Questions about Newton's third law

In summary, the conversation discusses the concept of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. The conversation delves into different situations where this law applies, such as a car driving and a person lifting weights, and the corresponding reaction forces that are involved. The concept of normal force is also discussed, as well as the difference between free fall and being on a surface. The conversation also clarifies that the gravitational force between a person and the Earth is mostly due to the Earth's mass, rather than the person's.
  • #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?
 
Science news on Phys.org
  • #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 :grumpy:
 
  • #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
 
  • #61
Byrgg said:
I'm not questioning his law or anything, I just need to understand a few things about it. I understand the basic concept of the law, for every force, there is an equal force in the oppostite direction.

But what provides the reaction force in the following situations?

A car driving, has an applied force forward, but what's the reaction force backwards? Also, in the same situation, what about friction? It applies a force backward, so what's the reaction force forward provided here?

Second, a person lifting weights, they apply a force upwards to lift the weight, what is the reaction force here provided by?

A car driving, has an applied force forward, but what's the reaction force backwards? Also, in the same situation, what about friction? It applies a force backward, so what's the reaction force forward provided here?

The reaction force is just the air resistance by air acting on the surface area of the moving car in the opposite direction to that of the car's motion as well as ground friction acting on the car's wheel, also in the opposite direction to that of the car's motion.

Second, a person lifting weights, they apply a force upwards to lift the weight, what is the reaction force here provided by?

Your hand is doing work by lifting the weights upwards. Since weight of the weights is acting downwards, the reaction force comes from your hand's upward movement when lifting the weights.
 
Last edited by a moderator:
<h2>1. What is Newton's third law?</h2><p>Newton's third law states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object will exert an equal force in the opposite direction on the first object.</p><h2>2. How does Newton's third law apply to everyday life?</h2><p>Newton's third law can be seen in many everyday situations, such as walking, throwing a ball, or even sitting in a chair. When you walk, your feet push against the ground, and the ground pushes back with an equal force, propelling you forward. When you throw a ball, your hand exerts a force on the ball, and the ball exerts an equal force in the opposite direction, causing it to move. When you sit in a chair, your body exerts a downward force on the chair, and the chair exerts an equal upward force on your body, keeping you from falling through.</p><h2>3. Can Newton's third law be violated?</h2><p>No, Newton's third law is a fundamental law of physics and cannot be violated. It is a natural consequence of the conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force.</p><h2>4. How does Newton's third law relate to the concept of forces?</h2><p>Newton's third law is closely related to the concept of forces. It explains that forces always occur in pairs, and that the two forces in a pair are equal in magnitude and opposite in direction. This means that forces cannot exist on their own; they always come in pairs.</p><h2>5. What is the difference between action and reaction in Newton's third law?</h2><p>In Newton's third law, action and reaction refer to the two forces in a pair. The action force is the force exerted by one object on another, while the reaction force is the force exerted by the second object on the first. Both forces are equal in magnitude and opposite in direction.</p>

1. What is Newton's third law?

Newton's third law states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object will exert an equal force in the opposite direction on the first object.

2. How does Newton's third law apply to everyday life?

Newton's third law can be seen in many everyday situations, such as walking, throwing a ball, or even sitting in a chair. When you walk, your feet push against the ground, and the ground pushes back with an equal force, propelling you forward. When you throw a ball, your hand exerts a force on the ball, and the ball exerts an equal force in the opposite direction, causing it to move. When you sit in a chair, your body exerts a downward force on the chair, and the chair exerts an equal upward force on your body, keeping you from falling through.

3. Can Newton's third law be violated?

No, Newton's third law is a fundamental law of physics and cannot be violated. It is a natural consequence of the conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force.

4. How does Newton's third law relate to the concept of forces?

Newton's third law is closely related to the concept of forces. It explains that forces always occur in pairs, and that the two forces in a pair are equal in magnitude and opposite in direction. This means that forces cannot exist on their own; they always come in pairs.

5. What is the difference between action and reaction in Newton's third law?

In Newton's third law, action and reaction refer to the two forces in a pair. The action force is the force exerted by one object on another, while the reaction force is the force exerted by the second object on the first. Both forces are equal in magnitude and opposite in direction.

Similar threads

Replies
127
Views
6K
  • Thermodynamics
Replies
2
Views
2K
Replies
8
Views
2K
Replies
14
Views
7K
Replies
7
Views
5K
  • Thermodynamics
Replies
4
Views
1K
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Replies
2
Views
1K
  • Thermodynamics
Replies
26
Views
2K
Back
Top