Four point masses are placed at the corners of a square of side 2 m as shown in the figure. Find the centre of mass of the system w.r.t. the centre of square.

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Solution

A=1k.g
B=2k.g
C=3k.g
D=4k.g
$X_{c} m = \frac{m_{1} \times x_{1} + m_{2} \times x_{2} + m_{3} \times x_{3} + m_{4} \times x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$X_{c} m = \frac{(1 \times 1) + (2 \times 1) + (3 \times - 1) + (4 \times - 1)}{1 + 2 + 3 + 4}$
$X_{c} m = \frac{1 + 2 - 3 - 4}{10} = \frac{- 4}{10}$
$X_{c} m = - 0.4$
$Y_{c} m = \frac{m_{1} \times y_{1} + m_{2} \times y_{2} + m_{3} \times y_{3} + m_{4} \times y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$Y_{c} m = \frac{(1 \times 1) + (- 1 \times 2) + (- 1 \times 3) + (4 \times 1)}{1 + 2 + 3 + 4}$
$Y_{c} m = \frac{1 - 2 - 3 + 4}{10} = \frac{0}{10}$
$Y_{c} m = 0$
The coordinate of center of mass (-0.4,0)

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Four point masses are placed at the corners of a square of side 2 m as shown in the figure. Find the centre of mass of the system w.r.t. the centre of square.

A=1k.gB=2k.gC=3k.gD=4k.gXcm=m1×x1+m2×x2+m3×x3+m4×x4m1+m2+m3+m4Xcm=(1×1)+(2×1)+(3×−1)+(4×−1)1+2+3+4Xcm=1+2−3−410=−410Xcm=−0.4Ycm=m1×y1+m2×y2+m3×y3+m4×y4m1+m2+m3+m4Ycm=(1×1)+(−1×2)+(−1×3)+(4×1)1+2+3+4Ycm=1−2−3+410=010Ycm=0The coordinate of center of mass (-0.4,0)