Center of Mass and Linear momentum Problem?

Question

A 2.53 kg particle has the xy coordinates (-1.36 m, 0.836 m), and a 3.24 kg particle has the xy coordinates (0.541 m, -0.112 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 4.11 kg particle such that the center of mass of the three-particle system has the coordinates (-0.515 m, -0.727 m)?

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Answer 1 · 2018-04-06T07:08:20

#"the (x,y) coordinates of the mass center of the other mass is:"#

#CM(-0.827,-2.173)#

#"let the mass center coordinates of the three-mass system be (x,y)."#

#"The mass center of the system can be calculated using the fallowing formula."#

#x=(Sigma m*x)/(Sigma m)=(m_1x_1+m_2x_2+m_3x_3+...)/(m_1+m_2+m_3+...)#

#y=(Sigma m*y)/(Sigma m)=(m_1y_1+m_2y_2+m_3y_3+...)/(m_1+m_2+m_3+...)#

#m_1=2.53 " "kg" "x_1=-1.36" "y_1=0.836#

#m_2=3.24 " "kg" "x_2=0.541" "y_2=-0.112#

#m_3=4.11 " "kg" "x_3= ?" "y_3= ?#

#m=m_1+m_2+m_3=2.53+3.24+4.11=9.88" "kg#

#-0.515=(-2.53*1.360+3.24*0.541+4.11*x_3)/(9.88)#

#-0.515=(-1.69+4.11*x_3)/(9.88)#

#-0.515*9.88+1.69=4.11*x_3#

#-3.398=4.11*x_3#

#x_3=(-3.398)/(4.11)#

#x_3=-0.827#

#-0.272=(2.53*0.836-3.24*0.112+4.11*y_3)/(9.88)#

#-0.727=(1.75+4.11*y_3)/(9.88)#

#-0.727*9.88-1.75=4.11*y_3#

#-8.933=4.11*y_3#

#y_3=(-8.933)/(4.11)#

#y_3=-2.173#

#"the (x,y) coordinates of the mass center of the other mass is:"#

#CM(-0.827,-2.173)#