Newton's third law (equal and opposite force)

[Chenkel]

Something about Newton's third law confuses me, when I hold my phone and I move it around it's velocity is changing, therefore because Newton's first law it's acted on by a force, and because Newton's second law, the force is directly proportional to the mass and acceleration of that object. When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction? (The hand is pushing on phone, so the phone is pushing on hand)

 

[haushofer]

Because the forces are opposite in direction.

If you stand on an accelerating platform, the platform exercises a force on you, hence you aaccelerate. As a 'reaction'/part of the interaction, you push onto the platform in opposite direction (increasing your weight).

Something similar goes for your scenario.

 

[Chenkel]

Because the forces are opposite in direction.

If you stand on an accelerating platform, the platform exercises a force on you, hence you aaccelerate. As a 'reaction'/part of the interaction, you push onto the platform in opposite direction (increasing your weight).

Something similar goes for your scenario.

If the platform is not moving then there's a down force on the platform (weight) and an equal and opposite reaction force keeping you from falling to the center of earth, this makes sense if the platform has a large mass and is bolted in place. But what happens if you have two objects with similar weight, like a phone and a hand, if the force is pushing the hand into the phone as the phone moves from right to left, what keeps the phone and the hand next to each other as they move from right to left, instead of flying in opposite directions? The force on the hand is to the left, the force on the phone is to the left, so what forces are we summing on the objects to get two forces to the left?

Suppose the object on the left has a force pushing on it equal to -2 Newton's, and the force on the second object (to the right of the first one) has a force of -8 Newton pushing on it, then if they have the same acceleration to the left then their masses differ, suppose the first objects mass 1kg, then the acceleration is -2m/s^2, then the second objects acceleration is the same, so it's mass is 4kg. What are the reaction forces in this example and why aren't these objects flying apart?

 

[PeroK]

Something about Newton's third law confuses me, when I hold my phone and I move it around it's velocity is changing, therefore because Newton's first law it's acted on by a force, and because Newton's second law, the force is directly proportional to the mass and acceleration of that object. When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction? (The hand is pushing on phone, so the phone is pushing on hand)

Because your hand is connected to your arm; your arm to your body; and, your body to the Earth; where you are effectively anchored and can resist small forces.

If you were floating in space and moved an object around, you would see the effect of the reaction force. Although, it would be better to do the experiment with something more massive than a phone.

Altenatively, if you throw something very heavy, then you are liable to fall over backwards from the recoil (reaction force).

 

[Lnewqban]

Something about Newton's third law confuses me, when I hold my phone and I move it around it's velocity is changing, therefore because Newton's first law it's acted on by a force, and because Newton's second law, the force is directly proportional to the mass and acceleration of that object. When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction? (The hand is pushing on phone, so the phone is pushing on hand)

Imagine you were trying to do the same about a 50 000 floating vessel, but standing on a floating small boat.

The fact that your hand is transferring kinetic energy to the phone proves the law: the phone must be pushing back or it would not move at all.

 

[Dale]

Something about Newton's third law confuses me, when I hold my phone and I move it around it's velocity is changing, therefore because Newton's first law it's acted on by a force, and because Newton's second law, the force is directly proportional to the mass and acceleration of that object. When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction? (The hand is pushing on phone, so the phone is pushing on hand)

It will help if you always state Newton’s 2nd law correctly: “because Newton's second law, the net force is directly proportional to the mass and acceleration of that object”.

Students often forget that the force in Newton’s 2nd law is the net force and the force in Newton’s 3rd law is a pair of forces individually acting on different objects.

When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction?

Because the force from the phone is not the only force acting on your hand. So although the 3rd law force is equal and opposite, the net force for the 2nd law is not.

 

[erobz]

"Theoretically" when you move the phone to the right, the Earth moves in the other direction, and because the Earth is tied to all other bodies of mass in the universe through gravitation and has moved to the left...( are we allowed to continue on this line of reasoning in PF? )

 

[Chenkel]

It will help if you always state Newton’s 2nd law correctly: “because Newton's second law, the net force is directly proportional to the mass and acceleration of that object”.

Students often forget that the force in Newton’s 2nd law is the net force and the force in Newton’s 3rd law is a pair of forces individually acting on different objects.

Because the force from the phone is not the only force acting on your hand. So although the 3rd law force is equal and opposite, the net force for the 2nd law is not.

So the net force includes all forces acting on the body, including reaction forces; this means that if my hand is pushing a ball to the left, there's a final force acting on the hand $F_{finalhand} = m_{hand}a$ and $F_{finalhand} = F_{push} - F_{ball}$ where $F_{push}$ is the pushing force of the hand, and $F_{ball}$ is the force on the ball, so the final force on the hand is the pushing force minus the reaction force. Let's say $m_{hand} = .5kg$ and the mass of the ball is $m_{ball} = .25kg$ let's also assume that the pushing force is to the left, so that means it's negative, and we can write $F_{push} = -1N$ then we can also write $F_{finalhand} = -1 - (.25*a)$ this is equal to the mass of the hand times the acceleration of the hand, by Newton's 2nd law; so we can write $F_{finalhand} = -1 - (.25*a) = m_{hand}a = (.5)a$ and solve for acceleration $(.75)a= -1$ so $a = -1.33 (\frac m {s^2})$ so both the hand and the ball will move to the left without them 'bouncing' off of each other.

Hopefully this makes sense!

 

[James Brown]

Something about Newton's third law confuses me, when I hold my phone and I move it around it's velocity is changing, therefore because Newton's first law it's acted on by a force, and because Newton's second law, the force is directly proportional to the mass and acceleration of that object. When my hand is accelerating my phone around why does my hand not bounce off the phone and move in the opposite direction? (The hand is pushing on phone, so the phone is pushing on hand)

Now, you are holding a phone and pushing it. Well I suppose you are pushing it horizontally, then will you hold your phone without your fingers? Definitely not. So when your fingers is holding your phone friction between your phone and your fingers will prevents the phone from bouncing back. If there is no frictional force, the one thay bounce is not your hand, it will be your phone. Also, your arm will be pushing your hand+phone which will prevent your hand from bouncing back.
The phone did exert a force on your hand, if you can feel it.
Also, from your explanation of Newton's 2nd law, the force is not directly proportional to mass, it is directly proportional to acceleration. You are like talking when force increase, mass AND acceleration increase.

 

[nasu]

So the net force includes all forces acting on the body, including reaction forces; this means that if my hand is pushing a ball to the left, there's a final force acting on the hand $F_{finalhand} = m_{hand}a$ and $F_{finalhand} = F_{push} - F_{ball}$ where $F_{push}$ is the pushing force of the hand, and $F_{ball}$ is the force on the ball, so the final force on the hand is the pushing force minus the reaction force. Let's say $m_{hand} = .5kg$ and the mass of the ball is $m_{ball} = .25kg$ let's also assume that the pushing force is to the left, so that means it's negative, and we can write $F_{push} = -1N$ then we can also write $F_{finalhand} = -1 - (.25*a)$ this is equal to the mass of the hand times the acceleration of the hand, by Newton's 2nd law; so we can write $F_{finalhand} = -1 - (.25*a) = m_{hand}a = (.5)a$ and solve for acceleration $(.75)a= -1$ so $a = -1.33 (\frac m {s^2})$ so both the hand and the ball will move to the left without them 'bouncing' off of each other.

Hopefully this makes sense!

You are adding forcese acting on the hand with forces acting on the ball. This is not the net force on the hand. Neither is on the ball. The net force on the hand is the sum of forces acting on the hand. Each force belongs to an "action-reaction" pair so saying "including the reaction forces" is useless. All forces are "reactions" to some "action". In your case, you have the hand interacting with two objects of interest for the question. It interacts with the ball and interacts with the arm (or just the rest of the body). Each one of these interaction will produce a force on the hand. Their sum is the net force on the hand.

The interaction hand-ball results in a force on the ball too. This is not part of the net force on the hand. The interaction with the rest of the body results in a force on the rest of the body. This is not part of the net force on the hand either.

Only the net force on the hand is equal to $m_{hand} \cdot a_{hand}$. If the ball is accelerated in forward direction (whatever this may be) the net force should be in forward direction.

 

[erobz]

If we isolate the ball, the force applied by the hand becomes an external force acting on the ball. The force applied to the ball from the hand is given by:

$x \rightarrow^+$

$$ F_1 = m_{ball} \ddot x $$

Using the same convention, isolating the hand, the reaction force from the ball becomes an external force on the hand. The force on the hand from the ball is given by:

$$ F_2 = -m_{ball} \ddot x $$

These forces are equal in magnitude, but opposite in direction.

$$ F_1 = -F_2$$

 

[Chenkel]

You are adding forcese acting on the hand with forces acting on the ball. This is not the net force on the hand. Neither is on the ball. The net force on the hand is the sum of forces acting on the hand. Each force belongs to an "action-reaction" pair so saying "including the reaction forces" is useless. All forces are "reactions" to some "action". In your case, you have the hand interacting with two objects of interest for the question. It interacts with the ball and interacts with the arm (or just the rest of the body). Each one of these interaction will produce a force on the hand. Their sum is the net force on the hand.

The interaction hand-ball results in a force on the ball too. This is not part of the net force on the hand. The interaction with the rest of the body results in a force on the rest of the body. This is not part of the net force on the hand either.

Only the net force on the hand is equal to $m_{hand} \cdot a_{hand}$. If the ball is accelerated in forward direction (whatever this may be) the net force should be in forward direction.

I believe there might be a misunderstanding, the net force on the hand is equal to all forces acting on the hand, in the example I gave this is what I did, so this should be consistent with your definition. The force of the hand plus the force from the ball onto the hand (reaction force) is what is used to get the 'net force.'

 

[nasu]

I believe there might be a misunderstanding, the net force on the hand is equal to all forces acting on the hand, in the example I gave this is what I did, so this should be consistent with your definition. The force of the hand plus the force from the ball onto the hand (reaction force) is what is used to get the 'net force.'

The force of the hand and the force on the hand are acting on different objets. The first one is not part of the net force acting on the hand. There is indeed a missunderstanding, still. There is only one force on the hand due to the hand-ball interaction. The pair of this force acts on the ball. The second force on the hand is due to the hand-rest of body interaction. The pair of this acts on the arm/body.
You should stop thinking about "reacting" forces. The forces come in pairs, both memebers on equal position. Look at what interacts with what. For each interaction you have two forces, acting on the two interacting objects.

 

[Chenkel]

The force of the hand and the force on the hand are acting on different objets. The first one is not part of the net force acting on the hand. There is indeed a missunderstanding, still. There is only one force on the hand due to the hand-ball interaction. The pair of this force acts on the ball. The second force on the hand is due to the hand-rest of body interaction. The pair of this acts on the arm/body.
You should stop thinking about "reacting" forces. The forces come in pairs, both memebers on equal position. Look at what interacts with what. For each interaction you have two forces, acting on the two interacting objects.

If what you're saying is true, then there should only be two forces for any given interaction of two objects, and I'm guessing these forces would point in opposite directions, because Newton's third law. If these are in opposite directions, wouldn't the ball and the hand move an infinitesimal amount away from each other, and stop accelerating because they're not touching anymore? How do we get both objects moving in the same direction, when the only forces acting on the objects are in opposite directions?

 

[Dale]

How do we get both objects moving in the same direction, when the only forces acting on the objects are in opposite directions?

You don’t get them accelerating in the same direction if the only force on each object is the force from the other. Of course, this is a different scenario than the hand and phone scenario since there are other forces involved there.

 

[Chenkel]

You don’t get them accelerating in the same direction if the only force on each object is the force from the other. Of course, this is a different scenario than the hand and phone scenario since there are other forces involved there.s

So there's an additional force on the hand that overcomes the force from the phone onto the hand, to get both objects moving in the same direction. What explains the additional force? Is the additional force the result of some interaction pair?

 

[Dale]

What explains the additional force?

The hand is attached to the wrist. The wrist can produce large forces on the hand.

Is the additional force the result of some interaction pair?

Yes, the force of the wrist on the hand is equal and opposite the force of the hand on the wrist

 

[Chenkel]

The hand is attached to the wrist. The wrist can produce large forces on the hand.

Yes, the force of the wrist on the hand is equal and opposite the force of the hand on the wrist

If the two objects are attached to each other, and one acts on the other, don't the forces cancel each other out? for example, if I was in a boat and I was pushing on one side of the boat while I'm inside it, I wouldn't be able to accelerate the boat.

 

[hmmm27]

If the two objects are attached to each other, and one acts on the other, don't the forces cancel each other out? for example, if I was in a boat and I was pushing on one side of the boat while I'm inside it, I wouldn't be able to accelerate the boat.

Well, that's wrong, innit. If you're inside a rowboat and push on say the inside of the front, you and your CG will move backwards and the boat's, forward. Acceleration all'round.

Pretty much immediately of course you run out of arm length and willingness to go for a swim, and the butt/boat interface decelerates both, so you're pretty much back where you started. with the system (boater & boat) retaining its initial CG.

In the cellphone/hand example, they are constantly trying to spring apart, but the wrist(etc) is keeping the force up, so they're stuck together, both compressed slightly depending on the springiness of each.

Decrease the acceleration of the hand, and the force between hand/cellphone decreases and both will decompress a bit. Stop acceleration completely and they will literally "spring apart", as almost unnoticeable as that would be.

 

[Dale]

if I was in a boat and I was pushing on one side of the boat while I'm inside it, I wouldn't be able to accelerate the boat.

You absolutely could accelerate the boat. What you cannot do is accelerate the center of mass of the boat+you system. The force between you and the boat is external to you and external to the boat. It is only an internal force for the boat+you system.

Similarly, the force of the wrist on the hand is external to the hand and can indeed accelerate the hand. It is internal to the whole body so it does not accelerate the whole body.

 

[nasu]

If what you're saying is true, then there should only be two forces for any given interaction of two objects, and I'm guessing these forces would point in opposite directions, because Newton's third law. If these are in opposite directions, wouldn't the ball and the hand move an infinitesimal amount away from each other, and stop accelerating because they're not touching anymore? How do we get both objects moving in the same direction, when the only forces acting on the objects are in opposite directions?

They do accelerate in opposite directions if these are the only forces. This is what you may have during an elastic collision. If this were the case (only intraction hand-ball but no hand-body interaction), the net forces on the two objects will be in opposite directions and they would accelerate in opposite directions. But even then, accelerating in opposite directions does not mean moving in opposite directions. They may move toghether for a while even though their acceerations are in opposite directions. Consider deformation of the two bodies in contact.

In the case of the a real hand (and not a un-atatched hand like in a horror movie) , the net force on the hand is in the same direction as the net force on the ball because the arm is pushing the hand with a larger force than the ball is pusing inopposite direction. Try to understand the concept of net force. It is essential to understand mechanics. The acceleartion of an object depends on the net force and not just one of the forces acting on it.

 

[Chenkel]

You absolutely could accelerate the boat. What you cannot do is accelerate the center of mass of the boat+you system. The force between you and the boat is external to you and external to the boat. It is only an internal force for the boat+you system.

Similarly, the force of the wrist on the hand is external to the hand and can indeed accelerate the hand. It is internal to the whole body so it does not accelerate the whole body.

I'm a little confused about external vs internal forces. From what I read, first you define your system, and then you define what forces are internal or external to the system. So external forces accelerate objects, and internal forces don't?

Also, so I understand the example better, is the force of the wrist on the hand a rotational force (torque) or a linear force?

If I'm sitting in the boat, and I push on the boat when I'm bolted to the boat, I imagine that wouldn't cause any acceleration to the boat, because I couldn't move (does this being bolted to the boat change which forces are internal, external?), but I suppose if I am not bolted to the boat, and I push on the wall of the boat, and it causes me to take a couple small steps back from the wall of the boat, it will also impart an acceleration to the boat.

 

[alan123hk]

Suppose a system consists of a person standing on the surface of the Earth with a mobile phone. When the person moves the phone, an observer in another inertial system sees the following.

Let the center of mass of the system be at point COM.

Both the distance and angle of the phone and the Earth relative to the COM point may change, but the COM remains in inertial motion without acceleration or deceleration.

The rotational speed of this COM-centered system may vary, but the angular momentum of the entire system is conserved.

But since the mass of the phone is negligible relative to the mass of the earth, all that is actually seen is the movement of the phone, and the Earth is not moving at all.

 

[Dale]

From what I read, first you define your system, and then you define what forces are internal or external to the system. So external forces accelerate objects, and internal forces don't?

Yes, that is correct. So in your OP, since the hand is your system the force from the wrist is an external force and can indeed accelerate the hand.

Also, so I understand the example better, is the force of the wrist on the hand a rotational force (torque) or a linear force?

Generally it is both.

If I'm sitting in the boat, and I push on the boat when I'm bolted to the boat, I imagine that wouldn't cause any acceleration to the boat, because I couldn't move (does this being bolted to the boat change which forces are internal, external?)

Considering you and the boat to be separate systems then being bolted to the bolt doesn’t change the fact pushing on the boat is an external force. What it does is add a second external force at the bolts. So now, when you push there is an external force at the point of the push and a second external force at the point of the bolts. That pair of external forces is equal and opposite so they prevent you from accelerating.

Note that the push force and the bolt force are equal and opposite, but not a 3rd law pair because they act in the same object (you). There is a 3rd law force acting on the boat at the point of the push and a different 3rd law force acting on the boat at the point of the bolts.

I suppose if I am not bolted to the boat, and I push on the wall of the boat, and it causes me to take a couple small steps back from the wall of the boat, it will also impart an acceleration to the boat

Yes, exactly. If the boat is very small, like a canoe, this can be significant and even dangerous.

 

[Chenkel]

I've been trying to gain some intuition on internal and external forces, has Walter Lewin or Richard Feynman covered these topics? I wonder if there are some good resources out there, that will give me a working knowledge of these forces, so I can better understand how to identify them, and their implications in physics problems.

 

[erobz]

I've been trying to gain some intuition on internal and external forces, has Walter Lewin or Richard Feynman covered these topics? I wonder if there are some good resources out there, that will give me a working knowledge of these forces, so I can better understand how to identify them, and their implications in physics problems.

Water Lewin definitely covers this in the MIT lecture series



I haven't watched it, but the title seems to cover the necessary topics

 

[Chenkel]

Water Lewin definitely covers this in the MIT lecture series



I haven't watched it, but the title seems to cover the necessary topics

I did watch that one, no coverage on internal vs external forces and the applications of defining a physical system.

 

[erobz]

how about this one?

 

[Chenkel]

how about this one?

I will check it out, thank you!

 

[erobz]

This one is more specific to your inquiry

 

[Dale]

I've been trying to gain some intuition on internal and external forces, has Walter Lewin or Richard Feynman covered these topics? I wonder if there are some good resources out there, that will give me a working knowledge of these forces, so I can better understand how to identify them, and their implications in physics problems.

First you define your system. Then any force between two different parts of the system is an internal force. Any force between the system and something outside the system is an external force.

 

[Chenkel]

First you define your system. Then any force between two different parts of the system is an internal force. Any force between the system and something outside the system is an external force.

I noticed that Walter Lewin gave an example of a star system, he said that the attraction of two stars inside the system is an example of internal forces, and the internal forces cancel each other out when computing the total force. That makes sense; when we sum up the forces of stars attraction we get a grand total of 0, so the overall momentum of the system hasn't changed. He went so far to say, that if the stars explode the total momentum hasn't changed, and also if they collide, the total momentum hasn't changed, is this because Newton's third law? I can see the total force being 0 for two objects colliding, but what about an explosion? Is an explosion purely pair wise interaction of particles, with equal and opposite force, and when they interact, the forces cancel each other out? If I push on one of the stars from outside the system with my hand, that would make an external force, so the total momentum would change for both the external system and the internal system. I hope I'm seeing things clearly.

The hand is attached to the wrist. The wrist can produce large forces on the hand.

Yes, the force of the wrist on the hand is equal and opposite the force of the hand on the wrist

So there's an external force to the ball+hand system causing the ball, and the hand, to move to the left, I'm wondering how this external force can be maintained continuously as I move my arm from right to left, is it easy to maintain an external force? What is going on physically to create a sustained external force?

 

[Dale]

Is an explosion purely pair wise interaction of particles, with equal and opposite force, and when they interact, the forces cancel each other out?

Yes. That is correct

If I push on one of the stars from outside the system with my hand, that would make an external force, so the total momentum would change for both the external system and the internal system.

Yes.

What is going on physically to create a sustained external force?

There isn’t a single general answer to that question. For your case, the force is a contact force between the wrist and the hand. The continuous (for a brief time) force is achieved by deformation of the body. This deformation allows the short-range contact forces to continue even though the hand is accelerating.

 

[PeroK]

Also, so I understand the example better, is the force of the wrist on the hand a rotational force (torque) or a linear force?

It's all those things. If it wasn't for the rest of your body, your hand would fall to the ground under gravity.

It's nonsensical to treat your hand as an isolated rigid object.

Your whole body is needed to move a phone, the skin on your hand is just the contact point.

 

[PeroK]

Also, you are confusing yourself by using something so light that you can ignore certain aspects of the mechanics. Instead of a phone have a heavy table. Now it should be more obvious that you need your feet planted firmly on the ground to move it.

You can injure your back lifting a heavy object. Why is that? Your back is not touching the table when you lift it.