Regarded as gm negative negative if m coincides with

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Unformatted text preview: oduced, GM • Equilibrium is unstable, Equilibrium unstable and • GM is regarded as GM negative negative • If M coincides with G, If coincides • Body is in neutral Body equilibrium equilibrium W = mg G B R=W θ θ G G M B1 Overturning moment M B1 x RW Determination of The Metacentric Height • Metacentric height of a vessel can be determined if the angle of tilt θ cause by moving a load P a know distance x across the deck is measured P • Overturning moment due to movement of load P = Px • If GM is the metacentric height, and • W = mg is the total weight of the vessel including P Righting moment = W x GM x θ θ x P P • For equilibrium in tilt position, the righting moment must equal to overturning moment, So that, W x GM x θ = Px Metacentric height, GM = Px/Wθ The true metacentric height is the value GM as θ 0 θ x P P Determination of The Position of The Metacentre Relative To The Centre of Buoyancy • Metacentric height GM determine the stability of a floating body • For a vessel of known shape and displacement, • Position of the centre of gravity G and buoyancy B is known • So GM can be found by knowing the distance BM GM = BM – BG O Small area a x O M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O • AC is the original waterline plane, and • B is the centre of buoyancy in equilibrium position • When the vessel is tilted through a small angle θ, • Centre of buoyancy will move to B’ (result in alteration in the shape of the displaced fluid) Small area a x O M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O • A’C’ is the waterline plane in the displaced position • For small angle of tilt, BM = BB’/θ Small area a x O • Movement of B to B’ is the result of removal of fluid volume AOA’ and addition of COC’ • Total weight of fluid displaced remains unchanged M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ Weight of wedge AOA’ = O Weight of wedge COC’ • a is a small area in the waterline plane at a distance x from the axis of rotation OO • It generates a volume when the vessel is tilted Small area a x O M A A’ Volume swept = DD’ x a = axθ out by a D D’ C’ O G B C B’ Upthrust, R = W θ O Summing all such volumes and, Multiply by the specific weight ρg of the liquid, Weight of = wedge AOA’ Small area a x x = AO ∑ ρgaxθ O x =0 M A Similarly, A’ x = CO Weight of = wedge COC’ ∑ ρgaxθ x =0 D D’ C’ O G B C B’ Upthrust, R = W θ O Since there is no chan...
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