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 density1025 kg/m. Its CG is 0.45m from the bottom. If a load of 3KN is placed on the top, find the maximumheight of the CG of this load above the bottom if the buoy is to remain in stable equilibrium.Solution:G = CG of buoyG1G1 = CG of load 3KNG' = Combined CG of load and buoy%DMh = depth of immersionG'В 3 СВ%3D1.2mOG =0.45mGВOG1=?hFinding depth of immersion (h)Weight of load and buoy = Weight of water displaced%3Dh = 0.53mOB = 0.53/2 = 0.265m0.38mBG' = OGʻ-OB = OGʻ-0.265G'M = MB-BG' = 0.38- OGʻ+0.265 = 0.645-OG'For stable equilibrium, G'M>0%3D%3D%3D0.645-OG'>0OGʻ < 0.645mTaking moments about O,3x10°xOG1 + 10.5x10°x0.45 = (3x10³+10.5x10³)x0.645OG1 = 1.3275m%3D

For checking the system is in stable or unstable equilibrium ,then tilted by a small angle then…

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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?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Section: Chapter Questions

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Concept explainers

Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its environment. This branch of physics keeps its roots in Ancient Greece in the form of the scripts of Aristotle and Archimedes.

Question

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

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