PROBLEMA \ref{492170}

Datos 

\(rA=4i+5j+4k\)
\(F1=100i-120j+75k\)
\(F2=-200i+250j+100k\)

Fórmula

\(M_{OT}=M_{o1}+M_{o2}\)
\(M_{O1}=\vec{rA}\ x\ \vec{F_1}\)
\(M_{o2}=\vec{rB}\ x\ \vec{F_2}\)

Desarrollo

\(M_{o1}=\vec{rA}\ x\ \vec{F_1}\)
\(i\left[\left(75\right)\left(5\right)-\left(-120\right)\left(3\right)\right]=735i\)
\(j\left[\left(4\right)\left(75\right)-\left(100\right)\left(3\right)\right]=0j\)
\(k\left[\left(4\right)\left(-120\right)-\left(100\right)\left(5\right)\right]=-20k\)
\(M_{o1}=735i-20k\)
\(M_{o2}=\vec{rA}\ x\ F_2\)
\(i\left[\left(5\right)\left(500\right)-\left(250\right)\left(3\right)\right]=1750i\)
\(j\left[\left(4\right)\left(100\right)-\left(-200\right)\left(3\right)\right]=1000j\)
\(k\left[\left(4\right)\left(250\right)-\left(-200\right)\left(5\right)\right]=2000k\)
\(M_{o2}=1750i+1000j+2000k\)

Solución

\(M_{oT}=2485i+1000j+1980k\ lb\ ft\)