1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pressure due to a perpendicular wind

  1. Nov 30, 2013 #1
    Sorry to ask another question :/
    1. The problem statement, all variables and given/known data
    Consider the change of momentum.
    A wind (of speed v) acts perpendicularly to a wall. Show that the pressure do to this wind acting of the wall is v2ρ, where ρ is air density.

    Hence show by suitable integration that the torque about the bottom edge of a vertical surface of height h and width x is given by 0.5v2ρLh2

    If a wall has height 2m, thickness 0.3m, mass 1200kg per metre length, what is the wind speed needed to topple this?
    Air density = 1.2 kgm-3





    The attempt at a solution
    Erm...
    I think P=F/A will be useful here.
    Mass will likely be substituted for ρ*V
    But I have no idea where to go :(
    Any tips appreciated
     
  2. jcsd
  3. Nov 30, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi LiamG_G! Welcome to PF! :smile:
    why?? that's what everybody's here for! :wink:
    as it says, consider the change of momentum …

    how much mass changes its momentum in time t ?

    (call the area "A", it'll cancel)
     
  4. Dec 1, 2013 #3
    The equations I would think would be useful are ρ=F/A, ρ=m/V, P=mv
    ρ= air density, P=momentum
    as m=ρV then I can substitute it into momentum giving p=ρVv.
    I just don't know what to do. It's most probably frustration blocking something simple.
    I'm not even sure how to describe the mass changing momentum with t.
    The only thing that comes to mind is:
    F=(ΔP)/t, F=ρA=(m/V)A
    (m/Ax)A=(ΔP)/t
    m=x(ΔP)/t
    but then I have no idea what to do with this, and I can't see how it relates :(
     
  5. Dec 1, 2013 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi LiamG_G! :smile:
    In time t, a block of air has reduced its speed from v to 0.

    What is the area of that block?

    What is the length of that block?

    And so what is the mass of that block?

    And what is the change of momentum of that block? :wink:
     
  6. Dec 1, 2013 #5
    What is the area of the block?
    Let area=A, representing the section of the wall impacted by the wind.

    The length of the block of air would be equal to vt, no? It makes sense that the amount of air hitting the wall is dependent of the speed of the air and the time taken.

    The mass would be equal to ρV, which is ρAvt.

    So then the momentum change would be v2ρAt

    p=F/A, and F=Δm/t so this leads to p=v2ρ
    Yay! I think that's it :)

    Thank you so much.


    Would you be able to help in the second part of the question?
    Hence show by suitable integration that the torque about the bottom edge of a vertical surface of height h and width x is given by 0.5v2ρLh2

    I would assume that I am to integrate with respect to h, so that the h becomes h2 and that would also bring the 0.5 into the equation. Although I don't see where this initial equation would come from.
    All I know for torque is that τ=I*α, with I=moment of inertia, and α=angular acceleration
     
  7. Dec 1, 2013 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    yes :smile:

    do you see how that came directly from the definition of force …

    force = rate of change of momentum (so impulse = force times time, which of course is good ol' Newton's second law)

    a lot of physics is simply using the appropriate basic equation (same with the second part! :wink: …)
    yes, it is integration, but since α = 0, it won't be τ=I*α, will it?

    it'll be the even more basic τ = Fh …

    slice the dam into horizontal slice of thickness dh …

    what do you get? :smile:
     
  8. Dec 1, 2013 #7
    Of course, thank you :)
    We determined before the F=ρAv2
    In this situation, A=Lh
    Considering a thin strip of height dh we can get to:
    τ=v2ρL∫0hhdh
    which leads to the equation needed; 0.5v2ρLh2

    With the third (and final :P ) part, I can get an equation for F in terms L and h from F=τ/h, leaving me with F=0.5v2ρLh
    My first thought was to set this equal to ma, to bring mass, but then this also brings acceleration into the mix, which I think(hope) is not necessary.
    My other thought immediately led to ρ being cancelled, but as the question tells to take the density of air as 1.2 kgm-3, cancelling out ρ can't be right.
     
  9. Dec 1, 2013 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Frankly, I don't understand the third part of the question …
    … we're not told anything about the foundations of the wall, so how can know the force needed to topple it?

    I suppose we could answer it on the basis that the wall has no foundations, and is just balancing there, waiting to be pushed over by any wind capable of tilting it to an angle of 0.3/2 :confused:
     
  10. Dec 1, 2013 #9
    Ah sorry, the question states 'a brick wall rests on flat ground' so I think we are supposed to assume that it has no foundations :)
     
  11. Dec 1, 2013 #10
    I have the same questions as Liam right now. Both trying to figure them out together over the phone :P
     
  12. Dec 1, 2013 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi SamQP! :smile:
    In that case, the basic equation to apply here is simply the (statics) equilibrium equation for torque

    there are three forces on the wall:
    i] its weight, mg
    ii] the wind
    iii] the reaction force (also mg) from the ground​

    the wall will start to topple (about the front edge of the wall) when the reaction force goes through the front edge of the wall! :wink:
     
  13. Aug 2, 2016 #12
    Hi,

    I have just come across this problem myself and the final part is confusing me. Do you have any further tips to get me started?

    Also, I don't understand why the reaction force would need to go through the front edge of the wall?

    Thanks, James
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted