Answer:
[tex]M_x = 2, M_y = 3[/tex]
Center of mass at [tex](\frac{3}{10},\frac{2}{10})[/tex]
Step-by-step explanation:
Given a set of objects [tex] m_1, \dots, m_k[/tex] located at the points [tex] P_1(x_1, y_1), \dots, P_k(x_k,y_k)[/tex] The moments Mx and My are calculated as follows.
[tex] M_x = \sum_{n=1}^k m_n\cdot y_n, M_y = \sum_{n=1}^k m_n\cdot x_n[/tex]
and the Center of mass is given at the coordinates [tex](\frac{My}{M}, \frac{Mx}{M})[/tex] where M is the total mass of the system.
We have that m1=3, m2=4 and m3=3, so M = 3+4+3 = 10.
We have that m1 is at point P1(2,-6), m2 at point P2(-3,2) and m3 at point P3 (3,4), then
[tex] M_x = \sum_{n=1}^k m_n\cdot y_n = 3\cdot(-6)+4\cdot(2)+3\cdot(4)=2[/tex]
[tex] M_y = \sum_{n=1}^k m_n\cdot x_n = 3\cdot(2)+4\cdot(-3)+3\cdot(3)= 3[/tex]
Then, the center of Mass is at [tex](\frac{3}{10},\frac{2}{10})[/tex]