**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*

*A rectangular barge of 10 m long and 5m wide, floating in SW at a draft of 3m, is being lifted out of the water by a heavy-lift crane. Find the load on the crane when the draft has reduced to 1m.*

* Solution:*

*Solution:*

**Area of rectangular barge = ( L x B)**

** = 10m x 5m,**

** RD = 1.025**

**Mass when draft 3m = (u/w volume at 3m) x (density)**

** = (10 x 5 x 3) x (1.025)**

** = 153.75t.**

**Mass after being lifted by the heavy crane =(u/w volume at 1m depth )x (density)**

** = (10 x5 x 1) x (1.0250**

** = 51.25 t**

**Hence ,Load taken by the crane = (153.75 – 51.25)**

** = 102.5 t**

*A rectangular box 2.4 m long, 1.2m wide and 0.8m high, floats in water of RD 1.012m at an even keel draft of 0.2m. find maximum mass of SW that can be poured in to it without sinking it.*

*Solution :*

*Solution :*

*Given :*

* L = 2.4m, B = 1.2m, H = 0.8m*

* RD = 1.012 , Depth = 0.2m*

*Mass of rectangular box = (L x B x H )x (1.012)*

* = (2.4 x 1.2 x 0.8) x (1.012)*

* = 2.33t*

*Mass of box even keel when draft is 0.2m*

* = (2.4 x 1.2 x 0.2) x(1.012)*

* = 0.583*

*Hence, Salt water that can poured is (2.33 – 0.58)*

* = 1.75 t.*

*A box – shaped vessel of 18450 t displacement in 150m long and 20 m wide. Find its draft in SW.*

* Solution:*

*Solution:*

*We know that:*

*Displacement = (U/w volume)x (density)*

* 18450 = ( 150 x 20x D) x (1.025)*

* D = ( 18450 ) / ( 150 x 20 x 1.025)*

* Draft = 6m*

*A box-shaped vessel 120m long and 15m wide is floating in DW of RD 1.005 at a draft of 5m. if her maximum permissible draft in SW in 6m , find how much cargo she can now load .*

*Solution:*

*Solution:*

*Area of box shape vessel = (Lx b)*

* = 120 x 15m,*

* RD = 1.005, Draft =5m*

* Mass = (u/w volume ) x (density)*

* = (120 x 15 x 5) x (1.005)*

* = 9045t .*

*Maximum cargo she can load = (120 x 15 x 6) x (1.025)= 11070 t*

*Cargo can be load = (11070 – 9045)*

* = 2025t.*

Solution No.4 is incorrect. It says the log floats on its horizontal axis thus we cant use the normal cylinder formula. We have to use the other formula ‘Area of a circle segment given height and radius”. Google it. Pls resolve and update.