A rubber ball filled with air has a diameter of 21.7 cm and a mass of 0.523 kg. What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool?
To find the force required to hold the ball in equilibrium immediately below the surface of water in a swimming pool, we need to consider the buoyant force acting on the ball.
The buoyant force is given by the formula: F_b = ρ * V * g,
where F_b is the buoyant force, ρ is the density of the fluid (in this case, water), V is the volume of the submerged object, and g is the acceleration due to gravity.
Since the ball is completely submerged in water, the volume of the ball that is submerged will be equal to its total volume. The volume of a sphere is given by the formula: V = (4/3) * π * r^3,
where V is the volume, π is a mathematical constant approximately equal to 3.14159, and r is the radius of the sphere.
In this case, the diameter of the ball is given as 21.7 cm, so the radius (r) will be half of that, or 10.85 cm (or 0.1085 m).
Using these values, we can calculate the volume of the ball: V = (4/3) * π * (0.1085)^3 ≈ 0.1658 m^3.
The density of water is approximately 1000 kg/m^3.
Now we can substitute these values into the equation for the buoyant force:
F_b = ρ * V * g = 1000 kg/m^3 * 0.1658 m^3 * 9.8 m/s^2 ≈ 1618.12 N.
Therefore, the force required to hold the ball in equilibrium immediately below the surface of water in a swimming pool is approximately 1618.12 Newtons (N).