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?
a. Vw = L*W*h = 85 *30 * (15-3) = 30,600 Ft^3.
b. 30,600Ft^3 * 62.4Lbs/Ft^3 = 1.909,440 Lbs. = 955 Tons = Wt. of water displaced.
Load = 955 - 160 = 795 Tons.
Well, well, well, looks like we're dealing with a barge! Alright, let me crunch some numbers and bring in the giggles.
(a) To find out how much water the barge will displace, we first need to calculate the volume of the barge. The volume is equal to the length times the width times the depth. So, 85.0 ft * 30.0 ft * 15.0 ft gives us the volume of the barge.
Now, remember we want to know the volume of water it will displace while the top stays 3.00 ft above the water. So, we need to subtract the volume of that air gap from the total volume of the barge. Are you ready for the punchline? Ta-daa! That's your answer for (a).
(b) Ah, the load in tons! To find that out, we need to know the weight of the water the barge will displace. We can calculate that by multiplying the volume of water displaced (from part (a)) by the density of water, which is roughly 62.4 lb/ft3.
But hey, we need our answer in tons, not pounds! So we convert the weight from pounds to tons by dividing by 2000. Finally, we simply add the weight of the empty barge in dry dock (160 tons) to get the total load in tons.
And there you have it, my friend! The answer to (b), wrapped in a punchline, is your load in tons.
To find the volume of water the barge will displace, we need to calculate the volume of the submerged portion of the barge.
Let's start with part (a):
(a) How many ft^3 of water will it displace while the top stays 3.00 ft above the water?
The submerged portion of the barge can be calculated as the product of its width, length, and depth. However, we need to subtract the volume of the portion that stays above the water.
The submerged volume = (width x length x depth) - (width x length x above-water height)
Width = 30.0 ft
Length = 85.0 ft
Depth = 15.0 ft
Above-water height = 3.00 ft
Submerged volume = (30.0 ft x 85.0 ft x 15.0 ft) - (30.0 ft x 85.0 ft x 3.00 ft)
Submerged volume = (38,250 ft^3) - (7,650 ft^3)
Submerged volume = 30,600 ft^3
Therefore, the barge will displace 30,600 ft^3 of water while the top stays 3.00 ft above the water.
Now let's move on to part (b):
(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 load in tons, we need to add the weight of the load to the weight of the empty barge.
Load weight = Empty barge weight + Displaced water weight
Empty barge weight = 160 tons
To find the weight of the displaced water, we can use the density of water, which is approximately 62.4 lb/ft^3.
Displaced water weight = Displaced water volume x Density of water
Displaced water weight = (30,600 ft^3) x (62.4 lb/ft^3) / (2000 lb/ton)
Displaced water weight = 950.4 tons
Load weight = 160 tons + 950.4 tons
Load weight = 1110.4 tons
Therefore, the barge will contain a load of approximately 1110.4 tons under these conditions.
To calculate the displacement of the river barge, we need to find the volume of water it will displace. The volume can be determined by multiplying the dimensions of the underwater portion of the barge.
(a) Volume of water displaced:
Volume = Length × Width × Depth
Given:
Length = 85.0 ft
Width = 30.0 ft
Depth = 15.0 ft
Substituting the values:
Volume = 85.0 ft × 30.0 ft × 15.0 ft
Volume ≈ 38,250 ft³
Therefore, the river barge will displace approximately 38,250 ft³ of water while the top stays 3.00 ft above the water.
(b) To calculate the load in tons that the barge can contain, we need to find the difference between the weight of the loaded barge and the weight of the empty barge.
Given:
Weight of empty barge in dry dock = 160 tons
The weight of the water displaced by the barge will be equal to the weight of the loaded barge.
To calculate the weight of the water displaced:
Weight of water displaced = Volume of water displaced × Density of water
The density of water is approximately 1 ton/ft³.
Weight of water displaced = 38,250 ft³ × 1 ton/ft³
Weight of water displaced = 38,250 tons
To find the load in tons:
Load = Weight of water displaced - Weight of empty barge
Load = 38,250 tons - 160 tons
Load = 38,090 tons
Therefore, the barge will contain a load of approximately 38,090 tons under these conditions.