# How a net force acts on 'Earth'?

• Kaneki123
In summary: N/m=0In summary, the net force on a ball is the force it applies to the air minus the force of the air on the ball. If there is a net force, the ball will accelerate.
Kaneki123
Hi, Newton's 2nd Law states that when a net force acts on a body, it causes acceleration in the body...This is well for forces acting on bodies in space...But what about on Earth??...On Earth when you apply force to an object, the air around it pushes back with same force!, So where is the concept of net force or 'unbalanced' force?...Kindly answer this question...

Kishor Kaphle
The 3rd (EDITED) law implies that the force you apply to the object is equal to the force the object applies to you, it's not equal to the force the air applies to the object.

To determine the motion of the object, we calculate the net force on the object. In this case, it's the force you apply to the object minus the force of the air on the object. If there is a net force, the object will accelerate.

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pixel said:
The 2nd law implies that the force you apply to the object is equal to the force the object applies to you, it's not equal to the force the air applies to the object.

To determine the motion of the object, we calculate the net force on the object. In this case, it's the force you apply to the object minus the force of the air on the object. If there is a net force, the object will accelerate.
Thanks for the quick reply...Suppose an object applies a force of 20N on the air particles...As a reaction, the air particles apply the same 20N force on the object...So where is the net force?...Should'nt the object be balanced by same force from both sides?

Kaneki123 said:
Hi, Newton's 2nd Law states that when a net force acts on a body, it causes acceleration in the body...This is well for forces acting on bodies in space...But what about on Earth??...On Earth when you apply force to an object, the air around it pushes back with same force!, So where is the concept of net force or 'unbalanced' force?...Kindly answer this question...
What do you mean by "the air pushes back with the same force"? That's only true in certain situations. Please give an example.

Kaneki123 said:
Thanks for the quick reply...Suppose an object applies a force of 20N on the air particles...As a reaction, the air particles apply the same 20N force on the object...So where is the net force?...Should'nt the object be balanced by same force from both sides?

Again, the motion of an object is determine by the net force on the object.

Kaneki123 said:
On Earth when you apply force to an object, the air around it pushes back with same force!
When I push on a basket-ball or a boomerang, air molecules move apart to let the object accelerate.

CWatters said:
What do you mean by "the air pushes back with the same force"? That's only true in certain situations. Please give an example.
I mean that when an object moves through the air, it comes in contact with air particles or the particles present in air...In other words it applies force to those particles...According to third law, should'nt the particles apply the same force back on the object...So the forces are 'balanced' in this situtation...For example, If a bullet is fired with 50N force, it applies the same force to air particles as it comes in contact with them...As a reaction, the air particles should apply the same force to the bullet.i.e.50 as action and reaction are equal...So there is a 'balance' of forces in this situation...Is'nt there?!

NascentOxygen said:
When I push on a basket-ball or a boomerang, air molecules move apart to let the object accelerate.
But they would still come in contact with the basketball of the boomerang...So should'nt there be an action-reaction between the objects and air molecules?

pixel said:
Again, the motion of an object is determine by the net force on the object.
Lets take an example of a ball...You throw the ball in the air with 10N force...The ball applies same 10N force to air molecules...Equal reaction causes a 10N force on the ball in OPPOSITE direction...(Taking gravity out of account)...SO there is a positive 10N force on ball and a negative 10N force on ball...So where is the net force?

Kaneki123 said:
But they would still come in contact with the basketball of the boomerang...So should'nt there be an action-reaction between the objects and air molecules?
It's almost too small for me to notice where I cite for these specific examples.

Kaneki123 said:
Lets take an example of a ball...You throw the ball in the air with 10N force...The ball applies same 10N force to air molecules...Equal reaction causes a 10N force on the ball in OPPOSITE direction...(Taking gravity out of account)...SO there is a positive 10N force on ball and a negative 10N force on ball...So where is the net force?
If the ball transferred all 10N to the air then there would be no net force on the ball, therefore it would have zero acceleration and would be moving at unchanging speed. When we observe this condition in a falling body we call it terminal velocity because there is [obviously] zero net force acting on the body.

Kaneki123 said:
Lets take an example of a ball...You throw the ball in the air with 10N force...The ball applies same 10N force to air molecules...Equal reaction causes a 10N force on the ball in OPPOSITE direction...(Taking gravity out of account)...SO there is a positive 10N force on ball and a negative 10N force on ball...So where is the net force?

The force between air and ball will be drag force and that is a function of how fast the ball is moving through the air. When the ball just starts to move the air will only push back with a force much less than the 10N. It won't be until the ball is moving at a considerable speed that the drag force will equal 10N.
You apply 10N to the ball. The air pushes back via drag (which has no relation to the force you are applying), and it is the difference in these forces that results in the net force.

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Third law is the most difficult for newcomer to understand. You are best to learn to draw vector diagrams of the forces.

On face value acceleration would be impossible due to the conundrum of the OP. The problem lies in what bodies you are assigning the forces to.

Vector diagrams will clarify your thinking.

Kaneki123 said:
Lets take an example of a ball...You throw the ball in the air with 10N force...The ball applies same 10N force to air molecules...Equal reaction causes a 10N force on the ball in OPPOSITE direction...(Taking gravity out of account)...SO there is a positive 10N force on ball and a negative 10N force on ball...So where is the net force?

You have to be careful here because when you "throw" a ball you are not continuously applying a force. Let's say instead that you are pushing an object with a continuous force. In your specific example there is no net force, so no acceleration. But if you increase the force on the object, at some point the air molecules cannot provide an equal force, so you could get acceleration. Note again that the 3rd law says:
• your force on the object = the object's force on you
• the object's force on the air = the air's force on the object
It does not say that your force on the object = the air's force on the object.

pixel said:
You have to be careful here because when you "throw" a ball you are not continuously applying a force. Let's say instead that you are pushing an object with a continuous force. In your specific example there is no net force, so no acceleration. But if you increase the force on the object, at some point the air molecules cannot provide an equal force, so you could get acceleration. Note again that the 3rd law says:
• your force on the object = the object's force on you
• the object's force on the air = the air's force on the object
It does not say that your force on the object = the air's force on the object.
Thanks for the answer...You said 'at some point the air molecules cannot provide an equal force'...Can you kindly state the reason as to why the air molecules would not be able to react with equal or same force?

Janus said:
The force between air and ball will be drag force and that is a function of how fast the ball is moving through the air. When the ball just starts to move the air will only push back with a force much less than the 10N. It won't be until the ball is moving at a considerable speed that the drag force will equal 10N.
You apply 10N to the ball. The air pushes back via drag (which has no relation to the force you are applying), and it is the difference in these forces that results in the net force.
You said ''When the ball just starts to move the air will only push back with a force much less than the 10N''...Why is that the air will push back with lesser force...Should'nt action-reaction be equal?As the ball moves through air, the ball applies 10 N force to air molecules it comes in contact with...So why is that the air molecules would not push back with same force?

NascentOxygen said:
If the ball transferred all 10N to the air then there would be no net force on the ball, therefore it would have zero acceleration and would be moving at unchanging speed. When we observe this condition in a falling body we call it terminal velocity because there is [obviously] zero net force acting on the body.
Why can't we observe the same phoenomenon to an object being pushed FORWARD (not upward)?

pixel said:
You have to be careful here because when you "throw" a ball you are not continuously applying a force. Let's say instead that you are pushing an object with a continuous force. In your specific example there is no net force, so no acceleration. But if you increase the force on the object, at some point the air molecules cannot provide an equal force, so you could get acceleration. Note again that the 3rd law says:
• your force on the object = the object's force on you
• the object's force on the air = the air's force on the object
It does not say that your force on the object = the air's force on the object.
You said''When the ball just starts to move the air will only push back with a force much less than the 10N''...Can't you state it like this..
Your force on object=The force the object applies to the air molecules=The force the air molecules apply BACK on the object...?

I think you are confusing three forces..

1) The force the air applies to the object (example = drag)
2) The force the object applies to the air
3) The force applied to the object (example = engine or rocket thrust).

1 and 2 are equal due to Newtons third law.

The net force acting on the object is the vector sum of all external forces (eg 1 and 3). Remember that Force 1 acts in the opposite direction to Force 3 so the sign of Force 1 is negative with respect to Force 3.

If the net force is >0 the object will accelerate
If the net force is <0 the object will decelerate
if the net force = 0 the object will maintain the current velocity.

DrewD
Kaneki123 said:
Your force on object=The force the object applies to the air molecules=The force the air molecules apply BACK on the object...?

No.

The force you apply to an object is not always the same as the force the air molecules apply back on the object.

The force you apply to an object is not always the same as the force object applies to the air molecules.

Perhaps see this diagram..

The red forces are equal (Newtons 3rd law)
The green forces are equal (Newtons 3rd law)
The red forces are not always the same as the green forces.

Kaneki123 said:
Thanks for the answer...You said 'at some point the air molecules cannot provide an equal force'...Can you kindly state the reason as to why the air molecules would not be able to react with equal or same force?

I only was responding to your example where you said the air molecules were applying the same force to the object as you were. As we are stating, this is not always the case, certainly it is not demanded by the 3rd law. So in general it is not true.

CWatters said:
No.

The force you apply to an object is not always the same as the force the air molecules apply back on the object.

The force you apply to an object is not always the same as the force object applies to the air molecules.
Thanks for the reply...It really clears up the confusion...One last thing...You said ''The force you apply to an object is not always the same as the force object applies to the air molecules''...Can you please elaborate as to why it is not same...Can you tell what are the 'factors' that cause this ''decrease''?

Kaneki123 said:
One last thing...You said ''The force you apply to an object is not always the same as the force object applies to the air molecules''...Can you please elaborate as to why it is not same...Can you tell what are the 'factors' that cause this ''decrease''?
It's not a "decrease". They are two completely different forces so there's no reason they have to be the same - if one is less than the other, it's not because the smaller one has been decreased, it's because it's smaller.

The force the object applies to the air molecules is equal to the force the air molecules apply to the object, which is the drag from air resistance. Meanwhile, you can push on the object with as much or as little force as you wish. If you apply more force than the air molecules the object will speed up; if you apply less force the object will slow down, and if you apply the exact same force then the object won't change its speed, like an airplane cruising at a constant speed.

Kaneki123 said:
Thanks for the reply...It really clears up the confusion...One last thing...You said ''The force you apply to an object is not always the same as the force object applies to the air molecules''...Can you please elaborate as to why it is not same...Can you tell what are the 'factors' that cause this ''decrease''?

Air is complicated because it comprises billions of individual molecules. To simplify the question consider a force Fa applied to two objects M1 and M2 that are touching each other as shown.

Newtons second law says that
Fa = (M1+M2) * a ..........(1)

where "a" is the acceleration of both. They both accelerate at the same rate because M1 is pushing M2 to the right.

So your question is about the green forces and why they don't equal the red forces.

F12 is the force M1 applies to M2
F21 is the force M2 applies to M1

Mass M2 is accelerating at "a" so the force F12 must be given by Newtons second law..

F12 = M2 * a ...........(2)

So comparing (1) and (2) you can see that the green force F12 is not equal to the red force Fa

For completeness..

The force F21 is given by Newtons third law so

F21 = F12

Once you understand CWatter's example, then consider this:

Take M1 alone and divide it into n thin sheets each with a mass of M1/n.
You are accelerating M1 with a force of Fa pushing on sheet 1 which is equal to M1 x a
Where M1 can also be expressed as the sum of all n of the sheets (n x M1/n)
Sheet 1 pushes on a mass equal to a mass equal to M1 minus the mass of sheet 1 or M1-M1/n, which is also being accelerated at a, so the force between sheet 1 and sheet 2 is (M1-M1/n) x a
Sheet 2 pushes on sheet 3 with a force of (M1-2M1/n) x a
Sheet 3 pushes on sheet 4 with a force of (M1-3M1/n) x a
...
Sheet n-1 pushes on sheet n with a force of( M1-(n-1)M1/n) x a = M1/n x a.

As you divide M1 into more sheets (n increases), the mass of each sheet (M1/n) decreases and the force between the last and second to last sheet decreases. As n approaches infinity, the force between the last two sheet approaches 0.

CWatters said:
Air is complicated because it comprises billions of individual molecules. To simplify the question consider a force Fa applied to two objects M1 and M2 that are touching each other as shown.

View attachment 103066

Newtons second law says that
Fa = (M1+M2) * a ..........(1)

where "a" is the acceleration of both. They both accelerate at the same rate because M1 is pushing M2 to the right.

So your question is about the green forces and why they don't equal the red forces.

F12 is the force M1 applies to M2
F21 is the force M2 applies to M1

Mass M2 is accelerating at "a" so the force F12 must be given by Newtons second law..

F12 = M2 * a ...........(2)

So comparing (1) and (2) you can see that the green force F12 is not equal to the red force Fa

For completeness..

The force F21 is given by Newtons third law so

F21 = F12
Okay...For sake of an example...lets say in space (or any other isolated system) if we 'throw' an object with some force and it collides with another object which was in rest, then if what you are saying is true then the first object should not come to a complete rest.i.e.it accelerates...Is this fact true even if the second object's mass is greater, smaller or equal to first object's mass?

Kaneki123 said:
Okay...For sake of an example...lets say in space (or any other isolated system) if we 'throw' an object with some force and it collides with another object which was in rest, then if what you are saying is true then the first object should not come to a complete rest.i.e.it accelerates...Is this fact true even if the second object's mass is greater, smaller or equal to first object's mass?
Let us not say we throw an object. Let us instead assume that an object exists already in motion. It does not matter how it came to have that motion. All that matters is that it is moving, that it has some mass and that, accordingly, it has some momentum. Saying that we throw it adds no useful information to the problem.

Now let us say that the moving object strikes another object that was at rest. That's fine.

Any acceleration in the second object will be matched by a corresponding acceleration in the first object. The ratio of the accelerations will be in inverse proportion to the ratio of the masses. That is what Newton's third law says. It amounts to a statement that the force between two objects always conserves momentum.

The first object may or may not stop. If it does stop, we know that the resulting momentum of the second object must be the same as the starting momentum of the first object. And vice versa. If the final momentum of the second object is equal to the initial momentum of the first object then the first object must stop.

Kaneki123 said:
Okay...For sake of an example...lets say in space (or any other isolated system) if we 'throw' an object with some force and it collides with another object which was in rest, then if what you are saying is true then the first object should not come to a complete rest.i.e.it accelerates...Is this fact true even if the second object's mass is greater, smaller or equal to first object's mass?

You are trying to apply what I said to a totally different situation. I assumed the two blocks start out in contact and a force is applied to one of them causing the pair to accelerate. No collision involved.

The situation you describe is quite different. For example what happens after the collision depends on the what the objects are made of. eg super bouncy balls behave differently to lumps of sticky clay. Usually you can easily work out what happens if they are one or the other. If they are somewhere in between then it can be a bit harder.

There are various possible outcomes to the situation you describe..

M1 can stick to M2 and they carry on together (eg bullet into a block of wood).
M1 can bounce off M2 and then they both continue in the same direction (Bowling ball and skittles)
M1 can bounce off M2 and return the way it came, while M2 accelerates away (Ball bouncing on the ground, M2 is the earth)
M1 stops and M2 continues (Two ball Newtons cradle).

It can be tricky to work out the forces involved during a collision so we frequently use other tricks to work out what happens to each object. Sometimes you can apply conservation of energy to the problem (total energy before = total energy after) sometimes not. You can always apply conservation of momentum.

jbriggs444
CWatters said:

You are trying to apply what I said to a totally different situation. I assumed the two blocks start out in contact and a force is applied to one of them causing the pair to accelerate. No collision involved.

The situation you describe is quite different. For example what happens after the collision depends on the what the objects are made of. eg super bouncy balls behave differently to lumps of sticky clay. Usually you can easily work out what happens if they are one or the other. If they are somewhere in between then it can be a bit harder.

There are various possible outcomes to the situation you describe..

M1 can stick to M2 and they carry on together (eg bullet into a block of wood).
M1 can bounce off M2 and then they both continue in the same direction (Bowling ball and skittles)
M1 can bounce off M2 and return the way it came, while M2 accelerates away (Ball bouncing on the ground, M2 is the earth)
M1 stops and M2 continues (Two ball Newtons cradle).

It can be tricky to work out the forces involved during a collision so we frequently use other tricks to work out what happens to each object. Sometimes you can apply conservation of energy to the problem (total energy before = total energy after) sometimes not. You can always apply conservation of momentum.
Okay...Can we just take it simple as to 'simple' objects...( no bouncy or rubbery stuff)...What i meant to say was...Like if a force of 10N is applied to an object and it continues in a straight line(of course after accelerating from rest)...And then it collides with the second object...Now if the first object applies a force of 10N to the second object then there is a reaction of 10N force on first object...Net force on first object becomes 0 and so it comes in rest or uniform velocity...But the same case is not observed with the object-air molecule relation...As you said so that the force on object is not neccesarily the force it applies to air molecule so that's how there is a net force on object...But if the reaction force on the first object is lesser than 10N than it should decelerate but still continue its motion...So it is really important for me to know that whether the reactional force in above scenario will be 10N or lesser than 10N?(Regardless of the second object's mass)

Kaneki123 said:
Okay...Can we just take it simple as to 'simple' objects...( no bouncy or rubbery stuff).
This does not bode well...
Now if the first object applies a force of 10N to the second object then there is a reaction of 10N force on first object...Net force on first object becomes 0 and so it comes in rest or uniform velocity...
How long is that 10N force applied? If we have simple objects which are perfectly rigid then the answer is... No time at all.
The resulting momentum change is... Zero.
The required mass of the object in order to change its state of motion in zero time with zero momentum change is... Zero.

Normally we do not talk about the force of impact. What we talk about instead is the quantity of momentum that is exchanged. This is the "impulse" associated with the collision. This eliminates the need to know how rigid the objects are..

So let's say that we have a head-on collision with an impulse of 10 kg m/sec. If the initially-moving object has a momentum equal to this, it stops. If it has a momentum greater than this, it keeps going. If it has a momentum less than this, it rebounds.

Kaneki123 said:
...if a force of 10N is applied to an object and it continues in a straight line(of course after accelerating from rest)...And then it collides with the second object...Now if the first object applies a force of 10N to the second object then there is a reaction of 10N force on first object...Net force on first object becomes 0...
This isn't how applied forces and collisions work. The fact that a force of 10N is applied to accelerate the object at first does not imply it applies a force of 10N on the second object. Nor do forces applied at different times (10N at first to accelerate the first object, 10N later when they collide) cancel each other out.

gmax137
russ_watters said:
This isn't how applied forces and collisions work. The fact that a force of 10N is applied to accelerate the object at first does not imply it applies a force of 10N on the second object. Nor do forces applied at different times (10N at first to accelerate the first object, 10N later when they collide) cancel each other out.
You said''The fact that a force of 10N is applied to accelerate the object at first does not imply it applies a force of 10N on the second object''...So is this a fact that the the force applied on first object is not neccessarily equal to the force it applies to the second object?...Because if it is not equal then ofcourse there would be a net force on first object...

Kaneki123 said:
You said''The fact that a force of 10N is applied to accelerate the object at first does not imply it applies a force of 10N on the second object''...So is this a fact that the the force applied on first object is not neccessarily equal to the force it applies to the second object?...Because if it is not equal then ofcourse there would be a net force on first object...

Right, there would be a net force on the first object. That's how an airplane increases its speed - the pilot pushes the throttle forward, the jet engines spool and and apply more force to the airplane, the force of the air on the airplane doesn't change so there's a net force on the airplane and it accelerates.

In general, the force applied to the first object and the force applied by the first object to the second object are completely different forces acting on completely different bodies. There's no reason why they should be the same.

CWatters and russ_watters

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