# Stability I – Chapter 3 Flotation Solutions

**A rectangular log of wood 8m long, 2m wide and 2m high floats in FW at a draft of 1.6m with one face of horizontal. Find its mass and RD.**

*Solution :*

*Solution :*

*Volume of rectangular log = (L x b x h)**= (8m x 2m x 2m)**= 32m ^{3}*

*Weight = (U/W volume ) x (Density of displaced water)**Weight = (8 x 2 x1.6) x(1)**= 25. 6t*

*RD = (mass / volume)**= 25.6 / 32**= 0.8t/m ^{3} .*

**A rectangular log of wood 5m x 1.6m x1.0m weight 6t and floats with it largest face horizontal. Find its draft in SW and its RD.**

*Solution :*

*Solution :*

*Volume of rectangular log = (L x B X H )**= (5m x 1.6m x 1.0m)*

*Weight of log = 6t**SW RD = 1.025*

*Weight = (u/w volume )x (Density of water displaced)**6 = (5 x 1.6 x D) x (1.025)**D = (6 /( 5 x 1.6 x1.025)**= 0.73m**Hence draft = 0.73m*

*As we know that**Density = (mass /volume)**= (6 / (5 x 1.6 x 1 )**=0.75 t/m ^{3}*

**A rectangular log 3m broad and 2m high floats with its breadth horizontal. If the density of the log is 0.7 tm-3, find its draft In water of RD 1.01.**

*Solution :*

*Solution :*

*Area of rectangular log = ( B X H ) = 3m x 2m**RD of log = 0.7 t/m ^{3}*

*Draft in water of RD 1.01 can be calculated as;*

*Density= (mass/volume)*

*0.7 = mass/ volume*

*Mass = volume x 0.7*

*Mass = (L x 3 x2) x ( 0.7)*

*= 4.2L t**Now mass = (U/w volume) at depth of D m x (1.01)**4.2L =( L x 3 x D )x( 1.01)**D = ( 4.2 / 3 x 1.01 )**= 1.386m.*

*Hence draft in water of RD 1.01 is 1.386m*

**A cylinder 2m in diameter and 10 m log floats in FW, with its axis horizontal, at draft of 0.6m. Find its mass.**

*Solution :*

*Solution :*

*Diameter of cylinder= D = 2m,**Radius = ( D / 2 )**= 1m**length = 10m, depth = 0.6m*

*Mass = (u/w volume) x (density of displaced water)*

*Weight = ( πr ^{2}h x density )*

*= (3.1416 x 1 x 1 x 0.6) x (1)*

*= 1.88t.***A barge of triangular cross section is 20m long, 12m wide and 6m deep. It floats in SW at a draft of 4m. Find its displacement.**

* Solution :*

*Solution :*

*Volume of triangular cross section = (L x B X H )**= (20 x 12 x6)*

*Depth = 4m*

*Here Displacement means = displacement by triangular cross section.**In triangle ABC, CG is perpendicular to AB**So triangle GBC and EFC are similar,**So by law of similar angle triangle**GB / EF = GC / EC**6 / EF = 6/ 4**EF = (6x 4)/6**= 4m*

*Now, DF = (2EF)**= 4×2**= 8m*

*Displacement = ( u/w volume ) x (1.025)**= (20 x 8 x 4) x (1.025)**= 656 t .*

**A cylindrical drum of 1.2m diameter and 2m height floats with it axis vertical in water of RD 1.016 at a draft of 1.4m. Find the maximum mass of lead shots that can be put in it with sinking it.**

*Solution:*

*Solution:*

*Radius = (d/2)**=(1.2m/2)**= 0.6m*

*H = 2m, D =1.4m , RD = 1.016*

*Mass of the cylinder = (r ^{2} h) x(density)*

*= (3.1416 x 0.6 x 0.6 x 2) x (1.016)*

*= 2.298t**Mass of the cylinder at 1.4m = 1.4 x r ^{2} h) x (density)*

*= 1.4 x (3.1416 x 0.6 x0 .6) x (1.016)*

*= 1.608 t**Hence ,the maximum mass of lead shots that can be put in it with sinking is (2.298 t – 1.608 t) = 0.69 t*