How can air displacement balance out forces in Newton's third law?

[jsmith613]

Newtons third law states:
"If object A acts on object B with force X, then object B will apply force X on object A" in other words "for every action there is an equal and opposite reaction".

Here are my few questions
1) If I apply pull a truck with force X and the truck pulls me with force X the truck SHOULD remain still but I can move it. With no resultant force, how is this possible. If the forces are not balanced, therefore, isn't Newtons law being disobeyed.

2) How can the Earth exert the correct upward force on every object. If I were to stand on the Earth (pulled down by gravity) the Earth pushes me up (again, by the force of gravity). If an object with the same volume as me, but with a different mass stood on the Earth the same thing would happen. How can the Earth apply the correct magnitude of force on every object

I am so confused.

Please explain

Thanks

 

[Doc Al]

Newtons third law states:
"If object A acts on object B with force X, then object B will apply force X on object A" in other words "for every action there is an equal and opposite reaction".

OK. Note that the 'action' and 'reaction' forces act on different objects.

Here are my few questions
1) If I apply pull a truck with force X and the truck pulls me with force X the truck SHOULD remain still but I can move it. With no resultant force, how is this possible. If the forces are not balanced, therefore, isn't Newtons law being disobeyed.

What determines whether the truck will accelerate or not is the net force on the truck, as described by Newton's 2nd law. If the force you exert is the only force acting on the truck, then it will start moving!

And if the truck's force on you were the only force acting on you, then you would start moving towards the truck. You'd pull each other together and get nowhere. That's what would happen if you and the truck were on a frictionless sheet of ice connected by a rope. You'd pull the rope, and then you and the truck would move towards each other.

What allows you to pull the truck and actually get somewhere is that there are other forces involved. For example, the ground could be exerting a friction force on your feet, allowing both you and the truck to move forward.

2) How can the Earth exert the correct upward force on every object. If I were to stand on the Earth (pulled down by gravity) the Earth pushes me up (again, by the force of gravity). If an object with the same volume as me, but with a different mass stood on the Earth the same thing would happen. How can the Earth apply the correct magnitude of force on every object

Careful here. Let's get the force pairs straight.

gravity: If the Earth exerts a certain downward gravitational force on you, then you exert an equal upward gravitational force on the earth. That's the nature of the gravitational force between two objects.

contact forces: The ground exerts an upward contact force (not gravitational!) that prevents you from falling through the floor, and you exert an equal downward contact force on the ground.

As long as what you are standing on is sufficiently strong to prevent you from pushing through it, then the upward contact force will be just enough to balance the downward gravitational force on you. You'll be in equilibrium. But step on a weak board that cannot support your weight and you'll start accelerating downward through the floor.

 

[jsmith613]

To DOC:

So what you are saying for Q1 is that:
- initially, when I start pulling on the tuck, the force I have on it is greater than the force it has on me (this, however, is only true if OTHER forces are involved). When the two forces are balanced, my resultant force = 0 so we remain at the same state of inertia as before. Correct?

For Q2, however, I am still confused. Here's why:
Firstly: Are we kept still when standing because of a Newton pair.

: If the Earth exerts a certain downward gravitational force on you, then you exert an equal upward gravitational force on the earth.

The ground exerts an upward contact force (not gravitational!) that prevents you from falling through the floor, and you exert an equal downward contact force on the ground.

If this statement is so, how do I exert the SAME gravitational force if my mass and acceleration is different to that of the earth. Furthermore, every other person has a different mass and acceleration to me, therefore the Earth should give into one of us.
The same question applies for the contact force.

Secondly: If two objects stand on the Earth's surface with the same mass and volume (assume it is literally on solid ground) how can the Earth provide the exact correct force to balance BOTH of our downward forces. Why is it not the same force no matter what the object. If this is the case then, should the Earth not push me up every time I step on it; gravity brings me down but the Earth should then push me up again.

Please explain

Thanks

 

[Xezlec]

So what you are saying for Q1 is that:
- initially, when I start pulling on the tuck, the force I have on it is greater than the force it has on me (this, however, is only true if OTHER forces are involved). When the two forces are balanced, my resultant force = 0 so we remain at the same state of inertia as before. Correct?

Not at all. The force you exert on the truck is always exactly equal and opposite to the force it has on you. The thing to understand here is that the concept of a force acting on you is not the same as the concept of a force acting on the truck. If you had two equal and opposite forces acting on you, you wouldn't move. Instead, what you have is one force acting on you and a different force acting on the truck and those forces happen to be equal and in opposite directions. Therefore, you and the truck are feeling pulled in opposite directions, and if that were the whole system, you would accelerate toward each other.

However, there are also other forces involved. There is a force between the ground in front of you and your feet, and that is pulling you and the ground toward each other horizontally (that's walking). The ground is big and therefore isn't going to move much. But you will move forward because that force is bigger than the force of the truck pulling you in the opposite direction.

If this statement is so, how do I exert the SAME gravitational force if my mass and acceleration is different to that of the earth.

Because that's how it works. The formula for gravitational force is G m1 m2 / r^2. The masses of the two objects are multiplied by each other, so it doesn't matter if you swap m1 and m2, you get the same force. Of course, applying a force to the Earth does not cause it to accelerate very much because it has such a large mass and F = m a (therefore a = F / m). Applying exactly the same amount of force to you would cause you to accelerate a lot more because your mass is a lot smaller than the Earth's.

If two objects stand on the Earth's surface with the same mass and volume (assume it is literally on solid ground) how can the Earth provide the exact correct force to balance BOTH of our downward forces.

Volume doesn't matter. Remember that the idea of a force on one object is not at all the same as the idea of a force on another object. Those are completely different concepts. Each object exerts a contact force -F on the Earth, and the Earth exerts a contact force F on each object. Therefore, each object feels a contact force of F, and the Earth feels a net contact force of -2F.

 

[saim_]

The forces are on different bodies that's why the resultant is not zero. You pull the truck and if your force is greater than its friction from the ground, the truck will moves towards you. The truck pulls you with the same force and if your friction from the ground is not equal to this force you will move towards the truck; you resist this force by pushing your feet down hard while pulling so you feet don't slip. Imagine pulling the truck while wearing roller blades; you'd be moving towards the truck much more than its moves towards you.

In case of gravity, Earth's pulls you down and you pull Earth up. This is one force pair. Now, if you hit the ground, due to its solid nature, the ground pushes you up with its contact force while you push the ground down with your contact force. So let's see the forces on both of you:

Forces on you: Gravity of Earth pulling you down, contact force from ground pushing you up; net force on you equals zero.

Forces on earth: Your gravity pulls the Earth up, your contact force with ground pushes the Earth down, net force on Earth equals zero.

 

[Doc Al]

To DOC:

So what you are saying for Q1 is that:
- initially, when I start pulling on the tuck, the force I have on it is greater than the force it has on me (this, however, is only true if OTHER forces are involved). When the two forces are balanced, my resultant force = 0 so we remain at the same state of inertia as before. Correct?

No, not correct.

First, the force you exert on the truck ALWAYS equals the force the truck exerts on you. That's Newton's 3rd law. All forces work this way.

Second, since those two forces act on different bodies, they never produce equilibrium directly. Forces can balance each other only if they act on the same body.

It is true that if no external forces act on you or the truck, then the center of mass of you and the truck cannot accelerate.

For Q2, however, I am still confused. Here's why:
Firstly: Are we kept still when standing because of a Newton pair.

No. You are kept still because the net force on you is zero. (Newton's 2nd law, not the 3rd.)

If this statement is so, how do I exert the SAME gravitational force if my mass and acceleration is different to that of the earth. Furthermore, every other person has a different mass and acceleration to me, therefore the Earth should give into one of us.

The gravitational attraction between two masses (you and the earth, say) is governed by Newton's law of gravity. The force that you exert on each other is the same, but the effect is not. If you jump off a ladder, you pull up the Earth with the same force that the Earth pulls down on you. Of course the effect is quite different! You are accelerated at 9.8 m/s^2; but the Earth is enormous and essential has zero acceleration due to the gravitational force you exert on it.

The same question applies for the contact force.

Whenever you push on something, it pushes back on you with the same force. Newton's 3rd law again.

Secondly: If two objects stand on the Earth's surface with the same mass and volume (assume it is literally on solid ground) how can the Earth provide the exact correct force to balance BOTH of our downward forces. Why is it not the same force no matter what the object. If this is the case then, should the Earth not push me up every time I step on it; gravity brings me down but the Earth should then push me up again.

Whenever two objects push against each other, they exert equal and opposite forces on each other. Newton's 3rd law again. Your weight causes you to push against the earth, the Earth pushes back equally. If two objects are on the surface, they each push against the ground with a different force (they weight different amounts); regardless, whatever force they push down with is always equal to the force the ground pushes up with.

 

[jsmith613]

A further two questions, therefore are as follows

1) Why does every force have an equal and opposite reaction (if possible, DONT use a mathematical formula to explain it because I would simply want to question why that formula works)

2)

Forces on earth: Your gravity pulls the Earth up

Does this concept work on the concent: f = ma or f = mg
So a = g = f / m.
Therefore I cause the Earth to accelerate upwards, when I stand on it at a = force applied / mass of earth. This would be an minute number.

 

[Doc Al]

A further two questions, therefore are as follows

1) Why does every force have an equal and opposite reaction (if possible, DONT use a mathematical formula to explain it because I would simply want to question why that formula works)

That's actually a rather deep question. It has to do with fundamental symmetry in nature and the fact that momentum is conserved. (I know that's not too helpful!)

2)


Does this concept work on the concent: f = ma or f = mg
So a = g = f / m.
Therefore I cause the Earth to accelerate upwards, when I stand on it at a = force applied / mass of earth. This would be an minute number.

Exactly. If you were in mid-air falling towards the earth, you could calculate the Earth's acceleration due to the gravitational force you exert on it just as you described: a = F/M. Since the mass of the Earth is huge, the acceleration of the Earth is incredibly minute and can be ignored. (Your acceleration, due to the Earth's gravitational force, is the usual g = 9.8 m/s^2; the same force acting on a much smaller mass gives a sizable acceleration.)

Just to nit-pick a bit: If you were standing on the earth, then you'd actually be exerting two forces on the earth. An upward gravitational force and an equal downward force from your feet on the ground. They balance out. The same is true for you: The Earth exerts a downward gravitational force on you and the ground pushes you up equally. That's why you don't accelerate when you're just standing still on the ground--the net force on you is zero.

 

[jsmith613]

Just to nit-pick a bit: If you were standing on the earth, then you'd actually be exerting two forces on the earth. An upward gravitational force and an equal downward force from your feet on the ground. They balance out. The same is true for you: The Earth exerts a downward gravitational force on you and the ground pushes you up equally. That's why you don't accelerate when you're just standing still on the ground--the net force on you is zero.

I assume, therefore, that both sets of forces are a Newton pair.

My question, therefore, is: If an airplane is flying above the Earth's surface (in the air) and the engines are NOT on, the plane will fall because the Earth causes it to accelerate down (there are no balanced forces - there is no contact force). Therefore, in order to remain airborne, the plane MUST push down enough air to counteract the acceleration due to gravity. If the plane pushes down the exact amount of air, the plane will not go up or down; if it pushes down more than necessary, the plane goes up; if it pushes down less than necessary, the plane falls.

Correct?

Concerning the plane, does it remain airborne in the following way: It pushes air down, thus displacing air. The air being displaced pushes up on the plane with the same force as the plane exerted on the air.

If this is not so, please explain why it happens.

Thank you.

 

[Pythagorean]

I assume, therefore, that both sets of forces are a Newton pair.

My question, therefore, is: If an airplane is flying above the Earth's surface (in the air) and the engines are NOT on, the plane will fall because the Earth causes it to accelerate down (there are no balanced forces - there is no contact force). Therefore, in order to remain airborne, the plane MUST push down enough air to counteract the acceleration due to gravity. If the plane pushes down the exact amount of air, the plane will not go up or down; if it pushes down more than necessary, the plane goes up; if it pushes down less than necessary, the plane falls.

Correct?

Concerning the plane, does it remain airborne in the following way: It pushes air down, thus displacing air. The air being displaced pushes up on the plane with the same force as the plane exerted on the air.

If this is not so, please explain why it happens.

Thank you.

Yes, that's sound logic, but I wouldn't say "the air being displaced pushes up on the plane". The displacement itself doesn't push the plane up; the displacement is a consequence of the air and the plane pushing on each other.

 

[Doc Al]

I assume, therefore, that both sets of forces are a Newton pair.

Just to be clear, the third-law pairs in that last example are:
- your gravitational force pulling up on the Earth and the Earth's gravitational force pulling down on you;
- your feet pushing down on the ground and the ground pushing up on your feet.

My question, therefore, is: If an airplane is flying above the Earth's surface (in the air) and the engines are NOT on, the plane will fall because the Earth causes it to accelerate down (there are no balanced forces - there is no contact force). Therefore, in order to remain airborne, the plane MUST push down enough air to counteract the acceleration due to gravity. If the plane pushes down the exact amount of air, the plane will not go up or down; if it pushes down more than necessary, the plane goes up; if it pushes down less than necessary, the plane falls.

Correct?

Concerning the plane, does it remain airborne in the following way: It pushes air down, thus displacing air. The air being displaced pushes up on the plane with the same force as the plane exerted on the air.

If this is not so, please explain why it happens.

Sounds good to me. For the air to support the plane, the plane must push down on the air, deflecting it.

 

[jsmith613]

Yes, that's sound logic, but I wouldn't say "the air being displaced pushes up on the plane". The displacement itself doesn't push the plane up; the displacement is a consequence of the air and the plane pushing on each other.

So the statement is correct, however, that air is being displaced with force x. To replace that air, more air rushes up with force -x, thus the two forces x + (-x) = 0, balance out.

Correct?

 

[Pythagorean]

So the statement is correct, however, that air is being displaced with force x. To replace that air, more air rushes up with force -x, thus the two forces x + (-x) = 0, balance out.

Correct?

yes, air is being displaces with force x, but I'm not so sure about the rest of your statement, as you haven't been terribly specific. From my understanding of what you've said, you're essentially forcing a three dimensional problem in one dimension the way you've formulated it. There's turbulent flows happening all over around the plane, so there's obviously not a net force of 0 on the air molecules themselves.

At some point, you'd have to specify whether you're talking about a molecule of air or a parcel of air. But unless you make the parcel really big, I doubt you'll have a sum of 0 for the forces on the air (because of turbulence).