Will the Unsecured Shed Roof Withstand the Storm Winds?

[vu10758]

Homework Statement

Someone is putting a small shed in their backyard. The roof is not nailed down so gravity alone is holding it down. The wind suddenly flows across the top of the roogf at 20 m/s. The air inside the shed is 1.01 x 10^15 Pa, the normal atmospheric pressure. The roof has an area of 16m^2 and a mass of 250 kg. Density of air is 1.29 kg/m^3. Will the roof fly off?

Homework Equations

The Attempt at a Solution

The answer is that the roof will fly off.

In order for this to be true, pressure inside plus gravity must be greater than pressure outside.

Pushing down I have Pressure_up = 1.01 x 10^5 Pa.

I know that Pressure_down = mg/A + P_wind

I know this must be less than 1.01 x 10^5

However, what is the formula for Pressure_wind? I know the faster the wind, the smaller the pressure. I know formulas such as p + .5*density*v^2 + density*g*y, but it seems like a higher velocity would yield a higher pressure with this formula. It shouldn't be that way. What am I doing wrong?

 

[AlephZero]

What is "p + .5*density*v^2 + density*g*y" equal to?

If you use the equation correctly, you will find higher velocity does give lower pressure.

 

[m2003]

Thank you for providing the necessary information to solve this problem. Based on the given information, we can use the formula for pressure, P = F/A, where F is the force exerted on the surface and A is the area of the surface. In this case, the force exerted by the wind on the roof would be the pressure of the wind (P_wind) multiplied by the area of the roof. Therefore, the formula for P_wind would be P_wind = F/A = (P_wind)(A).

To determine the pressure of the wind, we can use the Bernoulli's principle, which states that as the speed of a fluid (in this case, air) increases, its pressure decreases. This can be represented by the equation P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2, where P1 and P2 are the initial and final pressures, ρ is the density of the fluid, and v1 and v2 are the initial and final velocities, respectively. In this case, we can assume that the initial and final pressures are equal (since the air inside the shed is at atmospheric pressure) and solve for P_wind.

Using the given values, we can calculate the velocity of the wind by rearranging the equation to v2 = √((2(P1 - P2))/ρ), where P1 is the atmospheric pressure and P2 is the pressure inside the shed. Plugging in the values, we get v2 = √((2(1.01 x 10^5 - 1.01 x 10^15))/1.29) = 2.21 x 10^6 m/s.

This is a very high velocity, indicating that the wind is exerting a significant force on the roof. Therefore, the roof will likely fly off as the force of the wind is greater than the force of gravity holding it down. It is important to properly secure the roof to prevent this from happening.