Drag force reads

$$F_d = \frac{1}{2}\rho_F v^2 \mathcal{C} A$$

Where the letters have the usual meaning (respectively: fluid density, velocity relative to the fluid, drag coefficient, front area of the object).

When a sphere, for example, is left to sink in a jar of water, I expect then to meet two forces: weight and drag. For example if I want to find the velocity, I would obtain

$$v = \sqrt{\frac{2mg}{\rho_F \mathcal{C}A}}$$

However, it has been taught that the "correct" way to operate this kind of problem is to take into account weight, Archimede's law and Stokes' drag, respectively (with the correct sign according to the force's direction):

$$mg ~~~~~~~ -\rho_F V g ~~~~~~~ -6\pi \eta R v$$

Where $V$ is the volume of the sphere and $\eta$ is the viscosity of the fluid, $R$ is the sphere's radius.

Accordi to this, the velocity would be

$$v = \frac{2}{9} \frac{R^2 (\rho_S - \rho_F) g}{ \eta}$$

Clearly different from the previous one.

Now my question is: why is the first method wrong?


The first equation is not necessarily wrong.

When a sphere passes through a fluid at "low" speed the drag on the sphere is given by $6 \pi R v \eta$ and this is known as Stokes's law.
Derivation of this equation is difficult and requires a number of assumption to be made including that the fluid flow must be laminar.
You have used the equation in your second example of the sphere falling in water.
To decide whether or not this equation is applicable a dimensional parameter, Reynold's number $= \dfrac{\rho a v}{\eta}$ where $\rho$ is the density of the fluid and $a$ is a characteristic linear dimension which would be the radius of the sphere $R$ in this case, is used.
It has been found that for Stokes's law to be valid the Reynold's number should be less than one.

As the speed of the fluid relative to the object increases the inertia (density) of the fluid rather than the viscosity becomes relatively more and more important in determining the drag on an object moving through a fluid.

So the Stokes's law regime when the drag is proportional to the speed of the fluid relative to the object is superseded by the drag becoming proportional to the speed of the fluid relative to the object squared which is your first example.

  • $\begingroup$ This explains everything I needed to know! Thank you! $\endgroup$
    – Les Adieux
    Apr 9 '18 at 16:14
  • $\begingroup$ If you have time and will, could you please also check this other question of mine? physics.stackexchange.com/q/398708 $\endgroup$
    – Les Adieux
    Apr 9 '18 at 16:16

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