What force must be applied to hold a 6 dm3 ball under water?

| September 18, 2021 | education

V = 6 dm ^ 3 = 6 * 10 ^ -3 m ^ 3.
ρwater = 1000 kg / m ^ 3.
ρair = 1.2 kg / m ^ 3.
g = 9.8 m / s ^ 2.
F -?
Two forces act on a body immersed in water: the force of gravity Ft, directed vertically downward, and the buoyancy force of Archimedes Farkh, directed vertically upward.
The force of gravity is determined by the formula: Ft = m * g, where m is the mass of the body, g is the acceleration of gravity.
m = ρair * V.
Fт = ρair * V * g
The buoyancy force of Archimedes is determined by the formula: Farch = ρwater * g * V. Where ρwater is the density of the liquid in which the body is immersed, g is the acceleration of gravity, V is the volume of the immersed part of the body in the liquid.
F = Farch – Ft = ρwater * g * V – ρair * V * g.
F = 1000 kg / m ^ 3 * 9.8 m / s ^ 2 * 6 * 10 ^ -3 m ^ 3 – 1.2 kg / m ^ 3 * 9.8 m / s ^ 2 * 6 * 10 ^ -3 m ^ 3 = 58.73 N.
Answer: to keep the ball under water, you need to apply a force F = 58.73 N.



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