A flat-bottom river barge is 30.0 ft wide, 85.0 ft long, and 15.0 ft deep. (a) How many ft3 of water will it displace while the top stays 3.00 ft above the water? (b) What load in tons will the barge contain under these conditions if the empty barge weighs 160 tons in dry dock?
To find the volume of water displaced, we can calculate the volume of the rectangular prism formed by the dimensions of the barge and then subtract the volume of the part of the prism that is above the water level.
(a) The volume of the rectangular prism formed by the barge is given by the formula:
Volume = width × length × depth
Substituting the given values:
Volume = 30.0 ft × 85.0 ft × 15.0 ft
Volume = 38,250 ft³
Since the top of the barge stays 3.00 ft above the water, we need to subtract the volume of the prism with a 3.00 ft height. The volume of this prism is:
Volume = width × length × height
Volume = 30.0 ft × 85.0 ft × 3.0 ft
Volume = 7,650 ft³
Therefore, the volume of water displaced by the barge is:
Volume of water displaced = Volume - Volume above water
Volume of water displaced = 38,250 ft³ - 7,650 ft³
Volume of water displaced = 30,600 ft³
(b) To determine the load in tons, we need to consider the weight of the barge in dry dock and the weight of the water it displaces.
The weight of the load in tons is given by the formula:
Weight = Weight of empty barge + Weight of water displaced
The weight of the empty barge is given as 160 tons.
To find the weight of the water displaced, we need to calculate the mass of the water (in tons) and then convert it to weight. The mass of water is calculated using the density of water, which is approximately 1 ton/ft³.
Mass of water = Volume of water displaced × Density of water
Mass of water = 30,600 ft³ × 1 ton/ft³
Mass of water = 30,600 tons
Therefore, the weight of the water displaced is also 30,600 tons.
Weight = 160 tons + 30,600 tons
Weight = 30,760 tons
Therefore, the barge will contain a load of 30,760 tons under these conditions.