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The air of density $ν$ and moving with a velocity v strikes perpendicularly the inclined surface of area A and of a wedge kept on a horizontal surface. The mass of the wedge is m. Assuming that air molecules move along the surface of the wedge after collision. The minimum value of the coefficient of friction between the wedge and the ground so that the wedge does not move is (g is acceleration due to gravity)

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$\frac{ρA ν^{2} \sin θ}{mg + ρA ν^{2} \cos θ}$

$\frac{{ρAν}^{2}}{mg} \tan θ$

$\tan θ$

$\frac{{ρAv}^{2}}{mg + {ρAv}^{2} \cos θ}$

answer is A.

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Detailed Solution

$\begin{array}{l} N = mg + {ρAv}^{2} \cos θ \\ F = {ρAv}^{2} \sin θ = μN \Rightarrow μ = \frac{{ρAv}^{2} \sin θ}{mg + {ρAv}^{2} \cos θ} \end{array}$

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