find-the-center-of-gravity-with-respect-to-point-O-

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

• A big full square : S_1 = 10×10 = 100 cm^2 Ω_1 (5,5) • A small empty rectangle : S_2 = −4×2 = −8 cm^2 Ω_2 (8,3) • A small empty rectangle : S_3 = −4×2 = −8 cm^2 Ω_3 (8,7) S = S_1 +S_2 +S_3 = 100−8−8 = 84 cm^2 SOG^(→) = S_1 OΩ_1 ^(→) +S_2 OΩ_2 ^(→) +S_3 OΩ_3 ^(→) OG^(→) = ((100)/(84))(5,5)−(8/(84))(8,3)−(8/(84))(8,7) G(((31)/7),5)

$$\bullet\:\mathrm{A}\:\mathrm{big}\:\mathrm{full}\:\mathrm{square}\:: \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\mathrm{10}×\mathrm{10}\:=\:\mathrm{100}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{1}} \left(\mathrm{5},\mathrm{5}\right) \\ $$$$ \\ $$$$\bullet\:\mathrm{A}\:\mathrm{small}\:\mathrm{empty}\:\mathrm{rectangle}\:: \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:−\mathrm{4}×\mathrm{2}\:=\:−\mathrm{8}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{2}} \left(\mathrm{8},\mathrm{3}\right) \\ $$$$ \\ $$$$\bullet\:\mathrm{A}\:\mathrm{small}\:\mathrm{empty}\:\mathrm{rectangle}\:: \\ $$$$\mathrm{S}_{\mathrm{3}} \:=\:−\mathrm{4}×\mathrm{2}\:=\:−\mathrm{8}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{3}} \left(\mathrm{8},\mathrm{7}\right) \\ $$$$ \\ $$$$\mathrm{S}\:=\:\mathrm{S}_{\mathrm{1}} +\mathrm{S}_{\mathrm{2}} +\mathrm{S}_{\mathrm{3}} \:=\:\mathrm{100}−\mathrm{8}−\mathrm{8}\:=\:\mathrm{84}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{S}\overset{\rightarrow} {\mathrm{OG}}\:=\:\mathrm{S}_{\mathrm{1}} \overset{\rightarrow} {\mathrm{O}\Omega_{\mathrm{1}} }+\mathrm{S}_{\mathrm{2}} \overset{\rightarrow} {\mathrm{O}\Omega_{\mathrm{2}} }+\mathrm{S}_{\mathrm{3}} \overset{\rightarrow} {\mathrm{O}\Omega_{\mathrm{3}} } \\ $$$$\overset{\rightarrow} {\mathrm{OG}}\:=\:\frac{\mathrm{100}}{\mathrm{84}}\left(\mathrm{5},\mathrm{5}\right)−\frac{\mathrm{8}}{\mathrm{84}}\left(\mathrm{8},\mathrm{3}\right)−\frac{\mathrm{8}}{\mathrm{84}}\left(\mathrm{8},\mathrm{7}\right) \\ $$$$\mathrm{G}\left(\frac{\mathrm{31}}{\mathrm{7}},\mathrm{5}\right) \\ $$

Commented by john_santu last updated on 18/Aug/21

$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\Omega_{\mathrm{1}} \left(\mathrm{5},\mathrm{5}\right)\:. \\ $$

Commented by Olaf_Thorendsen last updated on 18/Aug/21

the square is 10×10. the center of gravity is (5,5).

$$\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\mathrm{10}×\mathrm{10}. \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{gravity}\:\mathrm{is}\:\left(\mathrm{5},\mathrm{5}\right). \\ $$

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