About moving through the air and Newton´s 3rd law

[Javier Mont]

How are we able to move trough the aire if the air exerts a force equal to the friction we use to push us through air?
I´ve had this problem in my mind since some days now. I did this picture:

url: https://www.kn3.net/60DAC45480AJPG.html

 

[mfb]

To start walking, you exert a larger force on the ground. To keep walking at constant speed, you need a force equilibrium, and you have that.

I don't understand where you need help.

 

[Javier Mont]

To start walking, you exert a larger force on the ground. To keep walking at constant speed, you need a force equilibrium, and you have that.

I don't understand where you need help.

I need heelp with that first larger force that you exert at the begining. If you apply a force on the ground to move, by 3rd law, the ground exerts the same large force as a friction on you, suppousedly moving you. I understand that. The problem is then with the air. Since you´re walking you have to push the air to make your way, but is that force of ´´pushing´´ the air the same that the ground exerts to move you? If so, then the air is pushing back on you with the same magnitud of force. So my question is: how is it posible to make your way through the air if the air is pushing back with the same force youre attempting to use to move, thus, canceling your moving?

 

[Bandersnatch]

I need heelp with that first larger force that you exert at the begining.

Take a look at the picture on the left, where you've marked green fs as equal to yellow fs.

Imagine instead of a man starting to walk, you have a rocket accelerating at full thrust, and instead of a whole atmosphere you have just one air molecule.

The engine acts on the rocket with some millions of Newtons. Does the rocket have to act on the air molecule with the same force to change its momentum (i.e. move it out of the way)?

 

[Javier Mont]

Take a look at the picture on the left, where you've marked green fs as equal to yellow fs.

Imagine instead of a man starting to walk, you have a rocket accelerating at full thrust, and instead of a whole atmosphere you have just one air molecule.

The engine acts on the rocket with some millions of Newtons. Does the rocket have to act on the air molecule with the same force to change its momentum (i.e. move it out of the way)?

So I was wrong! The force to push the air (fs green) is no the same used to push the body (fs yellow), right? The force used to push the air is one that gives the air particles the same speed as the body moving through them. Am I right?

 

[RedDelicious]

I believe I see your confusion. Whatever you come into contact with pushes back on you but isn't going to overcome the static friction holding you in place. Say you are pushing a crate across the floor, the force you need to exert on the crate overcome the static friction holding it in place will be less than the static friction required to move you in the opposite direction and so it moves forward but you don't move backward. In the case of the wall, you generally can't with your arms exert enough force to overcome neither the wall nor your static friction, and so you're both stuck in place. Air is similar to the crate, in that you can push it out of the way easily because of its light mass.

In contrast, on a low friction surface, you might be able to exert enough force to slip. If you wanted to move the crate on a completely frictionless surface, if you pushed off it, you would move backward as according to F = ma and it would move forward the same.

 

[Bandersnatch]

The force used to push the air is one that gives the air particles the same speed as the body moving through them. Am I right?

That's more or less it, if we keep in the back of our mind that the dynamics of moving through gas is more complicated than just giving it the same speed. You push it around, and there's turbulence, and what not.
In any case there will be some forces associated with moving air out of your way that you need to overcome in order to move yourself, in the same way as there are some frictional forces one needs to overcome in order to push a block of wood on a flat surface. But they're equal only if the motion is constant (incl. being at rest).
(I see RedDelicious gives a similar answer above)

 

[Javier Mont]

I believe I see your confusion. Whatever you come into contact with pushes back you but isn't going to overcome the static friction holding you in place. Say you are pushing a crate across the floor, the force you need to exert on the crate overcome the static friction holding it in place will be less than the static friction required to move you in the opposite direction. In the case of the wall, you generally can't with your arms exert enough force to overcome neither the wall or your static friction. Air is similar to the crate, in that you can push it out of the way easily because of its light mass.

In contrast, on a low friction surface, you might be able to exert enough force to slip. If you wanted to move the crate on a completely frictionless surface, if you pushed off it, you would move backward as according to F = ma and it would move forward the same.

So you´re meaning that fs green is not equal to fs yellow in the case of my picture? You mean then that the force I´m using to psuh through the air is less that the friction I´m using to walk so they don't cancel?

 

[Javier Mont]

That's more or less it, if we keep in the back of our mind that the dynamics of moving through gas is more complicated than just giving it the same speed. You push it around, and there's turbulence, and what not.
In any case there will be some forces associated with moving air out of your way that you need to overcome in order to move yourself, in the same way as there are some frictional forces one needs to overcome in order to push a block of wood on a flat surface. But they're equal only if the motion is constant (incl. being at rest).
(I see RedDelicious gives a similar answer above)

Thank you very much! For a moment I was in a loop thinking that the friction force was going to exert to every body making contact and thus to everything and cancelating their moves xD
Thank you!

 

[RedDelicious]

So you´re meaning that fs green is not equal to fs yellow in the case of my picture? You mean then that the force I´m using to psuh through the air is less that the friction I´m using to walk so they don't cancel?

No. They do not cancel out because these are two different interactions. The pushing yourself forward is what causes you to collide with the barrier in the first place. After you've accelerated yourself, you come into contact with barrier (air) which then exerts a force on you. These are not two equal and opposite forces acting on you the whole time, which is when they would cancel out. You continue moving forward because of inertia and these collisions with the air thereafter all work to slow you down, but don't cancel out the initial interaction, which is why your foot doesn't just freeze in the air even without static friction.

For example, when you throw a baseball, it will immediately hit an air molecule with force ~F and that air molecules exerts force ~F on it, but it continues moving through the air in the direction you threw it virtually unaffected because of inertia. The forces of the collisions with the air molecules acting on it from that time forward will all work to slow it down (drag), but clearly don't negate the initial interaction that propelled it forward.

 

[Bandersnatch]

No. They are equal but they do not cancel out because these are two different interactions. The pushing yourself forward is what causes you to collide with the barrier in the first place. After you've accelerated yourself, you come into contact with barrier (air) which then exerts a force on you. These are not two equal and opposite forces acting on you the whole time, which is when they would cancel out.

For example, when you throw a baseball, it will immediately hit an air molecule with force ~F and that air molecules exerts force ~F on it, but it continues moving through the air in the direction you threw it virtually unaffected because of inertia. The forces of the collisions with the air molecules acting on it from that time forward will all work to slow it down (drag), but clearly don't negate the initial interaction that propelled it forward.

I don't think it is correct given the context of the question.
What is being talked about in the quoted post is the equality of the force from the body hitting the molecules, and the reaction force these molecules exert on the body. I.e. (green) fs and R. These forces are equal and opposite, but applied to different bodies, so they don't cancel out, and due to different inertia they will result in different motions.
However, the question in post #8 was whether the forces marked (green) fs and (yellow) fs are equal when the body is accelerating, which they are not. You seemed to agree in post #6.

 

[Javier Mont]

However, the question in post #8 was whether the forces marked (green) fs and (yellow) fs are equal when the body is accelerating, which they are not. You seemed to agree in post #6.

Thanks again! Now I have it clear