P$\mathrm{.}2.(20$ points$)$ A $120$ lb $.$ force $(vec(F))$ is applied on the tube assembly at point $B$ through the cable $BC.$

a$)$ Determine the moment $(vec(M{)}_O)$ of the force $(vec(F))$ about point $O$ using Cartesian o $($S Vector Approach.

b$)$ Determine the magnitude of the moment $(vec(M{)}_{OA})$ of the force $(vec(F))$ about an axis extending between $O$ and $A.$

$F=12010$

ojoj$>$ojitek

A$)1i+4j+0k$

$|)=\frac{(B-C)i+(B-c)j+(B-c)K}{\sqrt[\phantom{2}]{i^2+j^2+k^2}}$

B$)1i+4i+2k$

$|vec(BC)||=\frac{(1-5)i+(4-0)j+(2-0)k}{?}=-4i+4j+2k$

$C$

A$)$

$F{U}_{BC}=120(-0\mathrm{.}666i+0\mathrm{.}666j+0\mathrm{.}333k)$

$=-79\mathrm{.}92i+79\mathrm{.}92j+39\mathrm{.}46k$

$B$

$\frac{r}{b}ar(0)=(B*0)$;$+(B-0)j+(B*0)K$

$=(1-0)i+(y-0)j+(0-0)k$

$=1i+4j+0k$

$2$

Part $(a)$

vec$(M{)}_0=-79\mathrm{.}92+79\mathrm{.}92j+39\mathrm{.}96k$

Part $(b)$

$|vec(M{)}_{OA}|=53\mathrm{.}278ft*1b$

$|MOA|=-0\mathrm{.}242((4-339\mathrm{.}95)-0)=-38\mathrm{.}68$

$=-0\mathrm{.}917(0-(139\mathrm{.}96))=36\mathrm{.}6$

$=01-1=0$

$|$ MOA $|=\sqrt[\phantom{2}]{{38\mathrm{.}68}^2+{36\mathrm{.}64}^2}$

:$53\mathrm{.}27$