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Chapter 2: Problem 137

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

Short Answer

Expert verified
The weight of the barge and its cargo (W) is 153600 lbs.

Step by step solution

01

Calculations for volume and weight

Since the barge sinks 0.25 ft more in river water than in seawater, the volume difference \( \Delta V \) between river water and sea water is: \( \Delta V = \text{area} \times \text{depth difference} = 40 \text{ft} \times 120 \text{ft} \times 0.25 \text{ft} = 1200 \text{ft}^3 \) . The additional weight that this volume of river water can support as compared to same volume of seawater is: \( \Delta W = \gamma \times \Delta V = 64 \text{lb/ft}^3 \times 1200 \text{ft}^3 = 76800 \text{lb} \).
02

Calculations for weight W

The weight of the barge and its cargo \( W \) is equal to the weight of the volume of the seawater displaced, which is the sum of the additional weight that river water can support plus the weight of the volume of seawater displaced: \( W = \mu V + \Delta W = 64 \text{lb/ft}^3 \times 1200 \text{ft}^3 + 76800 lbs = 153600 lbs \).

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Most popular questions from this chapter

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