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

Answer:

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

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:

$\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V$

$\Delta m = \rho_{w}\cdot \Delta V$

$\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)$

$\Delta m = 330000\,kg$

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

$\Delta m = \rho_{w}\cdot \Delta V$

$450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)$

The final submersion height is:

$h = 1.268\,m$

Answer 2

Answer:

a) m = 330000 kg = 330 tons

b) H3 = 1.268 meters

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