How much force is exerted on the roof?

  • Thread starter Thread starter lindsey4
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The discussion focuses on calculating the force exerted on a flat roof during a hurricane with wind speeds of 130 km/h. It highlights the need to apply Bernoulli's principle to determine the pressure difference between the air above the roof and the air inside the house. The dynamic pressure component of the Bernoulli equation is crucial for these calculations, specifically using the formula Pressure = Force/Area. The conversation also raises concerns about whether the roof can withstand the force, questioning if it will blow in or out if it fails. Understanding these principles is essential for accurately assessing the roof's structural integrity under extreme weather conditions.
lindsey4
Messages
3
Reaction score
0

Homework Statement


A hurricane wind blows across a 6.0mX15.0m flat roof at a speed of 130 km/h. Is the air pressure above the roof higher or lower than the air pressure inside the house? What is the pressure difference? How much force is exerted on the roof? If the roof cannot withstand the force, will it blow in or out?


Homework Equations



not sure

The Attempt at a Solution


I think you use something like Bernouli's principle but I really have no clue
 
Physics news on Phys.org
You are looking for the dynamic pressure part of the Bernoulli equation. Pressure=Force/area
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top