Find the buoyant force acting on a pine block 1 m long, 20 cm wide and 10 cm high with a density immersed in water 9 density of pine 400 kg \ m3 and water 1000 kg \ m3
l = 1 m.
a = 20 cm = 0.2 m.
h = 10 cm = 0.1 m.
ρс = 400 kg / m3.
ρw = 1000 kg / m3.
g = 9.8 m / s2.
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 = ρс * V = ρс * l * a * h.
Fт = ρс * l * a * h * g.
The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρw * g * V. Where ρ 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.
V = l * a * h.
Farch = ρv * g * l * a * h.
F = Farch – Ft = ρw * g * l * a * h – ρс * l * a * h * g = (ρw – ρс) * g * l * a * h.
F = (1000 kg / m3 – 400 kg / m3) * 9.8 m / s2 * 1 m * 0.2 m * 0.1 m = 1176 N.
Answer: the body will be pushed out of the water with a force of F = 1176 N.
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