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CHAPTER – 3 FLOTATION Chapter – 3 Flotation Exercise -3

Answer 1 Given:1. Dimensions of wooden log = 8m * 2m * 2m 2. Draft at FW = 1.6m RD of FW = 1.000

2m 1.6m 2m 8m

Volume displaced by log = L * B *draft Volume displaced by log = 8 * 2 * 1.6 Volume displaced by log = 25.6m3 RD of FW =

Mass of log Volume of water displaced

1.000 = Mass of log 25.6 25.6 t = mass of log Volume of log = L * B H Volume of log = 8 * 2 * 2 Volume of log = 32m3 Density of log = Mass of log Volume of log

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CHAPTER – 3 FLOTATION Density of log = 25.6t 32.0m3 Density of log = 0.8t/m3 1. Mass of log is 25.6t 2. Density of log is 0.8t/m3

Answer 2 Given:1. Dimensions of log = 5m * 1.6m * 1m 2. Mass of log = 6t

1m 1.6m 5m Volume of log = L * B * H Volume of log = 5 * 1.6 * 1 Volume of log = 8m3 Density of log = Mass of log Volume of log Density of log = 6 8 Density of log = 0.75 t/m3 RD = Density of log Density of FW RD = 0.75 1.0 RD = 0.75 Mass of SW displaced = 6t

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CHAPTER – 3 FLOTATION Density of SW =

Mass of log Volume displaced

1.025t/m3

= 6t . L * B * draft

1.025t/m3

= 6t . 5 * 1.6 * draft

Draft

= 6 . 5 * 1.6 * 1.025

Draft

= 0.732m

1. Draft in SW is 0.732m 2. RD of log is 0.75

Answer 3 Given:1. 2. 3. 4.

Breath of log = 3m Height of log = 2m RD of water = 1.01 Density of log = 0.7t/m3

3m

2m

Let length of log = l Let draft of log = d Density of log = Mass of log Volume of log 0.7 = Mass of log 3*2*l Page 3 of 13

CHAPTER – 3 FLOTATION 0.7 * 3 * 2 * l = mass of log Density of water= Mass of log Under water Volume 1.01

= 0.7 * 3 * 2 * l l*3*d

1.01 * 3 * l *d = 0.7 * 3 * 2 * l d

= 1.386m 1. Draft of log is 1.386m

Answer 4 Given:1. Diameter of cylinder 2. Length of cylinder 3. Draft

= 2m = 10m = 0.6 m

RD of FW = 1.000 NOTE : VOLUME OF ANY 3 DIMENSIONAL UNIFORM OBJECT IS GIVEN BY THE FORMULA BASE AREA * HEIGHT

2m A B 0.6m

E

C

D

Area of segment BDC = Area of sector ABDC – Area of triangle ABC AB = AD = AC = 1m (Radius of the circle) AE + ED = AD AE = AD – ED AE = 1m – 0.6m Page 4 of 13

CHAPTER – 3 FLOTATION AE = 0.4m In right angle triangle AEC (AE)2 + (EC)2 = (AC)2

(Pythagoras Theorem)

(0.4)2 + (EC)2 = (1)2 (EC)2 = (1)2 – (0.4)2 (EC)2 = 1 – 0.16 (EC)2 = 0.84 EC = √0.84 EC = 0.917m BE = EC (AE is perpendicular bisector on BC) BC = BE + EC BC = 0.917m + 0.917m BC = 1.834m Area of triangle ABC = 1 * Base * Height 2 Area of triangle ABC = 1 * BC * AE 2 Area of triangle ABC = 1 * 1.834 * 0.4 2 Area of triangle ABC = 0.3668m2 In triangle AEC Cos Ѳ = AE AC

(CosѲ = B) H

CosѲ = 0.4 1.0 Ѳ = Cos-1(0.4) Ѳ = 66.42° Similarly in triangle ABE, BAE = 66.42° BAC =

BAE + EAC

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CHAPTER – 3 FLOTATION BAC = 66.42° + 66.42° BAC = 132.84° Area of sector ABDC = Ѳ * ∏ * r2 360° Area of sector ABDC = 132.84° * 22 * (1)2 360° * 7 Area of sector ABDC = 1.1597m2 Area of segment BDC = Area of sector ABDC – Area of triangle ABC Area of segment BDC = 1.1597m2 – 0.3668m2 Area of segment BDC = 0.7929m2 Under water volume = Area of segment BDC * Height Under water volume = 0.7929 * 10 Under water volume = 7.929 m3 Density of FW =

Mass of log Under water volume of log

1 = Mass of log 7.929 1 * 7.929 = Mass of log 7.929 t = Mass of log Mass of log is 7.929 t

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CHAPTER – 3 FLOTATION Answer 5 Given:1. 2. 3. 4.

Length of triangular barge = 20m Breath of triangular barge = 12 m Depth of triangular barge = 06m Draft of triangular barge = 04m

G

12 m

B

C 6m

D

E

F

4m

A BG = GC (AG is perpendicular bisector on BC) BG = GC = 6m In triangle AGC and triangle AFE 1. 2. 3.

EAF = EAF AFE = AGC AEF = ACG

(Common angle) (Corresponding angles) (Corresponding angles)

Triangle AGC is similar to triangle AFE by AAA similarity. We know that sides of similar triangle are in equal ration, So AF = EF AG GC 4 = EF 6 6 4 * 6 = EF 6 Page 7 of 13

CHAPTER – 3 FLOTATION 4m = EF EF = FD

(AF is perpendicular bisector on DE)

4m = FD ED = EF + FD ED = 4m + 4m ED = 8m Area of triangle ADE = 1 * Base * Height 2 Area of triangle ADE = 1 * DE * FA 2 Area of triangle ADE = 1 * 8 * 4 2 Area of triangle ADE = 16m2 Density of SW = Mass of barge Underwater Volume of barge Density of SW = Mass of barge Underwater surface area of barge * length 1.025

= Mass of barge 16 *20

1.025 * 16 * 20 = Mass of barge 328 t

= Mass of barge

Answer 6 Given:1. 2. 3. 4.

Diameter of drum Height of drum RD of water Draft

= 1.2m = 2m = 1.016 = 1.4m

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CHAPTER – 3 FLOTATION

1.2m

1.4m

2m

Radius = Diameter 2 Radius = 1.2 2 Radius = 0.6m Height = Freeboard + Draft 2m = Freeboard + 1.4m 2m – 1.4m = Freeboard 0.6m = Freeboard Density of water

= Mass of lead Volume above water

Density of water

= Mass of lead ∏ * r2 * freeboard

1.016

= Mass of lead ∏ * r2 * 0.6

1.016 * 0.6 * ∏ * r2

= Mass of lead

1.016 * 0.6 * 22 * (0.6)2 = Mass of lead 7 1.016 * 0.6 * 22 * 0.6 * 0.6= Mass of lead 7 0.69 t

= Mass of lead

Maximum amount of lead mass that can be put in without sinking is 0.69 t Page 9 of 13

CHAPTER – 3 FLOTATION Answer 7 Given:1. 2. 3. 4.

Length of barge Breath of barge Initial draft Final draft

= 10m = 5m = 3m = 1m

RD of SW = 1.025

3m

1m

5m 10 m

Density of SW = Initial mass of barge Underwater Volume of barge 1.025

= Initial mass of barge L * B * draft

1.025

=

153.75t

= Initial mass of barge

Density of SW =

Initial mass of barge 10 * 5 * 3

Final mass of barge Underwater volume of barge

Density of SW = Final mass of barge L * B * draft 1.025

= Final mass of barge 10 * 5 * 1

1.025 * 10 * 5 * 1= Final mass of barge Page 10 of 13

CHAPTER – 3 FLOTATION 51.25 t

= Final mass of barge

Load on crane = change in displacement Load on crane = Initial displacement – Final displacement Load on crane = 153.75t – 51.25t Load on crane = 102.5t Load on crane is 102.5 t

Answer 8 Given:1. 2. 3. 4. 5.

Length of box Width of box Height of box RD of water Draft of box

= 2.4m = 1.2m = 0.8m = 1.012 = 0.2m

1.2 m 2.4 m 0.8 m

0.2 m

Freeboard of box

= Total height – Draft of box

Freeboard of box

= 0.8m – 0.2m

Freeboard of box

= 0.6m

RD of water

= Mass of SW to pour Volume above water

RD of water

= Mass of SW to pour L * B * Freeboard

1.012

= Mass of SW to pour 2.4m * 1.2m * 0.6m Page 11 of 13

CHAPTER – 3 FLOTATION 1.012 * 2.4 * 1.2 * 0.6 = Mass of SW to pour 1.749 t

= Mass of SW to pour

Maximum amount of SW that can be poured into without sinking is 1.749t

Answer 9 Given:1. Displacement of SW 2. Length of vessel 3. Breath of vessel

= 18450t = 150m = 20 m

18450 t

20 m 150 m

RD of SW

= 1.025

RD of SW

= Displacement Underwater Volume

RD of SW

= Displacement L * B * draft

1.025

= 18450 150 * 20 * draft

1.025 * 150 * 20 * draft = 18450 Draft

= 18450 1.025 * 150 * 20

Draft

= 6m

Draft of the vessel is 6m

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CHAPTER – 3 FLOTATION Answer 10 Given:1. 2. 3. 4. 5.

Length of vessel = 120m Breath of vessel = 15m RD of DW = 1.005 Draft = 5m Maximum permissible draft = 6m (in SW)

NOTE : FOR A BOX SHAPED VESSEL DENSITY IS INVERSELY PROPORTIONAL TO DRAFT Let initial draft in SW = x 5 = 1.025 X 1.005 5 * 1.005 = x 1.025 4.902m = x Maximum permissible draft (SW) = Initial draft in SW + Permissible sinkage 6m

= 4.902m + Permissible sinkage

RD of SW = Cargo to load Permissible sinkage of volume RD of SW = Cargo to load L * B * Permissible sinkage 1.025

= Cargo to load 120 * 15 * 1.098

1.025 * 120 * 15 * 1.098 = Cargo to load 2025.8 t = Cargo to load Cargo to load is 2025.8t

-o0o-

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