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    'Whenever two objects interact, the forces they exerton each other are equal and opposite.'

    Can anyone explain this statement to me, it's something that is often thrown around and I hear but I do not understand it. It's on the specification for AQA and I have no notes on it nor can I find any in my other books to explain it. I would appreciate anyone who can explain it to me!
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    (Original post by Aear)
    'Whenever two objects interact, the forces they exerton each other are equal and opposite.'

    Can anyone explain this statement to me, it's something that is often thrown around and I hear but I do not understand it. It's on the specification for AQA and I have no notes on it nor can I find any in my other books to explain it. I would appreciate anyone who can explain it to me!
    What are you exactly having difficulty with? The idea that they necessarily have to be in contact to exert forces of an equal magnitude but acted in different directions on each other?
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    (Original post by Aear)
    'Whenever two objects interact, the forces they exerton each other are equal and opposite.'

    Can anyone explain this statement to me, it's something that is often thrown around and I hear but I do not understand it. It's on the specification for AQA and I have no notes on it nor can I find any in my other books to explain it. I would appreciate anyone who can explain it to me!
    If object A exerts a force on object B, object B must exert a force on object A that is equal in magnitude to the other force, but in the opposite direction.
    ie. If A pushes B, then B pushes A with the same amount of force. If A pulls B, B pulls A with the same amount of force.

    (Original post by Mehrdad jafari)
    What are you exactly having difficulty with? The idea that they necessarily have to be in contact to exert forces of an equal magnitude but acted in different directions on each other?
    They don't need to be in contact. Consider gravity.
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    (Original post by morgan8002)
    If object A exerts a force on object B, object B must exert a force on object A that is equal in magnitude to the other force, but in the opposite direction.
    ie. If A pushes B, then B pushes A with the same amount of force. If A pulls B, B pulls A with the same amount of force.


    They don't need to be in contact. Consider gravity.
    Yeah, that's true! But for interactions acting at a distance, such a falling body under gravity, there is no normal reaction force as there would be in direct physical interactions, no? So, for example, for a body falling under gravity there is a force pulling the body but there isn't a force pushing the body in the opposite direction as there would in Direct interactions?
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    OP can probably ignore most of this.
    (Original post by Mehrdad jafari)
    Yeah, that's true! But for interactions acting at a distance, such a falling body under gravity, there is no normal reaction force as there would be in direct physical interactions, no? So, for example, for a body falling under gravity there is a force pulling the body but there isn't a force pushing the body in the opposite direction as there would in Direct interactions?
    You seem to be confusing two separate concepts.
    First N3
    For the object falling under gravity, there is a gravitational force acting on the object and the equal and opposite force from N3 is the force that the object is exerting on the Earth(the gravitational force pulling the Earth towards the object).

    For an object A in direct contact with object B(ignore gravity for now), for any normal reaction force that A exerts on B, the equal and opposite force from N3 is the normal reaction force acting on A(which has the same magnitude). Thus the two objects each have the same sized force acting on them.

    Now resolving components in equilibrium
    If we consider a more complicated example, eg. A person standing on the ground, there are two relevant forces acting on each body(the person and the Earth).
    The first is gravity. The Earth exerts a gravitational force on the person, pulling them towards it and the person exerts the same sized force on the Earth, pulling it towards them. These gravitational forces are an N3 pair.
    The second it the normal reaction force. The person exerts a normal reaction force in the Earth, pushing it away and the Earth the same sized force on the person, pushing them away. These normal reaction forces are an N3 pair.
    If we resolve vertically, we know that the person is in equilibrium. The upwards forces equal the downwards forces. So the gravitational force on the person is equal to the normal reaction force on the person, but in the opposite direction. Using the above N3 pairs, we know the same is true of the two forces on the Earth.
    The two forces on the person are equal in magnitude and opposite in direction but are not an N3 pair. They act on the same object and my statement of N3 above says a pair of N3 forces must act on different objects. Also, an N3 pair is two forces of the same fundamental type, these are not. The same can be said of the two forces on the Earth; they are equal in magnitude and opposite in direction, but are not an N3 pair.

    The laws of physics allow the weight to be larger than the normal reaction force(imagine a heavy mass on soft earth). In this case the normal reaction forces would not be equal in magnitude to the gravitational forces and the object would not be in equilibrium, but by N3, we know that the magnitude of the normal reaction force on the Earth is equal to the magnitude of the normal reaction force on the object(and the forces are in opposite directions) and that the magnitude of the gravitational force on the Earth is equal to the magnitude of the gravitational force on the object(and the forces are in opposite directions).

    edit: Thinking about it, a person standing on the surface of the Earth isn't in equilibrium near the equator because they're orbiting the Earth(so must have a net centripetal acceleration). Unless you stand at one of the poles, the normal reaction force is always smaller than the weight.
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    (Original post by morgan8002)
    OP can probably ignore most of this.


    You seem to be confusing two separate concepts.
    First N3
    For the object falling under gravity, there is a gravitational force acting on the object and the equal and opposite force from N3 is the force that the object is exerting on the Earth(the gravitational force pulling the Earth towards the object).

    For an object A in direct contact with object B(ignore gravity for now), for any normal reaction force that A exerts on B, the equal and opposite force from N3 is the normal reaction force acting on A(which has the same magnitude). Thus the two objects each have the same sized force acting on them.

    Now resolving components in equilibrium
    If we consider a more complicated example, eg. A person standing on the ground, there are two relevant forces acting on each body(the person and the Earth).
    The first is gravity. The Earth exerts a gravitational force on the person, pulling them towards it and the person exerts the same sized force on the Earth, pulling it towards them. These gravitational forces are an N3 pair.
    The second it the normal reaction force. The person exerts a normal reaction force in the Earth, pushing it away and the Earth the same sized force on the person, pushing them away. These normal reaction forces are an N3 pair.
    If we resolve vertically, we know that the person is in equilibrium. The upwards forces equal the downwards forces. So the gravitational force on the person is equal to the normal reaction force on the person, but in the opposite direction. Using the above N3 pairs, we know the same is true of the two forces on the Earth.
    The two forces on the person are equal in magnitude and opposite in direction but are not an N3 pair. They act on the same object and my statement of N3 above says a pair of N3 forces must act on different objects. Also, an N3 pair is two forces of the same fundamental type, these are not. The same can be said of the two forces on the Earth; they are equal in magnitude and opposite in direction, but are not an N3 pair.

    The laws of physics allow the weight to be larger than the normal reaction force(imagine a heavy mass on soft earth). In this case the normal reaction forces would not be equal in magnitude to the gravitational forces and the object would not be in equilibrium, but by N3, we know that the magnitude of the normal reaction force on the Earth is equal to the magnitude of the normal reaction force on the object(and the forces are in opposite directions) and that the magnitude of the gravitational force on the Earth is equal to the magnitude of the gravitational force on the object(and the forces are in opposite directions).

    edit: Thinking about it, a person standing on the surface of the Earth isn't in equilibrium near the equator because they're orbiting the Earth(so must have a net centripetal acceleration). Unless you stand at one of the poles, the normal reaction force is always smaller than the weight.
    I think I'm beginning to understand Newton's third law though I will need leave sometime for it to fully grasp it so that I can explain it a bit simpler, your explanation is too "heavy".

    In the case of a weight of an object being larger than the normal reaction force, I don't understand why that would be the case though. Surely if the object is at rest on the ground then the gravitational force would be equal to the normal reaction force caused by the electrostatic force of repulsion between the electrons of the surface of the two bodies as the two different fundamental forces would had to be equal for the object on the surface of the earth to be at rest.
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    (Original post by Mehrdad jafari)
    I think I'm beginning to understand Newton's third law though I will need leave sometime for it to fully grasp it so that I can explain it a bit simpler, your explanation is too "heavy".
    Once you understand it, find your own way to explain it to yourself. I find it easiest to just take it from the definition and think of examples where it applies and which forces are the pairs in those situations. You might find another way works better for you though.
    In the case of a weight of an object being larger than the normal reaction force, I don't understand why that would be the case though. Surely if the object is at rest on the ground then the gravitational force would be equal to the normal reaction force caused by the electrostatic force of repulsion between the electrons of the surface of the two bodies as the two different fundamental forces would had to be equal for the object on the surface of the earth to be at rest.
    You are correct in thinking that the weight and normal reaction force on the object would be balanced if the object was at rest on the surface of the Earth. However, no object is truly at rest on the surface of the Earth. They are orbiting the centre of the Earth. As they're going around in a circle(and a circle's gradient is constantly changing) their velocity must change(ie. they must accelerate so have a net force downwards). This is only a small net force, so doesn't usually need to be taken into account. If you haven't studied circular motion them don't think too deeply about this example. I just included it as another example that the contact force isn't required to equal the weight of an object.
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    (Original post by morgan8002)
    Once you understand it, find your own way to explain it to yourself. I find it easiest to just take it from the definition and think of examples where it applies and which forces are the pairs in those situations. You might find another way works better for you though.

    You are correct in thinking that the weight and normal reaction force on the object would be balanced if the object was at rest on the surface of the Earth. However, no object is truly at rest on the surface of the Earth. They are orbiting the centre of the Earth. As they're going around in a circle(and a circle's gradient is constantly changing) their velocity must change(ie. they must accelerate so have a net force downwards). This is only a small net force, so doesn't usually need to be taken into account. If you haven't studied circular motion them don't think too deeply about this example. I just included it as another example that the contact force isn't required to equal the weight of an object.
    I will, but thanks for your explanation.

    I see, I thought that was not related to your edited part of your post. It's true that the earth is constantly orbiting but considering that an object is somewhere near the equator then would its weight not be less than when when the earth wasn't rotating? Even in that case the normal reaction would be equal to the weight assuming that the object is not attached on the surface of the earth.
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    (Original post by Mehrdad jafari)
    I will, but thanks for your explanation.

    I see, I thought that was not related to your edited part of your post. It's true that the earth is constantly orbiting but considering that an object is somewhere near the equator then would its weight not be less than when when the earth wasn't rotating? Even in that case the normal reaction would be equal to the weight assuming that the object is not attached on the surface of the earth.
    The apparent weight would be less, but the actual force due to gravity(which I refer to as weight) would be the same. The apparent weight would be equal to the normal reaction force.

    The weight of O is given by F = \dfrac{Gm_Om_E}{r_E^2}.
    The apparent weight of O is given by F = \dfrac{Gm_Om_E}{r_E^2}-m_Or_E\omega^2, where \omega is the angular frequency of rotation of the Earth. (equation holds if on the equator, it becomes a little more complicated otherwise)
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    (Original post by morgan8002)
    The apparent weight would be less, but the actual force due to gravity(which I refer to as weight) would be the same. The apparent weight would be equal to the normal reaction force.

    The weight of O is given by F = \dfrac{Gm_Om_E}{r_E^2}.
    The apparent weight of O is given by F = \dfrac{Gm_Om_E}{r_E^2}-m_Or_E\omega^2, where \omega is the angular frequency of rotation of the Earth. (equation holds if on the equator, it becomes a little more complicated otherwise)
    I love how my simple question has turned into a full on physics debate with fun looking equations.

    I understand the theory of the statement, but not how it operates in the real world. If I push on a door, it pushes back on me, yet it still moves. This is one thing I do not understand. And while it is not mentioned, how can it exert a force on me? I think applications in the real world are what is confusing me in this topic.
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    (Original post by Aear)
    I love how my simple question has turned into a full on physics debate with fun looking equations.

    I understand the theory of the statement, but not how it operates in the real world. If I push on a door, it pushes back on me, yet it still moves. This is one thing I do not understand. And while it is not mentioned, how can it exert a force on me? I think applications in the real world are what is confusing me in this topic.
    It's a shame that you are required to memorise a few words so that you wouldn't have to understand it. How much people at GCSE level and below want to know the answers to these questions, adult students don't want.
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    (Original post by Aear)
    I love how my simple question has turned into a full on physics debate with fun looking equations.

    I understand the theory of the statement, but not how it operates in the real world. If I push on a door, it pushes back on me, yet it still moves. This is one thing I do not understand. And while it is not mentioned, how can it exert a force on me? I think applications in the real world are what is confusing me in this topic.
    When your hand pushes the door, the atoms in your hand get pressed against the atoms in the door. The atoms have electrons, which are negative, so when the atoms of your hand get pressed against the atoms of the door, there's an electric force between them. This obeys Newton's third law and if you have two atoms against each other, each atom will exert a force on the other equal to the force the other is exerting on it, but the two forces are in opposite directions(pushing each atom from the other one because like charges repel).
    The normal reaction force is just the sum of all of these small electric forces that act on one of the objects(eg. the sum of all the electric forces on the atoms in your hand). The normal reaction force on the door has the same size, but in the opposite direction(it's the sum of the same small electric forces, but this time the forces are all pointing in the opposite direction). The door moves because(ignoring friction etc.), the only force acting on it is the normal reaction force, pushing it away from you.

    I hope that made sense.
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    (Original post by morgan8002)
    When your hand pushes the door, the atoms in your hand get pressed against the atoms in the door. The atoms have electrons, which are negative, so when the atoms of your hand get pressed against the atoms of the door, there's an electric force between them. This obeys Newton's third law and if you have two atoms against each other, each atom will exert a force on the other equal to the force the other is exerting on it, but the two forces are in opposite directions(pushing each atom from the other one because like charges repel).
    The normal reaction force is just the sum of all of these small electric forces that act on one of the objects(eg. the sum of all the electric forces on the atoms in your hand). The normal reaction force on the door has the same size, but in the opposite direction(it's the sum of the same small electric forces, but this time the forces are all pointing in the opposite direction). The door moves because(ignoring friction etc.), the only force acting on it is the normal reaction force, pushing it away from you.

    I hope that made sense.

    That made a lot of sense, and it has helped, and I can see, I think, why the door moves, but why don't I move, is it because my mass is too large for the force to make me move?
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    (Original post by Aear)
    That made a lot of sense, and it has helped, and I can see, I think, why the door moves, but why don't I move, is it because my mass is too large for the force to make me move?
    Partially. Your mass is larger than(most) doors so you will experience a smaller acceleration than the door, so just a quick push doesn't change your velocity a lot, so you just slow down a little bit.
    If the door is particularly heavy(or experiences friction), then you'll probably lean into the door as you open it, pushing back on the floor with your feet. There will be a frictional force between the floor and you're feet. You'll then be roughly in equilibrium as you're transferring momentum between the door and the floor.
 
 
 
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