Chapter 4 - Buoyancy Stability.doc

CHAPTER 4 Buoyancy and Floatation

Think!!! Why did the Titanic Sink?

Why did the Titanic Sink? 

After it hit the iceberg, water began to fill the air filled compartments on the ship.



The added weight of the water, combined with the weight of the ship became greater than the buoyant force supporting the ship.



We all know what happened after that!

Density and buoyancy: An object that has a greater density than the fluid it is in, will sink. If its density is less than the fluid it will float.

A solid block of steel sinks in water. A steel ship with the same mass floats on the surface.

Liquids

F Buoyant Force

w

Bouyancy • Some object when placed in water, float while others sink. • Example : Pontoon • A fluid particle can be replaced by a body •

Pressure distribution over the boundary will not change

• Buoyancy force on the body equals to the weight of the replaced fluid particle •

Buoyancy force applied to the geometrical center of a submerged body

Floating bodies •

•

•

•

Buoyancy force acting on the submerged part of a floating body equal to body’s weight

Body weight is applied to body’s center of mass Buoyancy force is applied to the geometrical center of the submerged part A floating body is in equilibrium when the pair gravity- bouyancy does not produce moment

Who is Archimedes?

•

Archimedes (287-212 BC), pre-eminent Greek mathematician and inventor, who wrote important works on plane and solid geometry, arithmetic, and mechanics. – "Archimedes",Microsoft« Encarta« Encyclopedia 2001. ⌐ 1993-2000 Microsoft Corporation. All rights reserved

.

Archimedes Principle State Any object wholly or partly immersed in a fluid, is buoyed up by a force (up trust force) equal to the weight of the fluid displaced by the object.

The law •

Archimedes' Principle, law of physics that states that when an object is totally or partially immersed in a fluid, it experiences an up thrust equal to the weight of the fluid displaced.

Centre of Bouyancy The center of gravity of the displaced liquid is known as the center of buoyancy

Density and Buoyancy From Archimedes’s Principle: Buoyant Force = Weight of fluid displaced = mg (note : F = ma) = Vg (note:  = m) V Thus FB = Vg Where…… F = Buoyant Force or Up thrust B  = Density of fluid V = Volume of fluid displaced or the volume of the object that immersed in the fluid.

Submerged Body

Buoyant Force 

The magnitude of the buoyant force always equals the weight of the displaced fluid

F   gV  wfluid The buoyant force is the same for a totally submerged object of any size, shape, or density  The buoyant force is exerted by the fluid  Whether an object sinks or floats depends on the relationship between the buoyant force and the weight 

Center of Buoyancy is a center of gravity of the displaced liquid G – center of gravity B – Centre of Buoyancy H – Height of object d – depth of immersion

W

G d

B F

H

Figure below shows a wood block measuring 0.4m x 0.3m 0.2m is floating in the water. The wood relative density is 0.6. Determine the value of d

ρ

wood

xgxV

wood

=ρ

water

xgxV

waterdisplaced

600 x 9.81 x [0.4 x 0.3 x 0.2] = 1000 x 9.81 x [0.4 x 0.3 x d] 141.264 = 1177.2d d = 0.12 m

Example 1 A ship floating in the sea is known to displace 2200 metric tons of water. Calculate the volume of the 3 ship below the water level. (V= 2146 m ) 3 Sea water density = 1025 kg/m 3

m = 2200 x 1000 = 2200 x 10 kg ρ V

= m/V = 2200000 / 1025 3 = 2146 m

Example 2 A rectangular wooden block 100 cm x 50 cm x 40 cm is left floating in the water. If it is observed that a quarter of the volume of the block of wood lies above the water surface, calculate the mass of the wooden block. (m = 150 kg)

EXAMPLE 3 A block of wood 4 m long, 2 m wide and 1 m deep is floating horizontally in the water. If the density of wood is 700 kg/m3, calculate the: a)Mass of wooden block (5600 kg) 3

b)Volume of water displaced (5.6 m ) c)Depth of immersion, d (0.7 m) d)Position of center of buoyancy (0.35 m)

Solution Mass of wooden block ρ = m/V Vb = b x L x h =4 x 2 x 1 =

8m

3

Therefore; mb = ρb Vb = 700 (8) = 5600 kg

Volume of water displaced Vw = Vb Vw = mw/ρw = 5600 3 kg/1000kg/m 3 = 5.6 m Depth of Immersion, d Vi = b x L x d 5.6 = 4 x2xd d = 0.7 m

Centre of buoyancy B = d/2 = 0.7/2 = 0.35 m

EXAMPLE 4 An unloaded pontoon being used in a river estuary to transport construction materials has a mass of 15 tons. In plan the pontoon is 8 m long by 5 m wide. It is rectangular in section and has sides 1.5 m high. The water 3 is saline with density 1025 kg/m . Determine :a)

3

The volume of the water displaced (14.64 m )

b) Depth of immersion (0.366 m) c) The position of center of buoyancy (0.183 m)

EXAMPLE 5 An object weighing 100 N in air was found to weigh 75 N when fully immersed in water. If the object is in equilibrium, calculate the: a) Mass of the object (10.19 kg) -3

3

Volume of the object (2.55 x 10 m ) c) Relative density of the object (4.0) b)

Solution Weight in air = 100 N Weight in water = 75 N

Up trust Force, F = 100 – 75 = 25 N W = mg m = W/g = 100/9.81 = 10.19 kg Relative density Object density, ρ = m/V

Volume of object = Volume of water displaced Vb = Vw Therefore; F=ρgV 25 = 1000(9.81)Vw -3

Vw = 2.55 x 10 m

= 10.19/ 2.55 x 10 = 3996 kg/m3

3

Gs = ρb/ρw -3

= 3996/1000 = 4.0

Metacenter When ever a body floating in liquid is given small angular displacement, it starts oscillating about some point. This point, about which the body starts oscillating, it call Meta Centre. Metacenter – intersection of the line passing through the original Centre of buoyancy (B) and Centre of gravity of the body, and the vertical line through the new Centre of buoyancy.

Stability of Submerged or Floating Bodies Neutral Equilibrium -

Metacenter (M) is coincides with the center of gravity (G) of the body - GM=0 (neutral) -

A Ship in Equilibrium

If it occupies a new position and remains at rest in the new position when given small displacement.

Stable Equilibrium Stable equilibrium - If it returns back to its original position, when given a small angular displacement. Metacenter (M) is higher than the center of gravity (G) of the body - GM>0 (stable) -

Stable Equilibrium

Unstable Equilibrium Unstable equilibrium -

Unstable Equilibrium

If it does not return back to its original position and heels farther away, when given a small angular displacement.

- Meta center (M) is lower than the center of gravity (G) of the body - GM<0 (Unstable)

Step to calculate GM a) Calculate the depth of immersion, d d/H H

= Density of object/Density of fluid = height of object

b) Calculate the Height of Meta Center (GM) GM BM

= BM – BG = Ixx/Vd

G 4

d

Ixx Vd

– second moment of area (m ) 3 – Volume of displaced liquid (m )

BG

= OG – OB , which OG = H/2 and OB = d/2

B O

H

Exercise A floating pontoon is being used to transport construction materials. If it is known to displaced water if the pontoon floats on fresh water. 5.4 m wide, 12 m long and depth of immersion 1.5 m in fresh water. i. Mass of the pontoon if it floats in water a)

ii. Volume of the displaced water if the pontoon floats on fresh water. iii. Calculate the metacentric height GM iv. State the type of equilibrium

An object weighing 2703 N in air is immersed in water and recorded a weight of 1909 N. If the object equilibrium calculate the b)

i.

ii.

3

Volume of the object (0.081 m ) Relative density of the object (3.4)

c) Determine the metacentric height of a vehicle to transport pontoons that across a strait of sea water with a 3 density of 1150 kg/m . The pontoon measuring is 27 m in long, 19 m in wide and 9 m in high and the weight of the pontoon is 500 tones. 3

Vbody = 4617 m 3 Ρbody = 108.30 kg/m d = 0.85 m BG = 4.075 m 3

Vd = 436.05 m 4 Ixx= 15432.75 m BM = 35.392 m GM = 31.317 m

Determine the metacentric of a ferry across the Selat

Tebrau. The sea water density is 1029 kg/m3. The ferry dimension is 40 mx 15mx10m. The ferry mass is 700 tones metric.

Answer: Vd = 680.272 m3 Ixx = 11 250 m4 BM = 16.54 Density object = 116.67 kg/m3 d = 1.13 m BG = 4.435m GM = 12.105 m.

END OF SLIDE QUIZ AND TUTORIAL