OpenLLM-France/wikimedia huggingface dataset:
# Pokémon Crystal/Exploring Route 31
## Walkthrough

The path splits to the east and west, with the east featuring tall grass and the west featuring a paved path. The paved path is currently blocked, so head into the grass instead.
At the end of the grass lies the entrance to the Dark Cave. While it could technically be navigated, doing so in the pitch black darkness would be difficult. Hang near the entrance to catch the Pokémon Geodude if desired, but don't let the dim light of the cavern entrance leave your screen to make getting back out easier. Check out the item ball by the entrance to gain a potion.
Head west to find a berry tree that grows confusing curing bitter berries.

### Bug Catcher Wade

Head down south of the ledge to encounter Bug Catcher Wade. Before starting this battle, consider putting a weaker Pokemon in front to gain some experience in this relatively easy battle. An antidote may come in handy during or following this battle.
Wade starts with a level 2 male Caterpie. It knows tackle. Defeat it to earn 22 exp. He follows it up with an identical Caterpie.
Following that is his ace, a level 3 male Weedle. It knows the move poison sting, so consider using a poison type such as your own Weedle if you have one to safely dispatch it. Defeat it to earn 33 exp.
It's followed by one last Pokémon, a level 2 male Caterpie. Beat it to earn 22 more exp and 32. Talk to Wade again and he will ask for your phone number. Wade will call about bug catching contests, and will also call to offer you berries he found.

### Continuing to Violet City

Just south of Wade is a free pokéball. Head northwest to reach the gatehouse to Violet City, "The City of Nostalgic Scents".

## Pokémon

* Bellsprout - Morning
* Hoppip
* Pidgey
* Spinarak - Nite
# Statics/Method of Joints

The method of joints is a way to find unknown forces in a truss structure. The principle behind this method is that all forces acting on a joint must add to zero. If there were a net force, the joint would move.

## Example 1
### Question

Find the force in member BC of the truss pictured to the right.

### Answer

Using the method of joints, the force could be found by isolating the joint at either end of the member (joint B or C). Neither joint can be solved without further analysis; however, joint B can be solved if the force in member $AB$ and $BH$ is found.
To find force $AB$ analyze joint A. This joint has an external vertical force of 300N which must be countered by the members attached to the joint. Member $AE$ cannot possibly support any vertical load, otherwise it would not be loaded axially and the entire structure would no longer be a truss. If $AE$ has no load then member $AB$ is in 300N of tension.
When joint H is analyzed it is found that the force in members $BH$ and $HC$ must be zero. The reason why neither member can carry any load is that member $BH$ can only take a vertical load and member $HC$ can only take a horizontal load. In a real world application this structure might be useful if there was a load applied at joint $H$. Now joint $B$ can be analyzed.
The picture to the left shows the forces affecting joint B.
$\sum F_{y}=0=BH-BA+BC\cos(50)-BE\cos(50)$
$\sum F_{x}=0=BC\sin(50)+BE\sin(50)$

#### Substitution

From analysis of joint $A$
$\ BA=300N(Tension)$
From analysis of joint $H$
$\ BH=0N$
Put values for $BA$ and $BH$ into the equilibrium equations for joint B.
$\sum F_{y}=0=0-300N+BC\cos(50)-BE\cos(50)$
$\sum F_{x}=0=BC\sin(50)+BE\sin(50)$
$\ BC\sin(50)=-BE\sin(50)$
$\ BC=-BE$
Now $BC$ can be inserted in place of $-BE$ in $\sum F_{y}$, which gives:
$2BC\cos(50)-300N=0$
Finally, $BC$ can be solved for as follows:
$BC=233.4N$
# Mass Moments Of Inertia Of Common Geometric Shapes
## Slender Rod

$I_{x}=0$
$I_{y}=I_{z}={\frac {1}{12}}ml^{2}$

## Thin Quarter-Circular Rod

$I_{x}=I_{z}=mr^{2}({\frac {1}{2}}-{\frac {4}{\pi ^{2}}})$
$I_{y}=mr^{2}(1-{\frac {8}{\pi ^{2}}})$

## Thin Ring

$I_{x}=I_{y}={\frac {1}{2}}mr^{2}$
$I_{z}=mr^{2}$

## Sphere

$I_{x}=I_{y}=I_{z}={\frac {2}{5}}mr^{2}$

## Hemisphere

$I_{x}=I_{y}={\frac {83}{320}}mr^{2}$
$I_{z}={\frac {2}{5}}mr^{2}$

## Thin Circular Disk

$I_{x}=I_{y}={\frac {1}{4}}mr^{2}$
$I_{z}={\frac {1}{2}}mr^{2}$

## Rectangular Prism

$I_{x}={\frac {1}{12}}m\left(b^{2}+c^{2}\right)$
$I_{y}={\frac {1}{12}}m\left(a^{2}+c^{2}\right)$
$I_{z}={\frac {1}{12}}m\left(a^{2}+b^{2}\right)$

## Right Circular Cylinder

$I_{x}=I_{y}={\frac {1}{12}}m(3r^{2}+h^{2})$
$I_{z}={\frac {1}{2}}mr^{2}$

## Right Half Cylinder

$I_{x}={\frac {1}{12}}mh^{2}+mr^{2}({\frac {1}{4}}-{\frac {16}{9\pi ^{2}}})$
$I_{y}={\frac {1}{12}}mh^{2}+{\frac {1}{4}}mr^{2}$
$I_{z}=mr^{2}({\frac {1}{2}}-{\frac {16}{9\pi ^{2}}})$

## Thin Rectangular Plate

$I_{x}={\frac {1}{12}}mb^{2}$
$I_{y}={\frac {1}{12}}ma^{2}$
$I_{z}={\frac {1}{12}}m(a^{2}+b^{2})$

## Right Circular Cone

$I_{x}=I_{y}={\frac {3}{80}}m({4}{r^{2}}+h^{2})$
$I_{z}={\frac {3}{10}}mr^{2}$

## Right Tetrahedron

$I_{x}={\frac {3}{80}}m(b^{2}+c^{2})$
$I_{y}={\frac {3}{80}}m(a^{2}+c^{2})$
$I_{z}={\frac {3}{80}}m(a^{2}+b^{2})$

## further reading

* Statics/Geometric_Properties_of_Lines_and_Areas
* Mechanics/Rigid Bodies
# Statics/Introduction
## This Wikibook

This book was written at Wikibooks, a free online community where people write open-content textbooks. Any person with internet access is welcome to participate in the creation and improvement of this book. Because this book is continuously evolving, there are no finite "versions" or "editions" of this book. Permanent links to known good versions of the pages may be provided.

## Introduction

Statics is the branch of mechanics concerned with the study of forces and the effect of forces on a non-deformable, or rigid, system when the system is in a state of equilibrium.
This course is a crucial prerequisite for later areas, such as Dynamics and Properties of Materials. It utilizes principles of physics and calculus. It is fundamental in many different branches of engineering, from mechanical to civil engineering, and the principles of equilibrium, moment of inertia, and center of gravity will be revisited in more advanced fields. It is because an understanding of these topics is so crucial that statics does not cover a wide range of topics. Every problem will deal with some combination of two equations: the net forces being equal to zero, and/or net moments being equal to zero.

## Who is This Book For?

This book is for undergraduate students pursuing, or thinking about pursuing, a degree in engineering. As specified early, statics is needed in almost every field of engineering. It is not an introductory course, however. Statics does not attempt to reinvent the wheel, and is built firmly on the foundation of Physics. Newton's Laws are essential to statics. Mathematically, one should have a firm grip on vector algebra, though it is possible to solve many simple problems without the use of vectors. While it would benefit a student to understand calculus concepts, like the ability to calculate an area or a volume, most statics problems will involve simple geometries.

## How is this book organized

This book is organized to introduce concepts that will be later used in Dynamics and Properties of Materials. As such, it will begin by reviewing over vector mathematics and then introducing simple concepts, like Newton's Second Law, and then increasing the ways with which those concepts work together. Mathematically, this will involve increasing the number of forces, increasing the number of dimensions, and then expanding particle equilibrium to Rigid Body Equilibrium and Structural Equilibrium.
Finally, we will go over Moments of Inertia and Center of Mass. These concepts will be critical in higher level disciplines and a firm understanding will allow you to analyze more complex problems when the concepts of stress, strain, and acceleration are introduced.
# Statics/Method of Sections

The method of sections is another method to determine forces in members of a truss structure. In order to find unknown forces in using the method of sections, sections of the truss structure must be isolated. The net force on the entire isolated section must be zero since the isolated section does not move (if it did move it wouldn't be a statics problem). This method is often faster because instead of moving from joint to joint with the method of joints, the members of interest can be immediately isolated.

## Example 1
### Question

The truss pictured to the right is secured with pin mounts to the brown concrete block. A 100kg mass is hanging from the rightmost joint. For simplicity the weight of the trusses in the structure can be ignored. Calculate the compression in member A-B.

### Answer

This problem could be solved using the method of joints, but you would have to start at the joint above the weight then solve many other joints between the first joint and member A-B. The fifth joint solved would give member A-B's loading. It is easier using the method of sections.
The method of joints isolates a joint to find unknown forces. The method of sections is the same except an entire section is isolated. It should be obvious at this point that there cannot be any net force or moment on the section, if there was the section would move.
In order to find the stress in member A-B it is convenient to isolate the section of the truss as indicated by the image to the left. In problems where the method of joints was used, a joint was isolated by cutting surrounding members. Cutting is also used to isolate a section, three of the members of the truss were cut to isolate the section in this example. Because all cut members were part of a truss, all members supported an axial force of tension or compression. After being cut, all members must have the same axial force applied to support the section.
Equilibrium equations can now be written.
$\sum F_{x}=0=-T_{1}-T_{2}\cos(60)-T_{3}$
$\sum F_{y}=0=T_{2}\sin(60)-981N$
The equation summing forces in the Y direction only has one unknown because all cut members except A-B are horizontal.
$\ 981N=T_{2}\sin(60)$
$\ T_{2}=1130N$
Because $T_{2}$ is positive, member A-B is in 1130N of tension.
Quick links:Statics
# Centroids Of Common Shapes Of Areas And Lines
## Triangular Area

$Area={\frac {b*h}{2}}$

## Quarter Circular Area

$Area={\frac {\pi \ r^{2}}{4}}$

## Semicircular Area

$Area={\frac {\pi \ r^{2}}{2}}$

## Semiparabolic Area

$Area={\frac {2ah}{3}}$

## Parabolic Area

$Area={\frac {4ah}{3}}$

## Parabolic Spandrel

$Area={\frac {ah}{3}}$

## Circular Sector

$Area=\alpha \ r^{2}$

## Quarter Circular Arc

$Area={\frac {\pi \ }{2}}$

## Semi Circular Arc

$Area=\pi \ r$

## Arc Of Circle

$Area=2\alpha \ r$

# Area Moments Of Inertia of Common Geometric Shapes
## Rectangle

$I_{x}={\frac {1}{3}}bh^{3}$
$I_{y}={\frac {1}{3}}hb^{3}$
$I_{x'}={\frac {1}{12}}bh^{3}$
$I_{y'}={\frac {1}{12}}hb^{3}$

## Right Triangular Area

$I_{x}={\frac {1}{12}}bh^{3}$
$I_{y}={\frac {1}{4}}hb^{3}$
$I_{x'}={\frac {1}{36}}bh^{3}$
$I_{y'}={\frac {1}{36}}hb^{3}$

## Triangular Area

$I_{x}={\frac {1}{12}}bh^{3}$
$I_{x'}={\frac {1}{36}}bh^{3}$

## Circular Area

$J_{C}={\frac {\pi \ r^{4}}{2}}$
$I_{x'}=I_{y'}={\frac {\pi \ r^{4}}{4}}$

## Hollow circle

This is used for hollow cylinders where there is solid material between the outer and inner radius, but no material between the inner radius and the center, like a pipe's cross-section.
$I={\frac {\pi \ (r_{o}^{4}-r_{i}^{4})}{4}}$
$r_{o}$ is the outer radius $r_{i}$ is the inner radius

## Semicircular Area

$I_{x}=I_{y}={\frac {1}{8}}\pi \ r^{4}$
$I_{x'}=({\frac {\pi }{8}}-{\frac {8}{9\pi }})r^{4}$
$I_{y'}={\frac {1}{8}}\pi \ r^{4}$

## Quarter Circle

$I_{x}=I_{y}={\frac {1}{16}}\pi \ r^{4}$
$I_{x'}=I_{y'}=({\frac {\pi }{16}}-{\frac {4}{9\pi }})r^{4}$

# further reading

Statics/Geometric_Properties_of_Solids
# Statics/Moment of Inertia (contents)

The moment of inertia can be defined as the second moment about an axis and is usually designated the symbol I. The moment of inertia is very useful in solving a number of problems in mechanics. For example, the moment of inertia can be used to calculate angular momentum, and angular energy. Moment of inertia is also important in beam design.

## Shape moment of inertia for flat shapes

The area moment of inertia takes only shape into account, not mass.
    It can be used to calculate the moment of inertia of a flat shape about the x or y axis when I is only important at one cross-section.
$I_{x}=\int y^{2}\,dA$
$I_{y}=\int x^{2}\,dA$
$I_{z}=\int r^{2}\,dA$
Because, for flat shapes, $r^{2}=x^{2}+y^{2}$ the following is true $I_{z}=I_{x}+I_{y}$
The shape moment of inertia of the cross-section of a beam is used in Structural Engineering in order to find the stress and deflection of the beam.

## Shape moment of inertia for 3D shapes

For more shapes see Mass Moments Of Inertia Of Common Geometric Shapes.

## Mass moment of inertia

The mass moment of inertia takes mass into account. The mass moment of inertia of a point mass about a reference axis is equal to mass multiplied by the square of the distance from that point mass to the reference axis:
$I_{pointmass}=r^{2}m\,$
The metric units are kg*m^2.
The mass moment of inertia of any body of mass rotating around any axis is equal to the sum of the mass moment of inertia of each of the particles of that body:
$I_{m}=\sum r_{i}^{2}m_{i}$
Rather than adding up each particle individually, sometimes we can take a mathematical shortcut by integrating over all the particles:
$I_{m}=\int r^{2}\,dm$

## Radius of gyration

The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. If the mass of an object was 2kg, and the moment of inertia was $18kg*m^{2}$, then the radius of gyration would be 3m. In other words, if all of the mass was concentrated at a distance of 2m from the axis, then the moment of inertia would still be $18kg*m^{2}$. Radius of gyration is represented with a k.
$k={\sqrt {I/m}}\,$
The formula for the area radius of gyration replaces the mass with area.
$k={\sqrt {I/A}}\,$

## Parallel axis theorem

If the moment of inertia is known about an axis that runs through the center of mass, then the moment of inertia about any parallel axis is given by,
$I_{parallel-axis}=I_{center\,of\,mass}+md^{2}$
where d is the distance between the two axis of rotation.

## further reading

* Mechanics/Rigid Bodies
# Statics/Multiforce Members

There are many structural problems in statics that do not involve a truss. For example, a beam is not loaded axially. A beam is analyzed in terms of bending moments and shear. Many simple tools can be analyzed with statics.

## Example 1
### Question

The picture to the right shows a tool used to compress springs. In the tool's current state a person is squeezing the pliers with 40 Newtons applied to each handle to hold the spring(shown in orange) in a steady partially compressed state. The black circle is a pin (a pin can transfer forces, but cannot resist turning). The forces used to squeeze the pliers are vertical in addition to the forces applied on the spring.
Find the compressive force on the spring and the force on the pin.

#### Hint

Think of how the plier components can be disassembled and analyzed for equilibrium.

### Answer

If the pin in the question's picture was welded to the hole it fits in, the force from the spring could be any value; however, the pin is free to rotate (it can't resist a turning moment) so the picture to the left can be used for further analysis.
If moments are summed about the black pin (let's designate it "pin A"), the spring force(designated "F") can be solved for.
$\sum M_{A}=0=-6in*40N+2in*F$
$\ F=120N$
This means the spring is being compressed with 120N.
The force in the pin can be solved for with force equilibrium. Let's call the pin's $x$ force $P_{x}$ and the $y$ force $P_{y}$.
$\sum F_{x}=0=P_{x}$
$\sum F_{y}=0=40N+P_{y}+120N$
$\ P_{y}=-160N$
Since the equations were written in terms of the assumed force directions pictured and since $P_{y}$ is negative, the force acts in the opposite direction as shown in the picture.
The pin exerts a downward force on the handle shown in the left picture, but it exerts an upward force of the same magnitude on the other handle. This statement is very important and can be further explained by the picture to the right. The picture shows a dark gray and light gray rectangle. The light and dark gray corresponds to the previous pictures. The black rectangle represents the pin which goes through both handles as indicated with dashed lines. The dark handle is exerting an upward force of 160N on the pin and the light handle is exerting a downward force of 160N on the pin. Because the forces are equal and opposite the pin does not travel vertically. The force from the dark handle is transmitted through the pin to the light handle and the force from the light handle is transmitted through the pin to the dark handle. The pin is in 160N of shear.
# Statics/Newton's Laws and Equilibrium

They are not the law pertaining to Statics The laws are some bit different then this, They Are the sources of Statics

## Newton's First Law

Newton's first law states that an object that remains in uniform motion will remain in uniform motion unless it is acted upon by an external force. This also includes that an object at rest will remain at rest unless it is acted upon by an external force. When more than one force acts upon an object, the vector sum of these forces is the resultant force.
When the resultant force on an object is zero, it will remain at rest if it is at rest, or continue to move in a straight line at a constant velocity if it is in motion. There is no change in either the magnitude or direction of its velocity. That is, there is zero acceleration. This concept can also be applied to motion in any selected direction.
Consider an object moving along the x-axis. If no net force is applied to the object along the x-axis, it will continue to move along the x-axis at a constant velocity with no acceleration.We can extend this to the y- and z- axes.
In any system, unless the applied forces cancel each other out, that is, the resultant force is zero, there will be acceleration in the direction of the resultant force.The net force on such an object is zero

## Force Equilibrium

In static systems, where motion does not occur, the sum of the forces in all directions must always equal zero (otherwise, it's a dynamics problem). This concept can be represented mathematically with the following equations.

## Rotational Equilibrium

The concept also applies to rotational motion.
If the resultant moment about an axis is zero, the object will have no rotational acceleration about the axis. If the object is not spinning, it will not start to spin. If the object is spinning, it will continue to spin at the same constant angular velocity. Again, we can extend this to moments about the y-axis and the z-axis. This is represented mathematically with the following. ..............................................................................................................................................

## What happens if the sum of forces or moments is not zero?

If the object has a net moment not equal to zero, it will spin about whichever axes are not zero. If the object is acted upon by a net force not equal to zero, it will accelerate in whatever direction the net force is. Once the object moves it cannot be analyzed with statics; instead, the rules of dynamics take over. The acceleration of the object is governed by Newton's second law,
$\sum F=m\cdot a$
You will never need this equation in Statics.
# Structural Biochemistry/Reaction types

Types of chemical reactions

## Synthesis

In a synthesis reaction, two or more reactants combine to form one complex product. The general equation to represent this reaction is: A + B→AB

## Decomposition

The decomposition reaction is the opposite of the synthesis reaction. A complex compound breaks down into simpler substances. The general equation to represent this reaction is: AB→ A + B

## Dissociation

Dissociation looks like and can be mistaken for decomposition but they are very different reactions. When a compound dissociates, it breaks down into ions instead of atoms. The products are therefore charged. The general equation looks like: AB→ A$^+$ + B$^-$

## Single Replacement

A non-combined element replaces another in a compound and the element that it replaces has to be of the same group (e.g.metal replaces metal). The general equation: A + BC→ AC + B

## Double Replacement

In a double replacement, two compounds each trade one element of the same type over to the other. The general equation: AB + CD→ AC + BD

## Combustion

In a combustion reaction, a compound is combined with oxygen to produce carbon dioxide and water as products. Example: C$_n$H$_x$ + O$_2$→CO$_2$ + H$_2$O

## Acid-Base

The acid base reaction is a type of double replacement reaction. When the acid and base react, the H$^+$ ion reacts with the OH$^-$ ion to form H$_2$O (water). The products of this reaction are typically water and an ionic salt. HA + BOH→ H$_2$O + BA

## References

1. "Types of Reactions." ThinkQuest. Oracle Foundation, n.d. Web. 20 Nov. 2012. <http://library.thinkquest.org/2923/react.html>.
2. "TheSix Types of Chemical Reaction." The Six Types of Chemical Reaction. N.p., n.d. Web. 20 Nov. 2012. <http://misterguch.brinkster.net/6typesofchemicalrxn.html>.
# Structural Biochemistry/RecA Protein

RecA can be found in E. Coli and it is needed in order to repair and maintain DNA. A structural and functional homolog for Rec A can be found in every species and as a result, it is used as a method to exchange genetic information by homologous recombination through sexual reproduction. RecA has multiple functions, all of which are related to DNA repair. Its main function which has been most widely studied is that it facilitates DNA recombination in order to repair DNA double strand breaks. The RecA protein can catalyze one directional branch migration. This function makes it possible to complete recombination. Bacterial strains such as E. Coli that are lacking RecA are useful for cloning procedures.
# Structural Biochemistry/Receptor-mediated Endocytosis

Membranes must be able to separate or come together in order for the cells to absorb, transport, and release molecules. Receptor-mediated endocytosis is a common process in which cells take up molecules. In many cases, a large protein complex initially binds to a receptor located on the surface of the cell. Once the complex is bound to the receptor, specific proteins cause the membrane to invaginate. This ultimately causes the membrane to break off and fuse to form a vesicle.
This process is apparent when dealing with cholesterol in the blood (LDL). LDL binds to a specific integral membrane protein, LDL receptor. The LDL-LDL-receptor complex invaginates, therefore "breaking" the membrane. The LDL-LDL-receptor complex also separates such that LDL separates from the LDL recptor. The formed vesicle containing LDL fuses with a lysosome, which results in the degradation of LDL and the release of a cholesterol molecule.