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 calculate the volume of water displaced, we need to find the volume of the barge and subtract the volume of the portion above the water.
(a) Volume of the barge:
The volume of a rectangular prism is given by length x width x height. Therefore, the volume of the barge is:
Volume = 85.0 ft x 30.0 ft x 15.0 ft = 38,250 ft³
The volume of the portion above the water is calculated by finding the volume of a rectangular prism with dimensions 85.0 ft x 30.0 ft x 3.0 ft:
Volume above water = 85.0 ft x 30.0 ft x 3.0 ft = 7,650 ft³
Therefore, the volume of water displaced is:
Volume of water = Volume of barge - Volume above water
= 38,250 ft³ - 7,650 ft³
= 30,600 ft³
(b) To calculate the load in tons that the barge will contain under these conditions, we need to consider the weight of the empty barge, the weight of the water displaced, and the weight of the load.
Weight of the empty barge = 160 tons
Weight of the water displaced:
The weight of water is given by density x volume. For freshwater, the density is approximately 62.4 lb/ft³. To convert tons to pounds, we multiply by 2000.
Weight of water = Density of water x Volume of water
= 62.4 lb/ft³ x 30,600 ft³
= 1,907,040 lb
= 953.52 tons
Weight of the load = Weight of barge + Weight of water displaced - Weight of empty barge
= 160 tons + 953.52 tons - 160 tons
= 953.52 tons
Therefore, the load in tons that the barge will contain under these conditions is 953.52 tons.
To answer both parts of this question, we'll need to calculate the volume of the barge and then use that information to determine the displacement and load capacity.
(a) Volume of the Barge:
The barge can be considered as a rectangular prism with the given dimensions. So, we can calculate its volume by multiplying the width, length, and depth.
Volume = Width × Length × Depth
Volume = 30.0 ft × 85.0 ft × 15.0 ft
Volume = 38,250 ft³
(b) Displacement:
Since the top of the barge stays 3.00 ft above the water, the volume of water displaced will be the volume of the barge minus the volume of the space above the waterline.
Displacement = Volume of Barge - Volume above waterline
Displacement = 38,250 ft³ - (Width × Length × Above-water depth)
Displacement = 38,250 ft³ - (30.0 ft × 85.0 ft × 3.00 ft)
Displacement = 38,250 ft³ - 7,650 ft³
Displacement = 30,600 ft³
(c) Load Capacity:
To calculate the load capacity in tons, we need to consider the weight of the empty barge and the weight of the water it displaces.
Load Capacity = Weight of Barge + Weight of Displaced Water
Load Capacity = 160 tons + (Displacement × Density of Water)
Load Capacity = 160 tons + (30,600 ft³ × 62.4 lb/ft³) / (2000 lb/ton)
Load Capacity = 160 tons + 957.6 tons
Load Capacity = 1,117.6 tons
Therefore, the barge will displace 30,600 ft³ of water while the top stays 3.00 ft above the water. Also, it can carry a load of 1,117.6 tons under these conditions with an empty weight of 160 tons.