Question

Calculate the viscous force acting on a rain drop of diameter 2 mm, falling with a uniform velocity 4 m/s through air. The coefficient of viscosity of air is 1.8 x 10-5 Ns/m².

Answer 1

1.36 x 10-6 N

Answer 2

The viscous force acting on a spherical object moving through a fluid at low Reynolds number is given by Stokes' Law:

$F_v = 6 \pi \eta r v$

where:

$F_v$ is the viscous force $\eta$ is the coefficient of viscosity of the fluid $r$ is the radius of the spherical object $v$ is the velocity of the object

Given values:

Diameter of the rain drop, $d = 2$ mm Radius of the rain drop, $r = d/2 = 2/2 = 1$ mm $= 1 \times 10^{-3}$ m Uniform velocity of the rain drop, $v = 4$ m/s Coefficient of viscosity of air, $\eta = 1.8 \times 10^{-5}$ Ns/m²

Substitute these values into Stokes' Law formula:

$F_v = 6 \pi \times (1.8 \times 10^{-5} \text{ Ns/m²}) \times (1 \times 10^{-3} \text{ m}) \times (4 \text{ m/s})$ $F_v = (6 \times 1.8 \times 1 \times 4) \times \pi \times (10^{-5} \times 10^{-3})$ N $F_v = (10.8 \times 4) \times \pi \times 10^{-8}$ N $F_v = 43.2 \times \pi \times 10^{-8}$ N

Using the value of $\pi \approx 3.14159$:

$F_v \approx 43.2 \times 3.14159 \times 10^{-8}$ N $F_v \approx 135.7167648 \times 10^{-8}$ N $F_v \approx 1.357167648 \times 10^{-6}$ N

Rounding to 3 significant figures: $F_v \approx 1.36 \times 10^{-6}$ N.