Answer:
a) [tex]\bar x = \frac{d\cdot m_{3}+(d\cdot \cos 60^{\textdegree})\cdot m_{1}}{m_{1}+m_{2}+m_{3}}[/tex]
Step-by-step explanation:
a) Let be origin set in the left vertex of the triangle and the horizontal axis is parallel to the side between the right and left beads. The horizontal component of the center of mass is:
[tex]\bar x = \frac{d\cdot m_{3}+(d\cdot \cos 60^{\textdegree})\cdot m_{1}}{m_{1}+m_{2}+m_{3}}[/tex]
