Determine the Archimedean force acting on a 1 dm3 cork completely submerged in water.

Determine the Archimedean force acting on a 1 dm3 cork completely submerged in water. P plug = 240 kg / m3

We translate the dimension dm³ into m³ – 1 dm³ will mean 0.001 m³ (since 1 m³ contains 1000 dm³).

The buoyancy force acting on a body immersed in a liquid will be equal to the weight of the liquid displaced by it, or, in other words, the weight of the liquid that fills the volume occupied by the body immersed in the liquid.

Fout = Vρg, where V is the volume of the body, ρ is the density of the liquid (in our case, water, ρ of water – 1000 kg / m³), ​​g is the acceleration of gravity.

In our case, Fdiff = 0.001 m³ * 1000 kg / m³ * 9.8 m / s² = 1 * 9.8 kg * m / s² = 9.8 N.

The force of gravity “pushing” the plug down (or – its weight) will be – mg = Vρg – 0.001 m³ * 240 kg / m³ * 9.8 m / s² = 2.352 N.

As you can see from these values, the buoyant (Archimedean) force will be much greater than the weight of our plug, so it will never be able to sink.