$(27$ points$)$ Beam $AB$ holds a UTRGV sign that weighs $375$ lb and is supported by a steel guy wire BC and a pin at A$.$ To the nearest $\frac{1}{{16}^{\text{'}\text{'}}},$ determine the diameter of the wire BC if the failure normal stress is $7\mathrm{.}25$ ksi and the diametel of the double$-$shear pin $A$ if the failure shear stress is $5\mathrm{.}82$ ksi. Assume a factor of safety of $2\mathrm{.}15.$

${}_{\text{a}\text{l}\text{l}\text{o}\text{w}}=\frac{7\mathrm{.}25}{2\mathrm{.}15}=3\mathrm{.}372,{}_{\text{a}\text{l}\text{l}\text{o}\text{w}}=\frac{7\mathrm{.}25}{2\mathrm{.}15}=3\mathrm{.}372$

${?}_1{\displaystyle \underset{M}{\overset{?}{}}}=0$;

$0=(375)()-F_{BC}()$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=0,$

$(375)+A_y=0$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=0,$

$(375)-$

noizeg$1$qmo$)$ nolengT