Purpose: To demonstrate that the vector sum of the forces on a body in equilibrium is zero.

Some textbooks claim that the drag force is linearly proportional to the speed of a particle, some state the variation to be quadratic, while others indicate the drag to be linear at low speeds and become quadratic at higher speeds. This demonstration shows that, even at low speeds, the force is related quadratically to speed if the object has sufficient surface area that the flow around the edges is turbulent (corresponding to large Reynolds numbers). The transverse air speeds for such an object are not low.

(1) Linear drag force: If the drag force, D, is linear to speed, v, then D = kv leads to terminal speed
v_t = mg/k. In time, T, the object falls through height h = (mg/k)T. Thus, objects with masses m₁ and m₂ falling over the same time interval, T, drop distances h₁ and h₂ in direct proportion to their masses. That is,

(2) Quadratic drag force: A similar treatment using D = kv² leads to:

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