What force must be applied to hold a 5m3 steel plate in water?

Given:

V = 5 cubic meters – the volume of a steel plate immersed in water;

ro1 = 7800 kilograms per cubic meter – steel density;

ro = 1000 kilograms per cubic meter – the density of water.

It is required to determine F (Newton) – what force must be applied to keep the steel plate in water.

Find the mass possessed by the steel plate:

m = V * ro1 = 5 * 7800 = 39000 kilograms.

Then, to determine the strength, you must use the following formula:

F = F gravity – Farchimedes;

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

F = g * (m – ro * V);

F = 10 * (39000 – 1000 * 5) = 10 * (39000 – 5000) = 10 * 34000 = 340,000 Newtons (340 kN).

Answer: To keep the steel plate in water, you need to apply a force equal to 340,000 Newton (340 kN).