Question

organisms. If we have a sphere moving in a fluid of density ρ and viscosity μ, and the sphere has a radius R and speed r and drag coefficient Co, then the drag and viscous forces are given by these equations: 2. in this problem, we'll estimate which resistive force- viscous force or drag force-dominates for two different The Reynolds number is given by Re ()Rv. For air and water, we can use the densities and viscosities in the table below. We can also use the indicated drag coefficient for a sphere in each of these fluids Fluid Density p l viscosity μ 11.0 × 10' ko/m3-8 × 10-3 N·s/m2 Drag coefficient Co Air 2.0 x 105 N s/m20.20 water 10.25 Our first organism is the larva of a daphnia (a small crustacean). It lives in water and at certain times in its lifecycle it goes dormant and sinks passively to the bottom. We will model the larva as a sphere of radius 0.3 mm. When it's sinking passively, the larva travels at a speed of about 2 mm/s (assumed to be constant and straight down). a. Draw a free body diagram showing all the forces acting on the daphnia larva as it sinks. Label each of the forces. What is the net force on the larva? Use Newton's second law to write an equation relating all the forces on the larva. b.

Answer 1

(a) F_v is the viscous force, F_D is the drag force and since it is moving passively it has gravitational force mg downwards.

(b) Since the Larva is moving at a constant velocity downwards, the net force on Larva should be 0. By Newton's Laws we can write ( downward direction is -ve and upwards direction is +ve ).

$\Rightarrow mg = F_D + F_V$

(c) The Reynold's number can be calculated as,

$R_e = \frac{\rho R v}{\mu}$

The given medium is water, then substituting the values we get,

$R_e = \frac{\rho R v}{\mu} = \frac{(10^{3})(0.3 \times 10^{-3})(2 \times 10^{-3})}{0.8 \times 10^{-3}} = 0.75$

Therefore, the flow is laminar, Hence the viscous force will be more than drag vorce. (Viscous force will have more effect).

(d) The drag force is given by,

$F_D = \frac{1}{2}C_d\rho \pi R^2 v^2 =\frac{1}{2} (0.25) (10^3) \pi (0.3 \times 10^{-3})^2(2 \times 10^{-3})^2 = 1.41 \times 10^{-10} \,N$

The viscous force is given by,

$F_v = 6 \pi \mu Rv = 6 \pi (0.8 \times 10^{-3})(0.3 \times 10^{-3})(2 \times 10^{-3}) = 9.05 \times 10^{-9} \, N$

This is consistent with part (c).