Equilibrium of Floating Bodies

Introduction Phenomenon of Floatation for a body Either it sinks down Or floats on the liquid Body placed over a liquid is subjected to two forces: Gravitational force Upthrust of the liquid If gravitational force > upthrust , body will sink. If gravitational force < upthrust , body will float.

Archimedes’s Principle “Whenever a body is immersed wholly or partially in a fluid, it is buoyed up (i.e. lifted up) by a force equal to the weight of the fluid displaced by the body”

Buoyancy The tendency of a fluid to uplift a submerged body, because of the upward thrust of the fluid. Equal to the weight of the fluid displaced by the body. If the force of buoyancy is greater than the weight of the body, it will be pushed up till the weight of the fluid displaced is equal to the weight of the body and the body will float. If the force of buoyancy is less than the weight of the body, it will sink down.

Center of Buoyancy The center of buoyancy is the point, through which the force of buoyancy is supposed to act. It is always the center of gravity of the volume of the liquid displaced. Center of buoyancy is the center of area of the immersed section.

Example A uniform body 3m long ,2m wide and 1m deep floats in water. If the depth of immersion is 0.6m, what is the weight of the body? Solution : Length = 3m Width = 2m Depth = 1m Depth of immersion = 0.6m Volume of water displaced = 3 x 2 x 0.6 = 3.6 m3 Weight of body = Weight of water displaced Density = mass /volume Weight = 9.81 kN/ m3 x 3.6 m3 = 35.3kN

Example A block of wood 4m long,2m wide and 1m deep is floating horizontally in water. If density of the wood be 6.87 kN/m3, find the volume of water displaced and position of the center of buoyancy. Solution: Volume of water displaced Volume of Wooden Block = 4m x 2m x 1m = 8m3 Weight = Volume x Density = 8 x 6.87 = 55kN Volume of water displaced = Weight of block / Density of water = 55 / 9.81 = 5.6 m3 Position of Center of buoyancy Depth of immersion = Volume / Sectional Area = 5.6 / (4 x 2) = 0.7 m Center of Buoyancy = 0.7 / 2 = 0.35m from the base

Metacenter Whenever a body, floating in a liquid, is given a small angular displacement, it starts oscillating about some point. This point, about which the body starts oscillating is called metacenter. The metacenter may also be defined as the inter section of the line passing through the original center of buoyancy B and c.g., G of the body and the vertical line through the new center of buoyancy B1.

Metacentric Height The distance between the center of gravity of a floating body and the metacenter (i.e distance GM) is called metacentric height. The metacentric height of a floating body is a direct measure of its stability. More the metacentric height of a floating body, more it will be stable

Analytical Method for Metacentric Height

Analytical Method for Metacentric Height Ship is given clockwise rotation through a small angle (ϴ) in radians. Immersed section has now changed from acd1e1 from acde. Center of buoyancy has now changed from B to B1. Triangular wedge “aom” has come out of water whereas triangular wedge ocn has gone under water. Since volume of water displaced remains the same, there two triangular wedges must have equal areas. Triangular wedge has come out of water, thus decreasing force of buoyancy on left, tends to rotate vessel in anti-clockwise direction about O. Triangular wedge “ocn” has gone under water, thus increasing force of buoyancy on right, tends to rotate vessel in anti clockwise direction. These forces of buoyancy will form a couple, which will tend to rotate vessel in anticlockwise direction about O

Analytical Method for Metacentric Height Angle ϴ through which body is given rotation, is extremely small, ship may be assumed to rotate about M (i.e. Metacenter) l = Length of ship b= Breadth of ship ϴ = very small angle in radians through which ship is rotated V = Volume of water displaced by ship

Analytical Method for Metacentric Height 𝐵𝑀= 𝐼 𝑉 𝐵𝑀= 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑃𝑙𝑎𝑛𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 Metacentric Height ; GM = BM ± BG +ive sign used if G is lower than B -ive sign used if G is higher than B

Conditions of Equilibrium of a Floating Body A body is said to be in equilibrium, when it remains in steady state while floating in a liquid Stable Equilibrium Unstable Equilibrium Neutral Equilibrium

Conditions of Equilibrium of a Floating Body Stable Equilibrium A body is said to be in stable equilibrium, if it returns back to its original position, when given a small angular displacement. This happens when the metacenter (M) is higher than the center of gravity (G) of the floating body.

Conditions of Equilibrium of a Floating Body Un-Stable Equilibrium A body is said to be in stable equilibrium, if it does not return back to its original position and heels further away, when given a small angular displacement. This happens when the metacenter (M) is lower than the center of gravity (G) of the floating body.

Conditions of Equilibrium of a Floating Body Neutral Equilibrium A body is said to be in stable equilibrium, if it occupies a new position and remains at rest in this new position, when given a small angular displacement. This happens when the metacenter (M) coincides with the center of gravity (G) of the floating body.

Example A rectangular pontoon of 5m long, 2m wide and 1.2m deep is immersed in 0.8m in sea water. If density of sea water is 10 kN/m3, find the metacentric height of pontoon?