Question

A log is floating downstream. How would you calculate the drag force acting on it?

Answer 1

A log is floating downstream. How would you calculate the drag force acting on it?

Answer: There are two basic types of the log movement in the water.

1) if $\frac{\upsilon \cdot L \cdot \rho}{\mu} < 10$ , where $\upsilon$ is the velocity of log in the water, $m/s$ ; $L$ is the length or width of the log, $m$ ; $\rho$ is the water density, $\mathrm{kg/m^3}$ ; $\mu$ is the viscosity of water, $\mathrm{Pa \cdot s}$ .

In such conditions the drag force can be calculated as $F = k \cdot \mu \cdot L \cdot \upsilon$ , where $k$ is a dimensionless coefficient, which depends from the shape and streamlining of the log.

For example, if the log is a sphere with diameter $D$ , then $k = 3\pi \approx 9.42$ , and $F = 9.42 \cdot \mu \cdot D \cdot \upsilon$ . As you see, in this case the drag force is directly proportional to the speed of the log, $F \propto \upsilon$ .

2) if the log moves in such conditions, that $\frac{\upsilon \cdot L \cdot \rho}{\mu} > 100$ , then the drag force can be calculated as $F = C \cdot A \cdot \frac{\rho \cdot \upsilon^2}{2}$ , where $A$ is the maximal value of cross-sectional area of the log, $\mathfrak{m}^2$ ; $C$ is a dimensionless coefficient, which depends from the shape and streamlining of the log.

For example, if the log is a sphere with diameter $D$ , then $k = 0.44$ , and $F = 0.44 \cdot A \cdot \frac{\rho \cdot v^2}{2} = 0.44 \cdot \frac{\pi \cdot D^2}{4} \cdot \frac{\rho \cdot v^2}{2} = 0.173 \cdot \rho \cdot D^2 \cdot v^2$ .

As you see, in this case the drag force is directly proportional to the square of speed of the log, $F \propto v^2$ .