# Normal force cancels out the force of gravity?

Alright, so I know there will be a normal force pointing perpendicular to the surface and gravitational force, right?

Gravitational force = mg
But since normal force acts in "opposite direction to the gravitational force". Wouldn't normal force then cancel out the gravitational force? If it's so, wouldn't you just "fly" for a moment up, then come back? You'll come back because once you stop touching the ground, normal force would stop acting on you...

If an object rests on the ground, then angle is zero
cos(0) = 1.

The normal force is perpendicular to the surface. Sometimes that means it also acts opposite the gravitational force (like when the surface is horizontal), but not always.

Why would an object fly up if there is no net force accelerating it upwards? Consider the vertical direction: if the vertical component of the normal force and the force of gravity cancel out, then there will be no acceleration upwards.

I do agree that if for some reason the object found itself slightly above the surface, then there would be no normal force acting on it, though.

Why would an object fly up if there is no net force accelerating it upwards?
Yeah, sorry. But I guess, what it means is that only a slight push upwards would be necessary to get the object moving upwards, then, as the normal force stops acting on the object, then you'd need a stronger push upwards (that would account for the mass of the object), right?

So basically, for infinitesimal height you can move any object up with a slight push because normal force would act opposite to the force of gravity.

I do agree that if for some reason the object found itself slightly above the surface, then there would be no normal force acting on it, though.
What do you mean "slightly"? I thought that if an object doesn't touch any surface it wouldn't experience a normal force.

DaveC426913
Gold Member
Yeah, sorry. But I guess, what it means is that only a slight push upwards would be necessary to get the object moving upwards,
Why would it take only a slight push? Just because the net force is zero doesn't mean the forces themselves are not quite large. An object the size of a person is pushing on the ground with about 750N of force and the ground is pushing up with same. It'll take at least 750N of force to separate the two.

256bits
Gold Member
But since normal force acts in "opposite direction to the gravitational force". Wouldn't normal force then cancel out the gravitational force? If it's so, wouldn't you just "fly" for a moment up, then come back? You'll come back because once you stop touching the ground, normal force would stop acting on you...
Hello vizakenjack,

Your logic seems impeccable, but it does have a flaw that you have overlooked that throws a monkey wrench into your conclusion.
You cannot turn off gravity.

You are attracted to the earth, and the earth is attracted to you. This is whether you are standing on the ground or falling from a height.
In both cases you have a force of mg acting on you from the gravity of the earth.

On the ground, the surface you are standing on, such as a floor, stops you from falling towards the centre of the earth, by providing the normal force upwards. Similarily, you also provide a different normal force downwards from your feet onto the surface so the earth.

Do you think you should you re-work your conclusion?

Why would it take only a slight push?
Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.

Hello vizakenjack
...
Similarily, you also provide a different normal force downwards from your feet onto the surface so the earth.
Hi,
I don't get it, so an object provides a normal for to the ground too? uhm... well, what does it equal to? I mean, a normal force cancels out gravity
so what would an object's downward normal force cancel...? or what would be its magnitude?...

russ_watters
Mentor
Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.
A table has four legs. How much of the table's weight does each leg carry? If you add another leg, does the table launch itself into space or just sit there with a different weight distribution between the legs?

Next time you are on a bathroom scale, place it near the sink or another structure you can reach out and touch. Push down lightly on that structure. Do you launch yourself into the ceiling? What does the bathroom scale say?

Because normal force already cancels out the force of gravity for you. So you don't need to "overcome" force of gravity.

Consider an object sitting on the floor, weighing 10 newtons. 10 newtons of normal force act up on it and 10 newtons of gravity act down on it. Now you decide to grab it and pull up on it the tiniest bit, but not enough to lift it up. Let's say you pull it up by 1 newton. It's not enough to lift it off the ground. It now has 9 newtons of normal force acting up on it from the floor, and 1 newton of force pulling it up from you, and 10 newtons of gravity pulling down on it.

What happened and what is very important to remember is that the normal force changed! Don't think of it as "locked in" to the weight of the object. It doesn't always equal the weight of the object. It equals in magnitude the force pushed down upon the surface which could be more (say if you push down on an object) or less (if you pull up on it a bit, like in my example).

The object in my example won't start moving up until you pull up with at least 10 newtons of force.

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• vizakenjack
256bits
Gold Member
Hi,
I don't get it, so an object provides a normal for to the ground too? uhm... well, what does it equal to? I mean, a normal force cancels out gravity
so what would an object's downward normal force cancel...? or what would be its magnitude?...
Same thing as when you push on a wall, the wall pushes back. Both forces are equal in magnitude but opposite in direction.

Have you ever heard of Newton's Third Law - for every action there is a reaction? The two normal forces are an action-reaction pair and they cancel each other out.

The other action-reaction pair is the earth attracting you, and you attracting the earth, again both forces equal in magnitude but opposite in direction.

So this
"I mean, a normal force cancels out gravity" is not really correct.

The reason the normal force from the ground on you is equal to the attraction force of the earth on you by looking at the forces acting on your body. In this case we have the attractive force from gravity downwards, which we can say is W=mg, and the normal force N which acts upwards. Since you are not accelerating the net force is zero.
Fnet = W - N = 0, or
N=W

W is always mg ( at least close to the ground ).
If by chance you have a rope near you to pull on, the act of pulling will decrease the normal force from the ground. Only when your pull on the rope is equal to W, will you rise into the air. The normal force has not lifted you up, because by pulling on the rope you have decreased the normal steadily from mg to 0.

• vizakenjack
A.T.
Because normal force already cancels out the force of gravity for you.
Only if there are no other supporting forces. As soon you apply any upwards force the normal force drops by the same amount.

I don't get it, so an object provides a normal for to the ground too?
Yes, that's how you make footprints.

well, what does it equal to?
The two normal forces are equal but opposite. See Newtons 3rd Law.

I mean, a normal force cancels out gravity so what would an object's downward normal force cancel...?
The gravitational force that the objects exerts on the Earth.

sophiecentaur
Gold Member
2020 Award
It might help if you remember that, in every practical situation, there is always a small amount of movement involved - as if everything was actually composed of (very strong) springs. Forces are shared as the different parts of the structure deform a tiny bit and equilibrium is achieved. Adding a fifth leg to a table can produce any combination of force sharing, depending on the tiny details of strength and length of the leg. Think in terms of everything being made of sponge and you won't arrive at any apparently impossible situations in your thought experiments.

Yes, you're correct. When an object is touching the horizontal ground, the magnitude of the normal force and the gravitational force are equal. This is precisely why the object does not move. Now if the object is on a table, the perpendicular force exerted by the table on the object (normal) is equal to the gravitational force. If you exert a downward force on the object, the the normal will increase, as normal= gravitational force + your force.

If the object is on a slanted surface, then the magnitude of the normal force is not equal to the gravitational force. In fact, it's less. There's an angle between force and motion, so cos theta will have a value. This is why the object will slide down.

In mid air, the object does not experience a normal force, therefore it falls downwards.

What does the bathroom scale say?
Probably along the lines of: "wow, you're such an idiot that you can't figure out such a simple concept"
:(

The other action-reaction pair is the earth attracting you, and you attracting the earth, again both forces equal in magnitude but opposite in direction.
Earth attracts you - direction towards the earth's core, right? In other words, downwards.
So your attraction to the earth is ... upwards? Uhm... doesn't make sense...

the normal force changed!
Thanks, this really clarified it a bit. So does it mean, that if you pull a 10 newton object with 1 newton up, the normal force down would be 9 newtons and yours of course 1 newton up, and then they BOTH would cancel out the force of gravity?
I just don't kinda get it, why does floor have to cancel the force of gravity? Can't it just prevent an object from falling through it? Like support it and that's it. The gravity can keep acting on it. I saw the Cosmos with Neil Tyson, and there was an episode, in which it was explained about molecular ... attraction forces... dunno.
anyway, so the reason the ground would "push" an object upwards is because the molecules of the ground wouldn't like being bent down / squeezed by the object's mass so they would resist the object by wanting to remain as usual, rigid? Would this explain the reason a surface would act on an object with a normal force?

But what if ground doesn't cancel the force of gravity but simply prevents the object from falling down. It's impossible right? Because an object would act downwards on the area that it touches, right? So the area of the surface being touched would "resist" the object's downward force (due to gravity) ... Well, it makes sense then, but that only works if an object consists of molecules and atoms that have to resist...
what if an object was from a different universe... what if it didn't matter how hard you press on othe mysterious object? If it meant nothing to the object.. or whatever little components (like atoms) it's consisted of? Or,
Or,
what if it was possible to bent that mysterious object by pressing on it in downward direction... but what if the object's special atoms didn't care? Like "alright, fine, squeeze us all you want, we don't care, we don't want to go back to our original position". So the object could just keep getting whatever shape it can... I mean, there would be some sort of attraction forces within the mysterious object's atoms, but if they're broken... then it's fine... the "detached" atoms would simply reconnect with other atoms once they get closer to each other. So you could keep breaking bonds / forces of attraction between this mysterious object's atoms, and the atoms would simply get "reconnected" with other atoms, but the bonds between atoms wouldn't oppose your force...
For instance, if 2 newtons is needed to break a bond between two atoms of this mysterious object... then just 2 newtons is needed to break the bond, but there will be no resistance from the bond itself...
Oh wait, isn't it how it is in our universe already? It doesn't matter if the bonds "resist" whatever force that is trying to break that very bond, if it takes 2 newtons to break it, you'll break if you apply 2 newtons, regardless whether the bonds "resist" it or not, right?

Same thing as when you push on a wall, the wall pushes back.
So the wall is pushing back during the time you're pushing pushing the wall, right? Once you stop, the wall stops pushing back at you too, right? So there is no way to make an experience in which you suddenly stop pushing back but the wall continues pushing back at you for a brief moment? So what you're implying is that wall pushing back and you pushing the wall - these forces would stop acting on each other instantaneously?

Then another questions arises, if the wall is pushing back with equal magnitude to your force, how the heck do you manage to "move" the wall? Or let's say an object weights 10kg, would it mean the wall would ONLY push back with a force equal to 90 N? (10*9.8)??

If the object is on a slanted surface, then the magnitude of the normal force is not equal to the gravitational force. In fact, it's less. There's an angle between force and motion, so cos theta will have a value. This is why the object will slide down.
My problem is understanding why the surface would exert a normal force. For instance, molecular bonds/attraction would explain it (they don't want to be squeezed in, so they oppose an object in order to become "unsqueezed").

Assuming there is molecular attraction between the surface's molecules, wouldn't then, molecules inside the object (perpendicular block) would rush towards its end corner (circled in the picture above)? And if so, then the surface area that touches that end corner of the block.. would experience a stronger "push" downwards from the block (due to gravity), right? Or does the surface exert equally distributed normal force across all of its area that is affected by the block?

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A.T.
So your attraction to the earth is ... upwards?
Your attraction is towards you, obviously. That's what "attraction" means.

Then another questions arises, if the wall is pushing back with equal magnitude to your force, how the heck do you manage to "move" the wall?
If the wall moves, the ground pushes you more than the wall.

• vizakenjack
Mark44
Mentor
So your attraction to the earth is ... upwards?
A.T. said:
You attraction is towards you, obviously. That's what "attractions" means.
Yes, each body attracts the other, in forces that are oppositely directed.

Think about the earth and the moon. The earth exerts a gravitational force on the moon, and the moon likewise exerts a gravitational force on the earth.

My problem is understanding why the surface would exert a normal force.
Newton's third law of motion states: Every action has an equal and opposite reaction.
The weight of an object placed on the plane is the force exerted by it on the plane. So, the plane exerts an equal and opposite force, called the normal.
In the case of a slanted plane, the surface is not able to equal the gravitational force because like I said, there's an angle between force and motion.

Your attraction is towards you, obviously. That's what "attraction" means.
Uhm, what?
Okay let's get this straight. An object weighs 10kg, it rests on the ground.
How many forces are there?
Force due to gravity (10kg * g), right?
Then there is a "normal force" coming from the surface in upward direction, right?
But then, you also claim there is another force acting downwards... exerted by the object? What is this additional force that "leaves footprints" (when the object is a human)?

Yes, each body attracts the other, in forces that are oppositely directed.
So in the example above with an object that weighs 10 kg.
There are three forces:
The object's force due to gravity ↓
Surface's "normal force" acting on the object upwards ↑
The object attracts the earth to itself, so if it attracts the earth... then what would the direction of this force be? ↓ or ↑? On one hand, the object "pulls the earth" towards itself, so the force must be ↑ (towards the object), on the other... the object exerts the force to the earth and "tells" the earth that it "wants" it, then direction is towards the earth, down ↓
Although, I'm more inclined towards the former assumption, because an objects "pulls" the earth towards itself. So the direction should be towards the object... is it correct to say that the direction of such force is ↑?
...

A.T.
What is this additional force that "leaves footprints" (when the object is a human)?
Contact force, same but opposite as the normal force acting on the feet. See Newtons 3rd Law.

• vizakenjack
Contact force, same but opposite as the normal force acting on the feet.

So, in an example with an object that weighs 10 kg and rests on the ground, there are 3 forces present then?
Object's force downwards due to gravity.
Normal force exerted by the ground in upward direction.
Contact force between the object and the ground, the object exerts it in downwards direction.

Correct?

Mark44
Mentor
Uhm, what?
Okay let's get this straight. An object weighs 10kg, it rests on the ground.
How many forces are there?
Force due to gravity (10kg * g), right?
Then there is a "normal force" coming from the surface in upward direction, right?
But then, you also claim there is another force acting downwards... exerted by the object? What is this additional force that "leaves footprints" (when the object is a human)?
There are two forces, one up, one down. Each body (the object and the earth) attracts the other.
vizakenjack said:
So in the example above with an object that weighs 10 kg.
There are three forces:
No. There are two forces, as explained above. See Newton's Third Law (http://en.wikipedia.org/wiki/Newton's_laws_of_motion). The third law states that all forces exist in pairs.
vizakenjack said:
The object's force due to gravity ↓
Surface's "normal force" acting on the object upwards ↑
The object attracts the earth to itself, so if it attracts the earth... then what would the direction of this force be? ↓ or ↑? On one hand, the object "pulls the earth" towards itself, so the force must be ↑ (towards the object), on the other... the object exerts the force to the earth and "tells" the earth that it "wants" it, then direction is towards the earth, down ↓
Although, I'm more inclined towards the former assumption, because an objects "pulls" the earth towards itself. So the direction should be towards the object... is it correct to say that the direction of such force is ↑?
...

• vizakenjack
A.T.
So, in an example with an object that weighs 10 kg and rests on the ground, there are 3 forces present then?
No, there are 4 forces. Every interaction consists of a force pair, and you have two interactions: gravitational attraction and contact (electromagnetic repulsion).

There are two forces, one up, one down.
No, there are 4 forces.

So how many are really there?
Probably even more, if you also include the force of gravity from the black hole located at the center of our galaxy(?).

But only assuming earth as an isolated system.
I feel like, for simplicity sake it is said that there are only 2 forces acting on an object that weighs 10 kg and is resting on the ground, correct?

you have two interactions: gravitational attraction and contact (electromagnetic repulsion).
It is said here that an electromagnetic repulsion is the reason the surface "resists" and exerts a normal force on an object, correct?
What's the difference between a contact force and a normal force?

Mark44
Mentor
The contact forces that @A.T. mentioned keep the two objects separated. In force diagrams, these forces are usually ignored, at least that's my recollection from the physics courses I had long ago. The earth attracts the 10 kg object toward it, and the 10 kg object attracts the earth toward the object. The electromagnetic forces keep the objects from "welding" themselves together.

I'm not sure you realized it, @vizenjack, but the word "normal" in the term "normal force" isn't synonymous with "ordinary." The meaning here is "perpendicular," since the force is perpendicular to the ground. The

• vizakenjack
jbriggs444
Homework Helper
So how many are really there?
As many as you care to account for. Two obvious ones and their two third-law partners.

I feel like, for simplicity sake it is said that there are only 2 forces acting on an object that weighs 10 kg and is resting on the ground, correct?
Yes. Two forces acting on the object. Their third law partners act on other objects.
What's the difference between a contact force and a normal force?
A "contact" force is the force on an object from another surface or object that it is in direct contact with.

A "normal" force is the component of the contact force from a surface that is exerted perpendicular to the surface. In this context, "normal" means "perpendicular". [The component of a contact force that is tangential to the surface is usually frictional]

Two forces acting on the object. Their third law partners act on other objects.
Can you give an example of a third law partner in regards to a situation when a 10kg object rests on the ground?

Also, molecules inside the object (perpendicular block) would not "rush" towards the end corner (circled) because there is electrostatic repulsion between molecules in the block, correct? I mean, that makes sense, why would a molecule from the top attempt to go to the bottom (even if there is gravity pulling it downwards) if it simply CANNOT do so because other molecules would not allow it.. and even if you placed an extra molecule at the bottom, it would have become too "tight" at the bottom and all the molecules would have attempted to spread out so that the distance between molecules increased, right?

In other words, molecule from the top can't squeeze in-between other molecules beneath it and go like "uh, excuse me, I just want to go down, mind giving me way? thanks", right?

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