A glass is suspended from the dynamometer, filled with water to the brim. The dynamometer reading

A glass is suspended from the dynamometer, filled with water to the brim. The dynamometer reading is F1 = 3H. A small stone with a mass of mk = 100 g is lowered to the bottom of the glass, which turns out to be completely submerged in water. Determine the new dynamometer reading. The density of the stone is ρ = 2500 kg / m3.

The glass is filled to the brim with water and when the stone is immersed, a volume of water equal to the volume of the stone will be displaced.

Stone volume:

Vк = m / ρ, where m is the mass of the stone (m = 100 g = 0.1 kg), ρ is the density of the stone (ρ = 2500 kg / m3).

Vк = m / ρ = 0.1 / 2500 = 4 * 10-5 m3.

Spilled water mass:

mw = Vk * ρw, where ρw is the density of water (ρ = 1000 kg / m3).

mw = Vk * ρw = 4 * 10-5 * 1000 = 0.04 kg.

Dynamometer initial readings:

P = m * g = 3 H = 0.3 * 10.

Final indications:

P1 = m1 * g = (0.3 + 0.1 – 0.04) * 10 = 3.6 N.