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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

Commented by john_santu last updated on 18/Aug/21

Commented by Olaf_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 by john_santu last updated on 18/Aug/21

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

Commented by Olaf_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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