Math
posted by Catie on .
A hurricane wind blows across a 6.00 m times 15.0 m flat roof at a speed of 190 km/hr.
What is the pressure difference? Use 1.28 kg/m^3 for the density of air.
How much force is exerted on the roof?
If the roof cannot withstand this much force, will it "blow in" or "blow out"?
Help?

Use Bernoulli's equation:
P1+ρgh1+(1/2)ρv1² = P2+ρgh2+(1/2)ρv2²
where
P1=exterior pressure
P2=interior pressure
v1=exterior wind velocity
v2=interior wind velocity = 0
h1h2 is the negligible height differences
ρ=density of air (1.293 kgm^3)
which by neglecting ρg(h2h1), and substituting v1=0 we get
P2P1=(1/2)ρv1²
v1=190km/h=52.778m/s
ρ=1.293kg/m³=12.7N/m³
Suction force (blows out):
=area*pressure difference
=6m*15m*(1/2)*1.293kg/m³*(52.8m/s)²
=162,073 N
The above calculation of pressure assumes:
1. the pressure factor is 1 (which is not far for a flat roof).
See:
(Broken Link Removed)
2. the STP value of ρ=1.293kg/m³ instead of 1.28 as stipulated. You will need to make adjustments to get the correct answer.