass]] of the two bodies || <math>Gm_1m_2/r^2</math> where <math>G</math> is the gravitational constant, <math>m_1, m_2</math> are the masses and <math>r</math> is the distance between the centers of mass<br />
|-<br />
| [[Normal force]] || Bodies that share a surface of contact. || On each body, the force acts in the outward direction along the surface of contact. || The magnitude is determined to be such that the net relative acceleration along the direction perpendicular to the surface of contact is zero. It thus adjusts in response to the other forces being exerted on the bodies.<br />
|-<br />
| [[Static friction]] || Bodies that share a surface of contact. || On each body, the force acts in a direction parallel to the surface of contact, opposing the tendency of slippage of surfaces. || The magnitude is determined so that the net relative acceleration of surfaces along the surface of contact is zero. It thus adjusts in response to the other forces being exerted on the bodies. There is, however, an upper limit: the [[limiting coefficient of static friction]] times the [[normal force]] magnitude.<br />
|-<br />
| [[Kinetic friction]] || Bodies that share a surface of contact. || On each body, the force acts in a direction parallel to the surface of contact, opposing the actual direction of slippage of surfaces. || The magnitude is a fixed multiple of the [[normal force]], where the constant of proportionality is the [[coefficient of kinetic friction]] and is determined by the nature of the surfaces in contact.<br />
|-<br />
| [[Tension]] || An inextensible string and a body attached to one end of the string || The string pulls the body inward; the body pulls the string outward. || Determined from other constraints.<br />
|}<br />
<br />
===Concrete examples===<br />
<br />
{| class="sortable" border="1"<br />
! Situation !! Pair of bodies !! Action-reaction pair !! What does Newton's third law tell us?<br />
|-<br />
| Any object close to the surface of the earth || the object and the earth || The gravitational force exerted by the earth on the object, and the gravitational force exerted by the object on the earth || The two forces are equal in magnitude and opposite in direction. In other words, the "downward" force experienced by the object due to gravity has a corresponding "upward" force experienced by the earth. However, due to the huge mass of the earth, the resultant acceleration of the earth is too negligible to be noticed.<br />
|-<br />
| A block resting on a fixed horizontal floor || the block and the floor || the upward normal force exerted by the table on the block and the downward normal force exerted by the block on the floor || The two forces are equal in magnitude and opposite in direction. The normal force adjusts in magnitude to counteract other forces, in this case gravitational forces, so the block experiences no net acceleration and remains stable (by [[Newton's first law of motion]]). The floor does not accelerate downward either, presumably because whatever mechanism is fixing it is also generating forces that counteract the downward force exerted by the block.<br />
|-<br />
| The earth and the moon || the earth and the moon || The gravitational force exerted by the earth on the moon, and the gravitational force exerted by the moon on the earth. || The two forces are equal in magnitude and opposite in direction. The effect on the moon -- the moon orbiting the earth -- is more visible because the moon has less mass. The effect on the earth -- including tides -- is less salient because of the larger mass of the earth.<br />
|}<br />
<br />
==Misconceptions==<br />
<br />
===Misleading action-reaction formulation===<br />
<br />
The law is often stated as ''action and reaction are equal and opposite'' where "action" refers to one of the forces and "reaction" refers to the other force. However, this formulation is misleading because it suggests that one of the forces happens ''first'' and the other force happens in ''response'' to it. This is incorrect. The correct formulation is that both forces occur together and are a ''pair'', called an "action-reaction pair." Any physical phenomenon that causes a force also causes the corresponding reaction force.<br />
<br />
For instance, if I push a wall, the wall pushes me back. In terms of human intention, I might say that I was the "cause" of the pair of forces, so the force I exert is the "action" and the force exerted by the wall is the "reaction" force. However, as far as physics is concerned, the role of the two forces is completely symmetric.<br />
<br />
===Larger objects and larger forces===<br />
<br />
One of the common misconceptions surrounding Newton's third law is that the larger object must exert the larger force. There are two possible sources of this misconception:<br />
<br />
# The force exerted by the larger force has more of an ''effect'' on the smaller object than the force exerted by the smaller object. This is due to [[Newton's second law of motion]], which says that the magnitude of acceleration experienced due to a given force is inversely related to the mass.<br />
# It ''is'' true that if a smaller object were replaced by a larger object in a given setting, the larger object would exert more force than the smaller object. For instance, the forces exerted in a head-on collision of a light car and a heavy truck are greater than the forces exerted in a head-on collision of two cars. Newton's third law, in contrast, is comparing the forces between two objects ''within'' a given situation, rather than comparing across situations.<br />
<br />
===Normal "reaction" forces===<br />
<br />
Another common misconception is that forces that arise to cancel the effects of other forces are examples of Newton's third law. For instance, if a block is placed on a horizontal table, the table exerts an upward [[normal force]] on the block to counteract the downward gravitational force on the block.<br />
<br />
The normal force and gravitational force do ''not'' form an action-reaction pair and do ''not'' illustrate Newton's third law. The simplest way of seeing this is that both forces act ''on the same object''. Rather, the fact that they balance each other is due to [[Newton's first law of motion]], which causes the normal force to adjust in magnitude to cancel the downward force exerted due to gravity.<br />
<br />
Similar comments apply to [[static friction]] forces that arise to counteract external forces that would create a tendency for slipping.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Newton%27s_third_law_of_motion&diff=543Newton's third law of motion2012-08-18T00:15:36Z<p>Vipul: /* Examples */</p>
<hr />
<div>==Definition==<br />
<br />
'''Newton's third law of motion''' gives a relationship between the forces that two bodies exert on each other. It says that if body <math>A</math> exerts a [[force]] on body <math>B</math>, then body <math>B</math> exerts a force on body <math>A</math>, and further, that the forces are equal in magnitude and opposite in direction.<br />
<br />
An important corollary of this formulation is that every force on a body is exerted ''by'' something. In other words, it isn't possible for an object to feel a force that isn't being exerted by anything.<br />
<br />
==Examples==<br />
<br />
{| class="sortable" border="1"<br />
! Situation !! Pair of bodies !! Action-reaction pair !! What does Newton's third law tell us?<br />
|-<br />
| Any object close to the surface of the earth || the object and the earth || The gravitational force exerted by the earth on the object, and the gravitational force exerted by the object on the earth || The two forces are equal in magnitude and opposite in direction. In other words, the "downward" force experienced by the object due to gravity has a corresponding "upward" force experienced by the earth. However, due to the huge mass of the earth, the resultant acceleration of the earth is too negligible to be noticed.<br />
|-<br />
| A block resting on a fixed horizontal floor || the block and the floor || the upward normal force exerted by the table on the block and the downward normal force exerted by the block on the floor || The two forces are equal in magnitude and opposite in direction. The normal force adjusts in magnitude to counteract other forces, in this case gravitational forces, so the block experiences no net acceleration and remains stable (by [[Newton's first law of motion]]). The floor does not accelerate downward either, presumably because whatever mechanism is fixing it is also generating forces that counteract the downward force exerted by the block.<br />
|-<br />
| The earth and the moon || the earth and the moon || The gravitational force exerted by the earth on the moon, and the gravitational force exerted by the moon on the earth. || The two forces are equal in magnitude and opposite in direction. The effect on the moon -- the moon orbiting the earth -- is more visible because the moon has less mass. The effect on the earth -- including tides -- is less salient because of the larger mass of the earth.<br />
|}<br />
<br />
==Misconceptions==<br />
<br />
===Misleading action-reaction formulation===<br />
<br />
The law is often stated as ''action and reaction are equal and opposite'' where "action" refers to one of the forces and "reaction" refers to the other force. However, this formulation is misleading because it suggests that one of the forces happens ''first'' and the other force happens in ''response'' to it. This is incorrect. The correct formulation is that both forces occur together and are a ''pair'', called an "action-reaction pair." Any physical phenomenon that causes a force also causes the corresponding reaction force.<br />
<br />
For instance, if I push a wall, the wall pushes me back. In terms of human intention, I might say that I was the "cause" of the pair of forces, so the force I exert is the "action" and the force exerted by the wall is the "reaction" force. However, as far as physics is concerned, the role of the two forces is completely symmetric.<br />
<br />
===Larger objects and larger forces===<br />
<br />
One of the common misconceptions surrounding Newton's third law is that the larger object must exert the larger force. There are two possible sources of this misconception:<br />
<br />
# The force exerted by the larger force has more of an ''effect'' on the smaller object than the force exerted by the smaller object. This is due to [[Newton's second law of motion]], which says that the magnitude of acceleration experienced due to a given force is inversely related to the mass.<br />
# It ''is'' true that if a smaller object were replaced by a larger object in a given setting, the larger object would exert more force than the smaller object. For instance, the forces exerted in a head-on collision of a light car and a heavy truck are greater than the forces exerted in a head-on collision of two cars. Newton's third law, in contrast, is comparing the forces between two objects ''within'' a given situation, rather than comparing across situations.<br />
<br />
===Normal "reaction" forces===<br />
<br />
Another common misconception is that forces that arise to cancel the effects of other forces are examples of Newton's third law. For instance, if a block is placed on a horizontal table, the table exerts an upward [[normal force]] on the block to counteract the downward gravitational force on the block.<br />
<br />
The normal force and gravitational force do ''not'' form an action-reaction pair and do ''not'' illustrate Newton's third law. The simplest way of seeing this is that both forces act ''on the same object''. Rather, the fact that they balance each other is due to [[Newton's first law of motion]], which causes the normal force to adjust in magnitude to cancel the downward force exerted due to gravity.<br />
<br />
Similar comments apply to [[static friction]] forces that arise to counteract external forces that would create a tendency for slipping.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Newton%27s_third_law_of_motion&diff=542Newton's third law of motion2012-08-18T00:13:15Z<p>Vipul: Created page with "==Definition== '''Newton's third law of motion''' gives a relationship between the forces that two bodies exert on each other. It says that if body <math>A</math> exerts a [[..."</p>
<hr />
<div>==Definition==<br />
<br />
'''Newton's third law of motion''' gives a relationship between the forces that two bodies exert on each other. It says that if body <math>A</math> exerts a [[force]] on body <math>B</math>, then body <math>B</math> exerts a force on body <math>A</math>, and further, that the forces are equal in magnitude and opposite in direction.<br />
<br />
An important corollary of this formulation is that every force on a body is exerted ''by'' something. In other words, it isn't possible for an object to feel a force that isn't being exerted by anything.<br />
<br />
==Examples==<br />
<br />
{| class="sortable" border="1"<br />
! Situation !! Pair of bodies !! Action-reaction pair !! What does Newton's third law tell us?<br />
|-<br />
| Any object close to the surface of the earth || the object and the earth || The gravitational force exerted by the earth on the object, and the gravitational force exerted by the object on the earth || The two forces are equal in magnitude and opposite in direction. In other words, the "downward" force experienced by the object due to gravity has a corresponding "upward" force experienced by the earth. However, due to the huge mass of the earth, the resultant acceleration of the earth is too negligible to be noticed.<br />
|-<br />
| A block resting on a fixed horizontal floor || the block and the floor || the upward normal force exerted by the table on the block and the downward normal force exerted by the block on the floor || The two forces are equal in magnitude and opposite in direction. The normal force adjusts in magnitude to counteract other forces, in this case gravitational forces, so the block experiences no net acceleration and remains stable (by [[Newton' first law of motion]]). The floor does not accelerate downward either, presumably because whatever mechanism is fixing it is also generating forces that counteract the downward force exerted by the block.<br />
|}<br />
==Misconceptions==<br />
<br />
===Misleading action-reaction formulation===<br />
<br />
The law is often stated as ''action and reaction are equal and opposite'' where "action" refers to one of the forces and "reaction" refers to the other force. However, this formulation is misleading because it suggests that one of the forces happens ''first'' and the other force happens in ''response'' to it. This is incorrect. The correct formulation is that both forces occur together and are a ''pair'', called an "action-reaction pair." Any physical phenomenon that causes a force also causes the corresponding reaction force.<br />
<br />
For instance, if I push a wall, the wall pushes me back. In terms of human intention, I might say that I was the "cause" of the pair of forces, so the force I exert is the "action" and the force exerted by the wall is the "reaction" force. However, as far as physics is concerned, the role of the two forces is completely symmetric.<br />
<br />
===Larger objects and larger forces===<br />
<br />
One of the common misconceptions surrounding Newton's third law is that the larger object must exert the larger force. There are two possible sources of this misconception:<br />
<br />
# The force exerted by the larger force has more of an ''effect'' on the smaller object than the force exerted by the smaller object. This is due to [[Newton's second law of motion]], which says that the magnitude of acceleration experienced due to a given force is inversely related to the mass.<br />
# It ''is'' true that if a smaller object were replaced by a larger object in a given setting, the larger object would exert more force than the smaller object. For instance, the forces exerted in a head-on collision of a light car and a heavy truck are greater than the forces exerted in a head-on collision of two cars. Newton's third law, in contrast, is comparing the forces between two objects ''within'' a given situation, rather than comparing across situations.<br />
<br />
===Normal "reaction" forces===<br />
<br />
Another common misconception is that forces that arise to cancel the effects of other forces are examples of Newton's third law. For instance, if a block is placed on a horizontal table, the table exerts an upward [[normal force]] on the block to counteract the downward gravitational force on the block.<br />
<br />
The normal force and gravitational force do ''not'' form an action-reaction pair and do ''not'' illustrate Newton's third law. The simplest way of seeing this is that both forces act ''on the same object''. Rather, the fact that they balance each other is due to [[Newton's first law of motion]], which causes the normal force to adjust in magnitude to cancel the downward force exerted due to gravity.<br />
<br />
Similar comments apply to [[static friction]] forces that arise to counteract external forces that would create a tendency for slipping.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Mech:Error_log&diff=541Mech:Error log2012-04-19T23:55:55Z<p>Vipul: </p>
<hr />
<div>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 ''not'' require login, but you can optionally provide an email ID to be notified of the fix).<br />
<br />
We have just started logging errors. On this page, you will find information about all errors that persisted on the site for more than one month on any page. Note: This applies only to errors that were not fixed by December 18, 2011. Errors fixed before then are not being logged on this page.<br />
<br />
{| class="sortable" border="1"<br />
! Page !! Error description !! Version where it was introduced !! Version where it was corrected !! Why error? Why not caught? ("I" refers to [[User:Vipul|Vipul]])<br />
|-<br />
| [[pulley system on a double inclined plane]] || Had <math>\theta</math> in place of <math>\alpha_1, \alpha_2</math> || || [http://mech.subwiki.org/w/index.php?title=Pulley_system_on_a_double_inclined_plane&oldid=540 April 19, 2012] || didn't read page carefully after editing<br />
|-<br />
| [[sliding motion along an inclined plane]] || did not have square root on the formula for "speed at instant of return" in the "Given an initial speed upward" kinematics section || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=223 June 27, 2010] || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=537 January 15, 2012] || not intuitively clear enough<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Pulley_system_on_a_double_inclined_plane&diff=540Pulley system on a double inclined plane2012-04-19T19:56:01Z<p>Vipul: /* Basic components of force diagram */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
[[File:Pulleysystemondoubleinclinedplane.png|thumb|400px|right]]<br />
<br />
This article is about the following scenario. A fixed triangular wedge has two inclines <math>I_1</math> and <math>I_2</math> making angles <math>\alpha_1</math> and <math>\alpha_2</math> with the horizontal, thus making it a [[involves::double inclined plane]]. A [[involves::pulley]] is affixed to the top vertex of the triangle. A string through the pulley has attached at its two ends blocks of masses <math>m_1</math> and <math>m_2</math>, resting on the two inclines <math>I_1</math> and <math>I_2</math> respectively. The string is inextensible. The coefficients of static and kinetic friction between <math>m_1</math> and <math>I_1</math> are <math>\mu_{s1}</math> and <math>\mu_{k1}</math> respectively. The coefficients of static and kinetic friction between <math>m_2</math> and <math>I_2</math> are <math>\mu_{s2}</math> and <math>\mu_{k2}</math> respectively. Assume that <math>\mu_{k1} \le \mu_{s1}</math> and <math>\mu_{k2} \le \mu_{s2}</math>.<br />
<br />
We assume the pulley to be massless so that its moment of inertia can be ignored for the information below. We also assume the string to be massless, so that the tension values at the two ends of the string are equal in magnitude.<br />
<br />
==Summary of cases starting from rest==<br />
<br />
These cases will be justified later in the article, based on the force diagram and by solving the resulting equations.<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens qualitatively !! Magnitude of accelerations<br />
|-<br />
| <math>\! m_1\sin \alpha_1 - m_2\sin \alpha_2 > \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2</math> || <math>m_1</math> slides downward and <math>m_2</math> slides upward, with the same magnitude of acceleration || <math>\! a = g(m_1 \sin \alpha_1 - m_2 \sin \alpha_2 - \mu_{k1}m_1\cos\alpha_1 - \mu_{k2}m_2 \cos\alpha_2)/(m_1 + m_2)</math>.<br />
|-<br />
| <math>\! m_2\sin \alpha_2 - m_1\sin \alpha_1 > \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2</math> || <math>m_2</math> slides downward and <math>m_1</math> slides upward, with the same magnitude of acceleration || <math>\! a = g(m_2 \sin \alpha_2 - m_1 \sin \alpha_1 - \mu_{k1}m_1\cos \alpha_1 - \mu_{k2}m_2 \cos\alpha_2)/(m_1 + m_2)</math>.<br />
|-<br />
| <math>\! |m_1 \sin \alpha_1 - m_2 \sin \alpha_2| \le |\mu_{s1}m_1 \cos \alpha_1 + \mu_{s2}m_2 \cos \alpha_2|</math> || The system remains at rest || <math>0</math><br />
|}<br />
<br />
==Basic components of force diagram==<br />
<br />
There are ''two'' force diagrams of interest here, namely the force diagrams of the masses <math>m_1</math> and <math>m_2</math>.<br />
<br />
There are five candidate forces on each mass. We describe the situation for <math>m_1</math> below:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <math>m_1g</math> || [[gravitational force]] || unconditional|| <math>m_1g</math> where <math>m_1</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\alpha_1</math> with the normal to the incline and an angle <math>(\pi/2) - \alpha_1</math> with the incline <br />
|-<br />
| <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane.<br />
|-<br />
| <math>N_1</math> || [[normal force]] || unconditional|| adjusts so that there is no net acceleration perpendicular to the plane of the incline || outward normal to the incline <br />
|-<br />
| <math>f_{s1}</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline. At most equal to <math>\mu_{s1}N_1</math>. || The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull <math>m_1</math> down or pull it up.<br />
|-<br />
| <math>f_{k1}</math> || [[kinetic friction]] || sliding || <math>\mu_{k1}N_1</math> where <math>N_1</math> is the normal force || opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down.<br />
|}<br />
<br />
A similar description is valid for <math>m_2</math>:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <math>m_2g</math> || [[gravitational force]] || unconditional|| <math>m_2g</math> where <math>m_2</math> is the [[mass]] and <math>g</math> is the [[acceleration due to gravity]] || vertically downward, hence an angle <math>\alpha_2</math> with the normal to the incline and an angle <math>(\pi/2) - \alpha_2</math> with the incline <br />
|-<br />
| <math>T</math> || [[tension]] || unconditional || needs to be computed based on solving equations. || pulls the mass up the inclined plane.<br />
|-<br />
| <math>N_2</math> || [[normal force]] || unconditional|| adjusts so that there is no net acceleration perpendicular to the plane of the incline || outward normal to the incline <br />
|-<br />
| <math>f_{s2}</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline. At most equal to <math>\mu_{s2}N_2</math>. || The direction could be either up or down the incline, depending on whether the remaining forces create a net tendency to pull <math>m_2</math> down or pull it up.<br />
|-<br />
| <math>f_{k2}</math> || [[kinetic friction]] || sliding || <math>\mu_{k2}N_2</math> where <math>N_2</math> is the normal force || opposite direction of motion -- hence down the incline if sliding up, up the incline if sliding down. <br />
|}<br />
<br />
===Components perpendicular to the respective inclined planes===<br />
<br />
{{quotation|For more analysis of this part, see [[sliding motion along an inclined plane#Component perpendicular to the inclined plane]]}}<br />
<br />
For <math>m_1</math>, taking the component perpendicular to the inclined plane <math>I_1</math>, we get:<br />
<br />
<math>\! N_1 = m_1g \cos \alpha_1 \qquad (1.1)</math><br />
<br />
For <math>m_2</math>, taking the component perpendicular to the inclined plane <math>I_2</math>, we get:<br />
<br />
<math>\! N_2 = m_2g \cos \alpha_2 \qquad (1.2)</math><br />
<br />
Note that this part of the analysis is common to both the ''sliding'' and the ''no sliding'' cases.<br />
<br />
===Components along the respective inclined planes assuming no sliding===<br />
<br />
Note that the ''no sliding'' case has two subcases:<br />
<br />
* The system has a ''tendency'' to slide <math>m_1</math> down and <math>m_2</math> up, i.e., this is what would happen if there were no static friction. Thus, the static friction <math>f_{s1}</math> acts ''up'' the incline <math>I_1</math> and the static friction <math>f_{s2}</math> acts ''down'' the incline <math>I_2</math>.<br />
* The system has a ''tendency'' to slide <math>m_1</math> up and <math>m_2</math> down, i.e., this is what would happen if there were no static friction. Thus, the static friction <math>f_{s1}</math> acts ''down'' the incline <math>I_1</math> and the static friction <math>f_{s2}</math> acts ''up'' the incline <math>I_2</math>.<br />
<br />
Let's consider the first case, i.e., <math>m_1</math> down and <math>m_2</math> up. We get the equation for <math>m_1</math>:<br />
<br />
<math>m_1g\sin \alpha_1 = T + f_{s1} \qquad (2.1)</math><br />
<br />
and<br />
<br />
<math>0 \le f_{s1} \le \mu_{s1}N_1 \qquad (2.2)</math><br />
<br />
We get a similar equation for <math>m_2</math>:<br />
<br />
<math>T = m_2g\sin \alpha_2 + f_{s2} \qquad (2.3)</math><br />
<br />
and<br />
<br />
<math>0 \le f_{s2} \le \mu_{s2}N_2 \qquad (2.4)</math><br />
<br />
Add (2.1) and (2.3) and rearrange to get:<br />
<br />
<math>m_1g\sin \alpha_1 - m_2g\sin\alpha_2 = f_{s1} + f_{s2} \qquad (2.5)</math><br />
<br />
Plugging in (2.2) and (2.4) into (2.5), we get:<br />
<br />
<math>0 \le m_1g\sin \alpha_1 - m_2g\sin \alpha_2 \le \mu_{s1}N_1 + \mu_{s2}N_2 \qquad (2.6)</math><br />
<br />
Plugging in (1.1) and (1.2) into this yields:<br />
<br />
<math>0 \le m_1g\sin \alpha_1 - m_2g\sin \alpha_2 \le \mu_{s1}m_1g\cos \alpha_1 + \mu_{s2}m_2g\cos \alpha_2 \qquad (2.6)</math><br />
<br />
Cancel <math>g</math> from all sides to get:<br />
<br />
<math>0 \le m_1\sin \alpha_1 - m_2\sin \alpha_2 \le \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2 \qquad (2.7)</math><br />
<br />
This is the necessary and sufficient condition for the system to have a ''tendency'' to slide <math>m_1</math> down and <math>m_2</math> up, but to not in fact slide.<br />
<br />
Simialrly, in the other not sliding case (i.e., the system has a tendency to slide <math>m_2</math> down, <math>m_1</math> up), we get the following necessary and sufficient condition:<br />
<br />
<math>\! 0 \le m_2\sin \alpha_2 - m_1\sin \alpha_1 \le \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2 \qquad (2.8)</math><br />
<br />
Overall, for the no sliding case, we get the necessary and sufficient condition:<br />
<br />
<math>\! |m_1\sin \alpha_1 - m_2\sin \alpha_2| \le \mu_{s1}m_1\cos \alpha_1 + \mu_{s2}m_2\cos \alpha_2</math><br />
<br />
Note the following: in all these cases, it is ''not'' possible, using these equations, to determine the values of <math>T</math>, <math>f_{s1}</math> and <math>f_{s2}</math> individually.<br />
<br />
===Components along the respective inclined planes assuming sliding===<br />
<br />
We consider two cases:<br />
<br />
* <math>m_1</math> is sliding (and accelerating) down and <math>m_2</math> is sliding (and accelerating) up, so the force of kinetic friction <math>f_{k1}</math> acts ''up'' along <math>I_1</math> and the force of kinetic friction acts ''down'' along <math>I_2</math>.<br />
* <math>m_1</math> is sliding (and accelerating) up and <math>m_2</math> is sliding (and accelerating) down, so the force of kinetic friction <math>f_{k1}</math> acts ''down'' along <math>I_1</math> and the force of kinetic friction acts ''up'' along <math>I_2</math>.<br />
<br />
We first consider the <math>m_1</math> down, <math>m_2</math> up case. Let <math>a</math> denote the magnitude of acceleration for <math>m_1</math>. <math>a</math> is also equal to the magnitude of acceleration for <math>m_2</math>. We get:<br />
<br />
<math>\! m_1g \sin \alpha_1 - f_{k1} - T = m_1a \qquad (3.1)</math><br />
<br />
with<br />
<br />
<math>f_{k1} = \mu_{k1}N_1 \qquad (3.2)</math><br />
<br />
and<br />
<br />
<math>T - m_2g\sin \alpha_2 - f_{k2} = m_2a \qquad (3.3)</math><br />
<br />
with<br />
<br />
<math>f_{k2} = \mu_{k2}N_2 \qquad (3.4)</math><br />
<br />
Add (3.1) and (3.3), plug in (3.2),(1.1) and (3.4),(1.2) to get:<br />
<br />
<math>m_1g\sin \alpha_1 - m_2g\sin \alpha_2 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2 = (m_1 + m_2)a \qquad (3.5)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>a = \frac{m_1g\sin \alpha_1 - m_2g\sin \alpha_2 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2}{m_1 + m_2} \qquad (3.6)</math><br />
<br />
Since <math>a > 0</math> by our sign convention, this case is valid if:<br />
<br />
<math>m_1g\sin \alpha_1 - m_2g\sin \alpha_2 > \mu_{k1}m_1g\cos \alpha_1 + \mu_{k2}m_2g\cos \alpha_2 \qquad (3.7)</math><br />
<br />
and in particular:<br />
<br />
<math>m_1g\sin \alpha_1 > m_2g\sin \alpha_2 \qquad (3.8)</math><br />
<br />
The other case (<math>m_2</math> down, <math>m_1</math> up) occurs if the inequality sign is reversed, and we get, in that case, that:<br />
<br />
<math>a = \frac{m_2g\sin \alpha_2 - m_1g\sin \alpha_1 - \mu_{k1}m_1g\cos \alpha_1 - \mu_{k2}m_2g\cos \alpha_2}{m_1 + m_2} \qquad (3.9)</math></div>Vipulhttps://mech.subwiki.org/w/index.php?title=Mech:Error_log&diff=539Mech:Error log2012-01-16T03:27:02Z<p>Vipul: </p>
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We have just started logging errors. On this page, you will find information about all errors that persisted on the site for more than one month on any page. Note: This applies only to errors that were not fixed by December 18, 2011. Errors fixed before then are not being logged on this page.<br />
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! Page !! Error description !! Version where it was introduced !! Version where it was corrected !! Why error? Why not caught? ("I" refers to [[User:Vipul|Vipul]])<br />
|-<br />
| [[sliding motion along an inclined plane]] || did not have square root on the formula for "speed at instant of return" in the "Given an initial speed upward" kinematics section || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=223 June 27, 2010] || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=537 January 15, 2012] || not intuitively clear enough<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Mech:Error_log&diff=538Mech:Error log2012-01-16T03:26:17Z<p>Vipul: </p>
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{| class="sortable" border="1"<br />
! Page !! Error description !! Version where it was introduced !! Version where it was corrected !! Why error? Why not caught? ("I" refers to [[User:Vipul|Vipul]])<br />
|-<br />
| [[sliding motion along an inclined plane]] || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=223 June 27, 2010] || [http://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&oldid=537 January 15, 2012] || not intuitively clear enough<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&diff=537Sliding motion along an inclined plane2012-01-16T03:22:46Z<p>Vipul: /* Kinematics */</p>
<hr />
<div>{{perspectives}}<br />
{{mechanics scenario}}<br />
<br />
[[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.]]<br />
<br />
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.<br />
<br />
The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (whence, the plane is vertical).<br />
<br />
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.<br />
<br />
==Similar scenarios==<br />
<br />
The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.<br />
<br />
{| class="sortable" border="1"<br />
! Scenario !! Key addition/complication !! Picture<br />
|-<br />
| [[sliding motion along a frictionless inclined plane]] || <math>\mu_k = \mu_s = 0</math> || [[File:Blockonincline.png|100px]]<br />
|-<br />
| [[toppling motion along an inclined plane]] || The block can topple || [[File:Topplableblockonincline.png|100px]]<br />
|-<br />
| [[sliding motion for adjacent blocks along an inclined plane]] || Two blocks instead of one, with a normal force between them || [[File:Adjacentblocksonincline.png|100px]]<br />
|-<br />
| [[sliding-cum-rotational motion along an inclined plane]] || A sphere or cylinder on an inclined plane || [[File:Rotatoronincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[sliding motion along a frictionless circular incline]] || Incline is frictionless, but circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[sliding motion along a circular incline]] || Incline is circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[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]]<br />
|}<br />
<br />
==Basic components of force diagram==<br />
<br />
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).<br />
<br />
===The four candidate forces===<br />
<br />
Assuming no external forces are applied, there are four candidate forces on the block:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <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).<br />
|-<br />
| <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.<br />
|}<br />
<br />
Here is the force diagram (without components) in the ''no sliding'' case:<br />
<br />
[[File:Blockoninclinenoslidingforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding down'' case:<br />
<br />
[[File:Blockoninclineslidingdownforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding up'' case:<br />
<br />
[[File:Blockoninclineslidingupforcediagram.png|400px]]<br />
<br />
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]<br />
===Taking components of the gravitational force===<br />
<br />
{{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.}}<br />
<br />
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<br />
<br />
===Component perpendicular to the inclined plane===<br />
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component <math>mg\cos \theta</math> of gravitational force <math>mg</math> should cancel out normal force.]]<br />
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:<br />
<br />
<math>\! N = mg \cos \theta</math><br />
<br />
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.<br />
<br />
{{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.}}<br />
<br />
Some observations:<br />
<br />
{| class="sortable" border="1"<br />
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments<br />
|-<br />
| <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]].<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.<br />
<br />
===Component along (down) the inclined plane===<br />
<br />
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.<br />
:<br />
<br />
{| class="sortable" border="1"<br />
! 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|}<br />
==What about the inclined plane?==<br />
<br />
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?<br />
<br />
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).<br />
<br />
==Value of the acceleration function==<br />
<br />
===For a block sliding downward===<br />
<br />
As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Downaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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>.<br />
<br />
===For a block sliding upward===<br />
<br />
As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Upaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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:<br />
<br />
<math>\! a = g\sqrt{1 + \mu_k^2}</math><br />
==Behavior for a block initially placed at rest==<br />
<br />
===Statics and dynamics===<br />
<br />
If the block is initially placed gently at rest, we have the following cases:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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>.<br />
|}<br />
<br />
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.<br />
<br />
===Kinematics===<br />
<br />
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):<br />
<br />
{| class"sortable" border="1"<br />
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)<br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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> <br />
|}<br />
<br />
===Energy changes===<br />
<br />
==Given an initial speed downward==<br />
<br />
===Statics and dynamics===<br />
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>:<br />
<br />
* 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.<br />
* 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.<br />
<br />
===Kinematics===<br />
<br />
{{fillin}}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==Given an initial speed upward==<br />
<br />
===Statics and dynamics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
<br />
===Kinematics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken to reach highest point || <math>\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Displacement to highest point achieved (along incline) || <math>\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Vertical height to highest point achieved || <math>\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}</math><br />
|-<br />
| Horizontal displacement to highest point achieved || <math>\frac{u^2}{2g(\tan \theta + \mu_k)}</math><br />
|}<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken on return journey || <math>\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}</math><br />
|-<br />
| Speed at instant of return || <math>\! u \sqrt{\frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}}</math><br />
|}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==External links==<br />
<br />
===Instructional video links===<br />
<br />
See [[Video:Sliding motion along an inclined plane]].</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Mech:Error_log&diff=536Mech:Error log2011-12-20T00:49:34Z<p>Vipul: </p>
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We have just started logging errors. On this page, you will find information about all errors that persisted on the site for more than one month on any page. Note: This applies only to errors that were not fixed by December 18, 2011. Errors fixed before then are not being logged on this page.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Mech:Error_log&diff=535Mech:Error log2011-12-20T00:49:09Z<p>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 ''..."</p>
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<div>alpha</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=532Video:Sliding motion along an inclined plane2011-12-05T20:17:29Z<p>Vipul: /* Other coverage */</p>
<hr />
<div>{{foreign video warning}}<br />
<br />
==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) ||considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || considers the "sliding" case with [[kinetic friction]]<br />
|}<br />
<br />
===Other coverage===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| (embedding unavailable) || [http://mindbites.com/series/241-physics-the-forces-of-friction?lesson_id=4514 Friction and Inclines lecture by Thinwell available on Mindbites] -- only a previes is available for free, the lecture can be downloaded for a small price || full video || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=531Video:Sliding motion along an inclined plane2011-12-05T20:17:16Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>{{foreign video warning}}<br />
<br />
==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) ||considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || considers the "sliding" case with [[kinetic friction]]<br />
|}<br />
<br />
===Other coverage===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| (embedding unavailable) || [http://mindbites.com/series/241-physics-the-forces-of-friction?lesson_id=4514 Friction and Inclines lecture by Thinwell available on Mindbites] -- only a previes is available for free, the lecture can be downloaded for a small price || full video ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Template:Foreign_video_warning&diff=530Template:Foreign video warning2011-12-05T18:33:34Z<p>Vipul: Created page with "{{quotation|Some or all the videos linked to or embedded on this page were created by third parties independent of <tt>subwiki.org</tt> and its contributors. Thus, the copyright ..."</p>
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<hr />
<div>{{foreign video warning}}<br />
<br />
==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) ||considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || considers the "sliding" case with [[kinetic friction]]<br />
|}<br />
<br />
===Quick advanced coverage: Open Course Ware===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_a_frictionless_inclined_plane&diff=528Video:Sliding motion along a frictionless inclined plane2011-12-05T18:31:52Z<p>Vipul: </p>
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<div>{{foreign video warning}}<br />
<br />
==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_a_frictionless_inclined_plane&diff=527Video:Sliding motion along a frictionless inclined plane2011-12-05T18:25:45Z<p>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..."</p>
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<div>==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Template:Perspectives&diff=526Template:Perspectives2011-12-05T18:05:20Z<p>Vipul: </p>
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<div>{{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]]}}}}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Template:Perspectives&diff=525Template:Perspectives2011-12-05T18:04:07Z<p>Vipul: </p>
<hr />
<div>{{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]]}}}}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&diff=524Sliding motion along an inclined plane2011-12-05T18:03:12Z<p>Vipul: </p>
<hr />
<div>{{perspectives}}<br />
{{mechanics scenario}}<br />
<br />
[[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.]]<br />
<br />
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.<br />
<br />
The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (whence, the plane is vertical).<br />
<br />
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.<br />
<br />
==Similar scenarios==<br />
<br />
The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.<br />
<br />
{| class="sortable" border="1"<br />
! Scenario !! Key addition/complication !! Picture<br />
|-<br />
| [[sliding motion along a frictionless inclined plane]] || <math>\mu_k = \mu_s = 0</math> || [[File:Blockonincline.png|100px]]<br />
|-<br />
| [[toppling motion along an inclined plane]] || The block can topple || [[File:Topplableblockonincline.png|100px]]<br />
|-<br />
| [[sliding motion for adjacent blocks along an inclined plane]] || Two blocks instead of one, with a normal force between them || [[File:Adjacentblocksonincline.png|100px]]<br />
|-<br />
| [[sliding-cum-rotational motion along an inclined plane]] || A sphere or cylinder on an inclined plane || [[File:Rotatoronincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[sliding motion along a frictionless circular incline]] || Incline is frictionless, but circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[sliding motion along a circular incline]] || Incline is circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[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]]<br />
|}<br />
<br />
==Basic components of force diagram==<br />
<br />
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).<br />
<br />
===The four candidate forces===<br />
<br />
Assuming no external forces are applied, there are four candidate forces on the block:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <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).<br />
|-<br />
| <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.<br />
|}<br />
<br />
Here is the force diagram (without components) in the ''no sliding'' case:<br />
<br />
[[File:Blockoninclinenoslidingforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding down'' case:<br />
<br />
[[File:Blockoninclineslidingdownforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding up'' case:<br />
<br />
[[File:Blockoninclineslidingupforcediagram.png|400px]]<br />
<br />
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]<br />
===Taking components of the gravitational force===<br />
<br />
{{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.}}<br />
<br />
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<br />
<br />
===Component perpendicular to the inclined plane===<br />
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component <math>mg\cos \theta</math> of gravitational force <math>mg</math> should cancel out normal force.]]<br />
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:<br />
<br />
<math>\! N = mg \cos \theta</math><br />
<br />
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.<br />
<br />
{{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.}}<br />
<br />
Some observations:<br />
<br />
{| class="sortable" border="1"<br />
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments<br />
|-<br />
| <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]].<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.<br />
<br />
===Component along (down) the inclined plane===<br />
<br />
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.<br />
:<br />
<br />
{| class="sortable" border="1"<br />
! 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|}<br />
==What about the inclined plane?==<br />
<br />
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?<br />
<br />
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).<br />
<br />
==Value of the acceleration function==<br />
<br />
===For a block sliding downward===<br />
<br />
As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Downaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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>.<br />
<br />
===For a block sliding upward===<br />
<br />
As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Upaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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:<br />
<br />
<math>\! a = g\sqrt{1 + \mu_k^2}</math><br />
==Behavior for a block initially placed at rest==<br />
<br />
===Statics and dynamics===<br />
<br />
If the block is initially placed gently at rest, we have the following cases:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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>.<br />
|}<br />
<br />
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.<br />
<br />
===Kinematics===<br />
<br />
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):<br />
<br />
{| class"sortable" border="1"<br />
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)<br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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> <br />
|}<br />
<br />
===Energy changes===<br />
<br />
==Given an initial speed downward==<br />
<br />
===Statics and dynamics===<br />
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>:<br />
<br />
* 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.<br />
* 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.<br />
<br />
===Kinematics===<br />
<br />
{{fillin}}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==Given an initial speed upward==<br />
<br />
===Statics and dynamics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
<br />
===Kinematics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken to reach highest point || <math>\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Displacement to highest point achieved (along incline) || <math>\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Vertical height to highest point achieved || <math>\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}</math><br />
|-<br />
| Horizontal displacement to highest point achieved || <math>\frac{u^2}{2g(\tan \theta + \mu_k)}</math><br />
|}<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken on return journey || <math>\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}</math><br />
|-<br />
| Speed at instant of return || <math>\! u \frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}</math><br />
|}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==External links==<br />
<br />
===Instructional video links===<br />
<br />
See [[Video:Sliding motion along an inclined plane]].</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&diff=523Sliding motion along an inclined plane2011-12-05T18:02:35Z<p>Vipul: /* External links */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
[[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.]]<br />
<br />
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.<br />
<br />
The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (whence, the plane is vertical).<br />
<br />
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.<br />
<br />
==Similar scenarios==<br />
<br />
The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.<br />
<br />
{| class="sortable" border="1"<br />
! Scenario !! Key addition/complication !! Picture<br />
|-<br />
| [[sliding motion along a frictionless inclined plane]] || <math>\mu_k = \mu_s = 0</math> || [[File:Blockonincline.png|100px]]<br />
|-<br />
| [[toppling motion along an inclined plane]] || The block can topple || [[File:Topplableblockonincline.png|100px]]<br />
|-<br />
| [[sliding motion for adjacent blocks along an inclined plane]] || Two blocks instead of one, with a normal force between them || [[File:Adjacentblocksonincline.png|100px]]<br />
|-<br />
| [[sliding-cum-rotational motion along an inclined plane]] || A sphere or cylinder on an inclined plane || [[File:Rotatoronincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[sliding motion along a frictionless circular incline]] || Incline is frictionless, but circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[sliding motion along a circular incline]] || Incline is circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[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]]<br />
|}<br />
<br />
==Basic components of force diagram==<br />
<br />
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).<br />
<br />
===The four candidate forces===<br />
<br />
Assuming no external forces are applied, there are four candidate forces on the block:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <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).<br />
|-<br />
| <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.<br />
|}<br />
<br />
Here is the force diagram (without components) in the ''no sliding'' case:<br />
<br />
[[File:Blockoninclinenoslidingforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding down'' case:<br />
<br />
[[File:Blockoninclineslidingdownforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding up'' case:<br />
<br />
[[File:Blockoninclineslidingupforcediagram.png|400px]]<br />
<br />
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]<br />
===Taking components of the gravitational force===<br />
<br />
{{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.}}<br />
<br />
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<br />
<br />
===Component perpendicular to the inclined plane===<br />
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component <math>mg\cos \theta</math> of gravitational force <math>mg</math> should cancel out normal force.]]<br />
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:<br />
<br />
<math>\! N = mg \cos \theta</math><br />
<br />
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.<br />
<br />
{{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.}}<br />
<br />
Some observations:<br />
<br />
{| class="sortable" border="1"<br />
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments<br />
|-<br />
| <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]].<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.<br />
<br />
===Component along (down) the inclined plane===<br />
<br />
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.<br />
:<br />
<br />
{| class="sortable" border="1"<br />
! 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|}<br />
==What about the inclined plane?==<br />
<br />
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?<br />
<br />
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).<br />
<br />
==Value of the acceleration function==<br />
<br />
===For a block sliding downward===<br />
<br />
As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Downaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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>.<br />
<br />
===For a block sliding upward===<br />
<br />
As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Upaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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:<br />
<br />
<math>\! a = g\sqrt{1 + \mu_k^2}</math><br />
==Behavior for a block initially placed at rest==<br />
<br />
===Statics and dynamics===<br />
<br />
If the block is initially placed gently at rest, we have the following cases:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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>.<br />
|}<br />
<br />
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.<br />
<br />
===Kinematics===<br />
<br />
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):<br />
<br />
{| class"sortable" border="1"<br />
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)<br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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> <br />
|}<br />
<br />
===Energy changes===<br />
<br />
==Given an initial speed downward==<br />
<br />
===Statics and dynamics===<br />
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>:<br />
<br />
* 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.<br />
* 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.<br />
<br />
===Kinematics===<br />
<br />
{{fillin}}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==Given an initial speed upward==<br />
<br />
===Statics and dynamics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
<br />
===Kinematics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken to reach highest point || <math>\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Displacement to highest point achieved (along incline) || <math>\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Vertical height to highest point achieved || <math>\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}</math><br />
|-<br />
| Horizontal displacement to highest point achieved || <math>\frac{u^2}{2g(\tan \theta + \mu_k)}</math><br />
|}<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken on return journey || <math>\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}</math><br />
|-<br />
| Speed at instant of return || <math>\! u \frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}</math><br />
|}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==External links==<br />
<br />
===Instructional video links===<br />
<br />
See [[Video:Sliding motion along an inclined plane]].</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=522Video:Sliding motion along an inclined plane2011-12-05T18:01:56Z<p>Vipul: /* Basic coverage: Khan Academy videos */</p>
<hr />
<div>==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) ||considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || considers the "sliding" case with [[kinetic friction]]<br />
|}<br />
<br />
===Quick advanced coverage: Open Course Ware===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=521Video:Sliding motion along an inclined plane2011-12-05T18:01:39Z<p>Vipul: </p>
<hr />
<div>==Instructional video links==<br />
<br />
===Basic coverage: Khan Academy videos===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! <br />
|-<br />
| <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<br />
|-<br />
| <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. <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) ||considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || considers the "no sliding" case with [[static friction]]<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || considers the "sliding" case with [[kinetic friction]]<br />
|}<br />
<br />
===Quick advanced coverage: Open Course Ware===<br />
<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=520Video:Sliding motion along an inclined plane2011-12-05T17:58:45Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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) || || ||<br />
|-<br />
| <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) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}}</toggledisplay> || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=519Video:Sliding motion along an inclined plane2011-12-05T17:56:46Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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) || || ||<br />
|-<br />
| <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) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=BukTc4q9BMc}} || [http://www.khanacademy.org/video/correction-to-force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: correction to force of friction keeping the block stationary] || full video (0:51) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=iA7Thhnzc64}} || [http://www.khanacademy.org/video/force-of-friction-keeping-velocity-constant?playlist=Physics Khan Academy video: force of friction keeping velocity constant] || full video (9:09) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=518Video:Sliding motion along an inclined plane2011-12-05T17:51:09Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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) || || ||<br />
|-<br />
| <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) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=517Video:Sliding motion along an inclined plane2011-12-05T17:50:34Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Video link to correct start time (not necessarily on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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) || || ||<br />
|-<br />
| <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) || || || <br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || || full video (8:41) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=516Video:Sliding motion along an inclined plane2011-12-05T17:49:46Z<p>Vipul: </p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Video link to correct start time (not necessarily on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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) || || ||<br />
|-<br />
| <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) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=v8ODIMqbQ44}}</toggledisplay> || [http://www.khanacademy.org/video/force-of-friction-keeping-the-block-stationary?playlist=Physics Khan Academy video: force of friction keeping the block stationary] || full video (8:41) || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Video:Sliding_motion_along_an_inclined_plane&diff=515Video:Sliding motion along an inclined plane2011-12-05T17:44:22Z<p>Vipul: Created page with "==Instructional video links== Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video link..."</p>
<hr />
<div>==Instructional video links==<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Video link to correct start time (not necessarily on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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 this and more videos in the playlist on inclined planes) || || full video || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Template:Perspectives&diff=514Template:Perspectives2011-12-05T17:36:01Z<p>Vipul: </p>
<hr />
<div>{{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]]}}}}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Template:Perspectives&diff=513Template:Perspectives2011-12-05T17:35:10Z<p>Vipul: Created page with "{{quotation|'''ALSO CHECK OUT''': {{#ifexist:Questions:{{PAGENAME}}|Questions page (list of common doubts/questions/curiosities) <nowiki>|</nowiki>}}{{..."</p>
<hr />
<div>{{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]]}}}}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Sliding_motion_along_an_inclined_plane&diff=512Sliding motion along an inclined plane2011-12-05T17:18:57Z<p>Vipul: /* Instructional video links */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
[[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.]]<br />
<br />
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.<br />
<br />
The extremes are <math>\theta = 0</math> (whence, the plane is horizontal) and <math>\theta = \pi/2</math> (whence, the plane is vertical).<br />
<br />
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.<br />
<br />
==Similar scenarios==<br />
<br />
The scenario discussed here can be generalized/complicated in a number of different ways. Some of these are provided in the table below.<br />
<br />
{| class="sortable" border="1"<br />
! Scenario !! Key addition/complication !! Picture<br />
|-<br />
| [[sliding motion along a frictionless inclined plane]] || <math>\mu_k = \mu_s = 0</math> || [[File:Blockonincline.png|100px]]<br />
|-<br />
| [[toppling motion along an inclined plane]] || The block can topple || [[File:Topplableblockonincline.png|100px]]<br />
|-<br />
| [[sliding motion for adjacent blocks along an inclined plane]] || Two blocks instead of one, with a normal force between them || [[File:Adjacentblocksonincline.png|100px]]<br />
|-<br />
| [[sliding-cum-rotational motion along an inclined plane]] || A sphere or cylinder on an inclined plane || [[File:Rotatoronincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[sliding motion along a frictionless circular incline]] || Incline is frictionless, but circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[sliding motion along a circular incline]] || Incline is circular instead of linear (planar) || [[File:Blockoncircularincline.png|100px]]<br />
|-<br />
| [[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]]<br />
|-<br />
| [[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]]<br />
|}<br />
<br />
==Basic components of force diagram==<br />
<br />
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).<br />
<br />
===The four candidate forces===<br />
<br />
Assuming no external forces are applied, there are four candidate forces on the block:<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <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).<br />
|-<br />
| <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.<br />
|}<br />
<br />
Here is the force diagram (without components) in the ''no sliding'' case:<br />
<br />
[[File:Blockoninclinenoslidingforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding down'' case:<br />
<br />
[[File:Blockoninclineslidingdownforcediagram.png|400px]]<br />
<br />
Here is the force diagram (without taking components) in the ''sliding up'' case:<br />
<br />
[[File:Blockoninclineslidingupforcediagram.png|400px]]<br />
<br />
[[File:mgcomponentsforincline.png|thumb|300px|right|Component-taking for the gravitational force.]]<br />
===Taking components of the gravitational force===<br />
<br />
{{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.}}<br />
<br />
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<br />
<br />
===Component perpendicular to the inclined plane===<br />
[[File:Blockoninclineforcediagramnormalcomponents.png|thumb|300px|right|Normal component <math>mg\cos \theta</math> of gravitational force <math>mg</math> should cancel out normal force.]]<br />
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:<br />
<br />
<math>\! N = mg \cos \theta</math><br />
<br />
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.<br />
<br />
{{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.}}<br />
<br />
Some observations:<br />
<br />
{| class="sortable" border="1"<br />
! Value/change in value of <math>\theta</math> !! Value of <math>N</math> !! Comments<br />
|-<br />
| <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]].<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
For simplicity, we ignore the cases <math>\theta = 0</math> and <math>\theta = \pi/2</math> unless specifically dealing with them.<br />
<br />
===Component along (down) the inclined plane===<br />
<br />
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.<br />
:<br />
<br />
{| class="sortable" border="1"<br />
! 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|-<br />
| 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><br />
|}<br />
==What about the inclined plane?==<br />
<br />
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?<br />
<br />
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).<br />
<br />
==Value of the acceleration function==<br />
<br />
===For a block sliding downward===<br />
<br />
As we saw above, if the block is sliding downward, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta - \mu_k \cos \theta) = g\cos \theta(\tan \theta - \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Downaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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>.<br />
<br />
===For a block sliding upward===<br />
<br />
As we saw above, if the block is sliding above, the acceleration (downward positive) is given by:<br />
<br />
<math>\! a = g(\sin \theta + \mu_k \cos \theta) = g\cos \theta (\tan \theta + \mu_k)</math><br />
<br />
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.<br />
<br />
[[File:Upaccelerationintermsofinclineangle.png|400px]]<br />
<br />
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:<br />
<br />
<math>\! a = g\sqrt{1 + \mu_k^2}</math><br />
==Behavior for a block initially placed at rest==<br />
<br />
===Statics and dynamics===<br />
<br />
If the block is initially placed gently at rest, we have the following cases:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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>.<br />
|}<br />
<br />
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.<br />
<br />
===Kinematics===<br />
<br />
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):<br />
<br />
{| class"sortable" border="1"<br />
! !! <math>\! v</math> !! <math>\! t</math> !! <math>\! s</math> !! vertical displacement (call <math>\!h</math>) !! horizontal displacement (call <math>\! x</math>)<br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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><br />
|-<br />
| <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> <br />
|}<br />
<br />
===Energy changes===<br />
<br />
==Given an initial speed downward==<br />
<br />
===Statics and dynamics===<br />
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>:<br />
<br />
* 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.<br />
* 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.<br />
<br />
===Kinematics===<br />
<br />
{{fillin}}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==Given an initial speed upward==<br />
<br />
===Statics and dynamics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Case !! What happens<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|-<br />
| <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.<br />
|}<br />
<br />
===Kinematics===<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken to reach highest point || <math>\! t = \frac{u}{g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Displacement to highest point achieved (along incline) || <math>\frac{u^2}{2g \cos \theta (\tan \theta + \mu_k)}</math><br />
|-<br />
| Vertical height to highest point achieved || <math>\frac{u^2 \tan \theta}{2g (\tan \theta + \mu_k)}</math><br />
|-<br />
| Horizontal displacement to highest point achieved || <math>\frac{u^2}{2g(\tan \theta + \mu_k)}</math><br />
|}<br />
<br />
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:<br />
<br />
{| class="sortable" border="1"<br />
! Quantity !! Value<br />
|-<br />
| Time taken on return journey || <math>\! \frac{u}{g \cos \theta\sqrt{\tan^2 \theta - \mu_k^2}}</math><br />
|-<br />
| Speed at instant of return || <math>\! u \frac{\tan \theta - \mu_k}{\tan \theta + \mu_k}</math><br />
|}<br />
<br />
===Energy changes===<br />
<br />
{{fillin}}<br />
<br />
==External links==<br />
<br />
===Instructional video links===<br />
<br />
Note that these instructional videos are ''not'' part of Mech and the copyright on these is not with Mech contributors. Please visit the video links in the second column to get more information about the creators and context of the videos.<br />
{| class="sortable" border="1"<br />
! Video embedded in this page (click SHOW MORE to view) !! Video link (on course page) !! Video link to correct start time (not necessarily on course page) !! Segment !! Contextual information !! Transcript link !! Transcript segment <br />
|-<br />
| <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 this and more videos in the playlist on inclined planes) || || full video || || ||<br />
|-<br />
| <toggledisplay>{{#widget:YouTube|id=uZGbtK2KBoY}}</toggledisplay> || [http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed08.htm MIT OCW lecture on friction by Walter Lewin] || || 02:23 - 07:01 || Relates the [[limiting coefficient of static friction]] to the [[angle of friction]] to set the stage for some experimental demonstrations. || same as video link || It is fairly easy to measure ... to ...same kind of rubber.<br />
|}</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Normal_force&diff=511Normal force2011-12-04T01:35:27Z<p>Vipul: /* Work and energy */</p>
<hr />
<div>{{force type}}<br />
==Definition==<br />
<br />
{{quick phrase|[[quick phrase::force preventing objects from going into each other]], [[quick phrase::measure of how pressed together two bodies are]], [[quick phrase::component of contact force orthogonal to surface of contact]], [[quick phrase::component of contact force perpendicular to contact surface]], [[quick phrase::what's commonly mistaken for weight]]}}<br />
<br />
===Intensional definition===<br />
<br />
'''Normal force''', sometimes called '''normal reaction''', is the component of [[contact force]] between two bodies acting in a direction perpendicular to the plane of contact (or the tangent plane at the point of contact). It is in the outward direction on both bodies. The magnitudes of the normal force on both bodies are equal and the directions are opposite, in keeping with [[Newton's third law]].<br />
<br />
===Tendency-based definition===<br />
<br />
'''Normal force''' is the force exerted by two bodies on each other at a surface of contact that is intended to precisely counteract any tendency the bodies have of moving into each other.<br />
<br />
==Situational examples==<br />
<br />
{| class="sortable" border="1"<br />
! Example !! Direction of force on each body !! Magnitude of force !! Reason !! Picture<br />
|-<br />
| A block placed on a fixed horizontal surface. || Upward on the block, downward on the surface || Equal to the [[weight]] of the block, which is the product of the mass and acceleration due to gravity|| The net vertical force on the block should be zero || [[File:Blockonflatsurface.png|200px]]<br />
|-<br />
| A block placed on a fixed inclined plane with angle <math>\theta</math> with the horizontal: see [[sliding motion along a frictionless inclined plane]] and [[sliding motion along an inclined plane]] || Upward/outward on the block, downward/inward on the surface || The product <math>mg \cos \theta</math>, where <math>m</math> is the mass of the block, <math>g</math> is the acceleration due to gravity || The normal force must balance the component of gravitational force that is trying to send the block into the inclined plane. || [[File:Blockoninclineforcediagramnormalcomponents.png|200px]]<br />
|-<br />
| A person lifting a book placed on his horizontal palm || Upward on the book, downward on the palm || <math>m(g + a)</math> where <math>g</math> is the acceleration due to gravity and <math>a</math> is the upward acceleration. If the person is lifting the book with constant speed, it is just <math>mg</math>. If the speed is reducing, the normal force is less than <math>mg</math>. || Note that the normal force does ''not'' cancel <math>mg</math> but rather is determined so as to create the necessary acceleration. ||<br />
|-<br />
| Two blocks sliding down, adjacent to each other on an inclined plane. Assume that they do not get separated. || The block farther down exerts a normal force up the inclined plane on the other block, and the block further up exerts a downward force on the lower blocks. || {{fillin}}. Note that the normal force between the blocks is ''not'' set to ensure zero acceleration (hence, it does not balance out external forces in that direction), but it is set to ensure that the blocks accelerate equally, so that ''relative'' acceleration is still zero. || Obtained by solving system of equations || [[File:Adjacentblocksonincline.png|200px]]<br />
|-<br />
| A block on a circular incline, sliding down along the incline: see [[sliding motion along a frictionless circular incline]] and [[sliding motion along a circular incline]] || Upward/outward on the block, downward/inward on the incline surface || <math>mg \cos \theta - (mv^2/r)</math>, where <math>mg\cos \theta</math> is the component of gravitational force along the normal direction, and <math>mv^2/r</math> is the net centripetal force needed for the block to continue to stay on the circular incline. || Newton's second law. Note that here, <math>N</math> does ''not'' balance external forces in the normal direction, because we want the net force in the normal direction to be, not zero, but the centripetal force.<br />
|}<br />
<br />
==Confusion with Newton's third law==<br />
<br />
The normal force between two bodies exerted through a surface of contact depends on the other forces acting on these bodies that are creating a tendency for the bodies to move into each other. In fact, it adjusts itself in magnitude to ''precisely'' counteract the external forces along the normal direction. Hence, it is ''equal and opposite'' to these external forces and is sometimes called a ''reaction'' force.<br />
<br />
However, the normal force is '''not''' a ''reaction'' to the external force in the sense of [[Newton's third law]], even though it is (in many typical situations) equal and opposite to the external force.<br />
<br />
Rather, the normal forces that the two bodies exert on each other form an action-reaction pair. Here is an elaboration using the previous examples:<br />
<br />
{| class="sortable" border="1"<br />
! Example !! False interpretation of Newton's third law !! Correct interpretation of Newton's third law <br />
|-<br />
| A block placed on a fixed horizontal surface || The upward normal force exerted by the surface on the block forms an action-reaction pair with the downward gravitational force on the block || The upward normal force exerted by the surface on the block, and the downward normal force exerted by the block on the surface, form an action-reaction pair. The gravitational force exerted by the earth on the block and the gravitational force exerted by the block on the earth form another action-reaction pair. <br />
|-<br />
| A block placed on a fixed inclined plane with angle <math>\theta</math> with the horizontal || The upward/outward normal force exerted by the surface on the block forms an action-reaction pair with the component of gravitational force exerted on the block in the inward direction into the plane. || The upward/outward normal force exerted by the surface on the block forms an action-reaction pair with an inward/downward normal force exerted by the block on the surface.<br />
|}<br />
<br />
{{quotation|'''KEY POINT TO REMEMBER''': Two different forces acting on the same body ''cannot'' form an action-reaction pair. The two forces of an action-reaction pair must act on different bodies.}}<br />
<br />
==Units and dimensions==<br />
<br />
Normal force is a form of force and hence has the same dimensions and units as all forces do. The MLT dimensions are <math>MLT^{-2}</math> and the SI unit is <math>N</math> (Newton) or <math>kgms^{-2}</math>. Another unit typically used is <math>kgf</math>, which is the weight of <math>1kg</math> of material. Other units used include units obtained by replacing <math>kg</math> by other measures of weight, <math>m</math> by other measures of length, and <math>s</math> by other measures of time.<br />
<br />
==Related forces==<br />
<br />
* [[Friction]]: It is the component of contact force along the plane of contact.<br />
* [[Tension]]: Tension is a force exerted by a taut string on an object at its end, pulling the object inward.<br />
* [[Spring force]]: Spring force is a force exerted by a spring on an object attached at its end. When the spring is extended, the force is inward, analogous to [[tension]], while when the spring is compressed, this acts outward, analogous to [[normal force]].<br />
* [[Pressure]]: This is the normal force per unit area, or the differential of the normal force with respect to area. The normal force exerted through a surface of contact is the integral of the pressure with respect to area over the entire surface of contact.<br />
* [[Stress]]: This is force per unit area that tends to have a deforming effect, and the force could be arising from a normal force.<br />
<br />
==Subtleties==<br />
<br />
===Distribution over a flat surface of contact===<br />
<br />
The normal force acting through a surface of contact is the sum of forces acting on subparts of the surface. The magnitude of normal force per unit area is termed [[pressure]]. If we denote the pressure at a point on the surface of contact by <math>P</math>, then the total normal force can be given by integrating this pressure over the entire area of contact:<br />
<br />
<math>N = \int P \, dA</math><br />
<br />
where <math>dA</math> is the area element being integrated over. If the pressure is uniform through the surface of contact, then we have:<br />
<br />
<math>N = PA</math><br />
<br />
In general, the pressure need not be uniform over the entire surface of contact.<br />
<br />
===Torque due to normal force===<br />
<br />
{{further|[[Torque due to normal force]]}}<br />
<br />
Since the normal force is distributed over the entire surface of contact, different parts of it make different contributions to the torque. To determine the overall torque, we need to integrate, over the area of contact, the torque generated per unit area of contact.<br />
<br />
In many cases of interest, the torque due to normal force is zero. This happens when the normal force is uniformly distributed over the surface and the surfaces is symmetric about the normal line to the surface through the center of mass. For many of these situations, a simplification is made: the normal force is shown as a force with a single line of action, namely, the normal line through the center of mass, and this is used to conclude that the normal force is zero.<br />
<br />
===Normal force for a non-flat surface of contact===<br />
<br />
Consider a ball resting on/in a hemispherical surface of the same shape (so that there is a good fit). In this case, the surface of contact is not flat and there are different normal directions to different parts of the surface. The net direction of normal force in such situations is unclear. The normal force can be determined by the same integration as for a flat surface:<br />
<br />
<math>N =\int P d\overline{A}</math><br />
<br />
except that now the area element is treated as a vector pointing in the normal direction at the point of contact, and the integration is thus a vector integration.<br />
<br />
===Normal force for a hinge/point of contact===<br />
<br />
In some cases, the surface of contact is very small and, in the geometric approximation, is a point or a line rather than a surface. (In reality, it is still a surface, albeit one of very small area). The concept of normal force still makes sense, and it acts in the direction perpendicular to the tangent plane through the point of contact.<br />
<br />
In case of irregular surfaces, however,there may not be a well-defined tangent plane, and in this case, there may be some indeterminacy in the direction of normal force. The actual direction of normal force depends on the normal direction to the surface area as it is actually formed due to microscopic deformations. This becomes of significance in situations such as [[rolling friction]] and some examples of [[banking]].<br />
<br />
===Normal impulse===<br />
<br />
In [[head-on collision]]s, there is a huge normal force between the colliding bodies for a short interval of time. For such situations, it is not customary to calculate the value of the normal force over this short time interval, but the [[normal impulse]], which is the integral of this normal force over the time duration of the collision, is important.<br />
<br />
In [[oblique collision]]s, there is both a normal impulse and a [[frictional impulse]], which is the impulse due to frictional force. The frictional impulse is the [[coefficient of kinetic friction]] times the normal impulse.<br />
<br />
==Work and energy==<br />
<br />
Normal force can do work if the normal force causes motion, e.g., with one body ''pushing'' the other body in some direction. It does ''not'' do work if the normal force is balanced by other forces perpendicular to the plane of contact that result in no net motion in that direction.<br />
<br />
For instance, if a person lifts a book by placing it on the palm of her hand and moving her palm upward, then the normal force between the palm and the book causes the book to be lifted. The overall work done by the normal force is obtained by using the expression <math>\int F \cdot ds</math> where <math>F</math> is the normal force. Part of this work is converted to [[potential energy]], since it combats the gravitational force acting on the book, namely <math>\int mg \cdot ds = mgs</math> where <math>s</math> is the total height gained by the book. The rest of the work is converted into [[kinetic energy]], which is the energy that the book acquires because of its speed. This is given by <math>\int (F - mg) \cdot ds</math>.<br />
<br />
If a person pushes a block forward on a frictionless floor, the work done by the normal force is <math>\int F \cdot ds</math>, all of which is converted into the block's kinetic energy. If, however, the person pushes the block forward on a floor with friction, part of the work done, namely <math>\int \mu_k mg \cdot ds</math>, is dissipated in [[kinetic friction]], and the remaining is stored in the form of the block's kinetic energy.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Quiz:Sliding_motion_along_an_inclined_plane&diff=510Quiz:Sliding motion along an inclined plane2011-10-20T00:16:50Z<p>Vipul: Created page with "See sliding motion along an inclined plane for background information. ==Force diagram and acceleration analysis== <quiz display=simple> {Consider a situation where <math>..."</p>
<hr />
<div>See [[sliding motion along an inclined plane]] for background information.<br />
<br />
==Force diagram and acceleration analysis==<br />
<br />
<quiz display=simple><br />
<br />
{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)?<br />
|type="()"}<br />
- The magnitude of normal force equals <math>mg</math> and is independent of <math>\theta</math><br />
+ The magnitude of normal force decreases from <math>mg</math> to 0 as <math>\theta</math> increases from 0 to <math>\pi/2</math><br />
- The magnitude of normal force increases from 0 to <math>mg</math> as <math>\theta</math> increases from 0 to <math>\pi/2</math><br />
- 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><br />
- 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><br />
<br />
{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>?<br />
|type="()"}<br />
- <math>g(\sin \theta + \mu \cos \theta)</math><br />
- <math>g(\sin \theta - \mu \cos \theta)</math><br />
- <math>g(\cos \theta - \mu \sin \theta)</math><br />
+ <math>g(\cos \theta + \mu \sin \theta)</math><br />
<br />
</quiz></div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=509Banked turn at constant speed and constant height with no slipping2011-09-07T15:32:17Z<p>Vipul: /* Comparison of specific equations */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along the incline either up or down. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
{{quotation|'''AMBIGUOUS DIRECTION''': The direction of the static friction force <math>f_s</math> has a sense ambiguity, i.e., we are not sure whether it acts up the incline or down the incline. In order to avoid the ambiguity, we will assume a downward (inward) positive convention ''but'' allow <math>f_s</math> to have both positive and negative measurement.}}<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! |f_s| \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! |f_s| \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math> g \tan \theta - \mu_sg\le \frac{v^2}{r} \le g \tan \theta + \mu_sg\qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>\sqrt{\max \{ 0, rg (\tan \theta - \mu_s) \} } \le v \le \sqrt{rg(\tan \theta + \mu_s)} \qquad(2.4)</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v = \sqrt{rg \tan \theta}</math>. Note that the presence of static friction provides some leeway and allows a range of possible speeds rather than a unique single speed that works.<br />
<br />
===Two cases===<br />
<br />
We consider two cases:<br />
<br />
* <math>\tan \theta \le \mu_s</math>, so <math>0 \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In other words, the banking angle is less than or equal to the [[angle of friction]] <math>\arctan \mu_s</math>. In this case, the vehicle could be parked at the angle without downward slipping. There is thus no ''lower'' limit to the speed at which it can negotiate the turn. However, there is an ''upper'' limit of <math>\sqrt{rg(\tan \theta + \mu_s)}</math>.<br />
* <math>\tan \theta \ge \mu_s</math>, so <math>\sqrt{rg(\tan \theta - \mu_s)} \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In this case, there is both a minimum and a maximum speed needed to negotiate the turn at constant speed. The reason for a minimum speed is that if the vehicle is too slow, then it will slip downward because of gravity. To counteract that, a sufficiently high normal force must be applied which in turn causes a significant centripetal force which in turn causes a high centripetal acceleration, which is consistent only with a certain minimum speed because of <math>a = v^2/r</math>.<br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).<br />
<br />
===Comparison of specific equations===<br />
<br />
Contrast the equation for sliding motion along an inclined plane:<br />
<br />
<math>\! N = mg\cos \theta</math><br />
<br />
against the equation for banked turn with no slippage:<br />
<br />
<math>\! mg = N \cos \theta \implies N = \frac{mg}{\cos \theta} = mg \sec \theta</math><br />
<br />
In the case of sliding motion along an inclined plane, the normal force is smaller in magnitude than the gravitational force; in fact, it is <math>\cos \theta</math> times the gravitational force and thus successfully counteracts the component of gravitational force in the direction normal to the incline. The component along the incline, namely <math>mg\sin \theta</math>, is not canceled or affected by the normal force, though it could be partly or completely countered by the friction force. In other words, the vehicle is pressed against the surface ''less strongly'' than its weight. Thus, for instance, if you are going downhill in a vehicle, then the weight you exert on the seat is not your full weight, because part of your full weight is exerting a force on you to make you lurch forward.<br />
<br />
In the case of a banked turn the normal force is greater in magnitude than the gravitational force, so that it ''completely'' cancels the gravitational force in the pure vertical direction, and the ''remaining'' component serves the role of centripetal acceleration. In other words, the vehicle is pressed against the surface ''more strongly'' than its weight. Thus, when you are making a banked turn on a hillside, you are ''more'' strongly pressed to the vehicle.<br />
<br />
===How the stationary case of sliding motion along an inclined plane is a special case of this===<br />
<br />
The stationary case of sliding motion along an inclined plane can be viewed as a special case of this where <math>r = \infty</math>, i.e., there is no turn. We see that in this case, the only situation where the vehicle can maintain a constant height is if <math>\tan \theta \le \mu_s</math>.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=508Banked turn at constant speed and constant height with no slipping2011-09-07T15:29:57Z<p>Vipul: /* Two cases */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along the incline either up or down. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
{{quotation|'''AMBIGUOUS DIRECTION''': The direction of the static friction force <math>f_s</math> has a sense ambiguity, i.e., we are not sure whether it acts up the incline or down the incline. In order to avoid the ambiguity, we will assume a downward (inward) positive convention ''but'' allow <math>f_s</math> to have both positive and negative measurement.}}<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! |f_s| \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! |f_s| \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math> g \tan \theta - \mu_sg\le \frac{v^2}{r} \le g \tan \theta + \mu_sg\qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>\sqrt{\max \{ 0, rg (\tan \theta - \mu_s) \} } \le v \le \sqrt{rg(\tan \theta + \mu_s)} \qquad(2.4)</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v = \sqrt{rg \tan \theta}</math>. Note that the presence of static friction provides some leeway and allows a range of possible speeds rather than a unique single speed that works.<br />
<br />
===Two cases===<br />
<br />
We consider two cases:<br />
<br />
* <math>\tan \theta \le \mu_s</math>, so <math>0 \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In other words, the banking angle is less than or equal to the [[angle of friction]] <math>\arctan \mu_s</math>. In this case, the vehicle could be parked at the angle without downward slipping. There is thus no ''lower'' limit to the speed at which it can negotiate the turn. However, there is an ''upper'' limit of <math>\sqrt{rg(\tan \theta + \mu_s)}</math>.<br />
* <math>\tan \theta \ge \mu_s</math>, so <math>\sqrt{rg(\tan \theta - \mu_s)} \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In this case, there is both a minimum and a maximum speed needed to negotiate the turn at constant speed. The reason for a minimum speed is that if the vehicle is too slow, then it will slip downward because of gravity. To counteract that, a sufficiently high normal force must be applied which in turn causes a significant centripetal force which in turn causes a high centripetal acceleration, which is consistent only with a certain minimum speed because of <math>a = v^2/r</math>.<br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).<br />
<br />
===Comparison of specific equations===<br />
<br />
Contrast the equation for sliding motion along an inclined plane:<br />
<br />
<math>\! N = \mg\cos \theta</math><br />
<br />
against the equation for banked turn with no slippage:<br />
<br />
<math>\! mg = N \cos \theta \implies N = \frac{mg}{\cos \theta} = mg \sec \theta</math><br />
<br />
In the case of sliding motion along an inclined plane, the normal force is smaller in magnitude than the gravitational force; in fact, it is <math>\cos \theta</math> times the gravitational force and thus successfully counteracts the component of gravitational force in the direction normal to the incline. The component along the incline, namely <math>mg\sin \theta</math>, is not canceled or affected by the normal force, though it could be partly or completely countered by the friction force. In other words, the vehicle is pressed against the surface ''less strongly'' than its weight.<br />
<br />
In the case of a banked turn the normal force is greater in magnitude than the gravitational force, so that it ''completely'' cancels the gravitational force in the pure vertical direction, and the ''remaining'' component serves the role of centripetal acceleration. In other words, the vehicle is pressed against the surface ''more strongly'' than its weight.<br />
===How the stationary case of sliding motion along an inclined plane is a special case of this===<br />
<br />
The stationary case of sliding motion along an inclined plane can be viewed as a special case of this where <math>r = \infty</math>, i.e., there is no turn. We see that in this case, the only situation where the vehicle can maintain a constant height is if <math>\tan \theta \le \mu_s</math>.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=507Banked turn at constant speed and constant height with no slipping2011-09-07T15:29:12Z<p>Vipul: /* Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along the incline either up or down. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
{{quotation|'''AMBIGUOUS DIRECTION''': The direction of the static friction force <math>f_s</math> has a sense ambiguity, i.e., we are not sure whether it acts up the incline or down the incline. In order to avoid the ambiguity, we will assume a downward (inward) positive convention ''but'' allow <math>f_s</math> to have both positive and negative measurement.}}<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! |f_s| \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! |f_s| \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math> g \tan \theta - \mu_sg\le \frac{v^2}{r} \le g \tan \theta + \mu_sg\qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>\sqrt{\max \{ 0, rg (\tan \theta - \mu_s) \} } \le v \le \sqrt{rg(\tan \theta + \mu_s)} \qquad(2.4)</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v = \sqrt{rg \tan \theta}</math>. Note that the presence of static friction provides some leeway and allows a range of possible speeds rather than a unique single speed that works.<br />
<br />
===Two cases===<br />
<br />
We consider two cases:<br />
<br />
* <math>\tan \theta \le \mu_s</math>, so <math>0 \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In other words, the banking angle is less than or equal to the [[angle of friction]] <math>\arctan \mu_s</math>. In this case, the vehicle could be parked at the angle without downward slipping. There is thus no ''lower'' limit to the speed at which it can negotiate the turn. However, there is an ''upper'' limit of <math>\sqrt{rg(\tan \theta + \mu_s)}</math>.<br />
* <math>\tan \theta \ge \mu_s</math>, so <math>\sqrt{rg(\tan \theta - \mu_s)} \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In this case, there is both a minimum and a maximum speed needed to negotiate the turn at constant speed. The reason for a minimum speed is that if the vehicle is too slow, then it will slip downward because of gravity. To counteract that, a sufficiently high normal force must be applied which in turn causes a significant centripetal force which in turn causes a high centripetal acceleration, which is consistent only with a certain minimum speed because of math>a = v^2/r</math>. <br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).<br />
<br />
===Comparison of specific equations===<br />
<br />
Contrast the equation for sliding motion along an inclined plane:<br />
<br />
<math>\! N = \mg\cos \theta</math><br />
<br />
against the equation for banked turn with no slippage:<br />
<br />
<math>\! mg = N \cos \theta \implies N = \frac{mg}{\cos \theta} = mg \sec \theta</math><br />
<br />
In the case of sliding motion along an inclined plane, the normal force is smaller in magnitude than the gravitational force; in fact, it is <math>\cos \theta</math> times the gravitational force and thus successfully counteracts the component of gravitational force in the direction normal to the incline. The component along the incline, namely <math>mg\sin \theta</math>, is not canceled or affected by the normal force, though it could be partly or completely countered by the friction force. In other words, the vehicle is pressed against the surface ''less strongly'' than its weight.<br />
<br />
In the case of a banked turn the normal force is greater in magnitude than the gravitational force, so that it ''completely'' cancels the gravitational force in the pure vertical direction, and the ''remaining'' component serves the role of centripetal acceleration. In other words, the vehicle is pressed against the surface ''more strongly'' than its weight.<br />
===How the stationary case of sliding motion along an inclined plane is a special case of this===<br />
<br />
The stationary case of sliding motion along an inclined plane can be viewed as a special case of this where <math>r = \infty</math>, i.e., there is no turn. We see that in this case, the only situation where the vehicle can maintain a constant height is if <math>\tan \theta \le \mu_s</math>.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=506Banked turn at constant speed and constant height with no slipping2011-09-07T15:14:25Z<p>Vipul: </p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along the incline either up or down. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
{{quotation|'''AMBIGUOUS DIRECTION''': The direction of the static friction force <math>f_s</math> has a sense ambiguity, i.e., we are not sure whether it acts up the incline or down the incline. In order to avoid the ambiguity, we will assume a downward (inward) positive convention ''but'' allow <math>f_s</math> to have both positive and negative measurement.}}<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! |f_s| \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! |f_s| \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math> g \tan \theta - \m_sg\le \frac{v^2}{r} \le g \tan \theta + \mu_sg\qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>\sqrt{\max \{ 0, rg (\tan \theta - \mu_s) \} } \le v \le \sqrt{rg(\tan \theta + \mu_s)} \qquad(2.4)</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v = \sqrt{rg \tan \theta}</math>. Note that the presence of static friction provides some leeway and allows a range of possible speeds rather than a unique single speed that works.<br />
<br />
===Two cases===<br />
<br />
We consider two cases:<br />
<br />
* <math>\tan \theta \le \mu_s</math>, so <math>0 \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In other words, the banking angle is less than or equal to the [[angle of friction]] <math>\arctan \mu_s</math>. In this case, the vehicle could be parked at the angle without downward slipping. There is thus no ''lower'' limit to the speed at which it can negotiate the turn. However, there is an ''upper'' limit of <math>\sqrt{rg(\tan \theta + \mu_s)}</math>.<br />
* <math>\tan \theta \ge \mu_s</math>, so <math>\sqrt{rg(\tan \theta - \mu_s)} \le v \le \sqrt{rg(\tan \theta + \mu_s)}</math>: In this case, there is both a minimum and a maximum speed needed to negotiate the turn at constant speed. The reason for a minimum speed is that if the vehicle is too slow, then it will slip downward because of gravity. To counteract that, a sufficiently high normal force must be applied which in turn causes a significant centripetal force which in turn causes a high centripetal acceleration, which is consistent only with a certain minimum speed because of math>a = v^2/r</math>. <br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).<br />
<br />
===Comparison of specific equations===<br />
<br />
Contrast the equation for sliding motion along an inclined plane:<br />
<br />
<math>\! N = \mg\cos \theta</math><br />
<br />
against the equation for banked turn with no slippage:<br />
<br />
<math>\! mg = N \cos \theta \implies N = \frac{mg}{\cos \theta} = mg \sec \theta</math><br />
<br />
In the case of sliding motion along an inclined plane, the normal force is smaller in magnitude than the gravitational force; in fact, it is <math>\cos \theta</math> times the gravitational force and thus successfully counteracts the component of gravitational force in the direction normal to the incline. The component along the incline, namely <math>mg\sin \theta</math>, is not canceled or affected by the normal force, though it could be partly or completely countered by the friction force. In other words, the vehicle is pressed against the surface ''less strongly'' than its weight.<br />
<br />
In the case of a banked turn the normal force is greater in magnitude than the gravitational force, so that it ''completely'' cancels the gravitational force in the pure vertical direction, and the ''remaining'' component serves the role of centripetal acceleration. In other words, the vehicle is pressed against the surface ''more strongly'' than its weight.<br />
===How the stationary case of sliding motion along an inclined plane is a special case of this===<br />
<br />
The stationary case of sliding motion along an inclined plane can be viewed as a special case of this where <math>r = \infty</math>, i.e., there is no turn. We see that in this case, the only situation where the vehicle can maintain a constant height is if <math>\tan \theta \le \mu_s</math>.</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=505Banked turn at constant speed and constant height with no slipping2011-09-07T14:57:12Z<p>Vipul: /* Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along (up) the incline. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! f_s \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! 0 \le f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! 0 \le f_s \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math>\mu_sg \le \frac{v^2}{r} \le \mu_sg + g \tan \theta\qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>\sqrt{rg \tan \theta} \le v \le \sqrt{rg(\tan \theta + \mu_s)} \qquad(2.4)</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v \le \sqrt{rg \tan \theta}</math>.<br />
<br />
==Qualitative description of pair relationships==<br />
<br />
Equation (2.4) above gives a quantitative relationship governing the maximum possible speed of a banked turn for given values of radius of curvature, banking angle, and coefficient of static friction. We now discuss possible pair relationships between these four variables, and what they signify.<br />
<br />
{| class="sortable" border="1"<br />
! Variables whose relationship we are interested in studying !! Variables that we fix for studying this relationship !! Description of relationship !! Explanation/verbal formulation<br />
|-<br />
| <math>v</math> (maximum possible) and <math>r</math> (radius of curvature) || <math>\theta</math> and <math>\mu_s<math> || positive, <math>v \propto \sqrt{r}</math> || The larger the radius of curvature, the less curved (i.e. the more straightish) the curve, and the higher the speed that it is possible to negotiate. In the limit as <math>r \to \infty</math>, there is no turn at all and no limit on the value of <math>v</math>.<br />
|-<br />
| <math>v</math> (maximum possible) and <math>\theta</math> (banking angle) || <math>r</math> and <math>\mu_s</math> || positive, <math>v \propto \sqrt{\tan \theta}</math> ||<br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).</div>Vipulhttps://mech.subwiki.org/w/index.php?title=Banked_turn_at_constant_speed_and_constant_height_with_no_slipping&diff=504Banked turn at constant speed and constant height with no slipping2011-09-07T14:48:45Z<p>Vipul: /* Basic components of force diagram */</p>
<hr />
<div>{{mechanics scenario}}<br />
<br />
This article discusses a scenario where a vehicle with wheels is making a [[banked turn]] with a banking angle of <math>\theta</math> without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).<br />
<br />
==Basic components of force diagram==<br />
<br />
===The three candidate forces===<br />
<br />
{| class="sortable" border="1"<br />
! Force (letter) !! Nature of force !! Condition for existence !! Magnitude !! Direction <br />
|-<br />
| <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 <br />
|-<br />
| <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 <br />
|-<br />
| <math>f_s</math> || [[static friction]] || no sliding || adjusts so that there is no net acceleration along the incline || it is along (up) the incline. Note that if the vehicle's ''speed'' were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.<br />
|}<br />
<br />
==Equations==<br />
<br />
We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.<br />
<br />
It turns out that all the forces are in the plane spanned by the first two dimensions.<br />
<br />
The vehicle has no acceleration in the vertical direction, so [[Newton's first law of motion]] gives us that:<br />
<br />
<math>\! mg = N \cos \theta \qquad (1.1)</math><br />
<br />
The acceleration in the horizontal direction perpendicular to the motion of the vehicle is <math>v^2/r</math> where <math>v</math> is the speed and <math>r</math> is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then <math>r</math> is the radius of the circle.<br />
<br />
We thus get, by [[Newton's second law of motion]], that:<br />
<br />
<math>\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)</math><br />
<br />
We also have:<br />
<br />
<math>\! f_s \le \mu_sN \qquad (1.3)</math><br />
<br />
===Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn===<br />
<br />
Plugging (1.1) into (1.2), we get:<br />
<br />
<math>\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)</math><br />
<br />
Plugging (1.1) into (1.3), we have:<br />
<br />
<math>\! f_s \le \mu_smg/\cos \theta \qquad (2.2)</math><br />
<br />
Combining (2.1) and (2.2) we get:<br />
<br />
<math>\frac{v^2}{r} - g\tan \theta \le \mu_sg \qquad (2.3)</math><br />
<br />
Rearranging, we get:<br />
<br />
<math>v \le \sqrt{rg(\tan \theta + \mu_s)}</math><br />
<br />
In particular, in the ''frictionless'' case we would get <math>v \le \sqrt{rg \tan \theta}</math>.<br />
<br />
==Confusion with sliding motion along an inclined plane==<br />
<br />
The basic forces operating in the case of a banked turn are similar to the basic forces operating for [[sliding motion along an inclined plane]], yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.<br />
<br />
===Direction of acceleration and no acceleration===<br />
<br />
In the case of [[sliding motion along an inclined plane]], the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.<br />
<br />
In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the ''curve'' of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.<br />
<br />
===Choice of axes to take components===<br />
<br />
Once we have the force equations as ''vector'' equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.<br />
<br />
Based on our discussion above, we see that for the case of [[sliding motion along an inclined plane]], the best choice of axes is ''along the plane (up/down)'' (the direction of possible acceleration), ''perpendicular to the plane'' (the direction of the normal force and a component of the gravitational force), and ''along the plane (horizontal)'' (no forces, no acceleration).<br />
<br />
In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is ''horizontal, perpendicular to direct of motion'' (direction of acceleration -- centripetal), ''vertical'' (direction of gravitational force, other forces), and ''horizontal, along motion'' (no force or acceleration).</div>Vipul