Answer to Question #184383 in Mechanics | Relativity for Konoha

Question

Every few years, winds in Boulder, Colorado, attain sustained speeds of 58 m/s when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli equation on a roof having an area of 212 m²? Typical air density in Boulder is 48kg/m³, and the corresponding atmospheric pressure is 8.89×10⁴ N/m². 

Round your answer to 0 decimal places

Answer 1

Explanations & Calculations

• Refer to the attached figure to get follow the explanation better

• Considering streamlines from just above the roof area and a location farther outside from the roof height & applying Bernoulli's principal is the way to solve these kind of problems.
• As the streamlines get closer & compact above the roof velocity increases & the pressure decreases.
• Compared to the pressure of still air inside the house (as well as the farther location already chosen) this pressure is less which generates a pressure difference across the roof thereby imposing a force on it.
• When this force exceeds the limits, the roof is flown away.

• Applying the equation to a selected streamline measured from the ground level,

$\qquad\qquad \begin{aligned} \small P_1+\frac{\rho v_1^2}{2}+\rho gh_1 &=\small P_2+\frac{\rho v_2^2 }{2}+\rho gh_2\\ \small \pi+0+\rho gH&=\small P_1+\frac{\rho v^2}{2}+\rho gH\\ \small \pi-P&=\small \frac{\rho v^2 }{2}\\ \small &=\small \frac{48kgm^{-3}\times(58ms^{-1})^2}{2}\\ &=\small 80736\,Pa \end{aligned}$

• This is the pressure difference between the rooftop & the distance location but this is approximately the same pressure across the roof.
• Therefore, force on the roof can be calculated as follows,

$\qquad\qquad \begin{aligned} \small F&=\small \Delta P.A\\ &=\small 80736\,Nm^{-2}\times 212m^2\cdots(Pa=Nm^{-2}\\ &=\small \bold{1.712\times10^7}\,N \end{aligned}$