Questions about Newton's third law

Click For Summary

Discussion Overview

The discussion revolves around Newton's third law of motion, specifically exploring the nature of action and reaction forces in various scenarios, including a car in motion, a person lifting weights, and the concept of normal force. Participants seek clarification on the sources of reaction forces and how they relate to motion and acceleration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions what provides the reaction force when a car drives forward and how friction plays a role in this scenario.
  • Another participant suggests that the reaction force to the applied force of the car is provided by the ground, while the reaction force of friction is provided by the car itself.
  • There is a discussion about how the mass of the car and the Earth affects acceleration, with some participants pondering whether the car must have a lesser mass than the object providing the reaction force.
  • A participant raises the question of how normal force relates to gravity and whether it is the reaction force to gravity.
  • Concerns are expressed about the nature of forces during free fall, questioning how a reaction force exists when only gravity is acting on a falling object.
  • Some participants clarify that the normal force is not a third law pair with weight, as weight is the gravitational force exerted by the Earth on a person, while the reaction force is the gravitational force exerted by the person on the Earth.
  • There is a debate about the significance of a person's mass in relation to the gravitational field compared to that of the Earth, with some participants arguing that the gravitational field created by a person is negligible.

Areas of Agreement / Disagreement

Participants express various interpretations of Newton's third law, leading to multiple competing views on the nature of action and reaction forces. The discussion remains unresolved regarding the implications of mass and gravitational fields in these contexts.

Contextual Notes

Participants highlight limitations in their understanding of the concepts, particularly regarding the definitions of forces and the conditions under which they apply. There are unresolved questions about the relationship between mass, acceleration, and the effects of different forces.

  • #31
Well, I could try and simplify it to that... but...

I read up a bit on that drag stuff, doesn't seem too bad...

But if you can measure the drag on the car, that only gives you force applied to the car, and thus the force the car applies to the air.

I know doc al said that the air cannot be considered to have a point mass, but can't you at least estimate how much air you're dealing with here? I don't really need any complicated equations or anything yet, but I'm just curious as to how you could roughly estimate this, if the air is applying a force, surely there must be some mass you are dealing with?
 
Science news on Phys.org
  • #32
Someone? Please? Lend me hand here?
 
  • #33
The aerodyanmic drag is mostly due to acceleration of the air in the direction a car is moving. Assuming that the car isn't streamlined with a long tapered tail like a bonneville speed car, most of the drag occurs behind the car, because the car leaves a moving void behind it that creates a low pressure area that accelerates the air towards it. So this part of the drag is due to the old force = mass times acceleration.

Another component of aerodynamic drag is due to friction which increases the temperature of the air. I don't know how much of a part this plays in overall drag, but I suspect it's not much.

Trying to estimate the amount of air involved would be difficult, as there are different amounts of acceleration applied to the volume of air affected by a moving vehicle.

Regarding the air accelerating a car; if the car was very easy to move, then a strong breeze (headwind or tailwind) could move such a car.
 
Last edited:
  • #34
Ok, but I don't think my question has been answered... I know generally how drag works, but what mass of the air causes this? As said before, I know the air isn't reall a point mass, but if it's exerting a force, than, as I said, surely it has some mass?
 
  • #35
I think the problem isn't that there is no mass, but that it's ridiculously hard to calculate the mass that's involved. The air molecules, because they act like a fluid, work together. We don't know how many molecules are working together, so it's hard to estimate the mass that's involved.

You can't just break it up and say "Oh, this many cubic feet are affecting the drag." You also can't say "This many molecules are causing the friction." I'm not sure, other than an educated guess, you could ever find out how much mass is involved in something like that.
 
  • #36
Ok, now I think I understand a little better, but is it possible to estimate if the mass of the air causing the friction is greater than the mass of the car? That was basically my intention here, do we really have no idea what the mass is, or can we at least estimate it to be greater/less than the mass of the car?
 
  • #37
If you know the power output of the car and you know its top speed its pretty easy to estimate the drag force (and hence the mass of air) present when the car is traveling at its maximum speed. This is of course, ignoring any rolling resistance of the tyres, axles, drive staft, gearbox etc. But atleast it will give you an estimate.
 
  • #38
Err, if you have force, wouldn't you still need the acceleration to calculate mass? Or can the acceleration on the air from the car also be determined by the drag force?
 
  • #39
Byrgg said:
Err, if you have force, wouldn't you still need the acceleration to calculate mass? Or can the acceleration on the air from the car also be determined by the drag force?
If we assume that the car is a cuboid of defined volume and dimensions we could etimate the mass of air displaced.
 
  • #40
Go to a highway and measure approximately how far away from it you can stand, and still feel the breeze from passing cars. Then you can make at least a crude estimate of the volume of air that is affected by a passing car, and from that, the mass of the air.
 
  • #41
Hootenanny said:
If we assume that the car is a cuboid of defined volume and dimensions we could etimate the mass of air displaced.

jtbell said:
Go to a highway and measure approximately how far away from it you can stand, and still feel the breeze from passing cars. Then you can make at least a crude estimate of the volume of air that is affected by a passing car, and from that, the mass of the air.

Ok... which one is it here, if you know the volume of the car, is that equal to the volume of displaced air? And then knowing it's density, you could calculate the amount of displaced air? Actually, jtbell's answer is probably the same here, by crudely measuring as said... you could estimate the mass. But are the volume of the displaced air and the volume of the car the same? From that you could calculate(knowing the density of the air) the mass of the air...

But would you assume the car to be denser than the air? Probably, since most of what's in the car is solid...

Also...

Farsight said:
Oh look it up on Google. Air is circa 1.3kg per cubic metre, and a car is maybe six cubic metres, so we're talking about 18 pounds in weight.

How do you approximate the 18 pounds in weight here?

These will likely be my last questions in this thread by the way, once all of this is answered, I'll post a summary of what I've learned from this thread, and then ask if it's right.
 
  • #42
A very crude estimate of the air mass moved per unit time can be had by treating the car as a "plow" with a given area moving at a given speed, scooping out air as it goes. You can figure out the volume per second that the car "scoops" out of the way.
 
  • #43
Ok, what volume do you use per second then, the volume of the car? And I'm still curious about how the 18 lbs was estimated, unless that relates to this in some way...
 
  • #44
Byrgg said:
Ok, what volume do you use per second then, the volume of the car?
No. The rate at which volume is scooped out will equal Area x Speed. (Figure out why this is true.)

And I'm still curious about how the 18 lbs was estimated, unless that relates to this in some way...
What didn't you understand about this? Someone estimated the volume of a car and then calculated the mass & weight of a equal volume of air. (Not obvious that it's relevant.)
 
  • #45
How was the 18lbs calculated? They said something like 1.3kg/m cubed, and a car was about 6 cubic m... where did 18 come from?

Oh, and what area do you use then for the volume of air scooped out per unit time? Or is there a way to calculate this which you could show me, or even a link to a place where I could learn?
 
  • #46
Byrgg said:
How was the 18lbs calculated?

You're not from the USA or the UK, right? :smile:

One kilogram corresponds to about 2.2 pounds (mass). It's a standard conversion factor that every physics student in the USA knows (or should know). I'm not sure about the UK because they're officially on the metric system, but I think some non-metric units are still in common use.
 
  • #47
jtbell said:
You're not from the USA or the UK, right? :smile:

One kilogram corresponds to about 2.2 pounds (mass). It's a standard conversion factor that every physics student in the USA knows (or should know). I'm not sure about the UK because they're officially on the metric system, but I think some non-metric units are still in common use.
Sometimes we are required to convert between the two, however, it is something I severely dislike
 
  • #48
Ok... so then what was it? 1.3kg = 2.86lbs and then what? Weight was mentioned so I'm guessing that means multiplied by 9.8... which = 28.028... did I miss something??
 
  • #49
Byrgg said:
Ok... so then what was it? 1.3kg = 2.86lbs and then what? Weight was mentioned so I'm guessing that means multiplied by 9.8... which = 28.028... did I miss something??
We are taking six cubic meters of air, therefore; 6 x 1.3 x 2.2 = 17.16 lbs. Note, that pounds is a unit of mass not weight.
 
  • #50
Yeah... but it was said 18 lbs of weight... I guess this just fits in with the popular misuse of the word 'weight'.
 
  • #51
It's quite common to measure the "quantity" of something using either mass or weight, mixing units at will. It's OK in everyday usage, but not acceptable when doing physics.

Using standard units, a kilogram is a unit of mass. Near the Earth's surface that kilogram has a weight of about 9.8 Newtons, which is equivalent to about 2.2 pounds (force).
 
  • #52
2.2 lbs, force? Huh? I thought lbs for mass, what just happened?
 
  • #53
Byrgg said:
2.2 lbs, force? Huh? I thought lbs for mass, what just happened?
Pounds(lbs) is a unit of mass. However, pounds-force(lbf) is a unit of force or weight. One pound force is approximately 4 Newton's.
 
  • #54
The pound is used (confusingly, perhaps) as a measure both of force and of mass.

The standard pound (avoirdupois pounds) is a unit of mass defined in terms of the kilogram. (look it up)

But it's also commonly used as a unit of force in the imperial system of units, where mass is in slugs. The weight of one slug is (w = mg) 32 pounds (force).

Don't get hung up on this unit conversion trivia!

Personally, I stick with SI units. :wink:
 
  • #55
Pound-force? News to me, I've never heard of that, I guess that's what's meant when people use pounds when talking about pressure?

Approximately 4N is 1 lbf? How is that calculated?

1N = 1kg*m/s^2 I think... what calculation is used for getting lbf?
 
  • #56
Byrgg said:
Pound-force? News to me, I've never heard of that, I guess that's what's meant when people use pounds when talking about pressure?
Right. I'd say in physics or engineering it's more common to use pounds to measure force rather than mass. But you will almost always be able to tell by the context.

Approximately 4N is 1 lbf? How is that calculated?
I don't know what you mean by "calculated". This is a conversion factor between different systems of units. More accurately, 1 pound = 4.448 N.

1N = 1kg*m/s^2 I think... what calculation is used for getting lbf?
That's a definition. An analogous one for pounds-force would be:
1lbf = 1slug*ft/s^2.
 
  • #57
Oh, it uses feet as well, that's probably another thing I forgot to consider... so with ft and slug, the conversions work about to be about 4.448N? I think I've got it now
 
  • #58
Byrgg said:
Oh, it uses feet as well, that's probably another thing I forgot to consider... so with ft and slug, the conversions work about to be about 4.448N? I think I've got it now
Yeah, that is correct. You have to use imperial units for acceleration.
 
  • #59
To estimate the mass of air invovled, you need to know the amount of drag force and the average velocity of the air (assuming a no wind condition here) around a car, when the car is traveling at a specific velocity.

The amount of power due to drag is equal to the velocity of the car times the drag force of the air. For example, 1 horspower = speed in mph times force in lbs divided by 375 (conversion factor).

Using 1 second as a time factor, the work done by drag equals drag force times the distance the car moves in 1 second.

The amount of work done in 1 second will equal the amount of increase in energy, mostly kinetic, of the air in 1 second.

This is the tough part, you need to meausre or estimate the average velocity of the air affected by the car. Once this is estimated, then you can calculate the mass using the relationship between kinetic energy and mass e = 1/2 m v^2.

As an example, a 2006 Z06 Corvette, which has fairly low coeffiecient of drag, takes about 450 rear wheel horsepower to go 198mph. To make things easy, I'll assume that rolling resistance consumes about 20 hp, and that drag consumes about 430hp. This translates into a drag force of about 814lbs.

At 198 mph, the car is moving 290 feet / second. With a force of 814 lbs, the energy increase of the air is 236000 ft lbs. Assume that the average velocity of the diverted air is 30mph = 44 ft/sec, then the diverted mass per second is 244 lb mass per second or 118 cubic yards (2.06 lbs / cubic yard at sea level) of air per second. If the average velocity of the diverted air was 20mph = 29.3 ft / sec, then the diverted mass would be 550 lb mass per second or 267 cubic yards of air per second.

Need a second input here to confirm my math.
 
Last edited:
  • #60
An update to my previous post. How much of the air should be considered as diverted by a car passing through it? If you include the air that increased in velocity by 1/1000th of a mph, then the amount of involved is much larger than if you only considered air that was increased in velocity by 1mph.

Somewhat related is this picture of the Thrust SST creating an enormous shock wave. Since this is super sonic (or near super sonic) speeds, the way the air is affected is different than speeds well below supersonic, but it's a good example of how much air is affected in this case:

thrust1.jpg
 

Similar threads

Graduate Elastic collisions and Newton's third law
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Center of mass, Newton's third law, and rocket thrust
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Does Newton's Third law apply to torque/rotation?
  • · Replies 127 ·
5
Replies
127
Views
9K
High School Friction and Newton's laws in car braking systems
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Newton's Third Law: Action & Reaction Along the Line Connecting Two Objects
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Newton's third law in relation to field forces
  • · Replies 2 ·
Replies
2
Views
2K
High School Trouble understanding Newton's Third Law in Pulleys
  • · Replies 7 ·
Replies
7
Views
7K
Undergrad Newton's Third Law of a car and truck
  • · Replies 14 ·
Replies
14
Views
8K
High School Opposing forces and Newton's third law
  • · Replies 8 ·
Replies
8
Views
4K
High School Is Newton's Second Law Flawed? An Exploration of Zero Force in Space
  • · Replies 14 ·
Replies
14
Views
8K