Center of mass three object system is given by:
In x-direction
Xcm = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)
In y-direction
Ycm = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3)
Given m1 = m2 = m3 = 82.0 gm = m
m1 = mA = 82.0 g, (x1, y1) = (0, 0)
m2 = mB = 82.0 g, (x2, y2) = (10.2, 21.5)
m3 = mC = 82.0 g, (x3, y3) = (16.3, 15.4)
(Xcm, Ycm) = (?, ?)
Now
Xcm = (m*0 + m*10.2 + m*16.3)/(m + m + m)
Xcm = (0 + 10.2 + 16.3)/3
Xcm = 8.83 cm
Ycm = (m*0 + m*21.5 + m*15.4)/(m + m + m)
Ycm = (0 + 21.5 + 15.4)/3
Ycm = 12.3 cm
So, Xcm = 8.83 cm, Ycm = 12.3 cm
Part B. when two additional masses are added, then
Center of mass five object system is given by:
In x-direction
Xcm = (m1x1 + m2x2 + m3x3 + m4x4 + m5x5)/(m1 + m2 + m3 + m4 + m5)
In y-direction
Ycm = (m1y1 + m2y2 + m3y3 + m4y4 + m5y5)/(m1 + m2 + m3 + m4 + m5)
Given m1 = m2 = m3 = m4 = m5 = 82.0 gm = m
m1 = mA = 82.0 g, (x1, y1) = (0, 0)
m2 = mB = 82.0 g, (x2, y2) = (10.2, 21.5)
m3 = mC = 82.0 g, (x3, y3) = (16.3, 15.4)
m4 = mD = 82.0 g, (x4, y4) = (0, -24.7)
m5 = mE = 82.0 g, (x5, y5) = (0, 24.7)
(Xcm, Ycm) = (?, ?)
Now
Xcm = (m*0 + m*10.2 + m*16.3 + m*0 + m*0)/(m + m + m + m + m)
Xcm = (0 + 10.2 + 16.3 + 0 + 0)/5
Xcm = 5.3 cm
Ycm = (m*0 + m*21.5 + m*15.4 + m*(-24.7) + m*24.7)/(m + m + m + m + m)
Ycm = (0 + 21.5 + 15.4 - 24.7 + 24.7)/5
Ycm = 7.38 cm
So, Xcm = 5.3 cm, Ycm = 7.38 cm
Let me know if you've any query.
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