Stokes law and speed at equilibrium

Physics Asked by Vlad Uskar on November 27, 2020

It is stated that the equation leading to Stokes law describes an equilibrium state, such as the upwards force is equal to the downward force:

$$ 6pi reta v = frac43pi r^3 (d_1 – d_2)g $$

Now $v$ is velocity, and velocity at equilibrium is $0$, and this is the breaking point of my understanding. What is the nuance here?

2 Answers

(a) Your equation does not lead to Stokes's law but $incorporates$ Stokes's law. The law states that a sphere of radius $r$ moving at speed $v$ through a fluid of viscosity $eta$ experiences a resistive force given by $$F_text{res}=6 pi eta r v$$ and is in the opposite direction to the body's velocity. It holds provided that there is streamline flow of the liquid relative to the sphere.

(b) Your equation applies to when a body, falling through a fluid, has reached a constant velocity (provided that flow remains streamline). The condition for constant velocity is that upward and downward forces on the sphere are balanced.

The upward forces are the Stokes's law force and the Archimedean upthrust given by $$F_text{Arch}=tfrac43 pi r^3 d_2 g$$ in which $d_2$ is the fluid density. The downward force is the pull of gravity, $$F_text{grav}=tfrac43 pi r^3 d_1 g$$ in which $d_1$ is the density of the material of the sphere.

As we have said, when a steady velocity if reached $$F_text{res} + F_text{Arch} = F_text{grav}$$ Substituting from formulae above and re-arranging gives the equation you've quoted.

Correct answer by Philip Wood on November 27, 2020

The equilibrium state in this case just means that there isn’t a net force acting on the object—the net drag force on the object coming from Stokes’ law is equal and opposite to the gravitational force + buoyancy force on the object. As a result, it doesn’t accelerate, and its velocity is constant.

Answered by aghostinthefigures on November 27, 2020

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