What force must be applied to lift a cast iron part under water? Part volume 0.02 m3

Given:

V = 0.02 cubic meters – the volume of the cast iron part under water;

ro = 1000 kilograms / cubic meter – density of water;

ro1 = 7800 kilograms / cubic meter – density of cast iron (white cast iron).

It is required to determine F (Newton) – what force must be applied to lift the cast-iron part.

Find the mass of the cast iron part:

m = V * ro1 = 0.02 * 7800 = 156 kilograms.

We will assume that the part must be lifted evenly at a constant speed. Then, according to Newton’s first law:

F = F gravity – Farchimedes;

F = m * g – ro * V * g, where g = 10 Newton / kilogram (approximate value);

F = g * (m – ro * V);

F = 10 * (156 – 1000 * 0.02) = 10 * (156 – 20) = 10 * 136 = 1360 Newton.

Answer: To lift a cast iron part out of the water, you need to apply a force equal to 1360 Newton (1.36 kN).