A barge is floating in a harbor. When empty, it displaces V = 500 m^3 of water.

Question

A barge is floating in a harbor. When empty, it displaces V = 500 m^3 of water. Its geometry is shown in Figure 2 where h = 3 m, D = 21 m and L = 105 m. (a) What Is the mass of the barge? (b) If 1 times 10^6 kg of material is loaded on the barge, how much will the water line rise? (c) You are an engineer with the firm who manages these barges. Dock-workers begin loading 4 times 10^6 kg of material. Will it support this load? (d) What is the maximum load the barge will keep afloat? (e) The atmospheric pressure is p_a = 101325 Pa and so exerts a downward force of F_a = LDp_a = 2.23 times 10^8 N, much larger than the maximum load for this barge. Why doesn't the weight of the atmosphere sink the barge?

Explanation / Answer

solution:

archimedes principle tells that weight of liquid displaced by body is equal to weight of that body,hence bouyancy force is equal to weight of displaced fluid

Fb=Wwater

at equillibrium Fb=Wbody

2) for first case ,barge displaces V=500 m3 water

hence bouyancy force=Fb=density*V*G=1000*500*9.81=4905000 N

as Fb=Wb

Wb=m*g=4905000

mb=500000 kg

at that time volume of barge required to dipped in water

500=L*D*h1

h1=.2267 m

3) when additional load is added wieght of body turn s

wb=14715000 N

Wb=Ww=density*V2*9.81

V2=1500 m3

V2=D*L*h2

h2=.680272 m

hence rise in water level is=h2-h1=.6802-.2267=.4535 m

3) volume of barge is

V3=L*D*H=21*105*3=6615 m3

Ww=density*V3*9.81=64893150 N

Ww=Wb=Fb

Wb=64893150=mb1*9.81

mb1=6615000 kg

as mb1> 4*10^6 kg

we must allow them to load the material,but take care water line norms

4) hence maximum mass barge supported is

Mb1=6615000 kg

5) as barometric pressure is too high but it is acting on all entities on this earth ,water pressure at top is barometric pressure only acting upward and it is balancing the barometric pressure on barge and not let to sink

at topof water=Pw=Patm

at bottom of water=Pw=Patm+h*density*9.81

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