Pressure due to a perpendicular wind

[LiamG_G]

Sorry to ask another question :/

Homework Statement

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

 

[tiny-tim]

Hi LiamG_G! Welcome to PF!

Sorry to ask another question :/

why?? that's what everybody's here for!

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.

as it says, consider the change of momentum …

how much mass changes its momentum in time t ?

(call the area "A", it'll cancel)

 

[LiamG_G]

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 :(

 

[tiny-tim]

Hi LiamG_G!

I'm not even sure how to describe the mass changing momentum with t.

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? ​

 

[LiamG_G]

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?​

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

 

[tiny-tim]

… 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 :)

yes …

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! …)

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

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? ​

 

[LiamG_G]

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∫₀^hhdh
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.

 

[tiny-tim]

Frankly, I don't understand the third part of the question …

If a wall has height 2m, thickness 0.3m, mass 1200kg per metre length, what is the wind speed needed to topple this?

… 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

 

[SamQP]

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 :)

 

[SamQP]

I have the same questions as Liam right now. Both trying to figure them out together over the phone :P

 

[tiny-tim]

Hi SamQP!

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 :)

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!

 

[jsholliday7]

Hi SamQP! 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!

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