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Can anyone explain why you can take moment at O? Wouldn't all the forces be cancelled since all of them are vertical forces and they are in the line of action?

Can anyone explain why you can take moment at O? Wouldn't all the forces be cancelled since all of them are vertical forces and they are in the line of action?

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Can anyone explain why you can take moment at O? Wouldn't all the forces be cancelled since all of them are vertical forces and they are in the line of action?

19. A cylindrical buoy 1.8m in diameter, 1.2m high and weighing 10.5 KN floats in salt water of density 1025 kg/m. Its CG is 0.45m from the bottom. If a load of 3KN is placed on the top, find the maximum height of the CG of this load above the bottom if the buoy is to remain in stable equilibrium. Solution: G = CG of buoy G1 G1 = CG of load 3KN G' = Combined CG of load and buoy %D M h = depth of immersion G' В 3 СВ %3D 1.2m OG =0.45m G В OG1=? h Finding depth of immersion (h) Weight of load and buoy = Weight of water displaced %3D h = 0.53m OB = 0.53/2 = 0.265m 0.38m BG' = OGʻ-OB = OGʻ-0.265 G'M = MB-BG' = 0.38- OGʻ+0.265 = 0.645-OG' For stable equilibrium, G'M>0 %3D %3D %3D 0.645-OG'>0 OGʻ < 0.645m Taking moments about O, 3x10°xOG1 + 10.5x10°x0.45 = (3x10³+10.5x10³)x0.645 OG1 = 1.3275m %3D
Transcribed Image Text:19. A cylindrical buoy 1.8m in diameter, 1.2m high and weighing 10.5 KN floats in salt water of density 1025 kg/m. Its CG is 0.45m from the bottom. If a load of 3KN is placed on the top, find the maximum height of the CG of this load above the bottom if the buoy is to remain in stable equilibrium. Solution: G = CG of buoy G1 G1 = CG of load 3KN G' = Combined CG of load and buoy %D M h = depth of immersion G' В 3 СВ %3D 1.2m OG =0.45m G В OG1=? h Finding depth of immersion (h) Weight of load and buoy = Weight of water displaced %3D h = 0.53m OB = 0.53/2 = 0.265m 0.38m BG' = OGʻ-OB = OGʻ-0.265 G'M = MB-BG' = 0.38- OGʻ+0.265 = 0.645-OG' For stable equilibrium, G'M>0 %3D %3D %3D 0.645-OG'>0 OGʻ < 0.645m Taking moments about O, 3x10°xOG1 + 10.5x10°x0.45 = (3x10³+10.5x10³)x0.645 OG1 = 1.3275m %3D
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