What force must be applied to hold a stone in water, the weight of which in the air is 100N?

What force must be applied to hold a stone in water, the weight of which in the air is 100N? The density of the stone is 2600 kg / m3.

When the body hangs in water, we write the expression according to Newton’s second law, taking into account that the body is motionless:
T-m * g + Fa = 0
T = m * g-Fa
A buoyant force equal to the weight of the displaced fluid acts on a body immersed in a liquid, and is called the Archimedes force:
Fa = ρ * g * V, where ρ is the density of the liquid, g is the acceleration of gravity of a body raised above the Earth g = 9.8 m / s², V is the volume of the body.
Determine the volume of the body:
V = m / ρт, where ρт – body density, m – body mass.
We have ρт = 2600 kg / m³.
P = m * g
Let’s plug everything into the formula for determining the strength.
T = m * g-ρl * g * V = m * g-ρl * g * m / ρt = m * g * (1-ρl / ρt) = P * (1-ρl / ρt)
Let’s substitute the numerical values, taking into account that according to the reference book ρzh = 1000kg / m³:
T = P * (1-ρl / ρt) = 100 * (1-1000 / 2600) = 61.6 N.
Answer: the force to hold the stone is 61.6 N.