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\begin{document}

\title{Informe de soluci\selectlanguage{ngerman}ón de problemas sobre el momento de una fuerza~~~~\selectlanguage{english}}



\author[1]{Jessica Aide Quirino-Murillo }%
\affil[1]{Instituto Tecnológico Superior Zacatecas Occidente}%


\vspace{-1em}



  
  \date{May 14, 2024}


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\textbf{Instrucciones}~ Determine el momento de la(s) fuerza(s) en el
punto~\emph{O}~para cada uno de los problemas.

\par\null

\subsection*{~ ~Problema \#1}

{\label{992197}}\selectlanguage{english}
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\(rx=5ft+0.5+\cos30\selectlanguage{ngerman}º\)

\(ry=0.5ft\sin30º\)

\(fx=600\ lb\ \cos20\)

\(fy=600\ lb\sin20\)

\(M_{o\ }=\left(rx\ Fy-ry\ Fx\right)\)

\(=\left\{\left(5.43\right)\left(205.21\right)-\left(0.25\right)\left(563.81\right)\right\}\)

\(\left(1,114.29\right)-\left(140.95\right)\)

\(M_o\ 973.34\ MN\)

\par\null

\textbf{Problema \#2~}

\textbf{FORMULAS}

\textbf{\(M\)}\textsubscript{O\textbf{~}}\textbf{=~\(R\times F\ \left(3D\right)\)}

\(M\)\textsubscript{O~}=~\(\left(Rx-RyFx\right)\left(2D\right)\)\selectlanguage{english}
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Soluci\selectlanguage{ngerman}ón para hacer el~ calculo de M,N~ y~~\(\cos\ 45º\)~ tomamos R
Y F~

\(rx\ ry\ \ \ fx\ fy\)

\(ry\ =\ 3n\ +3m\ \cos\ 45º\)

\(para\ ry\)

\(3+\sin\ 45º\)

\(fx=500N\ \cos45º\)

\(fy=500N\ \sin\ 45º\)

\(M_O\ =\left(rx\ fy-ryfx\right)k\)

\(=\ \left(1060.66\right)N.M\)

\par\null

\textbf{Problema \#3}

Determinar el momento resultante de las fuerzas que estan en el punto 0
expresadas en el plano cartesiano.~\selectlanguage{english}
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\(M_o=f1=rxFF1=100i\ -120k+175k\)

\(f2=-200i+250j+100k\)

\(fr\ =-100i+130j+175k\)

\(r=4i+5j+3k\)

\(M_{o\ }=\left\{4i-100j5130k\right\}3175i\)

\(\left\{\left(5\text{*175}\right)-\left(130\text{*3}\right)\right\}=485i\ \ \ j\left(4\text{*175}\right)-\left\{3\text{*(-100)}\right\}=1000j\ \ k\left\{\left(4\text{*130}\right)-\left(100\text{*5}\right)\right\}\ =1020k\)

\(M_O\ =rx\ FF1=100i-120j+75k\ \ \ \ \ F2=200i+250j+100k\)

\(fr=-100i+130j+175k\)

\(r=4i+5j+3k\)

\(M_o=\left\{453-100i,\ 130j,175k\right\}=i\ \ \ \ \ \left\{\left(5\text{*175}\right)-\left(130\text{*3}\right)\right\}=485i\ \ \ \ \ j\left\{\left(4\text{*175}\right)3\text{*(-100)}\right\}=1000j\ \ \ k\ \left\{\left(4\text{*130}\right)-\left(5\text{*(-100)}\right)\right\}=1020k\)

\(M_o\ =485i-1000j+1020k\)\textbf{.}

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