By using Archimedes Principle the Barge will sit at a depth of 0.7203 m
What is Archimedes Principle?
The buoyancy of any floating object partially or totally submerged in a fluid can be computed thanks to Archimedes' principle. The weight of the thing alone exerts a downward force on it. The upward or buoyant force acting on the object is what is meant by the aforementioned Archimedes' principle. The difference between the buoyant force and the object's weight is therefore the net force acting on it. In the event that this net force is positive, the item will rise; in the event that it is negative, the object will sink; and in the event that it is zero, the object will be neutrally buoyant, remaining stationary.
Calculation:
The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:
density (salt water) = 1029 kg / [tex]m^3[/tex]
density (fresh water) = 1000 kg / [tex]m^3[/tex]
The amount of water (in mass) displaced by the barge should be equal in two waters.
mass displaced (salt water) = mass displaced (fresh water)
Since mass is also the product of density and volume,
therefore:
[density * volume] of salt water = [density * volume] of fresh water
First we calculate the amount of volume displaced in the harbour (salt water):
V = [tex]3.5 m\ X\ 24.0 m\ X\ 0.70 m[/tex]
V = [tex]58.8 m^3[/tex] of salt water
Plugging in the values into equation 1:
1029 kg / [tex]m^3[/tex] × 58.8 m^3 = 1000 kg/[tex]m^3[/tex]× Volume fresh water
Volume fresh water displaced = 60.5052 [tex]m^3[/tex]
Therefore the depth of the barge in the river is:
60.5052 [tex]m^3[/tex] = 3.5 m × 24.0 m × h
h = 0.7203 m
By using Archimedes Principle the Barge will sit at a depth of 0.7203 m
To refer more Archimedes Principle Problems visit:
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