Question #271861

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?


1
Expert's answer
2021-11-26T11:53:18-0500


The center of mass of the system is m1=12kg  m2=24kg  m3=36kgm_1=12kg \space\space m_2=24kg \space \space m_3=36kg a=0.12ma=0.12m

(X,Y)=(m1x1+m2x2+m3x3m1+m2+m3,m1y1+m2y2+m3y3m1+m2+m3)=\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})=


(m1(0)+m2(a)+m3(a2)m1+m2+m3,m1(0)+m2(0)+m3(3a2)m1+m2+m3)=\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})=


(m2(a)+m3(a2)m1+m2+m3,m3(3a2)m1+m2+m3)=\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})=


(240.12+360.0612+24+36,360.860.1212+24+36)\large(\frac{24*0.12+36*0.06}{12+24+36}, \frac{36*0.86*0.12}{12+24+36}) =(0.07m,0.05m)=(0.07m, 0.05m)



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