Archimedes’ principle states that the buoyant force on a submerged object is equal to the weight of the fluid it displaces. Our understanding of the relationship between pressure and depth can be used to demonstrate this principle.
The pressure on the top of the object is $P_1$, and the pressure on the bottom is $P_2$. This means that if the top and bottom of the object each have area $A$, the force on the top is $P_1A$ downward, while the force on the bottom is $P_2A$ upwards. The net force upwards on the object from the fluid is then,
$$F=P_2A-P_1A=\left(P_2-P_1\right)A$$
If the height of the object is $h$, then from the relationship between pressure and depth, $P_2 = P_1 + \rho_f gh$, so that,
$$F=\left(P_1 + \rho_f gh - P_1\right)A=\rho_f ghA = \rho_f g V = m_f g$$
where $m_f$ is the mass of the fluid which is displaced by the object (not the mass of the object). This is Archimedes’ Principle and the same goes for any shape of object. Thus the buoyancy force is given by the weight of the fluid displaced,
$$F_b=\rho_f gV_{\text{disp}}$$
The force due to gravity on the object is the object’s mass times gravity,
$$F_g = mg = \rho gV$$
The net force on the object must be zero if it is to be a situation of fluid statics such that Archimedes principle is applicable, and is thus the difference of the object’s weight and the buoyancy force,
$$F_T = 0 = F_g - F_b = mg - \rho_f gV_{\text{disp}}$$
If the buoyancy of an (unrestrained and unpowered) object exceeds its weight, it tends to rise, an object whose weight exceeds its buoyancy tends to sink.
Explore how the variations in the density and size of the object the boyancy.