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Wind and air pressure problem

  1. Apr 9, 2006 #1
    Any ideas how to start this problem (formulas or anything)?

    Just after its completion, the glass-walled skyscraper known as the John Hancock tower in Boston (see Figure P.64) was faced with an embarrassing problem: its windows kept falling out on windy days! The building soon became known as the "plywood palace," since at one point plywood covered a significant percentage of the exterior of the building as a temporary replacement for the lost windows. Needless to say, the 1.89 m x 2.86 m glass panes were quite a safety hazard as they cascaded to the pavement below. Fortunately, no casualties resulted from the flying panes. If a wind with a speed of 59.0 km/h is blowing parallel to the face of one of the windows, what is the magnitude of the net outward force on each window caused by the pressure difference between the inside and outside of the building?

    I have no clue how to calculate the pressure differences or the force that would be needed to knock the windows out.
     
  2. jcsd
  3. Apr 10, 2006 #2

    Andrew Mason

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    Assume the air pressure inside is the same as the air pressure outside with stationary air. Use Bernouilli's equation to determine the pressure drop caused by moving air:

    [tex]P + \frac{1}{2}\rho v^2 = P_{atm}[/tex]

    AM
     
  4. Apr 10, 2006 #3
    so P = 1.01E5 Pa + (1/2)*(1.168)*16.389 m/s ?
    how would the size of the window play into the problem then?

    thanks!
     
  5. Apr 10, 2006 #4
    actually, i meant:
    P = 1.01E5 Pa - (1/2)*(1.168)*16.389 m/s
    NOT
    P = 1.01E5 Pa + (1/2)*(1.168)*16.389 m/s

    but i'm still unsure as to how to factor in the size of the window...
     
  6. Apr 10, 2006 #5

    Hootenanny

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    How does one define pressure?

    -Hoot:smile:
     
  7. Apr 10, 2006 #6
    P=F/A so F=P*A

    but i tried this using the P = 1.01E5 Pa - (1/2)*(1.168)*16.389 m/s = 100990.4288 Pa
    and A = 1.89 m x 2.86 m = 5.4054 m^2
    so F = 545893 N, but this wrong

    am i still setting it up wrong, or do my numbers look wrong (is that not the value for air density i should use, etc?)
     
  8. Apr 10, 2006 #7

    Hootenanny

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    (1) You forgot to square the velocity.

    (2) The density of dry air at sea level is about 1.25 kg / m^3

    Number (2) shouldn't matter that much, but (1) will put you way off.

    -Hoot:smile:
     
  9. Apr 10, 2006 #8
    oh shoot, you're right, i did, thanks!

    so does this look right:
    P = 1.01E5 Pa - (1/2)*(1.25)*(16.389)^2 m/s = 100832 Pa
    and A = 1.89 m x 2.86 m = 5.4054 m^2
    so F = 545038 N ?

    it doesn't seem that off from the 545893 N i got before (within 1% of each other), so i am still unsure if i'm doing it right...
     
  10. Apr 10, 2006 #9

    Hootenanny

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    Your final calc looks correct to me.
     
  11. Apr 10, 2006 #10

    Andrew Mason

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    The larger the area, the larger the net force: [itex]F = (P_{in} - P_{out})A[/itex]

    Also, the density of air is 1.29 kg/m^3 at standard temperature and pressure.

    The net force is due to this pressure difference. So the net outward force on the window is:

    [tex]\Delta P A = 173 N/m^2 \times 5.15 m^2 = 890 N[/tex]

    This is the equivalent of the weight of a man pulling outward on each window.

    AM
     
    Last edited: Apr 10, 2006
  12. Apr 10, 2006 #11
    is P(in)=1atm=1.01E5 Pa?
    and is P(out)=1.00826E5 Pa ?

    if so then would the solution be:
    (1.01E5 - 1.00826E5)*(5.4054)=936.454 N
    ?
     
  13. Apr 10, 2006 #12
    I see where you got 173 N/m^2, but shouldn't Area=1.89m*2.86m=5.4054m^2 (not 5.15) or is this a dif number? If this is a mistake, then the answer should be about 936N, right?

    What do you mean by this? How much does an average man weigh, or how do you know what force he could pull out with? I am just curious about this part.

    Thanks!!
     
  14. Apr 10, 2006 #13

    Hootenanny

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    Sorry, for misleading you kelly. I missed the difference in pressure.
     
  15. Apr 10, 2006 #14

    Andrew Mason

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    Right.



    A 95.5 kg. person weights 936 N. So that gives you some idea of the force pushing outward on the window.

    AM
     
  16. Apr 2, 2009 #15
    I'm doing the same question, the second part asks
    "Compare the magnitude of this force to the magnitude of the weight of a 72.2 kg insurance salesperson."

    the force in my case (slightly different numbers) was 1012 N. Any help is appreciated. thanks
     
  17. Apr 2, 2009 #16
    Compare the magnitude of this force to the magnitude of the weight of a 72.2 kg insurance salesperson. And it says (times weight of the salesman) near the answer box....I am unsure what this means. Again my force was 1012 N.
    thank you
     
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