L14 - BUOYANCY

# The centroid of the waterline plane m a a d d c o g b

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Unformatted text preview: ge in displacement, x = AO ρgθ Small area a x = CO ∑ ax = ρgθ ∑ ax x =0 x x =0 O ∑ ax = 0 But ∑ ax is the first moment of area of the waterline plane about OO, therefore the axis OO must pass through the centroid of the waterline plane M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O Distance BB’ can now be calculated, Couple produced by the movement of the wedge AOA’ to COC’ must be equal to the couple due to the movement of R from B to B’ Small area a x O M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O Small area a Moment about OO of the weight of fluid = ρgaxθ x x swept out by area a x O Total moment due to = ρgθ ∑ ax2 altered displacement M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O second moment of Putting ∑ ax2 = I = area of waterline plane about OO Small area a x Total moment due to = ρgθI altered displacement Moment due to = R x BB’ movement of R = ρgV x BB’ Where, V = volume of liquid displaced O M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O Small area a ρgV x BB’ = ρgθI x BB’ = θI / V O giving BM = BB’/θ BM = I/V The distance BM is known as the metacentric radius M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ O Small area a Substitute BM = I/V x Thus, GM = BM – BG GM = I/V – BG O M A A’ D D’ C’ O G B C B’ Upthrust, R = W θ...
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