F = m* a please explain

[Boltzmannbrain]

2 hours ago, Lorentz Jr said:

Other way around: F causes the m to a.

Yes, I agree.

The OP, as I understand it, is confused about what F = ma means regarding the scenario given.  It reads as if the air is not accelerating and therefor can't cause a force.  But like you say, F cause m to a.  And really, I suppose that the air is accelerating, in the negative direction, from the force being applied to it from the wall.

It might make more sense for the OP to know that F_wind = - F_wall , which is from Newton's 3rd law.   

Edited February 12, 2023 by Boltzmannbrain

[sethoflagos]

3 hours ago, joigus said:

Mmmm. Sorry, I didn't see your factor 1/2.

I don't think conservation of (kinetic) energy will give you the right answer here. There's no reason to assume collisions are elastic. OTOH, momentum is always conserved at a macroscopic level. That's why I think Swansont's suggestion is the most secure foundation for this kind of reasoning. Does that make sense?

I'm sorry Josh, I missed the last bit.

I took a little time out from the dog-piling to check my facts. I read the OP as 'What is the stagnation pressure acting on a wall normal to a headwind'. The response I gave is  simply textbook fluid mechanics and is correct given the stated assumptions. I find it strange for once being called on to defend the Bernoulli equation, but then lots of things seem strange to me these days.

Both viewpoints expressed comply with conservation of momentum. The key difference is that I reduce momentum incrementally into a dynamic pressure field while others drop all the incoming momentum at the face of the wall.

Now I have an easy way of propelling the fluid up and over the wall: it's accelerated up and away by the pressure field built with the momentum of the incoming stream.

Others seem to have stationary gas stuck to the wall and no source of energy to move it away. 

I look forward to some imaginative explanations.

 

[studiot]

15 hours ago, Rocks said:

If a gust of wind blows at constant speed, it can demolish a wall - so what is the mass of the wind (where do you start measuring the length of the wind)? And isn't the acceleration 0?

So what is the force of the wind?

 

Perhaps you would like to elaborate on this wall of yours, and what you actually want to know.

 

The dam wall I was thinking of is 90metres high and 500 metres wide.

 

Wind loading analysis can be very complicated and what I offered was the beginning of it to try to connect the quantities you mentioned without calculus, for understanding purposes.

Winds are normally specified by a general incident pressure and a host of shape and other factors that control the wind regime.

The resultant load is of course  forces and/ or moments. I say moment because it may be that for a brick wall the wind pressure can cause unwanted tension in the brickwork, which cannot support tension.

 

[joigus]

I think I got a reasoning that recovers the 1/2 factor that @sethoflagos posited from Bernouilli's equation and, by now, I'm totally convinced is right. In the general situation --but still idealised, as we are assuming an infinite wall, and a constant velocity field pointing right in the image, we'd have something like,

An ensemble of particles that's equivalent as to momentum --and K.E.-- transfer is,

 

This would give a velocity field and a density that are consistent with the continuity equation iff 1/2 the molecules are going right, and the other 1/2 is going left --but with half the velocity. Otherwise, we wouldn't be getting an overall velocity field going right with velocity v, as 2v-v=v, while v-v=0.

So the density \( \rho \) must --on average-- be made up of 1/2 the molecules going right --and eventually hitting the wall--, and the other 1/2 going left --and therefore never hitting it again. Although we still have \( \rho\triangle V=\rho Sv\triangle t \), this gives us twice the molecules that are going to hit the wall within time \( \triangle t \). Therefore, I should have written,

\[ \triangle p=\frac{1}{2}v\triangle m \]

which gives \( \frac{1}{2}\rho v^{2} \) for the pressure.

In the more realistic case, it is certainly true that molecules moving wildly in all kinds of oblique directions would contribute to the --constant-- density, but this is made up for in the simplistic model by assuming that they're not there, and the average leftward motion is totally carried by particles moving perpendicularly to the wall. 

Note for @Boltzmannbrain on aspects already mention by other members: Newton's equations of motion are valid thoughout, of course. But there are cases in which thinking in terms of momentum transfer is more useful. This is, I think, one of them.

Please tell me whether you agree/disagree with this --admittedly-- drastically simplified model.

 

 

[Boltzmannbrain]

2 hours ago, joigus said:

 

Note for @Boltzmannbrain on aspects already mention by other members: Newton's equations of motion are valid thoughout, of course. But there are cases in which thinking in terms of momentum transfer is more useful. This is, I think, one of them.

Please tell me whether you agree/disagree with this --admittedly-- drastically simplified model.

 

 

Yeah, I think you nailed it.  I was looking around and I saw something that reminded me what I learnt a long time ago.  F = ma actually means a net force = ma.  And net force is a change in momentum over a change in time.  See, 5.3 Newton’s Second Law – University Physics Volume 1 (ucf.edu) (see lower part of webpage for reference).

So I think momentum is crucial to this problem too.  In that case, I think the issue that the OP had with force is that the air particles with seemingly no acceleration (even though there is acceleration) needs more than just F = ma to explain what is happening.  So, momentum, as you pointed out, is one of the major factors here.

Edited February 13, 2023 by Boltzmannbrain

[joigus]

5 minutes ago, Boltzmannbrain said:

Yeah, I think you nailed it.

There is the possibility that I made a mistake. Let's see what other people have to say.  

Now that I think about it, it's not so clear the the continuity equation is satisfied at the point of hitting the wall. \( 2v\rho \) mass per unit area per unit time is hitting the wall, while only \( v\rho \) mass per unit area per unit time is bouncing off. There seems to be a mismatch.

Qualitatively, it's clearly all about transfer of momentum, which in this case is better looked upon in terms of mass flow and individual momentum transfer per particle. Getting the constants right is another matter.

I'll probably have to think about it some more when I get the chance.

Anyway, it's just fun thinking about these problems.

[studiot]

I don't wish to enter a person slanging match with anybody.

 

So lets clear up the stagnation pressure once and for all.

 

Quote

The pressure at the stagnation point is higher than teh pressure in the undisturbed stream by by the product of teh fluid density and the undisturbed kinetic energy per unit mass of fluid.

I have indicated the relevent theory and the factor of 1/2.

 

So Seth's formula is the overpressure above normal stram pressure, which has not been included

[sethoflagos]

2 hours ago, joigus said:

There is the possibility that I made a mistake. Let's see what other people have to say.  

Firstly a word about deriving the result from Newton's second law. Maybe you missed the details of my earlier post: 

On 2/12/2023 at 6:20 PM, sethoflagos said:

One problem with this approach is that the system is steady state so there is no clear timelime to integrate over.

However we can proceed along the lines of:

dP/dx = -p dv/dt = - p v dv/dx 

hence dP = - p v dv

If we keep things simple and ignore air compressibility and elevation changes we get on integration

P₁ - P₀ = pv^2 / 2 

Which is a form of the Bernoulli equation.

 

Please read the first calculation line carefully. I begin with rate of change of momentum, just as @swansont recommends, but since I can't integrate that directly I use the chain rule to recast the ODE into a simple form I can integrate. This gives the correct answer.

Attempting crude numerical integration of the raw momentum equation via left Riemann sum as we saw yesterday gave a 100% overestimate, to which I suggested you try the middle Riemann sum. 

23 hours ago, sethoflagos said:

No worries. Just adjust the first line of your calc to dL = (v/2) dt and I think everything else clicks into place.

You are right to maintain some doubt over the validity of this procedure. It gives the correct solution but only by accident. You're working with at least one assumption that is profoundly unphysical.

I made a little explanatory sketch of how I see air flow over a wall. (A small one - let's keep it simple)

Interactions between the wind and wall are mediated by a clockwise rotating prism of air that acts as a ramp to accelerate the incoming airstream up and over the wall.  

It is clear that the main flow retains much of its horizontal momentum throughout - it doesn't dump it all into wall in fact it mustn't (continuity). But it isn't total momentum that is in the equation - it's rate of change of momentum that counts.

So what happens inside the prism. Wind shear transfers momentum into the angled face accelerating it up to near bulk velocity. And that momentum is transferred via an exchange of angular momentum with a flow parallel to the wall rather than linear momentum from direct impingement. A similar effect happens on the ground providing the necessary reaction to generate upwards momentum.

Almost none of the main flow gets anywhere near the wall.

Anyway, this model works for me and is pretty close to what I've seen from wind-tunnel tests of blast walls and the like. Hope it helps someone out there, though I suspect the OP is long gone after yesterday's sorry show. 

 

 

Edited February 13, 2023 by sethoflagos

[joigus]

6 hours ago, studiot said:

I don't wish to enter a person slanging match with anybody.

Neither do I. And I can assure you I won't.

5 hours ago, sethoflagos said:

You are right to maintain some doubt over the validity of this procedure. It gives the correct solution but only by accident. You're working with at least one assumption that is profoundly unphysical.

Absolutely. I would even go as far as to say that there are several assumptions that are unphysical in my picture. In my 'defence,' my attempt was not meant to illustrate how real wind behaves. A spatially-confined gust of wind that's hitting a wall obviously does not correspond to my simpleminded 'model' of particles coming from -infinity and bouncing off an infinite wall, back to -infinity, while keeping completely parallel to each other. It's just an illustration of how a continuous supply of particles hitting a wall at a certain velocity will transfer momentum per unit area, but you have to give up on F=ma, not because it's any the less true, but because that relation is not the one that's useful. What's useful is transfer of momentum per unit area per unit time.

Unfortunately, my model cannot be made consistent with the continuity equation, which I now realise after having thought about it for a while --even if it gives the right result by tinkering with the numbers. I think Studiot is thinking of a slightly different setting than you are, with no ground to sustain the wind so as to produce eddies, as you have pictured. But maybe he will be willing to ellaborate on that.

I do believe @Boltzmannbrain has probably understood the main point, without getting bogged down in finer details of zero-velocity areas, stationary currents circulating, and the like, of which I'm --partially, at least-- guilty. The take-home lesson is: A continuous supply of particles at speed v, and constant density d, hitting a wall, will exert a force on a wall that's more simply expressed as proportional to dv². The calculation of the dimensionless coefficient being a matter of correctly applying fluid mechanics. What's going on is transfer of momentum.

Shall we leave it at that? I am happy with that, at least, for the time being. But if you want to discuss more, I'll be happy to take a back sit and learn --or refresh my memory-- a little more.

[sethoflagos]

58 minutes ago, joigus said:

Shall we leave it at that? I am happy with that, at least, for the time being. But if you want to discuss more, I'll be happy to take a back sit and learn --or refresh my memory-- a little more.

Let's let this thread die a well deserved death. The well has been poisoned by those unable to yield ground in their turf wars.

[studiot]

9 hours ago, joigus said:

Neither do I. And I can assure you I won't.

Absolutely. I would even go as far as to say that there are several assumptions that are unphysical in my picture. In my 'defence,' my attempt was not meant to illustrate how real wind behaves. A spatially-confined gust of wind that's hitting a wall obviously does not correspond to my simpleminded 'model' of particles coming from -infinity and bouncing off an infinite wall, back to -infinity, while keeping completely parallel to each other. It's just an illustration of how a continuous supply of particles hitting a wall at a certain velocity will transfer momentum per unit area, but you have to give up on F=ma, not because it's any the less true, but because that relation is not the one that's useful. What's useful is transfer of momentum per unit area per unit time.

Unfortunately, my model cannot be made consistent with the continuity equation, which I now realise after having thought about it for a while --even if it gives the right result by tinkering with the numbers. I think Studiot is thinking of a slightly different setting than you are, with no ground to sustain the wind so as to produce eddies, as you have pictured. But maybe he will be willing to ellaborate on that.

I do believe @Boltzmannbrain has probably understood the main point, without getting bogged down in finer details of zero-velocity areas, stationary currents circulating, and the like, of which I'm --partially, at least-- guilty. The take-home lesson is: A continuous supply of particles at speed v, and constant density d, hitting a wall, will exert a force on a wall that's more simply expressed as proportional to dv². The calculation of the dimensionless coefficient being a matter of correctly applying fluid mechanics. What's going on is transfer of momentum.

Shall we leave it at that? I am happy with that, at least, for the time being. But if you want to discuss more, I'll be happy to take a back sit and learn --or refresh my memory-- a little more.

Most definitely a great summary. +1

We have all missed points here which is why the wind loading design codes offer exactly this.

For example CP3, Chapter 5 in the UK offers

1) From the map of standardised wind speeds and modifying tables choose the design wind speed, V,  appropriate to the location.  V is in m/s

2) Calculate the 'dynamic pressure' , q  from the equation   q = 0.613V²  N/m²

 

The coefficient 0.613 is not dimensionless but enjoys suitable units of conversion.
 

This will yield a pressure similar to Seth's calculation, but for different reasons from either his or mine.

 

The history of wind loading in UK structural engineering really starts in 1879 with the Tay Bridge disaster.

It is interesting to note that in those days responsibility for meteorolgy rested with The Astronomer Royal, then Sir George Airy.

The bridge designer, Sir Thomas Bouch wrote to Airy, requesting a pressure loading figure and received a written reply of 10 pounds per square foot

But he hastily revised this fugure to 120 pounds per square foot, following the disaster.

Bouch, however collected all the blame.

Subsequent to this a whole series of experiments were carried out at the National Physical Laboratory relating loading to building shape and size during the later 1880s and 1890s.

This led to the CP3 figure, originally in imperial units along with a bunch of modifying factors.

 

8 hours ago, sethoflagos said:

Let's let this thread die a well deserved death. The well has been poisoned by those unable to yield ground in their turf wars.

 

So I disagree that there is nothing worth discussing here, so long is it is not in such inflammatory language.

The mechanics of this must be Newtonian so we have all missed points here, myself included and we can all benefit from acknowledging the contributions from all concerned (including those not mentioned in this post) as no one was completely right or completely wrong.

 

 

[sethoflagos]

1 hour ago, studiot said:

For example CP3, Chapter 5 in the UK offers

1) From the map of standardised wind speeds and modifying tables choose the design wind speed, V,  appropriate to the location.  V is in m/s

2) Calculate the 'dynamic pressure' , q  from the equation   q = 0.613V²  N/m²

So my posts were 'wrong' because I failed to address all the safety factors the building codes include to cover for topological features such as wind funnelling? 

You'll notice that the expression for dynamic pressure is the same as the one I gave calculated at a standard air density @ 15 ⁰C of 1.226 kg/m³. This standard doesn't appear to think the equation I gave is wrong. Unlike some other codes, this one factors the safety margin into the tabulated design wind speeds rather than use a crude drag coefficient.

Ask yourself why the engineering codes use the incompressible form of the Bernoulli equation for a fluid that is demonstrably compressible.

Noting that we can rearrange it in the form P₁/P₀ = 1 + p₀ v₀² / 2 P₀

We can compare with the compressible form: P₁/P₀ = (1 + (k-1) / k . p₀ v₀² / 2 P₀)^(k /(k-1))

So are the engineering protocols 'wrong' by your exactling standards? 

Or are they just simpler and conservative. Engineers are usually ok with that.

Sorry for having to point out the fundamental flaws in your thinking earlier in the thread. I did try to do it as sensitively as possible - by addressing @joiguswho can handle differences of opinion in an adult manner. But you spat your dummy out anyway. You owe me and the other contributors to this thread an apology - but I don't suppose in a million years that we'll ever get one.

Edited February 14, 2023 by sethoflagos

[studiot]

36 minutes ago, sethoflagos said:

So my posts were 'wrong' because I failed to address all the safety factors the building codes include to cover for topological features such as wind funnelling? 

I didn't say that was the reason. It wasn't.

And I didn't say they were wrong full stop at any point, not a courtesy to extended to me.

I did say they were not completely correct.

I also indicated that others in this thread made some good points as well as some less good ones.

 

Yet you singled me out in most of your posts for uncouth personal comments.

So yes I will remain a 'baby' if that is the way adults behave these days.

[sethoflagos]

24 minutes ago, studiot said:

Yet you singled me out in most of your posts for uncouth personal comments.

Where?

[studiot]

So rather than leave this thread on a sour note, I think I will post more of my thoughts on the actual subject.

Firstly the stubborn factor of 1/2.

I don't know why the code writes incorporated it but I do know thet acknowledge that the real wind velocity variest from near zero close to the ground to a maximum near the top of a wall.

So taking the max velocity and multiplying it by some factor less than one seems reasonable as an average.

The code also describes finding and using the centre of pressure for the actual application of teh wind load to the wall.

 

Now a couple of other members have pointed out Newton's third law say that

If the wind applies a force F to the wall then the wall applies an equal but opposite force F to the wind.

 

So where does that leave my analysis ?

Over cautious, but safe in engineering terms.

 

Modern codes also stress that they are starting point average simple solutions with no complicating factors.

And that the individual situations should take account of special circumstances, using wind tunnels if necessary.

 

What lessons does that leave us with ?

Well the wind stream is taken as steady and continuous.

This poses difficulties for the idea that a region of higher density builds up near the wall as this cannot continue indefinitely.

Eventually as much air must be leaving the vicinity as arrives.

 

So what we can say is the the presence of the wall alters the stream pattern which is best represented as a pressure averaged over the whole face of the wall.

This pressure is then converted to a representative force, applied at a suitable point.

 

This point of application is important because Rocks wondered how the wind could blow over a wall.

For a wall this collected force applies a moment about the wall foundation.

As the wind changes this can cause rocking or oscillation.

Such motion can cause the wall to shift over time on its foundation sometimes sufficiently to become unstable.

[sethoflagos]

Okay, let's have another go at this, but taken in short (hopefully) simple steps.

What do we know with certainty?

We know the incoming wind gets over the wall in its entirity, there is no mass transfer in or out.

We know that it gets over the wall very quickly (no time for meaningful heat transfer) and no external work is done on it. Therefore there is negligible energy exchange with the environment.

By some means the incoming flow horizontal flow is deflected upwards at an angle we can call A. 

By conservation of energy horizontal velocity becomes v_x = v cosA, and vertical velocity v_y = v sinA

By conservation of momentum the rate of change of momentum with the ground per unit area = p v² sinA 

Ditto the rate of change of momentum with the wall per unit area = p v² (1 - cosA)

By conservation of energy (I think, help me out here), it is necessary that (1-cosA) = cosA hence A = pi/3 radians (60^o) 

P_wall - P₀ =  0.500 p v² 

Pground - P₀ =  0.866 p v²

[Tutoroot]

Newton's second law is often stated as F=ma, which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a). This means the more mass an object has, the more force you need to accelerate it. And the greater the force, the greater the object's acceleration.

[studiot]

18 hours ago, sethoflagos said:

Okay, let's have another go at this, but taken in short (hopefully) simple steps.

What do we know with certainty?

Great stuff.

18 hours ago, sethoflagos said:

We know the incoming wind gets over the wall in its entirity, there is no mass transfer in or out.

Yes, agreed,  but over or around please.

 

18 hours ago, sethoflagos said:

We know that it gets over the wall very quickly (no time for meaningful heat transfer) and no external work is done on it. Therefore there is negligible energy exchange with the environment.

Agreed.

So the question you asked before of where does the energy arise for it to leave does not arise. It always has the energy it arrives with.

That of course implies the the wind never actually stops ie V is never actually zero.

So my full on jet is really an absolute upper bound, which is why I said it was a very simplistic safe option.

18 hours ago, sethoflagos said:

By conservation of energy horizontal velocity becomes v_x = v cosA, and vertical velocity v_y = v sinA

OK, this is the start of a form of analysis.

18 hours ago, sethoflagos said:

By conservation of energy horizontal velocity becomes v_x = v cosA, and vertical velocity v_y = v sinA

By conservation of momentum the rate of change of momentum with the ground per unit area = p v² sinA 

If the rate of change of momentum has a value other than zero it implies a force is acting on something.

So what do you think is applying a force to what please ?

 

 

[sethoflagos]

20 hours ago, sethoflagos said:

By conservation of energy (I think, help me out here), it is necessary that (1-cosA) = cosA hence A = pi/3 radians (60^o)

I think I've got this now into a form that doesn't involve some non-physical model.

Deflecting the airstream upwards requires, as shown above, a force with h,v components of -p v² (1 - cosA), p v² sinA 

By Newton's 3rd Law that generates an equal and opposite reaction force of p v² (1 - cosA), - p v² sinA 

If we can reflect this reaction force back in the direction it came it will become its own source.

I picture the means of transmission as a pressure wave (there are other valid pictures)

To achieve the reflection necessary for a stable, steady-state system the wave must reflect off both the ground and the wall. It doesn't matter in which order but if say part of the wave 'bounces' over the wall, or strikes the ground ahead of the air deflection zone the momentum leakage will work to alter the angle of deflection of the airflow.

A little study of the geometry shows that the wave path between ground and wall must be parallel to the deflected airstream ie 

21 hours ago, sethoflagos said:

(1-cosA) = cosA hence A = pi/3 radians (60^o)

This angle is self-maintaining as discussed above. As the wind picks up from still conditions (no air ramp), the growing pile up of air against the wall will naturally and very quickly construct its own 60⁰ ramp and maintain it in dynamic equilibrium.

[sethoflagos]

On 2/15/2023 at 6:35 PM, sethoflagos said:

P_wall - P₀ =  0.500 p v² 

Pground - P₀ =  0.866 p v²

Just to tidy up, I forgot to adjust the ground load for the reduced area of ground relative to wall, so that figure is wrong. It needs to be divided by cot (60) (ie multiplied by 3^0.5).

I make the triangle of forces to be as shown in the sketch:

 

 

Edited February 17, 2023 by sethoflagos

[sethoflagos]

I think it all balances up but I'd be grateful if someone checked it over.

[studiot]

I'm not sure about extensive calculations on the wind regime since we are really interested in the effect of the wind on the wall, rather than the other waya round.

So let us start from two proportionalities, since proportion is the simplest variation mathematically.

 

Assuming the force of applied by the wind is proportional to the square of the speed, V, with constant, b

F = bV²

And assuming that the wind speed is proportional to the height above the ground. h

V=ah

Substituting for F yields and combining constants.

F = ba²h² = ch²

 

This leads to the total horizontal shear force due to the wind at the base being

[math]Shear = \int_0^h {Fdh = } c\int_0^h {{h^2}dh = } c\left[ {\frac{{{h^3}}}{3}} \right]_0^h = c\frac{{{h^3}}}{3} = \frac{{c{h^3}}}{3}[/math]

 

That is the shear force is proportional to the cube of the windspeed

But the reaql kicker comes with the moment induced at the base of the wall, which is seen to be proportional to the fourth power of the height.

[math]Moment = \int_0^h {hFdh = } c\int_0^h {h{h^2}dh = } c\left[ {\frac{{{h^4}}}{4}} \right]_0^h = \frac{{c{h^4}}}{4}[/math]

 

 

Edited February 17, 2023 by studiot