An iron nut was lowered into a vessel with mercury. Will the nut sink?

univerkov | March 29, 2021 | education

Two forces will act on the nut in opposite directions:

1. The force of gravity: Fт = mg – directed downward.

2. The buoyancy force of Archimedes FА– is directed upwards.

FA = ρрgV (ρр – density of mercury, V – nut volume)

We find the volume of the nut:

V = m / ρс.

Substitute the expression for volume into the formula for the Archimedes force:

FA = (ρрgm) / ρс = (ρр / ρс) gm.

We see that FA differs from Ft by the coefficient (ρр / ρс). This coefficient is greater than 1, since the density of mercury is greater than that of steel.

FA> Fт.

Answer: The buoyancy force of Archimedes is greater than the force of gravity. The nut will not sink.

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