If wind blows at 39 m/s over a house,

  • Thread starter Carnivroar
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In summary, Bernouilli's principle states that when the speed of air is greater, the pressure is smaller. The pressure is smaller when the flat roof is at ground level, so the wind blows over the roof at a faster speed and the pressure is less than the pressure at the roof when the wind is blowing at a slower speed. The wind blows over the roof at a speed of v2=39 m/s and the pressure is p2=p1-1/2 (rho) v22. The pressure means the force exerted to unit surface. So the net force on the roof is F=PA-P2A=(P1-P2)A=[P1-(P1-1/2
  • #1
Carnivroar
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1
what is the net force on the roof if its area is 320 m^2 and is flat?

I don't even know where to start.
 
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  • #2
What class are you taking?
 
  • #3
LawrenceC said:
What class are you taking?

physics 1

i think i have to use bernoulli's formula but i don't know how
 
  • #4
What does Bernouilli's principle state?

ehild
 
  • #5
ehild said:
what does bernouilli's principle state?

Ehild

1/2ρv12 + p1 = 1/2ρv22 + p2
 
  • #6
Carnivroar said:
1/2ρv12 + p1 = 1/2ρv22 + p2

Correct. If you look at the equation you see that the pressure is smaller when the speed of the air is greater. The speed of air is zero below the roof in the building, v1=0 and the pressure below the roof is just the usual p1=1 atm. The wind blows over the roof, its speed is v2=39 m/s. p2 =p1 - 1/2 (rho) v22.
The pressure means the force exerted to unit surface. The air inside the building exerts F1=p1A upward force to the flat roof of area A. The air outside exerts p2A downward force to the roof. What is the net force then?

ehild
 
  • #7
This is the answer I'm supposed to get

3.1×10^5

or 310000

Let me try it the way you said

Since V1 = 0 the first term goes away

P1 = 1/2 ρ 39^2 + P2

(1.225 = density of air)

101325 - 1/2 * 1.225 * 1521 = P2

101325 - 931.6125 = 100393.3875

100393.3875 * A = Fnet

100393.3875 * 320 = 32125884

Not right.

What am I doing wrong?
 
  • #8
Carnivroar said:
This is the answer I'm supposed to get

3.1×10^5

or 310000

Let me try it the way you said

Since V1 = 0 the first term goes away

P1 = 1/2 ρ 39^2 + P2

(1.225 = density of air)

101325 - 1/2 * 1.225 * 1521 = P2

101325 - 931.6125 = 100393.3875

100393.3875 * A = Fnet

100393.3875 * 320 = 32125884

Not right.

What am I doing wrong?

For one thing you should really put units on stuff. For another thing you are only calculating the force on the roof exerted by the air outside the house. There's another force exerted on the roof by the air inside the house that isn't moving. They are in opposite directions.
 
  • #9
You calculated the force of the blowing air, but you need the net force. Read my previous post.

ehild
 
  • #10
Oh, okay. Got it now. The problem with this question is that it doesn't give you the value for air density, so even a slight difference in the value you chose will make a difference in the big numbers. With the 1.225 I used above I was getting ~290000 then I used 1.29 and got the ~310000 I needed.

Thanks!
 
  • #11
You get more accurate result if you express the force symbolically and pug in the data at the end. The net force is
F=PA-P2A=(P1-P2)A=[P1-(P1-1/2 ρv2)]A=(1/2 ρv2)A
No need to calculate the pressure.
The density of air depends quite much on the temperature and hight above sea level.
It should have been given in the problem.

ehild
 

Related to If wind blows at 39 m/s over a house,

1. What is the speed of the wind blowing over the house?

The speed of the wind blowing over the house is 39 m/s.

2. Is 39 m/s considered a strong wind?

Yes, 39 m/s is considered a strong wind. It falls under the category of "strong gale" according to the Beaufort Wind Scale.

3. How does wind speed affect the house?

The higher the wind speed, the greater the force applied on the house. This force can cause damage to the structure and potentially lift or displace objects on or around the house.

4. Can the wind blowing at 39 m/s cause a tornado?

No, a tornado is a specific type of weather phenomenon that requires specific atmospheric conditions and wind patterns. Wind blowing at 39 m/s alone is not enough to cause a tornado.

5. How can the wind speed over the house be measured?

The wind speed over the house can be measured using an anemometer, which is a device that measures wind speed and direction. It consists of cups that spin in the wind, and the speed is calculated based on the rotation of the cups.

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