Question

A barge floating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the

bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully

loaded with coal the bottom of the barge is located H; = 1.05 m below the surface. Part (a) Find the mass of the coal in kilograms.

Numeric : A numeric value is expected and not an expression.

m1 =

Part (b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge? Numeric : A numeric value is expected and not an expression.

Answer 1

a)
`\Delta m = 330000\,kg`, b)
`h = 1.268\,m`

Step-by-step explanation:

a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:

`\Delta m \cdot g = \rho_(w)\cdot g \cdot \Delta V`

`\Delta m = \rho_(w)\cdot \Delta V`

`\Delta m = \left(1000\,(kg)/(m^(3)) \right)\cdot (550\,m^(2))\cdot (1.05\,m-0.45\,m)`

`\Delta m = 330000\,kg`

b) The submersion height is found by using the equation derived previously:

`\Delta m = \rho_(w)\cdot \Delta V`

`450000\,kg = \left(1000\,(kg)/(m^(3))\right)\cdot (550\,m^(2))\cdot (h-0.45\,m)`

The final submersion height is:

`h = 1.268\,m`

Answer 2

a) m = 330000 kg = 330 tons

b) H3 = 1.268 meters

Step-by-step explanation:

Given:-

- The density of fresh-water, ρ = 1000 kg/m^3

- The base area of the rectangular prism boat, A = 550 m^2

- The height of the boat, H = 2.0 m ( empty )

- The bottom of boat barge is H1 = 0.45 m of the total height H under water. ( empty )

- The bottom of boat barge is H2 = 1.05 m of the total height H under water

Find:-

a) Find the mass of the coal in kilograms.

b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge?

Solution:-

- We will consider the boat as our system with mass ( M ). The weight of the boat "Wb" acts downward while there is an upward force exerted by the body of water ( Volume ) displaced by the boat called buoyant force (Fb):

- We will apply the Newton's equilibrium condition on the boat:

Fnet = 0

Fb - Wb = 0

Fb = Wb

Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V1 ). The expression of buoyant force (Fb) is given as:

Fb = ρ*V1*g

Where,

V1 : Volume displaced when the boat is empty and the barge of the boat is H1 = 0.45 m under the water:

V1 = A*H1

Hence,

Fb = ρ*A*g*H1

Therefore, the equilibrium equation becomes:

ρ*A*g*H1 = M*g

M = ρ*A*H1

- Similarly, apply the Newton's equilibrium condition on the boat + coal:

Fnet = 0

Fb - Wb - Wc = 0

Fb = Wb + Wc

Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V2 ). The expression of buoyant force (Fb) is given as:

Fb = ρ*V2*g

Where,

V2 : Volume displaced when the boat is filled with coal and the barge of the boat is H2 = 1.05 m under the water:

V2 = A*H2

Hence,

Fb = ρ*A*g*H2

Therefore, the equilibrium equation becomes:

ρ*A*g*H2 = g*( M + m )

m+M = ρ*A*H2

m = ρ*A*H2 - ρ*A*H1

m = ρ*A*( H2 - H1 )

m = 1000*550*(1.05-0.45)

m = 330000 kg = 330 tons

- We will set the new depth of barge under water as H3, if we were to add a mass of coal m = 450,000 kg then what would be the new depth of coal H3.

- We will use the previously derived result:

m = ρ*A*( H3 - H1 )

H3 = m/ρ*A + H1

H3 = (450000 / 1000*550) + 0.45

H3 = 1.268 m