Particle settling in water at terminal velocity: which force balance condition is correct for a rigid sphere? (Assume steady fall in a quiescent liquid with constant properties.)

Introduction:
This question tests core fluid–particle mechanics: how forces balance on a particle falling through a liquid at its terminal (steady) settling velocity. At terminal velocity the acceleration is zero, so the net force must be zero. Understanding which forces are present and how they sum is foundational for designing thickeners, classifiers, and sedimentation tanks.

Given Data / Assumptions:

• Spherical rigid particle falling in water.
• Steady fall at terminal velocity (no acceleration).
• Quiescent, homogeneous fluid; constant properties.
• Forces considered: weight (W), buoyancy (B), and drag (D).

Concept / Approach:
At terminal velocity, Newton’s second law gives the algebraic sum of forces equal to zero. Taking downward as positive, weight acts downward, buoyancy upward, and drag opposite to motion (upward for a falling particle). The correct balance is W = B + D. Equivalently, D = W − B (drag equals the apparent weight).

Step-by-Step Solution:
Identify forces: W downward; B upward; D upward.Apply equilibrium at terminal state: sum of vertical forces = 0.Write balance: W − B − D = 0.Rearrange: W = B + D.

Verification / Alternative check:
In Stokes regime (very small Reynolds number), the analytical solution gives D = 3πμ d v_t. At terminal velocity D precisely equals W − B, confirming the general balance.

Why Other Options Are Wrong:

• Buoyancy = weight + drag: would imply buoyancy exceeds weight, contradicting settled motion.
• Drag = buoyancy + weight: would require drag to exceed weight, unphysical for downward fall.
• Drag = weight: ignores buoyancy; only true if fluid density were negligible.
• Apparent weight = 0: at terminal velocity the apparent weight equals drag, not zero.

Common Pitfalls:
Omitting buoyancy in liquids, or confusing instantaneous force equality with time-averaged relations. Also, students sometimes assume terminal velocity requires drag = weight, which is only valid in gases of negligible density.

Final Answer:
Weight = buoyancy + drag