What force must be applied to keep a stone in the water whose weight in the air is 100N and the density is 2600KG * m3.

Given:

P = 100 Newton is the weight of the stone in the air;

ro = 2600 kilograms per cubic meter – the density of the stone;

ro1 = 1000 kilograms per cubic meter – the density of water.

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

Let’s find the volume of the stone:

V = m / ro = P / (ro * g), where g = 10 Newton / kilogram is an approximate value;

V = 100 / (10 * 2600);

V = 100/26000 = 0.004 m3 (the result has been rounded to one thousandth).

Then the buoyancy force acting on the stone will be equal to:

A = ro1 * g * V = 1000 * 10 * 0.004 = 4 * 10 = 40 Newtons.

According to Newton’s first law, the force will be equal to:

F = P – A = 100 – 40 = 60 Newtons.

Answer: to keep a stone in water, you need to apply a force equal to 60 Newtons.