Question

A river barge, whose cross section is approximately rectangular, carries a load of grain.  The barge is 28 ft wide and 90 ft long.  When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain is 7 ft.  Determine:  (a) the unloaded weight of the barge, and (b) the weight of the grain.

Answer 1

Concepts and Reason

A river barge is a flat-bottomed boat, built mainly for river and canal transport of heavy goods. Due to the load carried by it, it sinks in water by a certain amount. The total weight of the river barge (self-weight plus load carried with it) is equal to the weight of water displaced by it.

By knowing the volume of the barge and by marking the depth of submergence, it is possible to know the total load carried by it.

Fundamentals

The weight of the barge without the grain can be calculated as,

${W_b} = \gamma {V_b}$

Here, the weight density is $\gamma$ and the submerged volume of the barge without the grain is ${V_b}$.

The submerged volume of the barge without the grain is,

${V_b} = Ah$

Here, the area of the barge is A and the depth of the barge without the grain is h.

The weight of the barge with the grain can be calculated as,

${W_{bg}} = \gamma {V_{bg}}$

Here, the submerged volume of the barge with grain is ${V_{bg}}$.

The submerged volume of the barge with the grain is,

${V_{bg}} = A{h_g}$

Here, the depth of the barge with the grain is ${h_g}$.

(a)

Calculate the submerged volume of the barge without the grain.

${V_b} = Ah$

Here, the area of the barge is A and the depth of the barge without the grain is h.

Substitute $\left( {28 \times 90} \right){\rm{ f}}{{\rm{t}}^2}$ for A and 5 ft for h.

$\begin{array}{c}\\{V_b} = \left( {28 \times 90} \right) \times 5\\\\ = 12600{\rm{ f}}{{\rm{t}}^3}\\\end{array}$

Calculate the weight of the barge without the grain.

${W_b} = \gamma {V_b}$

Here, the weight density is $\gamma$.

Substitute $62.4{\rm{ lb/f}}{{\rm{t}}^3}$ for $\gamma$ and $12600{\rm{ f}}{{\rm{t}}^3}$ for ${V_b}$.

$\begin{array}{c}\\{W_b} = 62.4 \times 12600\\\\ = 786240{\rm{ lb}}\\\end{array}$

(b)

Calculate the submerged volume of the barge with the grain.

${V_{bg}} = A{h_g}$

Here, the depth of the barge with the grain is ${h_g}$.

Substitute $\left( {28 \times 90} \right){\rm{ f}}{{\rm{t}}^2}$ for A and 7 ft for h.

$\begin{array}{c}\\{V_{bg}} = \left( {28 \times 90} \right) \times 7\\\\ = 17640{\rm{ f}}{{\rm{t}}^3}\\\end{array}$

Calculate the weight of the barge with the grain.

${W_{bg}} = \gamma {V_{bg}}$

Here, the submerged volume of the barge with grain is ${V_{bg}}$.

Substitute $62.4{\rm{ lb/f}}{{\rm{t}}^3}$ for $\gamma$ and $17640{\rm{ f}}{{\rm{t}}^3}$ for ${V_{bg}}$.

$\begin{array}{c}\\{W_b} = 62.4 \times 17640\\\\ = 1100736{\rm{ lb}}\\\end{array}$

Calculate the weight of the grain.

${W_g} = {W_{bg}} - {W_b}$

Substitute 786240 lb for ${W_b}$ and 1100736 lb for ${W_{bg}}$.

$\begin{array}{c}\\{W_g} = 1100736 - 786240\\\\ = 314496{\rm{ lb}}\\\end{array}$

Ans: Part a

The unloaded weight of the barge is 786240 lb.

Part b

The weight of the grain is 314496 lb.