A2-2. Forces in Fluids: Drag and Buoyancy
Buoyancy (Upthrust)
When an object is fully or partially immersed in a fluid (a liquid or a gas), it experiences an upward force. This happens because fluid pressure increases with depth; the pressure pushing up on the bottom of the object is greater than the pressure pushing down on the top.
Archimedes' Principle: The buoyant force (upthrust) acting on an object is exactly equal to the weight of the fluid that the object displaces.
- $\rho$: Density of the fluid ($\text{kg/m}^3$)
- $V$: Volume of the displaced fluid ($\text{m}^3$)
- $g$: Gravitational field strength ($\text{N/kg}$)
⚠ Floating vs. Sinking
If an object is floating in equilibrium, the upward buoyant force perfectly balances the downward weight ($F_b = W$). The fraction of the object's volume that is submerged is equal to the ratio of the object's density to the fluid's density ($\dfrac{\rho_{object}}{\rho_{fluid}}$).
Fluid Friction (Drag) & Stoke's Law
As an object moves through a fluid, it must push fluid particles out of the way, experiencing a resistive drag force acting exactly opposite to its direction of motion. Drag heavily depends on the object's speed, its cross-sectional area, and the fluid's viscosity (thickness).
Stoke's Law: For a small, perfectly spherical object moving relatively slowly through a viscous fluid, the drag force can be precisely calculated as:
- $\eta$: Fluid viscosity ($\text{Pa}\cdot\text{s}$)
- $r$: Radius of the sphere ($\text{m}$)
- $v$: Velocity of the sphere ($\text{m/s}$)
Terminal Velocity
If an object is dropped from a high altitude, it initially accelerates downward due to gravity. As its velocity $v$ increases, the upward drag force $F_D$ also increases. Eventually, the upward forces (Drag + Buoyancy) perfectly balance the downward Weight.
At this exact moment, the net force becomes zero ($\Sigma F = 0$), acceleration drops to zero, and the object falls at a constant, maximum speed known as terminal velocity.
Example 1: Terminal Velocity of a Falling Sphere
Problem: A small spherical steel ball bearing of radius $r = 2.0 \times 10^{-3}\text{ m}$ and density $\rho_s = 7800\text{ kg/m}^3$ is dropped into a tall cylinder of oil. The oil has a density of $\rho_o = 900\text{ kg/m}^3$ and a dynamic viscosity of $\eta = 0.050\text{ Pa}\cdot\text{s}$. Calculate the terminal velocity of the ball bearing as it falls through the oil. (Assume $g = 9.81\text{ m/s}^2$)
Solution:
- a) Set up the force balance equation:
At terminal velocity, the net force is zero. The downward weight ($W$) is perfectly balanced by the upward buoyant force ($F_b$) and drag force ($F_D$).
$$W = F_b + F_D$$ - b) Substitute the formulas for each force:
Weight of the sphere: $W = m_s g = (\rho_s V)g = \rho_s \left(\dfrac{4}{3}\pi r^3\right)g$
Buoyant force: $F_b = \rho_o V g = \rho_o \left(\dfrac{4}{3}\pi r^3\right)g$
Stoke's Drag: $F_D = 6\pi \eta r v_t$ - c) Rearrange to solve for terminal velocity ($v_t$):
$$6\pi \eta r v_t = \rho_s \left(\dfrac{4}{3}\pi r^3\right)g - \rho_o \left(\dfrac{4}{3}\pi r^3\right)g$$
$$6\pi \eta r v_t = \dfrac{4}{3}\pi r^3 g (\rho_s - \rho_o)$$
$$v_t = \dfrac{2 r^2 g (\rho_s - \rho_o)}{9 \eta}$$ - d) Plug in the given values:
$$v_t = \dfrac{2(2.0 \times 10^{-3})^2 (9.81)(7800 - 900)}{9(0.050)}$$
$$v_t = \dfrac{2(4.0 \times 10^{-6})(9.81)(6900)}{0.45} = \dfrac{0.5415}{0.45} \approx \mathbf{1.2\text{ m/s}}$$