Question 1075922
*[illustration F4.png].
.
.
.
Summing the forces in the z direction,
{{{R[1]+R[2]+R[3]=500}}}
Summing the moments about the x axis,
{{{500*C-R[3]*D+R[2]*D=0}}}
Summing the moments about the y axis,
{{{500*E-R[2]*F-R[3]*F=0}}}
Find {{{C}}} using the law of cosines to find {{{A}}},
{{{cos(A)=(900^2+625^2-325^2)/(2(900)(625))}}}
{{{cos(A)=73/75}}}
{{{A=13.26}}}
So then 
{{{30-A=16.74}}}
So then,
{{{C=625sin(16.74)}}}
{{{C=625(0.2880)}}}
{{{C=180.0}}}{{{mm}}}
and
{{{E=625cos(16.74)}}}
{{{E=598.5}}}{{{mm}}}
and
{{{F=900cos(30)}}}
{{{F=779.4}}}{{{mm}}}
So then,
Summing about x,
{{{500*180-R[3]*450+R[2]*450=0}}}
{{{450(R[3]-R[3])=500*180}}}
{{{R[3]-R[2]=(500*180)/450}}}
1.{{{R[3]-R[2]=200}}}
and summing about y,
{{{500*E-R[2]*F-R[3]*F=0}}}
{{{500(598.5)=(R[2]+R[3])(779.4)}}}
2.{{{R[2]+R[3]=383.9}}}
So then adding eqs. 1 and 2,
{{{R[3]-R[2]+R[2]+R[3]=200+383.9}}}
{{{2R[3]=583.9}}}
{{{R[3]=291.97}}}{{{N}}}
Then,
{{{291.97-R[2]=200}}}
{{{R[2]=91.97}}}{{{N}}}
and,
{{{R[1]+91.97+291.97=500}}}
{{{R[1]=116.06}}}{{{N}}}