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 by the barge, we can calculate the difference in volume between the barge and the barge with water at the given height:
(a) Volume of water displaced:
The volume of the barge can be calculated by multiplying its length, width, and height:
Volume of barge = Length x Width x Height = 85.0 ft x 30.0 ft x 15.0 ft = 38,250 ft³
The volume of the barge with water at the given height can be calculated by adding the extra height to the height of the barge and then multiplying by the length and width:
Volume of barge with water = Length x Width x (Height + Extra height) = 85.0 ft x 30.0 ft x (15.0 ft + 3.0 ft) = 85.0 ft x 30.0 ft x 18.0 ft = 46,350 ft³
Therefore, the volume of water displaced by the barge is:
Volume of water displaced = Volume of barge with water - Volume of barge
Volume of water displaced = 46,350 ft³ - 38,250 ft³ = 8,100 ft³
(b) Load in tons:
To calculate the load in tons, we need to know the weight of one cubic foot of water and the weight of the empty barge.
The weight of water varies with its temperature and salinity. At room temperature and average salinity, the weight of water is approximately 62.4 pounds per cubic foot.
1 ton = 2000 pounds.
To convert from pounds to tons, divide the weight by 2000.
Weight of water displaced = Volume of water displaced x Weight of water
Weight of water displaced = 8,100 ft³ x 62.4 lb/ft³ = 505,440 lb
Load in tons = (Weight of water displaced + Weight of empty barge) / 2000
Load in tons = (505,440 lb + 320,000 lb) / 2000 = 825.22 tons
Therefore, the barge will displace 8,100 ft³ of water while the top stays 3.00 ft above the water, and the load in tons will be approximately 825.22 tons under these conditions if the empty barge weighs 160 tons in dry dock.