Physics Book - User contributions [en]

Torque vs Work

2016-11-27T23:40:29Z

Qfaber3:

[[File:Wvt003.jpg|300px|thumb|right]]

When it comes to our universe, there still remains much that we do not understand. That we we do know and understand, we measure in relation to other ideas or subjects, in terms of abstracts we call units. In modern physics, and most mathematical applications we relate and measure matter and interactions in terms of these units. However, two such descriptions having the same units does necessarily connotate a representation of the same idea or measurement. Here, we will examine one such occurrence in modern physics, specifically the difference between torque and work when both are expressed as Newton*meters.

==The Newton-Meter --> Vector vs Scaler==

In modern physics, there a two concepts that, while describing different ideas, happen to utilize the same units, these units being:

<math> Newton * meters </math>
<math> (N * m) </math>

===Mathematical Model: The Concepts===

[[File:Wvt002.jpg|250px|thumb|left|Torque on an Object]]

The first of these concepts is '''torque''', which is a measurement of the moment that a force causes in a system. A moment is also known as a system's tendency of a force to cause rotation.

<math> \vec{\tau} = \vec{r}_cm X \vec{F} </math>

[[File:Wvt001.jpg|300px|thumb|right|Work on an Object]]

<i>where '''<math>\tau</math>''' is the vector cross product of the force vector and position vector relative to the center of mass</i>

The second concept is '''work''', which is the measurement of a force acting over a distance on a given system.

<math> W = \vec{F}*\vec{r} </math>

<i>where '''W''' is the scaler dot product of the force vector and change in position vector</i>

Both of these concepts are measured using the units of <math> Newton * meters </math> , however torque is a vector value while work is a scaler value. This is a key point to keep in mind.

Previously, in Physics, when we addressed the energy principle and energy in general, we describe the measurement of energy and change in energy using the units:

<math> Joules </math>
<math> (J) </math>

As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.

<math> W = ∆K = K_{final} - K_{initial} </math>

<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>

So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a scaler measurement, while torque is a vector representation. They must be measure individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.

===Nuances in Work===

According to it's definition, work is a force acting over a given distance. So what about forces acting on an object that does not move? Let's look at an example. Take this block sitting on the earth and being acted on by a man who tries to lift it.

[[File:Wvt004.jpg|300px|middle]]

Let the mass of the block be <math> 5 kg </math> and the force of the man pulling up be <math> 20 N </math>.

We know: <math> F = mg </math> <i>where '''F''' is the force of gravity, '''m''' is the mass of the object, and '''g''' is the acceleration due to gravity</i>

Force due to man: <math> F_{man} = 20 N </math>

Force due to gravity: <math> F_{grav} = (5 kg) * (-9.8 m/(s^{2}) = -49 N </math>

Net force: <math> F_{net} = 20 N + (-49 N) = -29 N </math>

However, we know that the block will not move through the earth due to the force of the earth acting on it, and so the block will stay in place. Since the block does not move there is no change in position. Applying this to our formula for work:

Work by man: <math> W_{man} = \vec{F}_{man} * \vec{r}_{block} = <0, 20, 0> N * <0, 0, 0> m = (0*0)Nm + (20*0)Nm + (0*0)Nm = 0 Nm </math>

So even though the man applied a force to the block, no work was done because the block did not move. It a very important to realize that it doesn't matter how large the force applied to an object is, if the object does not move, no work is done.

In this situation: <math> W = ∆K = 0Nm </math>

===Nuances in Torque===

Like work, torque also has some interesting situations that are important to take note of. Consider the following example where a wheel is being acted on by a force that is directly toward the center of mass of the wheel.

[[File:Wvt005.jpg|300px|middle]]

Let the radius of the wheel be <math> 2 m </math> and the force applied be <math> 15 N </math>.

Using these values, let's take the cross product of position cross force to find the torque cause by the force on the wheel.

Torque: <math> \vec{\tau} = \vec{r}_{cm} X \vec{F} = <2, 0, 0> m X <-15, 0, 0> N </math>

<math> \vec{\tau} = (2 m * -15 N)(iXi) + (2 m * 0 N)(iXj) + (2 m * 0 N)(iXk) + (0 m * -15 N)(jXi) + (0 m * 0 N)(jXj) + (0 m * 0 N)(jXk) + (0 m * 0 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>

<math> \vec{\tau} = (-30 Nm)(0) + (0 Nm)(k) + (0 Nm)(-j) + (0 Nm)(-k) + (0 Nm)(0) + (0 Nm)(i) + (0 Nm)(j) + (0 Nm)(-i) + (0 Nm)(0) </math>

<math> \vec{\tau} = <0, 0, 0> Nm </math>

So, even though a force was applied at a distance relative to the center of mass of the wheel, there was no net torque because the force was acting straight through the center of mass of the wheel. This is important to remember as any force acting through an objects center of mass may be disregarded as having any effect on the torque and therefore does not change the angular momentum of the object.

==Example Problems==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple Torque===

If the center of mass of the spool of thread is at the origin, determine the torque caused by the force as shown below.

[[File:Wvt006.jpg|300px|middle]]

Solution:

<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>

<math> \vec{\tau} = <0, 4, 0> m X <-2, 0, 0> N </math>

<math> \vec{\tau} = (0 m * -2 N)(iXi) + (0 m * 0 N)(iXj) + (0 m * 0 N)(iXk) + (4 m * -2 N)(jXi) + (4 m * 0 N)(jXj) + (4 m * 0 N)(jXk) + (0 m * -2 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>

<math> \vec{\tau} = (0 Nm)(0) + (0 Nm)(k) + (0 Nm)(-j) + (-8 Nm)(-k) + (0 Nm)(0) + (0 Nm)(i) + (0 Nm)(j) + (0 Nm)(-i) + (0 Nm)(0) </math>

<math> \vec{\tau} = <0, 0, 8> N*m </math>

===Middling Work===

In the system below, determine the net work that is accomplished by the two forces.

[[File:Wvt007.jpg|300px|middle]]

Solution:

<math> W = \vec{F}*\vec{r} </math>

<math> W_{1} = <30, 0, 0> N * <15, 0, 0> m </math>

<math> W_{1} = (30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>

<math> W_{1} = 450 Nm + 0 Nm + 0 Nm </math>

<math> W_{1} = 450 Nm </math>

<math> W_{2} = <-30, 0, 0> N * <15, 0, 0> m </math>

<math> W_{2} = (-30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>

<math> W_{2} = -450 Nm + 0 Nm + 0 Nm </math>

<math> W_{2} = -450 Nm </math>

<math> W_{net} = W_{1} + W_{2} </math>

<math> W_{net} = 450 Nm - 450 Nm </math>

<math> W_{net} = 0 N*m </math>

===Difficult Torque===

In the complex system below, determine the given force is applied at point '''A'''. Determine the net torque about the origin due to the force.

[[File:Wvt008.jpg|300px|middle]]

Solution:

<math> \vec{r} = <12, -2, 6> m </math>

<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>

<math> \vec{\tau} = <12, -2, 6> m X <2, 1, 7> N </math>

<math> \vec{\tau} = (12 m * 2 N)(iXi) + (12 m * 1 N)(iXj) + (12 m * 7 N)(iXk) + (-2 m * 2 N)(jXi) + (-2 m * 1 N)(jXj) + (-2 m * 7 N)(jXk) + (6 m * 2 N)(kXi) + (6 m * 1 N)(kXj) + (6 m * 7 N)(kXk) </math>

<math> \vec{\tau} = (24 Nm)(0) + (12 Nm)(k) + (84 Nm)(-j) + (-4 Nm)(-k) + (-2 Nm)(0) + (-14 Nm)(i) + (12 Nm)(j) + (6 Nm)(-i) + (42 Nm)(0) </math>

<math> \vec{\tau} = <-20, -72, 16> N*m </math>

==Connectedness==

===Connected to My Own Interests?===

While I am more interested in digital and electronic designs, I love to tinker and create with mechanical devices. In order to make a device efficient and able to function without breaking, you must understand how work is done and the strain that torque can put on a device.

===Connected to Electrical Engineering?===

While torque and work are not integral parts of electrical engineering, it is very important to understand the two concepts because they do have the potential to alter the design of something I might construct. In this day and age, everything is electronic and most modern tools even use digital instructions. It's important when designing a mechanical tool to understand torque in order to determine the best avenue through which to run wire and to know how much extra give to allow to wire to move if the tools workload necessitates movement.

===Connected to Industry===

The industrial applications for both torque and work are enormous. From power tools, to vehicles, to simple machines. If one can do more wore with less energy, or apply greater torque with smaller forces, it greatly increases the workload that an individual can accomplish. So, yes, torque and work are quite important in industry.

==Origin==

===Torque===
The idea of torque first originated with Aristotle and his work with levers.

===Work===
Energy was also initially a notion put forth by Aristotle, but the concept of work was something later derived by several scientists and mathematicians, including Joule and Carnot.

== See also ==

===Further reading===

[http://www.physicsbook.gatech.edu/Work Wiki - Work]<div></div>
[http://www.physicsbook.gatech.edu/Torque Wiki - Torque]<div></div>
[http://www.physicsbook.gatech.edu/The_Energy_Principle Wiki - Energy Principle]<div></div>

===External links===
[https://www.youtube.com/watch?v=MHtAK1rMa1Y Explanation of Torque]<div></div>
[http://science360.gov/obj/video/9112c778-48e4-4b75-a09b-2f2d2404da12/science-nfl-football-torque Torque in Football - Cool!]<div></div>
[http://torquenitup.weebly.com/torque-problems.html Torque Example Problems]<div></div>

==References==

[http://www.wiley.com//college/sc/chabay/ Matter & Interactions 4th Edition]<div></div>
[http://physics.stackexchange.com/ Physics Stack Exchange]<div></div>
[https://en.wikipedia.org/wiki/Torque Wikipedia Torque]<div></div>

jderemer3

[[Category: Angular Momentum]]

Gyroscopes

2016-11-27T23:40:05Z

Qfaber3:

'''claimed by Quincy Faber Fall 2016'''

An explanation by Ansley Marks

A [https://en.wikipedia.org/wiki/Gyroscope gyroscope] is a device containing a wheel or disk that is free to rotate about its own axis independent of a change in direction of the axis itself. Since the spinning wheel persists in maintaining its plane of rotation, a [https://www.youtube.com/watch?v=ty9QSiVC2g0 gyroscopic effect] can be observed.

[[File: gyro.gif]]

==The Main Idea==

Although insignificant looking and seemingly uninteresting when still, gyroscopes become a fascinating device when in motion and can be explained using the angular momentum principle. Gyroscopes come in all different forms with varying parts. The main component of a gyroscope is a spinning wheel or a disk mounted on an axle. Typically gyroscopes contain a suspended rotor inside three rings called gimbals. In order to ensure that little torque is applied to the inside rotor, the gimbals are mounted on high quality bearing surfaces, allowing free movement of the spinning wheel in the middle. These types of gyroscopes with multiple gimbals are useful for stabilization because the wheels can change direction without affecting the inner rotor. If the spinning axle of a gyroscope is placed on a support, then a complex motion can be observed. The motion of a gyroscope will be modeled and explained further on in this page.

===A Mathematical Model===

When the spinning axis of a gyroscope is placed on a support, a gyroscopic effect is observed. The gyroscope bobs up and down--nutation--and rotates about the support--precession. For the sake of simplifying the mathematical equations for a gyroscope's motion, [http://dictionary.reference.com/browse/nutation nutation] (the upwards and downwards movement of the rotor) will be ignored. We will only look at the [http://dictionary.reference.com/browse/precession precession] motion of the gyroscope.

[[File:gyropic1.png|200px|thumb|left|A gyroscope processing in the x,z plane with the y-axis positioned upwards along the vertical support.]]

To start off with, the gyroscope's rotor rotates about its own axis with an angular velocity of ω and has a moment of inertia ''I''. Thus, the rotational angular momentum of the rotor can be modeled as:

'''Lrot,r = Iω'''

Where the rotational angular momentum points horizontal to the rotor.

The Lrot,r will always change direction as the rotor rotates about the support. The rotor processes about the support with an angular velocity Ω, which is constant in magnitude and direction.

If Ω is known, then the velocity of the center of mass of the rotor device can be derived using the following relationship:

'''Ω = ''V''cm/''r'''''

Where ''r'' is equal to the distance from the support to the center of mass of the rotor device. The linear momentum of the gyroscope is then Ω''P''. [[File:figure11.611.png|200px|thumb|left|A view of the gyroscope from the side with all the forces labeled.]]

Since the rotor is processing about the support, there must be a perpendicular force ''f'' exerted by the support such that Ω''P'' = ''f'', where ''P'' is equal to ''M(Ωr)''. Thus, ''f'' = ''Mr''<math>Ω^2</math>.

There is also a translational angular momentum of the rotor processing about the support. This can be modeled by finding the magnitude of the position vector crossed with the momentum.

'''Lsupport = |''R'' x ''P'' | '''

Since the direction of the rotational angular momentum of the rotor around the support is constantly changing direction, the rate of change of the rotational angular momentum can be written as:

'''LrotΩ'''

Thus the only remaining element that is needed to complete the Angular Momentum Principle is the torque. The torque is equal to the distance from the support to the center of mass of the rotor, ''r'', multiplied by the force exerted, which is the mass times gravity. Therefore, since the change in rotational angular momentum is ''LrotΩ'', that must be equal to ''τCM''. By setting the two equations equal to each other, the angular momentum can be isolated to one side. This yields the following result:

'''Ω = τCM/Lrot = ''r''Mg/''I''ω '''

==Real World Examples==

===Magnetic Resonance Imaging===

A good analogy for the way that a Magnetic Resonance Imaging (MRI) works is a gyroscope. To start off with a little background, the way that an MRI works is that all the hydrogen atoms in your body are aligned by using strong magnetic fields. Once these hydrogen atoms are aligned, similar to how a compass's needle is aligned, radio waves can be sent into the body and signals are created from the way the photons emit the radio waves.

Identical to gyroscopes, the hydrogen nucleus rotates about its own axis at a particular frequency. The strength and direction of the magnetic field can effect the direction and angular speed of these rotating protons in the nuclei. By controlling the direction and rotation speed, the location of the hydrogen nucleus can be deduced and thus helping the process of creating images.

===Aviation===

Gyroscopes offer two functions in aircraft. The first is rigidity in space, which means that the gyroscope will resist any attempt to change the direction of the axis. This is useful when the plane makes turns, throwing the plane off its natural horizontal. The gyroscope acts as a "fake horizon", which orientates the plane back to its natural position. The technical term for this gyroscope contraption is an attitude indicator.

The second function of gyroscopes in planes is precession. In this context, this means that any perpendicular force applied to a gyroscope's axis of rotation will manifest itself 90° further along the axis of rotation. This property is useful because the amount of banking before a turn can be determined. If the gyroscope was oriented with the longitudinal axis of the plane, then only the rate of turn could be determined instead of the amount of bank on the aircraft.

==Connectedness==
This topic is interesting because gyroscopes have held the fascination of pretty much anyone that has ever seen one in motion including myself. Although the explanation that I gave was a simplified version of a gyroscope which only processes and doesn't nutate, there are many other complex mathematical models of the complicated motion of gyroscopes. Many papers and even books have been written on the subject of gyroscopes, and they have baffled nobel prize winners such as Niels Bohr and famous physicists alike. Gyroscopes are connected to my major because they are huge in industrial manufacturing of numerous materials. We use some sort of gyroscope in our everyday lives from cars to airplanes and other mechanical equipments.

==History==

Gyroscopes have been around for nearly 200 years. The first person to discover the gyroscope was Johann Bohnenberger in 1817 at the University of Tubingen. However, Bohnenberger was not credited with the discovery of the gyroscope. The French scientist Jean Bernard Leon Foucault (1826-1864) coined the term "gyroscope" and ended up with being credited for the discovery of a gyroscope. Thanks to his experiments with the gyroscope, they started to become mainstream and studied by many other physicists. In the early 20th century, gyroscopes were first used in boats and eventually in aircraft. Gyroscopes have been modified and tweaked to suit many purposes that are widely used today mainly as stabilizers.

== See also ==

===Further reading===

Compass and Gyroscope: Integrating Science and Politics for the Environment

Mathematical model for gyroscope effects: http://scitation.aip.org/content/aip/proceeding/aipcp/10.1063/1.4915651

YouTube video on gyroscope procession: https://www.youtube.com/watch?v=ty9QSiVC2g0

Wikipedia: https://en.wikipedia.org/wiki/Gyroscope

===External links===

https://www.youtube.com/watch?v=ty9QSiVC2g0

http://dictionary.reference.com/browse/precession

http://dictionary.reference.com/browse/nutation

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

==References==

Oxford Dictionaries: http://www.oxforddictionaries.com/us/definition/american_english/gyroscope

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

Wikipedia: https://en.wikipedia.org/wiki/Gyroscope

Gyroscope History: http://www.gyroscopes.org/history.asp

Science Learning: http://sciencelearn.org.nz/Contexts/See-through-Body/Sci-Media/Video/So-how-does-MRI-work

Quora: https://www.quora.com/What-the-function-of-gyroscopes-in-airplane

[[Category: Angular Momentum]]

Torque vs Work

2016-11-23T18:41:38Z

Qfaber3:

[[File:Wvt003.jpg|300px|thumb|right]]

Claimed by Quincy Faber (11/23/16)

When it comes to our universe, there still remains much that we do not understand. That we we do know and understand, we measure in relation to other ideas or subjects, in terms of abstracts we call units. In modern physics, and most mathematical applications we relate and measure matter and interactions in terms of these units. However, two such descriptions having the same units does necessarily connotate a representation of the same idea or measurement. Here, we will examine one such occurrence in modern physics, specifically the difference between torque and work when both are expressed as Newton*meters.

==The Newton-Meter --> Vector vs Scaler==

In modern physics, there a two concepts that, while describing different ideas, happen to utilize the same units, these units being:

<math> Newton * meters </math>
<math> (N * m) </math>

===Mathematical Model: The Concepts===

[[File:Wvt002.jpg|250px|thumb|left|Torque on an Object]]

The first of these concepts is '''torque''', which is a measurement of the moment that a force causes in a system. A moment is also known as a system's tendency of a force to cause rotation.

<math> \vec{\tau} = \vec{r}_cm X \vec{F} </math>

[[File:Wvt001.jpg|300px|thumb|right|Work on an Object]]

<i>where '''<math>\tau</math>''' is the vector cross product of the force vector and position vector relative to the center of mass</i>

The second concept is '''work''', which is the measurement of a force acting over a distance on a given system.

<math> W = \vec{F}*\vec{r} </math>

<i>where '''W''' is the scaler dot product of the force vector and change in position vector</i>

Both of these concepts are measured using the units of <math> Newton * meters </math> , however torque is a vector value while work is a scaler value. This is a key point to keep in mind.

Previously, in Physics, when we addressed the energy principle and energy in general, we describe the measurement of energy and change in energy using the units:

<math> Joules </math>
<math> (J) </math>

As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.

<math> W = ∆K = K_{final} - K_{initial} </math>

<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>

So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a scaler measurement, while torque is a vector representation. They must be measure individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.

===Nuances in Work===

According to it's definition, work is a force acting over a given distance. So what about forces acting on an object that does not move? Let's look at an example. Take this block sitting on the earth and being acted on by a man who tries to lift it.

[[File:Wvt004.jpg|300px|middle]]

Let the mass of the block be <math> 5 kg </math> and the force of the man pulling up be <math> 20 N </math>.

We know: <math> F = mg </math> <i>where '''F''' is the force of gravity, '''m''' is the mass of the object, and '''g''' is the acceleration due to gravity</i>

Force due to man: <math> F_{man} = 20 N </math>

Force due to gravity: <math> F_{grav} = (5 kg) * (-9.8 m/(s^{2}) = -49 N </math>

Net force: <math> F_{net} = 20 N + (-49 N) = -29 N </math>

However, we know that the block will not move through the earth due to the force of the earth acting on it, and so the block will stay in place. Since the block does not move there is no change in position. Applying this to our formula for work:

Work by man: <math> W_{man} = \vec{F}_{man} * \vec{r}_{block} = <0, 20, 0> N * <0, 0, 0> m = (0*0)Nm + (20*0)Nm + (0*0)Nm = 0 Nm </math>

So even though the man applied a force to the block, no work was done because the block did not move. It a very important to realize that it doesn't matter how large the force applied to an object is, if the object does not move, no work is done.

In this situation: <math> W = ∆K = 0Nm </math>

===Nuances in Torque===

Like work, torque also has some interesting situations that are important to take note of. Consider the following example where a wheel is being acted on by a force that is directly toward the center of mass of the wheel.

[[File:Wvt005.jpg|300px|middle]]

Let the radius of the wheel be <math> 2 m </math> and the force applied be <math> 15 N </math>.

Using these values, let's take the cross product of position cross force to find the torque cause by the force on the wheel.

Torque: <math> \vec{\tau} = \vec{r}_{cm} X \vec{F} = <2, 0, 0> m X <-15, 0, 0> N </math>

<math> \vec{\tau} = (2 m * -15 N)(iXi) + (2 m * 0 N)(iXj) + (2 m * 0 N)(iXk) + (0 m * -15 N)(jXi) + (0 m * 0 N)(jXj) + (0 m * 0 N)(jXk) + (0 m * 0 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>

<math> \vec{\tau} = (-30 Nm)(0) + (0 Nm)(k) + (0 Nm)(-j) + (0 Nm)(-k) + (0 Nm)(0) + (0 Nm)(i) + (0 Nm)(j) + (0 Nm)(-i) + (0 Nm)(0) </math>

<math> \vec{\tau} = <0, 0, 0> Nm </math>

So, even though a force was applied at a distance relative to the center of mass of the wheel, there was no net torque because the force was acting straight through the center of mass of the wheel. This is important to remember as any force acting through an objects center of mass may be disregarded as having any effect on the torque and therefore does not change the angular momentum of the object.

==Example Problems==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple Torque===

If the center of mass of the spool of thread is at the origin, determine the torque caused by the force as shown below.

[[File:Wvt006.jpg|300px|middle]]

Solution:

<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>

<math> \vec{\tau} = <0, 4, 0> m X <-2, 0, 0> N </math>

<math> \vec{\tau} = (0 m * -2 N)(iXi) + (0 m * 0 N)(iXj) + (0 m * 0 N)(iXk) + (4 m * -2 N)(jXi) + (4 m * 0 N)(jXj) + (4 m * 0 N)(jXk) + (0 m * -2 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>

<math> \vec{\tau} = (0 Nm)(0) + (0 Nm)(k) + (0 Nm)(-j) + (-8 Nm)(-k) + (0 Nm)(0) + (0 Nm)(i) + (0 Nm)(j) + (0 Nm)(-i) + (0 Nm)(0) </math>

<math> \vec{\tau} = <0, 0, 8> N*m </math>

===Middling Work===

In the system below, determine the net work that is accomplished by the two forces.

[[File:Wvt007.jpg|300px|middle]]

Solution:

<math> W = \vec{F}*\vec{r} </math>

<math> W_{1} = <30, 0, 0> N * <15, 0, 0> m </math>

<math> W_{1} = (30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>

<math> W_{1} = 450 Nm + 0 Nm + 0 Nm </math>

<math> W_{1} = 450 Nm </math>

<math> W_{2} = <-30, 0, 0> N * <15, 0, 0> m </math>

<math> W_{2} = (-30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>

<math> W_{2} = -450 Nm + 0 Nm + 0 Nm </math>

<math> W_{2} = -450 Nm </math>

<math> W_{net} = W_{1} + W_{2} </math>

<math> W_{net} = 450 Nm - 450 Nm </math>

<math> W_{net} = 0 N*m </math>

===Difficult Torque===

In the complex system below, determine the given force is applied at point '''A'''. Determine the net torque about the origin due to the force.

[[File:Wvt008.jpg|300px|middle]]

Solution:

<math> \vec{r} = <12, -2, 6> m </math>

<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>

<math> \vec{\tau} = <12, -2, 6> m X <2, 1, 7> N </math>

<math> \vec{\tau} = (12 m * 2 N)(iXi) + (12 m * 1 N)(iXj) + (12 m * 7 N)(iXk) + (-2 m * 2 N)(jXi) + (-2 m * 1 N)(jXj) + (-2 m * 7 N)(jXk) + (6 m * 2 N)(kXi) + (6 m * 1 N)(kXj) + (6 m * 7 N)(kXk) </math>

<math> \vec{\tau} = (24 Nm)(0) + (12 Nm)(k) + (84 Nm)(-j) + (-4 Nm)(-k) + (-2 Nm)(0) + (-14 Nm)(i) + (12 Nm)(j) + (6 Nm)(-i) + (42 Nm)(0) </math>

<math> \vec{\tau} = <-20, -72, 16> N*m </math>

==Connectedness==

===Connected to My Own Interests?===

While I am more interested in digital and electronic designs, I love to tinker and create with mechanical devices. In order to make a device efficient and able to function without breaking, you must understand how work is done and the strain that torque can put on a device.

===Connected to Electrical Engineering?===

While torque and work are not integral parts of electrical engineering, it is very important to understand the two concepts because they do have the potential to alter the design of something I might construct. In this day and age, everything is electronic and most modern tools even use digital instructions. It's important when designing a mechanical tool to understand torque in order to determine the best avenue through which to run wire and to know how much extra give to allow to wire to move if the tools workload necessitates movement.

===Connected to Industry===

The industrial applications for both torque and work are enormous. From power tools, to vehicles, to simple machines. If one can do more wore with less energy, or apply greater torque with smaller forces, it greatly increases the workload that an individual can accomplish. So, yes, torque and work are quite important in industry.

==Origin==

===Torque===
The idea of torque first originated with Aristotle and his work with levers.

===Work===
Energy was also initially a notion put forth by Aristotle, but the concept of work was something later derived by several scientists and mathematicians, including Joule and Carnot.

== See also ==

===Further reading===

[http://www.physicsbook.gatech.edu/Work Wiki - Work]<div></div>
[http://www.physicsbook.gatech.edu/Torque Wiki - Torque]<div></div>
[http://www.physicsbook.gatech.edu/The_Energy_Principle Wiki - Energy Principle]<div></div>

===External links===
[https://www.youtube.com/watch?v=MHtAK1rMa1Y Explanation of Torque]<div></div>
[http://science360.gov/obj/video/9112c778-48e4-4b75-a09b-2f2d2404da12/science-nfl-football-torque Torque in Football - Cool!]<div></div>
[http://torquenitup.weebly.com/torque-problems.html Torque Example Problems]<div></div>

==References==

[http://www.wiley.com//college/sc/chabay/ Matter & Interactions 4th Edition]<div></div>
[http://physics.stackexchange.com/ Physics Stack Exchange]<div></div>
[https://en.wikipedia.org/wiki/Torque Wikipedia Torque]<div></div>

jderemer3

[[Category: Angular Momentum]]