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\usepackage{graphicx}
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%  \usepackage[nomarkers]{endfloat}
% \let\MYoriglatexcaption\caption
% \renewcommand{\caption}[2][\relax]{\MYoriglatexcaption[#2]{#2}}
%\fi
% endfloat.sty was written by James Darrell McCauley and Jeff Goldberg.
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% *** Do not adjust lengths that control margins, column widths, etc. ***
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% correct bad hyphenation here
\hyphenation{op-tical net-works semi-conduc-tor}

% Use this header.tex file for:
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%  - Example: \usepackage{xspace}
% 2. global macros available in all document blocks
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% NOT allowed in the main document, and definitions will be local to the current block.



\begin{document}
%
% paper title
% can use linebreaks \\ within to get better formatting as desired
\title{practica 1}
%
%
% author names and IEEE memberships
% note positions of commas and nonbreaking spaces ( ~ ) LaTeX will not break
% a structure at a ~ so this keeps an author's name from being broken across
% two lines.
% use \thanks{} to gain access to the first footnote area
% a separate \thanks must be used for each paragraph as LaTeX2e's \thanks
% was not built to handle multiple paragraphs
%
%
%\IEEEcompsocitemizethanks is a special \thanks that produces the bulleted
% lists the Computer Society journals use for "first footnote" author
% affiliations. Use \IEEEcompsocthanksitem which works much like \item
% for each affiliation group. When not in compsoc mode,
% \IEEEcompsocitemizethanks becomes like \thanks and
% \IEEEcompsocthanksitem becomes a line break with idention. This
% facilitates dual compilation, although admittedly the differences in the
% desired content of \author between the different types of papers makes a
% one-size-fits-all approach a daunting prospect. For instance, compsoc
% journal papers have the author affiliations above the "Manuscript
% received ..."  text while in non-compsoc journals this is reversed. Sigh.

\author{Lizett Cardiel% <-this % stops a space
\IEEEcompsocitemizethanks{\IEEEcompsocthanksitem Lizett Cardiel is with Instituto Tecnológico Superior Zacatecas Occidente\protect\\
% note need leading \protect in front of \\ to get a newline within \thanks as
% \\ is fragile and will error, could use \hfil\break instead.
\IEEEcompsocthanksitem }}% <-this % stops a space
%\thanks{Manuscript received \today;}


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% these prevent an unwanted space from occurring between the last author name
% and the end of the author line. i.e., if you had this:
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%                     ^------------^------------^----Do not want these spaces!
%
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% instance, "\textbf{A} \textbf{B}" will typeset as "A B" not "AB". To get
% "AB" then you have to do: "\textbf{A}\textbf{B}"
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% that ends a line with a % and do not let a space in before the next \thanks.
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% you are supposed to have spaces between the names. For what it is worth,
% this is a minor point as most people would not even notice if the said evil
% space somehow managed to creep in.



% The paper headers
\markboth{Computer Science Journal of IEEE,~Vol.~1, No.~1, 2013}%
{Lizett Cardiel : practica 1}
% The only time the second header will appear is for the odd numbered pages
% after the title page when using the twoside option.
%
% *** Note that you probably will NOT want to include the author's ***
% *** name in the headers of peer review papers.                   ***
% You can use \ifCLASSOPTIONpeerreview for conditional compilation here if
% you desire.



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% outside of the main text columns and, therefore, unlike regular IEEE
% journals, the available text space is not reduced by their presence.
% If you want to put a publisher's ID mark on the page you can do it like
% this:
%\IEEEpubid{0000--0000/00\$00.00~\copyright~2007 IEEE}
% or like this to get the Computer Society new two part style.
%\IEEEpubid{\makebox[\columnwidth]{\hfill 0000--0000/00/\$00.00~\copyright~2007 IEEE}%
%\hspace{\columnsep}\makebox[\columnwidth]{Published by the IEEE Computer Society\hfill}}
% Remember, if you use this you must call \IEEEpubidadjcol in the second
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% compsoc under a non-conference mode.


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% \fi
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1.- calcular la soluci\selectlanguage{ngerman}ón a los siguientes ejercicios

2.- obtener el punto entre los vectores \(\left(7i,-11j\right)\ y\ \left(13i,-17j\right)\)

3.-obtener el punto cruz entre los vectores~\(\left(7i,-11j\right)\ y\ \left(13i,-17j\right)\)

4.-calcule el angulo entre los vectores~\(\left(7i,-11j\right)\ y\ \left(13i,-17j\right)\)

\par\null

primeramente obtendremos el punto de los vectores que se hace por los
componentes del mismo exponente

1.- (7)(13)+(11)(17) = 91+187=278

después haremos lo del punto cruz que hace con los puntos cruzados como
ya la mencionábamos anteriormente

2.-(7)(-17)+(13)(11) = -119+143=24

finalmente con la siguiente formula sacaremos el angulo de los vectores
pero para eso primeramente necesitamos sacar el cuadrado de los vectores
que nos proporcionaron ~\(\left(^{7i^{2^{ }},-11j^2}\right)\) y ~\(^{\left(13i^2,-17j^2\right)}\) y queda
de la siguiente manera

\subsection*{\(\left(\left|a\right|\right)\) =~\(\sqrt{49+121}\)
=~\(\sqrt{170}\)~}

{\label{882636}}

\subsection*{}

{\label{882636}}

\subsection*{}

{\label{882636}}

\subsection*{\(\left(\left|b\right|\right)\) =~\(\sqrt{169+289}\)
=~\(\sqrt{458}\)~}

{\label{882636}}

\subsection*{}

{\label{882636}}

\subsection*{\texorpdfstring{\(\theta\)=\(\cos^{-1}\)\(\left(\frac{\vec{a.\vec{b}}}{\left|a\right|\ \left|b\right|}\right)\)=~\(\left(\frac{278}{\sqrt{170\ \sqrt{458}}}\right)\)=\emph{\textbf{4.93}}}{\(\theta\)=\(\cos^{-1}\)\(\left(\frac{\vec{a.\vec{b}}}{\left|a\right|\ \left|b\right|}\right)\)=~\(\left(\frac{278}{\sqrt{170\ \sqrt{458}}}\right)\)=4.93}}

{\label{131957}}

\subsection*{}

{\label{131957}}

\subsection*{}

{\label{131957}}

~y graficado nos queda así.

\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/tareaa/tareaa}
\caption{{GRAFICACION DE LOS VECTORES
{\label{593555}}%
}}
\end{center}
\end{figure}

\section*{PROBLEMA \#2}

{\label{607498}}

Determine el momento de la fuerza al rededor del punto 0 desprecie el
grosor del miembro

\par\null

Solucion:

sacaremos la fuerza del momento.

\(\vec{Mo}=F_xF_y\)~

a continuaci\selectlanguage{ngerman}ón sacaremos los valores de~\(F_{x\ }\ Y\ F_y\)

\(\vec{f_x}=0.341+0.141j\)

\(\vec{f_y}=50NCos60i-50Sin60j\)

\(Cos60=\frac{x}{50N}\ \ \ \ \ \ \ X=50NCos60\)

\par\null

después realizaremos una tabla para sacar el valor de K.

~ ~ I ~ ~ ~ ~ ~ ~J ~ ~ ~ K

0.341 ~0.141 ~0 ~ ~ ~ ~ ~ ~ ~ ~ ~A continuacion nos queda asi
~\(=K\left[\left(0.341\right)\left(-43.3\right)-\left(25\right)\left(0\right)\left(0\right)\right]\)

-20 ~ ~ ~ -43.3 ~ 0 ~ ~ ~ ~ ~y esto es igual al resultado que
es~\(K=-11.2\ n.m\)

\section*{PRACTICA \#3}

{\label{350703}}

Determine el momento de la fuerza del punto 0.

\par\null

Solucion:

sacaremos la fuerza del momento.

\(\vec{Mo}=F_XF_y\)

a continuación sacaremos los valores de r y F del angulo de 20°

\(F_x=\)\(600lbCos20=563.8\)

\(\vec{F_y}=600lbSin20=205.21\)

ahora del angulo de 30°

\(F_x=5ftCos30-.5Cos60=4.08ft\)

\(F_y=5ftSin60=2.93ft\)

\par\null

A continuación sacaremos el resultado de las~\(F_{x\ }\ Y\ F_y\)

\(Mo=\left(205.11\right)\left(2.93\right)+\left(563.8\right)\left(2.93\right)=2489.2\ ft\)

\par\null

a continuación pondré las imágenes de los vectores que hemos realizado~

\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/2/2}
\caption{{Vectores
{\label{768501}}%
}}
\end{center}
\end{figure}

\selectlanguage{english}
\FloatBarrier
\end{document}

