An undersea research centre I spherical with an external diameter of 5.20m. The mass of the chamber when occupied is 74400kg. It is anchored to the sea bottom by a cable. What is
a.) The buoyant force on the chamber?
b.) The tension in the cable?
To solve these problems, we need to use the principles of buoyancy and tension.
a) To find the buoyant force on the chamber, we can use Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. In this case, the fluid is water.
1. Find the volume of the chamber: Since the research center is spherical, we can use the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the sphere.
Given that the diameter of the chamber is 5.20 m, the radius (r) can be calculated by dividing the diameter by 2:
r = 5.20 m / 2 = 2.60 m
Now, we can calculate the volume:
V = (4/3)π(2.60 m)^3 = (4/3)π(17.576 m^3) ≈ 73.575 m^3
2. Calculate the weight of the fluid displaced: Since the chamber is submerged in water, we need to calculate the weight of the water displaced by the chamber.
The density of water is approximately 1000 kg/m^3, so we can calculate the weight using the formula: weight = density * volume * gravity, where gravity is approximately 9.8 m/s^2.
Weight of water displaced = (1000 kg/m^3) * (73.575 m^3) * (9.8 m/s^2) ≈ 716167 N
Therefore, the buoyant force on the chamber is approximately 716167 N.
b) To find the tension in the cable, we need to consider the forces acting on the chamber.
The tension in the cable must balance the weight of the chamber and the buoyant force. Therefore, the tension in the cable is equal to the sum of the weight of the chamber and the buoyant force.
Weight of the chamber = mass * gravity
Weight of the chamber = 74400 kg * 9.8 m/s^2 ≈ 728160 N
Tension in the cable = Weight of the chamber + Buoyant force
Tension in the cable = 728160 N + 716167 N ≈ 1447327 N
Therefore, the tension in the cable is approximately 1447327 N.
To find the answers to the given questions, we need to use some concepts from physics related to buoyancy and tension.
a.) The buoyant force on the chamber:
The buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the object is the undersea research center.
To calculate the buoyant force, we need to determine the volume of water displaced by the research center. Since the center is spherical, we can use the formula for the volume of a sphere.
The formula for the volume of a sphere is:
V = (4/3)πr³
Given that the external diameter of the research center is 5.20m, we can calculate the radius (r) using half of the diameter:
r = 5.20m / 2 = 2.60m
Now substitute the radius into the volume formula:
V = (4/3)π(2.60m)³ = 71.99m³
The buoyant force (F_b) can be calculated by multiplying the volume of water displaced by the research center by the density of water (ρ_water) and the acceleration due to gravity (g).
The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.8 m/s².
F_b = V * ρ_water * g
F_b = 71.99m³ * 1000 kg/m³ * 9.8 m/s²
Thus, the buoyant force on the chamber is approximately:
F_b ≈ 705,044 N (rounded to the nearest whole number)
b.) The tension in the cable:
The tension in the cable can be determined by analyzing the forces acting on the chamber. When the chamber is being held stationary, the tension in the cable is equal to the downward force acting on the chamber combined with the buoyant force exerted on it.
Tension in the cable (T) = Weight of the chamber (m * g) + Buoyant force (F_b)
Given that the mass of the chamber is 74,400 kg and the acceleration due to gravity is 9.8 m/s²:
T = 74,400 kg * 9.8 m/s² + 705,044 N
Thus, the tension in the cable is approximately:
T ≈ 1,427,256 N (rounded to the nearest whole number)
the volume of the sphere is 4/3 pi r^3 = 589 m^3
The buoyant force is the weight of the water displaced by the sphere.
mass = volume * density
The tension is the buoyant force minus the weight of the sphere.