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There was a similar question asked earlier but I did not really understand the answer, so I'll try again:

So I have this object we will call it object A and we have a floor, which we will name B. Object A exerts a gravitational force $F_g$ on the floor B, see picture below. At the same time object A has a normal force, which comes to my first question, is the normal force exerted from the floor to object A or is the normal force from object A in itself?

enter image description here

We also have according to Newton's third law a counteracting force from the gravitational force $F_g$ which comes to my second question, this counteracting force is not the same force as the normal force is it? And what is the force that cancels the counteracting gravitational force, would it not accelerate upwards because of the net force it has from the counteracting gravitational force.

I think the core problem I have is differentiating the normal force and Newton's third law, all help is much appreciated!

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    $\begingroup$ The answer is simple: action and reaction act on different bodies. $\endgroup$ – Anurag B. Apr 24 '18 at 13:12
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The term normal reaction is used when a force is at right angles to a surface and "reacting" to the presence of one object touching another object.
Such forces are set up to prevent one object passing through another object and are produced by each of the objects being (slightly) compressed when they are in contact.

Remember that when you apply Newton's third law:

  • The forces must act on different objects
  • The forces must be of the same type either non contact (gravitational, electrostatic, etc) or contact (objects touching).

In your example of static equilibrium the confusion might arise because you are dealing with four forces all of which have the same magnitude:

  • Gravitational attraction on object A due to object B (ground and Earth), $mg$, which is downwards.
  • Gravitational attraction on object B due to object A, $mg$, which is upwards.

These two forces are a Newton's third law pair.

  • Contact force on object A due to object B, $mg$, which is upwards and is called the normal reaction on object A due to object B.
  • Contact force on object B due to object A, $mg$, which is downwards and is called the normal reaction on object B due to object A.

These two forces are a Newton's third law pair.

The gravitation attraction on object A due to object B and the normal reaction on object A due to object B are two equal magnitude forces acting in opposite directions but are not a Newton's third law pair because:

  • The forces are both acting on object A.
  • One is a non-contact force and the other is a contact force.
  • If object A was not touching object B the gravitational force would still exist but the contact force would not.

The net force on object A (gravitational and contact due to B) is zero and the net force on object B (gravitational and contact due to A) is zero

Update with a diagram I produced earlier.

enter image description here

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  • $\begingroup$ On your second bullet, why is the contact force pointing upwards? Contact force on object B due to object A, mg, which is upwards and is called the normal reaction on object B due to object A. $\endgroup$ – Iram Haque Jun 1 '17 at 15:21
  • $\begingroup$ @IramHaque Thanks for your comment. It is an error because of cutting and pasting and not editing correctly that hopefully I have now corrected. $\endgroup$ – Farcher Jun 1 '17 at 15:27
  • $\begingroup$ Alright thanks, is it something like this if i draw it here dl2.pushbulletusercontent.com/fOWVl2qU2LyectaDIFoliEXJYvZomFj4/… I can paste it on pastebin if that's needed $\endgroup$ – Iram Haque Jun 1 '17 at 15:31
  • $\begingroup$ @IramHaque Your diagram shows the four forces correctly. The upward contact force you have a question mark by is correct. Perhaps I would show the contact forces acting a the surfaces of the block and the Earth? $\endgroup$ – Farcher Jun 1 '17 at 15:42
  • $\begingroup$ But later when adding them, it's as if they would be at the center of mass right? $\endgroup$ – Iram Haque Jun 1 '17 at 15:47
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Yes, you are confused. Right off the bat, object A does not exert a gravitational force on B. And I'm confused by what you mean by "counteracting force".

When doing these analyses, be sure to define your system, and focus on one system a time. Always keep in mind that a force has an object and an agent. An object cannot apply a force on itself, to answer one of your questions. Object A has two forces on it: gravity (the agent is the Earth) and the normal force (the agent is B). Since object A is at rest, there is no acceleration. Thus Newton's second law tells us $$m_Aa_A = 0 = N_{A,B} + F_{A, \mathrm{Earth}}$$ $$N_{A,B} = -F_{A, \,\mathrm{Earth}}$$ Note that this is a consequence of Newton's second law.

Newton's third law tells us that the force of A on B is equal in magnitude but opposite in direction to the force of B on A Each object exerts a normal force on the other. $$ N_{A,\,B} =-N_{B,\,A}$$ and the force of gravity due to the Earth on A is equal and opposite to the force of gravity due to A on the Earth. $$F_{A, \,\mathrm{Earth}} =-F_{\mathrm{Earth},\, A}$$

The second law is a statement of how a single object responds to the forces on it. The third law is a statement of the relationship between forces on two objects.

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Here goes:

There are two Newton's Third Law pairs of forces here:

Pull of (whole of) Earth on object = – Pull of object on (whole of) Earth

Contact force of floor on object = – Contact force of object on floor

So Newton's Third Law gives no relationship between the (gravitational) pull of the Earth on the object (call it W) and the upward push (call it F) of the floor on the object.

Indeed, W and F aren't necessarily equal and opposite. For example, if the object falls on to the floor from a height, there will be a short period during the impact when F is of greater magnitude than W.

However, after a short time, F will become equal and opposite to W. We know this because the object will remain at rest on the floor, and therefore has no resultant force on it.

You may not wish to read what follows. It goes more deeply into what's happening, but it's not usually explained in textbooks, and you may find it too weird...

How can it possibly come about that F is equal and opposite to W? Remember that it does NOT follow from Newton's Third Law. What's even stranger is that an object of twice the mass (and therefore acted on by a W of twice the magnitude) must also experience an equal and opposite F on it – unless the floor gives way! How does the floor manage to supply a force that is always equal and opposite to W? The answer has already been hinted at: "unless the floor gives way" wasn't entirely facetious… Even if the floor doesn't actually give way it deforms a little. The heavier the object the more the floor deforms, and the object moves downwards a little. The more the floor is deformed the greater the upward force it exerts on the object (just as a spring exerts a greater force the more it is extended). The object achieves equilibrium when the floor is deformed enough for F to be equal and opposite to W. [You probably won't see the floor deforming; we hope the floorboards are thick enough not to deform very much, but there are methods (e.g involving interference of light) which will show the deformation clearly.]

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  • $\begingroup$ So contact force in this case is normal force? $\endgroup$ – Iram Haque May 29 '17 at 18:47
  • $\begingroup$ Yes, here "contact" force is a normal force. The perpendicular component of a contact force is a normal force. Recall that normal is a synonym for perpendicular. The horizontal component of a contact force could be, for example, friction. I'm sure there are special cases and exceptions ... but I can't think of any at the moment. $\endgroup$ – garyp May 29 '17 at 20:10
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Assume B is the earth.

A exerts a force on B, (Gravity) and B exerts the same opposite force on A (Gravity). At the point where A hits the surface and stops accelerating, B exerts a second force on A, A exerts the second equal and opposite force on B.

This second force is the source of confusion. It is equal and opposite to gravity because it happens to stop both objects accelerating (for known reasons). It acts upwards on A and downwards on B.

The confusion perhaps arises from the treatment of gravity and normal as forces that come in pairs. They do not, a normal force can act on A separately to B, but as garyp has said, forces always come in pairs. In this case, the normal force is provided by B. The normal force can be anything, it does not have to equal gravity. It is deduced here to be gravity because the object A and B are not accelerating.

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This can get a bit confusing as there are sometimes distinguishable forces broken into two categories actual and effective. The summation of the actual forces must equal the summation of the effective forces. In this case the normal force is the effective force applied on the block from the floor. This force is Newtons third law. It is the equal and opposite force exerted on the block from the floor. This counter force is essentially created also from newtons second law, F=ma where a is gravity.

The normal force will always act perpendicular to the surface and equal mg.

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  • $\begingroup$ Note that the normal force does not always equal $mg$. $\endgroup$ – garyp May 29 '17 at 18:07
  • $\begingroup$ But will always equal a component of mg apologies $\endgroup$ – Brian Dibble May 29 '17 at 18:22
  • $\begingroup$ Again, not always true. I guess my main issue here is that normal forces have no relationship at all to gravity. You don't say that they do, but you do leave open room for misinterpretation. $\endgroup$ – garyp May 29 '17 at 20:06
  • $\begingroup$ You got a misconception here. The normal force needn't always be equal to mg. For example : Consider a block of mass $ m $ inside a lift accelerating upward and let $ N $ be the normal force by the lift on the block. Then for the block, $ N - mg = ma $ where $ a $ is the upward acceleration of the lift ( hence that of the block ). Or consider a bike of moving on a circular arc. Here too, the difference between the cosine component of mg and the normal force by the arc on the bike is the centripetal force required by the bike to move aver the arc. $\endgroup$ – Physicpsycho May 30 '17 at 15:29

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