# F = m* a please explain

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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?

Edited by Rocks
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The air definitely experiences an acceleration, that's at least part of what causes the pressure on the wall, but calculating the amount of force is complicated. It's a problem in fluid dynamics, and where you have to start analyzing it will depend on what the source of the wind is. Air will slow down and be pushed upward, and there will be a chaotic vortex pattern that flows over the wall. High pressure on the wind side, low pressure on the other side, and the wall may be toppled by the torque of the wind acting in a region near the top of the wall and above the base.

Edited by Lorentz Jr
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59 minutes 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?

The mass of wind is not a well-defined concept as far as I know.  Wind is better characterized by its velocity, pressure, and density.

edit: x-post with @Lorentz Jr; good answer, see above

Edited by Ghideon
x-post
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2 hours ago, Rocks said:

isn't the acceleration 0?

No, the acceleration is not 0. The air in the wind moves with the constant velocity before it gets to the wall. At this time, its acceleration is 0. But when it hits the wall, the air cannot continue moving as before because it cannot go through the wall.

It starts slowing down before it hits the wall, because of the air in front of it, and eventually it stops moving toward the wall. So, its velocity changes during this time. Change in velocity per time is acceleration.

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2 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?

Good morning, Rocks and welcome.

Yes the wind speed may be constant and force does indeed equal mass times acceleration.

But whilst speed is a scalar it is not the same as velocity, which is a vector as is force and acceleration.

So the equation Force  = mass x acceleration is a vector equation.

I hope you know that vectors have both magnitude (the scalar part) and direction and that these two quantities can change indpendently of one and other.

So the acceleration (at the wall) is not zero because there is a change of direction.

To see how this works you need to consider yet another vector quantity called momentum, which undergoes significant magnitude change at the wall.

The force arises as Newton has another equation Force = rate of change of momentum.

Please let us know where you are studying momentum so we can finish off an appropriate reply, to show you how to calculate the force from the speed and mass of the wind.

The other replies are true, but their effects are not really relevent to your question.

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Why is pressure connected to acceleration? A column of air also has pressure, vertically from gravity but I think there is horizontal pressure too. Plus isn't the wind compressed from the wind behind it, and compressed air obviously produces pressure.

Okay so if the speed is constant maybe the air isn't compressed, but isn't there any horizontal pressure anyway? If the air was more rarified on one side I bet a window, or something thin, would fall to that side.

studiot I finished highschool a long time ago, things I didn't understand back then still come back to me ocassionally though.

I remember studying momentum m*v, it is also connected to mass.

Edited by Rocks
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19 minutes ago, Rocks said:

Why is pressure connected to acceleration? A column of air also has pressure, vertically from gravity but I think there is horizontal pressure too. Plus isn't the wind compressed from the wind behind it, and compressed air obviously produces pressure.

studiot I finished highschool a long time ago, things I didn't understand back then still come back to me ocassionally though.

I remember studying momentum m*v, it is also connected to mass.

If you work it through F = ma becomes F = density x area x velocity^2 / 2. Divide by area and you get the pressure rise acting on the windward side of the wall. On the downwind side the reverse happens, air is accelerated back up to wind speed creating a partial vacuum. The nett force on the wall is 2F (give or take).

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It may help to note that Newton’s second law is F = dp/dt (p is momentum). This becomes F = ma if the mass is constant.

So one can view a force as the rate of change of momentum. As studiot points out, there is a change in direction, meaning there is a change in momentum. Thus, a force.

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2 hours ago, Rocks said:

Why is pressure connected to acceleration? A column of air also has pressure, vertically from gravity but I think there is horizontal pressure too. Plus isn't the wind compressed from the wind behind it, and compressed air obviously produces pressure.

Okay so if the speed is constant maybe the air isn't compressed, but isn't there any horizontal pressure anyway? If the air was more rarified on one side I bet a window, or something thin, would fall to that side.

studiot I finished highschool a long time ago, things I didn't understand back then still come back to me ocassionally though.

I remember studying momentum m*v, it is also connected to mass.

Smashing. I will try to work with that ans see how we can get to Seth's formula.

First I would like to make it absolutely clear that

The force on the windward face of the wall depends only on the winf at that face.
It does not depend in any way on what happens on the other side of the wall.
It may even be that the pressure and force on the other side of the wall is very much greater, for example if the wall is a dam wall.
The windard force will be the same whatever happens in all cases.

The wind blows horizontally for convenience and I would like to consider a round tube of air in this wind with an area of 1 metre2.
The reason for choosing a round tube will become clear.

Fig1 shows this tube and fig 2 shows how the airstream in the tube impacts on the wall.

OK so the wind is travelling horizontally along this tube with steady speed v until it hits the wall face.

Splat.  Except that is is a continuous process.

So the air must be diverted sideways along the wall face.
If this did not happen then there would be a pile up of air at the wall face as more air continued to move along the tube.
This is called continuity and we therefore have the equation

Volume or mass of air arriving at the wall face  =  volume or mass of air spreading out along the wall face

Now mass arriving per second = volume x density arriving per second = cross sectional area x length of tube arriving per second x density

And the length of tube arriving per second is the speed of the air flow (speed = distance/time).
And we have set the area of cross section to 1.

So we have mass per second = 1 * v * d      Where d is the density.

You have mentioned that momentum = mass time velocity

So the momentum per second = mass per second  x velocity = v *d*v = v2d  units of momentum.

We have also specified a horizontal wind so all the momentum is horizontal. The vertical momentum is zero as is the sideways momentum.

But at the wall fact the horizontal speed suddenly bcomes zero so the horizontal momentum suddenly becomes zero.
That is all the horizontal momentum is destroyed or lost.

Now you may have heard of the principle of conservation of energy ?

There is also a principle of conservation of momentum. This says that momentum is also conserved, unless there is a force in which case the force is equal to the change of momentum per second.

The change of momentum per second = (v2d - 0) so this equals the force acting on every square metre of wall the wind blows agains.

I have tried to show this diagramatically in Figs 2 and 3.

This also shows why I have used a circular tube.
The vertical and sideways momenta start off as zero and no force acts so they must remain as zero.
This happens as you can see in figs 2 an 3 by symmetry there is always a balance of momenta that sums to zero ##### Share on other sites

3 hours ago, Rocks said:

Why is pressure connected to acceleration? A column of air also has pressure, vertically from gravity but I think there is horizontal pressure too. Plus isn't the wind compressed from the wind behind it, and compressed air obviously produces pressure.

Okay so if the speed is constant maybe the air isn't compressed, but isn't there any horizontal pressure anyway? If the air was more rarified on one side I bet a window, or something thin, would fall to that side.

studiot I finished highschool a long time ago, things I didn't understand back then still come back to me ocassionally though.

I remember studying momentum m*v, it is also connected to mass.

I think @swansont's idea of rate of change of momentum is the simplest way to think about wind exerting a force of a wall. You have a stream of air molecules hitting the wall and rebounding. The harder the wind blows, the more of these you have in unit time. F = ma is also F = d (mv) /dt i.e. rate of change of momentum with time.

So faster wind means more momentum change in unit time, which means greater force.

Pressure is just force per unit area.

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Wait everyone.  From what I understand F = ma doesn't mean that ma causes F.  It is just that F is equivalent to ma.

And the air would not be the F or ma; the wall is.  The wall's acceleration is how we would find the force caused by the wind.

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1 hour ago, exchemist said:

I think @swansont's idea of rate of change of momentum is the simplest way to think about wind exerting a force of a wall. You have a stream of air molecules hitting the wall and rebounding. The harder the wind blows, the more of these you have in unit time. F = ma is also F = d (mv) /dt i.e. rate of change of momentum with time.

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.

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22 minutes ago, sethoflagos said:

Which is a form of the Bernoulli equation.

Bernoulli is an energy equation.

Unless the wall moves, which we are not considering here, there is no energy loss by the wind 1/2mv2 is conserved.

Energy, pressure and volume are of course all scalars.

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1 hour ago, exchemist said:

I think @swansont's idea of rate of change of momentum is the simplest way to think about wind exerting a force of a wall. You have a stream of air molecules hitting the wall and rebounding. The harder the wind blows, the more of these you have in unit time. F = ma is also F = d (mv) /dt i.e. rate of change of momentum with time.

Agreed. You can get Seth's 'hydrodynamic' formula,

5 hours ago, sethoflagos said:

F = ma becomes F = density x area x velocity^2 / 2. Divide by area and you get the pressure rise acting on the windward side of the wall.

from Swansont's suggestion. Consider some amount of time $$\triangle t$$. With @studiot's suggestion of 1 second, the maths is slightly simplified. This defines a reference layer of wind.

$\triangle l=v\triangle t$

$\triangle V=Sv\triangle t$

Every $$\triangle t$$ seconds the corresponding layer of wind will transfer momentum to the wall by an amount,

$\triangle p=v\triangle m$

where,

$\triangle m=\rho\triangle V=\rho Sv\triangle t$

And the resulting pressure would be,

$P=\frac{F}{S}=\frac{v\triangle m}{S\triangle t}=\frac{\rho Sv^{2}\triangle t}{S\triangle t}=\rho v^{2}$

As Studiot pointed out, momentum parallel to the walls is canceled out. While the air molecules bouncing back cannot transfer momentum back to the wind, so to speak, as we're assuming the gust of wind to be constant --it has reached some static regime.

I forgot to mention my $$\rho$$ is @studiot's d --the constant density.

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21 minutes ago, studiot said:

Bernoulli is an energy equation.

Unless the wall moves, which we are not considering here, there is no energy loss by the wind 1/2mv2 is conserved.

Of course Bernoulli is an energy equation - that's what you get when you integrate a force balance.

If kinetic energy is conserved as you say, then where does the work necessary to compress the air come from?

These things do happen very quickly but they are not instantaneous. The momentum exchange does not occur at constant velocity v. It occurs gradually over the velocity range v to zero. On average that's v/2.

This is why your results are double what they should be. You need to integrate over a valid path.

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6 minutes ago, sethoflagos said:

Of course Bernoulli is an energy equation - that's what you get when you integrate a force balance.

If kinetic energy is conserved as you say, then where does the work necessary to compress the air come from?

These things do happen very quickly but they are not instantaneous. The momentum exchange does not occur at constant velocity v. It occurs gradually over the velocity range v to zero. On average that's v/2.

This is why your results are double what they should be. You need to integrate over a valid path.

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?

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11 minutes ago, joigus said:

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

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

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24 minutes ago, sethoflagos said:

Of course Bernoulli is an energy equation - that's what you get when you integrate a force balance.

If kinetic energy is conserved as you say, then where does the work necessary to compress the air come from?

These things do happen very quickly but they are not instantaneous. The momentum exchange does not occur at constant velocity v. It occurs gradually over the velocity range v to zero. On average that's v/2.

This is why your results are double what they should be. You need to integrate over a valid path.

Every fluid element is travelling at the winf speed v, until it is brought to rest by the wall surface.

Therefore every element has the full mass x velocity momentum

There is no averaging factor of 1/2 for momentum, only kinetic energy.

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10 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?

To make it easier to imagine fundamentally, we can make the air one big point particle and the wall one big point particle too.

What is happening?

The air (let's call it a gigantic oxygen molecule O2) travels towards the wall particle (let's call it a mix of molecules called W) and as it gets closer and closer there is a repulsive force (from electrons that surround both O2 and W, but that is not important), an equal and opposite reaction occurs.  Eventually, if the wall breaks, that means that the W molecule/wall gave and accelerated in the direction of the O2.  Or, if the wall does not break, then something had to give, namely the Earth.  Ever so sightly the Earth had to accelerate and gain momentum (this is from the law of conservation of momentum: the O2 must transfer its momentum).

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17 minutes ago, studiot said:

Every fluid element is travelling at the winf speed v, until it is brought to rest by the wall surface.

What a very curious idea.

27 minutes ago, studiot said:

Why exactly are inelastic impacts relevant to the discussion?

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8 hours ago, Rocks said:

Why is pressure connected to acceleration?

It's Newton's 3rd law, the action-reaction law. A force is required to slow down the air, and therefore the air must exert an equal and opposite force on the wall. Pressure is force per unit area.

@studiot's comment about the air's direction changing is partially right: Air can't just slow down and not go anywhere, because it already occupies all the volume in the region in front of the wall. So it curves upward (and doesn't slow down as much overall) before becoming turbulent around the top of the wall (at least if the wall is free-standing).

3 hours ago, Boltzmannbrain said:

Wait everyone.  From what I understand F = ma doesn't mean that ma causes F.  It is just that F is equivalent to ma.

Other way around: For solid objects, F causes the m to a. In the case of air being redirected by the wall, F = dp/dt = d(mv)/dt = v (dm/dt), i.e. the force from the wall keeps slowing down the horizontal velocity component of more and more incoming air, while the ground increases its vertical component. The wall and the ground can't do work on the air because they're not moving, so the overall effect is redirection.

3 hours ago, Boltzmannbrain said:

And the air would not be the F or ma; the wall is.  The wall's acceleration is how we would find the force caused by the wind.

Generally speaking, there will usually be an ma for the air but not much for the wall. A brick wall won't go much of anywhere (i.e. not more than a few millimeters) until the force on it creates a torque that snaps it in two near the base. The force on the wall is caused by a combination of things: A distant(-ish) source of static high pressure, loss of momentum by a temporary gust of wind (that's the ma), and the net force can be further increased by a decrease in the pressure on the other side.

50 minutes ago, Boltzmannbrain said:

as [the air] gets closer and closer there is a repulsive force

You can think of the incoming air as riding up a sort of curved "ramp" that's made of pressurized air already in place near the lower regions of the wall. The microscopic picture is more complicated, but this is a reasonable macroscopic approximation.

It's the same reason the nose of an ideal aerodynamic shape is rounded instead of pointy: At subsonic speeds, there's always a small amount of fluid that's already accelerated to the object's speed and repels incoming fluid more or less smoothly. Pressure is transmitted through air at the speed of sound, so this doesn't happen above that speed. That's why the noses of supersonic jets are pointy.

Edited by Lorentz Jr
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36 minutes ago, sethoflagos said:

What a very curious idea.

Why exactly are inelastic impacts relevant to the discussion?

?

?   Why indeed I don't recall mentioning them.

You don't seem to have anything to say about the bog standard fluds calculation.

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17 minutes ago, studiot said:

?   Why indeed I don't recall mentioning them.

That was I --speaking a la Gandalf. By the way, I still think @swansont's argument is the ticket here.

@Lorentz Jr is right though. Any realistic situation will be riddled with turbulence. Still, you can concoct a non-turbulent system for which the energy/momenta analysis is equivalent.

Crudely speaking, that's because you can picture all the molecules hitting the wall at speed 2v, and all of them bouncing off perpendicularly (to the wall) at speed -v, the wall absorbing the deficit of kinetic energy, and the flow of particles still being a stationary flow which on average has velocity v towards the wall.

Momentum would be conserved, and the continuity equation wouldn't suffer in the least.

Edited by joigus
minor correction
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8 minutes ago, studiot said:

Why indeed I don't recall mentioning them.

The calculations you presented specifically state 'The jet strikes the plate and does not rebound but spreads sideways over the surface of the plate. The momentum normal to the plate is destroyed.' Sounds pretty inelastic to me.

23 minutes ago, studiot said:

You don't seem to have anything to say about the bog standard fluds calculation.

It seems quite credible. Just inappropriate to the OP. In particular there is no mention of the jet being slowed by an adverse pressure gradient prior to impact. In fact no mention of pressure at all. Which at least then avoids having to explain away the infinite pressure necessary to bring a jet to a halt instantaneously.

I provided my own simplified calculation of 'the force of the wind' some dozen or so posts back. Perhaps you could tell me where exactly the mistake is?

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15 minutes ago, sethoflagos said:

The jet strikes the plate and does not rebound but spreads sideways over the surface of the plate. The momentum normal to the plate is destroyed.' Sounds pretty inelastic to me.

Inelastic refers to KE, not momentum. A wall effectively has infinite mass, being anchored to the earth. Up until it breaks apart.

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