Answer
a) $m_{2}$ = $2$
b) $m_{3}$ = $10$
Work Step by Step
$M_{x}$ = $m_{1}(0)+m_{2}(0)+m_{3}(4)$ = $4m_{3}$
$M_{y}$ = $m_{1}(-1)+m_{2}(3)+m_{3}(0)$ = $3m_{2}-m_{1}$
the total mass of the system is $m_{1}+m_{2}+m_{3}$
so the coordinates of the center of mass are
$(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(\frac{3m_{2}-m_{1}}{m_{1}+m_{2}+m_{3}}, \frac{4m_{3}}{m_{1}+m_{2}+m_{3}})$
a)
$3m_{2}-m_{1}$ = $0$
$3m_{2}-6$ = $0$
$m_{2}$ = $2$
b)
$\frac{4m_{3}}{m_{1}+m_{2}+m_{3}}$ = $2$
$\frac{4m_{3}}{6+4+m_{3}}$ = $2$
$4m_{3}$ = $20+2m_{3}$
$m_{3}$ = $10$
