Four bodies of masses 2, 3, 5 and 8,kg are pl | vedclass

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Question

The Co-ordinates of the corners of the square are

$(0,0),(2,0),(2,2),(0,2) .$ Hence,

$\mathrm{X}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{

Vedclass · Accepted Answer

$\left( {\frac{8}{9},\frac{{13}}{9}} \right)$

Vedclass · Answer

The Co-ordinates of the corners of the square are

$(0,0),(2,0),(2,2),(0,2) .$ Hence,

$\mathrm{X}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}+\mathrm{m}_{4} \mathrm{x}_{4}}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}+\mathrm{m}_{4}}$

$=\frac{2 	imes 0+3 	imes 2+5 	imes 2+8 	imes 0}{2+3+5+8}=\frac{16}{18}=\frac{8}{9} \mathrm{m}$

$\mathrm{Y}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{y}_{1}+\mathrm{m}_{2} \mathrm{y}_{2}+\mathrm{m}_{3} \mathrm{y}_{3}+\mathrm{m}_{4} \mathrm{y}_{4}}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}+\mathrm{m}_{4}}$

$=\frac{2 	imes 0+3 	imes 0+5 	imes 2+8 	imes 2}{2+3+5+8}=\frac{26}{18}=\frac{13}{9} \mathrm{m}$

$	herefore$Co-ordinates of the centre of mass $=\left(\frac{8}{9}, \frac{13}{9}ight)$

Four bodies of masses $2, 3, 5$ and $8\,kg$ are placed at the four corners of a square of side $2\,m$ as shown. The position of $CM$ will be

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