Another explanation is that the stones move only when the water dumped on the playa during a storm freeze into a large, thin sheet of ice. The stones are trapped in place in the ice. Then, as air flows across the ice during a wind, the air-drag forces on the ice and stones move them both, with the stones gouging out the trails. The magnitude of the air-drag force on this horizontal “ice sail” is given by ${\mathbf{D}}_{\mathbf{ice}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}{\mathbf{C}}_{\mathbf{ice}}{\mathbf{rA}}_{\mathbf{ice}}{\mathbf{v}}^{\mathbf{2}}$, whereice is the drag coefficient$\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{103}\mathbf{)}$, r is the air density$\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{)}$,${\mathbf{A}}_{\mathbf{ice}}$is the horizontal area of the ice, and vis the wind speed along the ice. Assume the following: The ice sheet measures$\mathbf{400}\mathbf{}\mathbf{m}$by$\mathbf{500}\mathbf{}\mathbf{m}$ by$\mathbf{4}\mathbf{.}\mathbf{}\mathbf{0}\mathbf{}\mathbf{mm}$and has a coefficient of kinetic friction of $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}$with the ground and a density of$\mathbf{917}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. Also assume that 100 stones identical to the one in Problem 8 are trapped in the ice. To maintain the motion of the sheet, what are the required wind speeds

(a) near the sheet and

(b) at a height of$\mathbf{10}\mathbf{}\mathbf{m}$?

(c) Are these reasonable values for high-speed winds in a storm?

• The dimension of ice is 400 m by 500 m by 4 mm.
• The density of ice is$917{\text{\hspace{0.17em}kg/m}}^{\text{3}}$.
• The air-drag force is:$\mathrm{D}=4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{\rho A}}_{\mathrm{ice}}{\mathrm{v}}^2$.
• The drag coefficient,${\mathrm{C}}_{\mathrm{ice}}=2\times {10}^{-3}$ .
• The air density,$\mathrm{\rho }=1.21{\text{\hspace{0.17em}kg/m}}^{\text{3}}$.
• The number of stones is 100.
• The mass of each stone is 20 kg.
• The kinetic friction coefficient between ice and ground, ${\mathrm{\mu }}_{\mathrm{k}}=0.10$.

The problem deals with the relation between air-drag force and frictional force. Air-drag force is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. The frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. Here equate the air-drag force to the frictional force to solve this question.

Formula:

Air drag force is given by,

$\mathrm{D}=4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{\rho A}}_{\mathrm{ice}}{\mathrm{v}}^2$

Frictional force is

$\mathrm{F}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mg}$

The mass of ice:

$\begin{array}{l}{\mathrm{m}}_{\mathrm{ice}}=917{\text{\hspace{0.17em}kg/m}}^{\text{3}}\times (400\text{\hspace{0.17em}m}\times 500\text{\hspace{0.17em}m}\times 0.004\text{\hspace{0.17em}m})\\ {\mathrm{m}}_{\mathrm{ice}}=7.34\times {10}^5\text{\hspace{0.17em}kg}\end{array}$

The mass of ice with 100 stones:

$\begin{array}{l}\mathrm{m}={\mathrm{m}}_{\mathrm{ice}}+100\times 20\text{\hspace{0.17em}kg}\\ \mathrm{m}=7.34\times {10}^5\text{\hspace{0.17em}kg}+2000\text{\hspace{0.17em}kg}\\ \mathrm{m}=7.36\times {10}^5\text{\hspace{0.17em}kg}\end{array}$

To calculate the velocity of wind, equate air-drag force to frictional force:

$\begin{array}{c}\mathrm{D}=\mathrm{F}\\ 4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{\rho A}}_{\mathrm{ice}}{\mathrm{v}}^2={\mathrm{\mu }}_{\mathrm{k}}\mathrm{mg}\\ \mathrm{v}=\sqrt{\frac{{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mg}}{4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{\rho A}}_{\mathrm{ice}}}}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{\left(0.10\right)\times (7.36\times {10}^5\text{\hspace{0.17em}kg})\times (9.8{\text{\hspace{0.17em}m/s}}^{\text{2}})}{4\times (2\times {10}^{-3})\times (1.21{\text{kg/m}}^{\text{3}})\times (400\text{\hspace{0.17em}m}\times 500\text{\hspace{0.17em}m})}}\\ \mathrm{v}=19\text{\hspace{0.17em}m/s}\times \frac{3600\text{\hspace{0.17em}s}}{1\text{hr}}\times \frac{\text{1\hspace{0.17em}km}}{\text{1000\hspace{0.17em}m}}\\ \mathrm{v}=\text{69\hspace{0.17em}km/h}\end{array}$

Thus, the required speed is 69 km/h.

Doubling our previous result, we find the required speed to be$139\text{\hspace{0.17em}km/h}$ .

Thus, the speed will be 139 km/h.

The result is reasonable for storm winds. A category-5 hurricane has speeds on the order of $2.6\times {10}^2\text{m/s}$.

Thus, we can say that these values for high-speed winds in a storm are reasonable.