Buoyant Force

When we talk about buoyancy, the first thing we think about is that objects tend to weigh less when submerged in a fluid. For example, a rock that could be lifted with difficulty underwater would be almost impossible to move outside of it. This phenomenon is explained by the appearance of a force called buoyancy.

When an object is submerged in a fluid, an upward buoyant force appears, equal to the weight of the liquid displaced by the submerged body. For this reason, all bodies submerged in a fluid seem to lose part of their weight, as there is a difference in forces between different regions of the body according to its depth. Thus, the buoyant force will be:

F_{buoy} = F_2 - F_1

Since P=F/A and P=\rho g h, we can infer that F=\rho g A h, where \rho is the fluid density, h the depth, A the area of the surface where the pressure is applied, and g the gravitational acceleration. With this, the forces exerted on the top and bottom faces are given by:

F_1 = \rho g A h_1

F_2 = \rho g A h_2

Therefore, we have:

\begin{array}{rl} F_{\text{buoyancy}} &= \rho g A h_2 - \rho g A h_1 \\ \\ &=\rho g A \underbrace{h_2 - h_1}_{\Delta h} \\ \\ &= \rho gV \\ \\ & =\text{Weight of the displaced fluid volume} \end{array}

This is what is known as Archimedes’ principle.

EXAMPLE: A 70[kg] rock lies at the bottom of a lake. If its volume is 3\cdot 10^4 [cm^3], what force is needed to lift it?

SOLUTION:

The buoyant force on the rock will be:

\begin{array}{rl} F_{\text{buoyancy}} &= \rho_{\text{water}} g V_{rock} \\ \\ &= 10^3 \left[\dfrac{kg}{m^3}\right] \cdot 9.81\left[\dfrac{m}{s^2}\right] \cdot 3 \cdot 10^4 [cm^3] \\ \\ &= 10^3 \left[\dfrac{kg}{m^3}\right] \cdot 9.81\left[\dfrac{m}{s^2}\right] \cdot 3 \cdot 10^4 \left[\dfrac{m}{100}\right]^3 = 294[N] \end{array}

While the weight force of the rock is:

F_{\text{weight}} = m_{\text{rock}}g = 70[kg] \cdot 9.81 \left[\dfrac{m}{s^2}\right]=686[N]

Therefore, to lift the rock underwater, a force of F = 686[N] - 294[N] = 392[N] will suffice. Underwater, this rock can be lifted with almost half the force you would need outside of it.

Buoyancy and Archimedes’ Principle

Archimedes’ principle helps us understand why some objects float when submerged in certain fluids, such as wood in water. In general, an object floats in a fluid if the medium’s density is greater than that of the object, and it will float until a portion emerges above the interface. The body will rise until it reaches an equilibrium position. How can we calculate the portion of the body that emerges above the fluid? It’s easy to calculate.

If we equate the weight force with the buoyant force, we can calculate what portion of the body will emerge above the interface. The reasoning is as follows:

\begin{array}{rl} & F_{\text{weight}} = F_{\text{buoyancy}}\\ \\ \equiv & m_{\text{object}} g = m_{\text{submerged}} g \\ \\ \equiv & \rho_{\text{object}}V_{\text{object}} g = \rho_{\text{submerged}} V_{\text{submerged}} g \\ \\ \equiv & \dfrac{\rho_{\text{object}}}{\rho_{\text{submerged}}} = \dfrac{V_{\text{submerged}}}{V_{\text{object}} } = \text{Percentage of the body submerged} \end{array}

EXAMPLE: A simple model assumes a continent as a solid block of rock (with a density =2800[kg/m^3]) floating on the surrounding Earth’s mantle (with a density =3300[kg/m^3]). Assuming the continent has an average thickness of 35[km], calculate the average height that emerges above the mantle.

SOLUTION:

The percentage of the body submerged will be:

Submerged percentage \displaystyle = \frac{\rho_{body}}{\rho_{fluid}}

Therefore, the percentage of the body that is not submerged and that emerges above the mantle will be:

Emerging percentage \displaystyle = 1 - \frac{\rho_{body}}{\rho_{fluid}} = 1 - \frac{2800}{3300} \approx 0,15 = 15\%

Since the average thickness is 35[km], what emerges on average will be 15\% 35[km]\approx 5.3 [km].