Question

three particles of the respective masses m1-12kg,m2-24kg, and m3-36kg formed an equilateral triangle of side length a=120cm.If we locate m1 at the origin of the xy-plane and put m2 to right of m1 at right axis.What is the approximate coordinate s of the center of the mass of the system?

Answer 1

The center of mass of the system is $m_1=12kg \space\space m_2=24kg \space \space m_3=36kg$ $a=0.12m$

$\large(X,Y)=(\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3},\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3})=$

$\large(\frac{m_1(0)+m_2(a)+m_3(\frac{a}{2})}{m_1+m_2+m_3},\frac{m_1(0)+m_2(0)+m_3(\frac{\sqrt{3}a}{2})}{m_1+m_2+m_3})=$

$\large(\frac{m_2(a)+m_3(\frac{a}{2})}{m_1+m_2+m_3},\frac{m_3(\frac{\sqrt{3}a}{2})}{m_1+m_2+m_3})=$

$\large(\frac{24*0.12+36*0.06}{12+24+36}, \frac{36*0.86*0.12}{12+24+36})$ $=(0.07m, 0.05m)$