Physics Book - User contributions [en]

Gauss's Flux Theorem

2016-11-26T18:11:56Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> \theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.
==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T18:07:39Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> \theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.
==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

File:Flux diagram.png

2016-11-26T17:56:41Z

Patrycjakot:

Gauss's Flux Theorem

2016-11-26T17:53:20Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> \theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T16:57:39Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> \theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T16:57:20Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T16:56:48Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and <math> /theta</math> is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T16:13:06Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model (as shown below) used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} </math> is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T15:52:08Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, \hat{n} is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction and angle of an electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T15:35:35Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, \hat{n} is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction of the electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder. This use of Gauss's Law is especially useful in the field of Physics due to its aid in simplifying the calculations (as opposed to using Coulomb's Law) in finding the total charge inside a closed surface. Furthermore, the use of electric flux in Gauss's Flux Theorem assists in studying and understanding Faraday's Law and its applications to objects such as electrical generators, transformers, and inductors.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T15:24:05Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, \hat{n} is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

===A Computational Model===
Gauss's Flux Theorem is able to be visualized by observing the direction of the electric field traveling through a given surface.

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes. It can be observed that using Gauss's Law can assist in finding the derivation of the electric field for objects such as a sphere, sheet of charge, and cylinder.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics. The use of Gauss's Law aids in simplifying the calculations ( as opposed to using Coulomb's Law) in finding the total charge inside a closed surface.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T14:34:23Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, \hat{n} is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T14:31:35Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q\_(inside)}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, <math> \hat{n} <math> is the unit normal vector, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T14:26:00Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q_\inside}{\varepsilon_0} = \oint_C E\bullet dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

==Examples==

===Easy===
===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<'''Bold text'''math>\Phi_E = 2.573</math> Volt Meters

===Middling===
===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Difficult===
===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T14:21:03Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{\Sigma q_\inside}{\varepsilon_0} = \oint_C E\bullet dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

==Examples==

===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<math>\Phi_E = 2.573</math> Volt Meters

===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Gauss's Flux Theorem

2016-11-26T14:00:11Z

Patrycjakot:

Claimed by Brenda Dang '''Claimed by Patrycja Kotowska (Fall 2016)'''

This is a page about Gauss's Flux Theorem. A work in progress by [[Special:mypage|Jeff Patz]]

==The Main Idea==

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field.

===A Mathematical Model===

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

<math>\Phi_E = \frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

<math>\Phi_E = EAcos(\theta)</math>

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

==Examples==

===One Surface and Uniform Electric Field===

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

[[File:GL-Ex-1.S.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>, fill out the known values, which in the case is all values needed.

<math>\Phi_E = (400)(\pi0.05^2)cos(35)</math>

<math>\Phi_E = 2.573</math> Volt Meters

===Multiple Surfaces and Uniform Electric field===

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

[[File:Ex-2-2.JPG]]

Using the simplified version of Gauss's Law because the electric field is uniform: <math>\Phi_E = EAcos(\theta)</math>

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

<math>\Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)]</math>

<math>\Phi_E = 0</math>

===Non-Uniform Electric Field===

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

[[File:GL-Ex-3.JPG]]

For this problem we can utilize the equality of <math>\frac{Q}{\varepsilon_0} = \oint_C E\bullet dA</math>

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

<math>\frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2</math>

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

<math>q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3}</math>

<math>q = Q\frac{r_1^3}{R^3}</math>

<math>E = Q\frac{r_1}{4\pi R^3}</math>

==Connectedness==
Most of the applications of Gauss's Law are finding the electrical fields of different shapes.
This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

==History==

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

== See also ==

===Further reading===

===External links===

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

==References==

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf

[[Category:Which Category did you place this in?]]

Compression or Normal Force

2016-04-17T16:41:58Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math> where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 1 and block 2 and find the parallel component of the net force, <math>F_n</math> for block 2.

====Answer====
The free body is as shown below:

[[File:Free-body-diagrams.jpg|center|Example 3]]

Looking at the definition of the normal force, this type of contact force is always perpendicular to the surface at which it touches. Because of this definition, there is no parallel component of the net force, only a perpendicular component is present. There is however, a parallel component to the gravitational force on the block. Yet in this situation, the normal force is equal to the perpendicular component of the gravitational force, <math> F_n = M_2gcos(\alpha)</math>. The image for Example 2 can be used as a reference.

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

http://www.texample.net/tikz/examples/free-body-diagrams/

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T16:30:15Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math> where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 1 and block 2 and find the parallel component of the net force, <math>F_n</math> for block 2.

====Answer====
The free body is as shown below:

[[File:Free-body-diagrams.jpg|center|Example 3]]

Since the question is asking for the parallel component of the net force, it would be helpful to draw a triangle that depicts the trigonometric quantities <math>sin(\alpha)</math> and <math> cos(\alpha)</math> (see Example 2 image for reference). The component that is parallel to the surface on which block 2 is moving on is <math>M_2gsin(\alpha)</math>. This shows that the parallel component of the net force is, <math> F_n = M_2gsin(\alpha)</math>.

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

http://www.texample.net/tikz/examples/free-body-diagrams/

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T16:29:39Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math> where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 1 and block 2 and find the parallel component of the net force, <math>F_n</math> for block 2.

====Answer====
The free body is as shown below:

[[File:Free-body-diagrams.jpg|center|Example 3]]

Since the question is asking for the parallel component of the net force, it would be helpful to draw a triangle that depicts the trigonometric quantities <math>sin(\alpha)</math> and <math> cos(\alpha)</math> (see Example 2 image for reference). The component that is parallel to the surface on which block 2 is moving on is <math>M_2gsin(\alpha)</math>. This shows that the parallel component of the net force is, <math> F_n = M_2gsin(\alpha)</math>.

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T16:27:54Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math> where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 1 and block 2 and find the parallel component of the net force, <math>F_n</math> for block 2.

====Answer====
The free body is as shown below:

[[File:Free-body-diagrams.jpg|center|Example 3]]

Since the question is asking for the parallel component of the net force, it would be helpful to draw a triangle that depicts the trigonometric quantities <math>sin(\alpha)</math> and <math> cos(\alpha)<math/> (see Example 2 image for reference). The component that is parallel to the surface on which block 2 is moving on is <math>M_2gsin(\alpha)</math>. This shows that the parallel component of the net force, <math> F_n</math>, is equal to <math>M_2gsin(\alpha)</math>.

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

File:Free-body-diagrams.jpg

2016-04-17T16:19:18Z

Patrycjakot:

File:NormalForceQuestion3.png

2016-04-17T16:16:37Z

Patrycjakot:

Compression or Normal Force

2016-04-17T16:15:03Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math> where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 1 and block 2 and find the parallel component of the net force, <math>F_n</math> for block 2.

====Answer====
The free body is as shown below:

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T16:12:36Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

===Difficult===
In a pulley system, block 1 of mass <math>M_1</math> is hanging over the edge of an inclined plane at an angle, <math>(\alpha)</math>where it is attached to block 2 of mass <math>M_2</math>. Block 2 is sliding upwards on the ramp as block 1 is slowly moving downwards. Draw a free body diagram for block 2 and find the parallel component of the net force, <math>F_n</math>.

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:54:16Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:53:26Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

'''Answer:'''
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

''Answer:''
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:52:16Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:51:10Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F_n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:50:58Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math>F-n</math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface. Understanding information about the normal force on an object can also help to finding the magnitude of other forces present in a system, such as the force of friction.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:48:48Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface.

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object situated on an angle is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 m/s^2)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
<math>m = 14 kg</math>
<math>g = 9.8 m</math>
<math>(\theta) = 48</math>

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:46:43Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface.

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = (\mu)F_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 N/kg)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
m = 14 kg
g = 9.8 N/kg
theta = 48

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:46:26Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface.

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = \muF_n </math>
:<math>= (\mu)mg \cos(\theta))</math>
:<math> (\mu) </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1),a7 kilogram block rests on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is <math>9.8 m/s^2</math>.

====Answer====
Using the provided free body diagram, the net force acting on the block is equal to the normal force minus the gravitational force. Since the object is at rest, the net force is equal to zero. This allows the normal force to be set equal to the gravitational force generate the formula <math>F_n = mg</math>. Plugging in the given values into the equation will yield the correct results.
Given:
<math>m = 7kg </math>
<math>g = 9.8 m/s^2</math>
<math> F_n = mg</math>
<math>F_n = (7 kg)(9.8 N/kg)</math>
<math> = 68.6 N</math>

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
m = 14 kg
g = 9.8 N/kg
theta = 48

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's thrid laws of motion extends to today's uses due to the high prevalence of the normal force in real world situations, such as landing a rocket on the moon.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-17T00:07:49Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move along surfaces with various angles. This results in the normal force being present in both the direction parallel and perpendicular to the surface.

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force is dependent on the positioning of the system, however, the most commonly used formula for normal force is represented below:
:<math>F_n = mg </math>
:<math>F_n </math> is the normal force of the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = \muF_n </math>
:<math>= \mumg \cos(\theta))</math>
:<math> \mu </math> = coefficient of static or kinetic friction
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>

==Examples==

The examples below represent different situations in which the net force changes due to the placement of the system.

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1), there is a 7 kilogram block resting on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is 9.8 N/kg.

====Answer====
Fn = mgcos(theta)
In this example, theta will be 90 since the normal force is perpendicular to the surface that it is resting on, thus 90 degrees.
Now we can plug into our formula:
m = 7kg
g = 9.8 N/kg
<math> F_n = mg</math>
Fn = (7 kg)(9.8 N/kg)
= 68.6 N

===Middling===
[[File: InclinedPlane.jpg|200px|right|Example 2]]

A 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline using the accompanying image to the right.
The force of gravity is <math>9.8 m/S^2</math>.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:

Given:
m = 14 kg
g = 9.8 N/kg
theta = 48

By analyzing the image, it is seen that the cosine trig identity correlates to the direction in which the normal force is perpendicular to the surface. Multiplying the mass of the block and force of gravity by the cosine of the stated angle, the perpendicular normal force will be found.

<math>F_n = mgcos(\theta)</math>
<math>F_n = (14 kg)(9.8 m/s^2)cos(48)</math>
<math> = 91.805 N</math>

==Connectedness==
This force is very crucial to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why objects do not keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when drawing free body diagrams, such as for construction and architectural purposes.

==History==

[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's three laws of motion extend to today's uses due to the important of understanding how much

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

File:InclinedPlane.jpg

2016-04-16T23:28:12Z

Patrycjakot:

File:FnInclinedPlane.png

2016-04-16T23:25:39Z

Patrycjakot:

Compression or Normal Force

2016-04-16T23:25:25Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move at various angles, therefore resulting in the normal force to

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force at an angle is:
:<math>F_n = mg \cos(\theta) </math>
:<math>F_n </math> is the normal force present in the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>
:<math>\theta</math> is the angle of the incline (

The normal force of an object (such as a block) on an inclined plane must be separated into parallel and perpendicular components. This results in two separate equations for the normal force on the object.

:Perpendicular component of normal force:
:<math>F_n = mgcos(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

:Parallel component of normal force (with no friction present):
:<math>F_n = mgsin(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = \mu_k F_n </math>
:<math>= \mu_k (mg \cos(\theta))</math>

==Examples==

Time for a few examples:

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1), there is a 7 kilogram block resting on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is 9.8 N/kg.

====Answer====
Fn = mgcos(theta)
In this example, theta will be 90 since the normal force is perpendicular to the surface that it is resting on, thus 90 degrees.
Now we can plug into our formula:
m = 7kg
g = 9.8 N/kg
theta = 90

Fn = (7 kg)(9.8 N/kg)cos(90)
= 68.6 N

===Middling===
[[File: SecondExampleNorm.gif|right| Example 2]]

In Example 2 (the figure to the right), the 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline.
The force of gravity is 9.8 N/kg.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:
Given:
m = 14 kg
g = 9.8 N/kg
theta = 48
Using trigonometry, it is seen that the cosine trig identity correlates to
Fn = mgcos(theta)
Fn = (14 kg)(9.8 N/kg)cos(48)
= 91.805 N

==Connectedness==
This force is pretty important to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why we don't keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when creating any free body diagram or any construction.

==History==
[[Media:]]
[[File:SirNewton16.jpg|right|thumb|Sir Isaac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Isaac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's three laws of motion extend to today's uses due to the important of understanding how much

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

https://en.wikipedia.org/wiki/Friction

[[Category: Contact Interactions]]

File:SirNewton16.jpg

2016-04-16T23:22:47Z

Patrycjakot: Patrycjakot uploaded a new version of "File:SirNewton16.jpg"

File:SirNewton16.jpg

2016-04-16T23:20:47Z

Patrycjakot:

Compression or Normal Force

2016-04-16T23:15:59Z

Patrycjakot:

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force is often equal to the weight of the object, however, objects can move at various angles, therefore resulting in the normal force to

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force at an angle is:
:<math>F_n = mg \cos(\theta) </math>
:<math>F_n </math> is the normal force present in the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>
:<math>\theta</math> is the angle of the incline (

The normal force of an object (such as a block) on an inclined plane must be separated into parallel and perpendicular components. This results in two separate equations for the normal force on the object.

:Perpendicular component of normal force:
:<math>F_n = mgcos(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

:Parallel component of normal force (with no friction present):
:<math>F_n = mgsin(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = \mu_k F_n </math>
:<math>= \mu_k (mg \cos(\theta))</math>

==Examples==

Time for a few examples:

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1), there is a 7 kilogram block resting on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is 9.8 N/kg.

====Answer====
Fn = mgcos(theta)
In this example, theta will be 90 since the normal force is perpendicular to the surface that it is resting on, thus 90 degrees.
Now we can plug into our formula:
m = 7kg
g = 9.8 N/kg
theta = 90

Fn = (7 kg)(9.8 N/kg)cos(90)
= 68.6 N

===Middling===
[[File: SecondExampleNorm.gif|right| Example 2]]

In Example 2 (the figure to the right), the 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline.
The force of gravity is 9.8 N/kg.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:
Given:
m = 14 kg
g = 9.8 N/kg
theta = 48
Using trigonometry, it is seen that the cosine trig identity correlates to
Fn = mgcos(theta)
Fn = (14 kg)(9.8 N/kg)cos(48)
= 91.805 N

==Connectedness==
This force is pretty important to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why we don't keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when creating any free body diagram or any construction.

==History==
[[Media:]]
[[File:Newton12.jpg|right|thumb|Sir Issac Newton]]

The normal force a direct application of Newton's Third Law of Motion. Sir Issac Newton was a famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed many of his theories about physics and stated his three laws of motion. The application of Newton's three laws of motion extend to today's uses due to the important of understanding how much

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:

https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:

https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

https://en.wikipedia.org/wiki/Philosophiæ_Naturalis_Principia_Mathematica

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

[[Category: Contact Interactions]]

Compression or Normal Force

2016-04-16T21:45:13Z

Patrycjakot: I am correcting the equations and defining variables.

Claimed by Hemanth Koralla

[[File:Normal force.png|right|thumb|''FN'' represents the '''normal force''']]

==Compression of Normal Force==
The compression or also commonly known as the normal force, <math> F_n\ </math>, is a simple fundamental concept that must be understood before attempting any contact force problem. First, it is important to understand that the normal force is NOT a kind of fundamental force such as the electric or gravitational force. It is just a force used to describe the interaction between atoms. As hinted by the name, this force simply points in the perpendicular or "normal" direction to the surface(s) that it is in contact with. The magnitude of this normal force often equal to the weight of the object, however,

Understanding information about the normal force on an object can also help . The approximate force of friction is the static friction coefficient multiplied by this normal force.

==== Equations ====
The formula for calculating the normal force at an angle is:
:<math>F_n = mg \cos(\theta) </math>
:<math>F_n </math> is the normal force present in the system
:<math>m</math> is the mass of the system
:<math>g</math> is a constant value for the force due to gravity and is equivalent to 9.8 <math>m/s^2</math>
:<math>\theta</math> is the angle of the incline (

The normal force of an object (such as a block) on an inclined plane must be separated into parallel and perpendicular components. This results in two separate equations for the normal force on the object.

:Perpendicular component of normal force:
:<math>F_n = mgcos(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

:Parallel component of normal force (with no friction present):
:<math>F_n = mgsin(\theta)</math>
:<math>m</math> is the mass of the object
:<math>g</math> is the force of gravity and is equivalent to <math>9.8 m/s^2</math>
:<math>\theta</math> is the angle of the incline

The formula for calculating the force of friction using this normal force when an object is on an inclined plane is:
:<math>F_f (static) = \mu_k F_n </math>
:<math>= \mu_k (mg \cos(\theta))</math>

==Examples==

Time for a few examples:

===Simple===

[[File: ExampleNorm1.png|left|thumb|Example 1]]

On the image to the left (Example 1), there is a 7 kilogram block resting on top a flat table. Find the normal force, <math> F_n\ </math>, that is being exerted on the block.
The force of gravity is 9.8 N/kg.

====Answer====
<math>F_n = mgcos(\theta) </math>
In this example, theta will be 90 since the normal force is perpendicular to the surface that it is resting on, thus 90 degrees.
Now we can plug into our formula:
m = 7kg
g = 9.8 N/kg
theta = 90

Fn = (7 kg)(9.8 N/kg)cos(90)
= 68.6 N

===Middling===
[[File: SecondExampleNorm.gif|right| Example 2]]

In Example 2 (the figure to the right), the 14 kg block is lying stationary on a plane that is inclined 48 degrees. Find the normal force perpendicular to the block on the incline.
The force of gravity is 9.8 N/kg.

====Answer====
Because this problem asks for the normal force on an inclined plane, it is important to break up the normal force into components:
Given:
m = 14 kg
g = 9.8 N/kg
theta = 48
Using trigonometry, it is seen that the cosine trig identity correlates to
Fn = mgcos(theta)
Fn = (14 kg)(9.8 N/kg)cos(48)
= 91.805 N

==Connectedness==
This force is pretty important to understand because it explains the fundamental reason for why gravity doesn't keep pulling us down and explains this not-so-apparent force that is acting on an object. It helps explain why we don't keep falling into the Earth. The application of the normal force can be seen in many industrial situations. It must be taken into account when creating any free body diagram or any construction.

==History==

[[File:Newton12.jpg|right|thumb|Sir Issac Newton]]

The normal force a direct application of Newton's Third Law of Motion. As you probably know, Sir Issac Newton was famous scientist from England who roamed the Earth from 1643 to 1727. His work in math and physics set the stage for many of the principles and theories that we have today. His main work, the Philosophiae Naturalis Principia Mathematic, discussed the many of his theories about physics and stated his three laws of motion. You can find even more information about him under the Notable Scientists tab.

== See also ==

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Types-of-Forces

https://www.physics.uoguelph.ca/tutorials/fbd/FBD3.htm

==References==
Information:
https://en.wikipedia.org/wiki/Normal_force

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html

Images:
https://en.wikipedia.org/wiki/Normal_force#/media/File:Friction_relative_to_normal_force_(diagram).png

http://www.fromoldbooks.org/Aubrey-HistoryOfEngland-Vol3/pages/vol3-401-Sir-Isaac-Newton/

http://www.sparknotes.com/physics/dynamics/newtonapplications/section1.rhtml

[[Category: Contact Interactions]]