Continuation of Problem 8. Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical ${}^{\mathbf{20}\mathbf{kg}}$stone, which presents to the wind a vertical cross-sectional area of${}^{\mathbf{0}\mathbf{.}\mathbf{040}\mathbf{}{\mathbf{m}}^{\mathbf{2}}}$and has a drag coefficient C of$\mathbf{0}\mathbf{.}\mathbf{80}$. (a) In kilometers per hour, what wind speed${}^{\mathbf{V}}$along the ground is needed to maintain the stone’s motion once it has started moving? Because winds along the ground are retarded by the ground, the wind speeds reported for storms are often measured at a height of${}^{\mathbf{10}\mathbf{}\mathbf{m}}$. Assume wind speeds are${}^{\mathbf{2}\mathbf{.}\mathbf{00}}$ times those along the ground. (b) For your answer to (a), what wind speed would be reported for the storm? (c) Is that value reasonable for a high-speed wind in a storm?

Mass of the stone,$M=20\mathrm{kg}$

Area of cross section,$A=0.040{\mathrm{m}}^2$

Drag coefficient,$\mathrm{C}=0.8$

Air density,$p=1.21\mathrm{kg}/{\mathrm{m}}^3$

Coefficient of kinetic friction,${\mu }_k=0.80.$

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium.

Formula:

The terminal speed is given by

$v_t=\sqrt{\frac{2F_g}{CpA}}$

Where C is the drag coefficient,${}^p$is the fluid density, A is the effective cross-sectional area, and${}^{F_g}$is the gravitational force.

Free body diagram of the stone:

Kinetic frictional force must act on the stone during its motion.

To find its magnitude, use Newton’s 2nd law along y direction.

Stone is not moving along y.So, the acceleration is 0, it gives,

$$\begin{gathered} \sum _{}{F}_y=0 \\ N-mg=0 \\ N=mg \end{gathered}$$

Kinetic frictional force is defined as,

$$\begin{gathered} f_k={\mu }_{k}N \\ ={\mu }_k\left(mg\right) \\ =\left(0.80\right)\left(\left(20\right)\left(9.81\right)\right) \\ =156.96\mathrm{N} \end{gathered}$$

In this case,${}^{F_k}$is nothing but the drag force. Thus,

Where D is the drag force. Then, by using the equation,

$D=\frac{1}{2}CpA{V}^2,$

expression for speed of the wind can be written as,

$$\begin{gathered} v=\sqrt{\frac{2F}{CpA}} \\ =\sqrt{\frac{2\left(157\mathrm{N}\right)}{\left(0.80\right)\left(1.21\mathrm{kg}/{\mathrm{m}}^3\right)\left(0.040{\mathrm{m}}^2\right)}} \\ =90\mathrm{m}/\mathrm{s} \\ =3.2\times {10}^2\mathrm{km}/\mathrm{h} \end{gathered}$$

Hence,the wind speed V along the ground is needed to maintain the stone’s motion once it has started moving is${}^{3.2\times {10}^2\mathrm{km}/\mathrm{h}}$.

Doubling our previous result, the reported speed is found to be${}^{6.5\times {10}^2\mathrm{km}/\mathrm{h}}$.

The result is not reasonable for a terrestrial storm. A category 5 hurricane has speeds

on the order of${}^{2.6\times {10}^2\mathrm{m}/\mathrm{s}}$.