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Question Number 151096 by john_santu last updated on 18/Aug/21

$$$$ find the center of gravity with respect to point O\n

Commented byjohn_santu last updated on 18/Aug/21

Commented byOlaf_Thorendsen last updated on 18/Aug/21

$$\bullet \mathrm{A}\mathrm{big}\mathrm{full}\mathrm{square}:$$ $${\mathrm{S}}_1=10\times 10=100{\mathrm{cm}}^2$$ $${\mathrm{\Omega }}_1(5,5)$$ $$$$ $$\bullet \mathrm{A}\mathrm{small}\mathrm{empty}\mathrm{rectangle}:$$ $${\mathrm{S}}_2=-4\times 2=-8{\mathrm{cm}}^2$$ $${\mathrm{\Omega }}_2(8,3)$$ $$$$ $$\bullet \mathrm{A}\mathrm{small}\mathrm{empty}\mathrm{rectangle}:$$ $${\mathrm{S}}_3=-4\times 2=-8{\mathrm{cm}}^2$$ $${\mathrm{\Omega }}_3(8,7)$$ $$$$ $$\mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3=100-8-8=84{\mathrm{cm}}^2$$ $$$$ $$\mathrm{S}\overrightarrow{\mathrm{OG}}={\mathrm{S}}_1\overrightarrow{\mathrm{O}{\mathrm{\Omega }}_1}+{\mathrm{S}}_2\overrightarrow{\mathrm{O}{\mathrm{\Omega }}_2}+{\mathrm{S}}_3\overrightarrow{\mathrm{O}{\mathrm{\Omega }}_3}$$ $$\overrightarrow{\mathrm{OG}}=\frac{100}{84}(5,5)-\frac{8}{84}(8,3)-\frac{8}{84}(8,7)$$ $$\mathrm{G}(\frac{31}{7},5)$$

Commented byjohn_santu last updated on 18/Aug/21

$$\mathrm{how}\mathrm{to}\mathrm{get}{\mathrm{\Omega }}_1(5,5).$$

Commented byOlaf_Thorendsen last updated on 18/Aug/21

$$\mathrm{the}\mathrm{square}\mathrm{is}10\times 10.$$ $$\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{gravity}\mathrm{is}(5,5).$$

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