# Why normal force is not equal to gravity?

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• user079622
user079622
We learned that normal force equal in magnitude as mg and opposite direction, it is reaction force to gravity.
1.If normal force is not equal gravity, isnt this violate newton 3. law action=reaction?
2. If gravity is higher than normal force , this system has net Force non zero,it means accelerate in dircetion of higher force ,which is not true?
3.Professor say it must be higher otherwise we will not move with earth?
It can be equal, when rotate stone with rope,stone pull your arm with same force as arm pull stone,so not need to be higher..

The normal force is not the "reaction" to gravity as in a third law pair. The force pairs in Newton third law are always due to same interaction, between the same two objects. The pair of gravitational attraction on the object is the gravitational attraction on the Earth. Normal forces describe a different interaction which is not even gravitational. And no, the normal force can have any value, depending on the situation. Does not have to be equal to the weight of the object.

russ_watters
nasu said:
The normal force is not the "reaction" to gravity as in a third law pair. The force pairs in Newton third law are always due to same interaction, between the same two objects. The pair of gravitational attraction on the object is the gravitational attraction on the Earth. Normal forces describe a different interaction which is not even gravitational. And no, the normal force can have any value, depending on the situation. Does not have to be equal to the weight of the object.
In static normal force has equal magnitude as mg.

It is not true that gravity must be higher than normal f. otherwise we wil not move with earth

What you are saying does not seem to make any sense. I am NOT going to watch a 17 minute video just to see if you understood him correctly. At what time-point does he say that?

user079622 said:
In static normal force has equal magnitude as mg.
Just because two forces happen to be equal but opposite in some cases, doesn't mean they are a Newton's 3rd Law pair.

user079622 said:
We learned that normal force equal in magnitude as mg and opposite direction, it is reaction force to gravity.
Ugh, this is not true on several counts.

First, the normal force is not always equal in magnitude as ##m\vec g##. For example, if you are jumping off the ground then the normal force can be larger than ##m\vec g## and if you are in free fall the normal force is zero.

Second, the normal force is in the direction perpendicular to the surface. It is only opposite the direction of gravity if the surface is horizontal.

Third, it is not the reaction force to gravity. It is a reaction force to the contact force that is pushing on the surface.

russ_watters and user079622
phinds said:
What you are saying does not seem to make any sense. I am NOT going to watch a 17 minute video just to see if you understood him correctly. At what time-point does he say that?
16:23

Dale said:
Ugh, this is not true on several counts.

First, the normal force is not always equal in magnitude as ##m\vec g##. For example, if you are jumping off the ground then the normal force can be larger than ##m\vec g## and if you are in free fall the normal force is zero.

Second, the normal force is in the direction perpendicular to the surface. It is only opposite the direction of gravity if the surface is horizontal.

Third, it is not the reaction force to gravity. It is a reaction force to the contact force that is pushing on the surface.
N=mg-mv2/r....... where g is moon gravitation acc.

If I want find normal force exerted on man feet on dark side of moon, do I need take in calculation revolution and rotation, because they both have their own mv2/r ?

Or will be normal force different if moon only revolve around earth and not rotate around itself?

A.T. said:
Just because two forces happen to be equal but opposite in some cases, doesn't mean they are a Newton's 3rd Law pair.
Why we cant say mg=mv2/r ..gravity is centripetal force in case for earth, just like friction is centripetal force in car turn?
Profesor say centripetal force is not real force, real forces are gravity and normal force.
I dont agree that normal foce is real force, that force dont even exist...it is math construction.
Real force are gravity ,electromagnet force and strong and weak force..

Motore and phinds
user079622 said:
I dont agree that normal foce is real force, that force dont even exist...it is math construction.
The normal force is the name given to a force that a surface exerts on an object away from it. Why do you walk down the stairs instead of jumping out the top floor of a building which is faster? Because, regardless of whether you believe that the normal force exists or not, if you jumped out the window you would be injured (or worse) by the normal force exerted on you by the ground when you land . Its origin is electromagnetic and arises by the electrons in your body being repelled by the electrons in the ground.

Dale
user079622 said:
I dont agree that normal foce is real force, that force dont even exist...it is math construction.
Who taught you that nonsense?

kuruman said:
The normal force is the name given to a force that a surface exerts on an object away from it. Why do you walk down the stairs instead of jumping out the top floor of a building which is faster? Because, regardless of whether you believe that the normal force exists or not, if you jumped out the window you would be injured (or worse) by the normal force exerted on you by the ground when you land . Its origin is electromagnetic and arises by the electrons in your body being repelled by the electrons in the ground.
System has non zero net force but not accelereting in direction of higher force.
How is this possible?

user079622 said:
N=mg-mv2/r....... where g is moon gravitation acc.
Here you appear to be considering an object that is at rest on the surface at the equator of a body that is rotating with a surface velocity of ##v## and which has a radius of ##r##.

Let us do some back of the envelope estimation...

The moon rotates about 28 times more slowly than the Earth (rotational period, not surface velocity). It has a radius about 1/4 of the Earth's. One handy formulation for centrifugal acceleration is ##F = \frac{4\pi^2r}{t^2}##. That is two factors of 28 and one factor of 4 weaker than the Earth's centrifugal acceleration.

The Earth's centrifugal acceleration [plus the deviation in the same direction from the shape of the geoid] is in the neighborhood of 0.5 percent. That is an over-estimate.

The Moon's gravitational force is about 1/6 that of the Earth.

So we would expect the percentage deviation due to the moon's centrifugal acceleration to be 0.5 times 6 divided by 28 divided by 28 divided by 4. My calculator says that is 0.001 percent.

user079622 said:
If I want find normal force exerted on man feet on dark side of moon, do I need take in calculation revolution and rotation, because they both have their own mv2/r ?
No. To a first approximation, both the moon and the man are accelerating in the direction of the Earth at the same rate. So you can ignore the moon's orbital velocity in its path around the Earth and count rotation only. Of course, rotation is also negligible as estimated above.

To a second approximation, there would be a small tidal correction because the man is farther from the Earth than the moon's center of gravity. Tidal gravity from the Earth is stretching them apart. Then too, the moon is not a rigid sphere. So the moon stretches a bit. It gets complicated. The precise normal force of a hypothetical man on the very center of the dark side of the moon is of little practical importance.

user079622 said:
System has non zero net force but not accelereting in direction of higher force.
How is this possible?
It is possible if the system is going around in a circle. The Earth experiences a non-zero net force of attraction in the direction of the Sun. Its acceleration is in the direction of the Sun but it does not fall into the Sun. You are confusing acceleration with direction of motion.

jbriggs444 said:
No. To a first approximation, both the moon and the man are accelerating in the direction of the Earth at the same rate. So you can ignore the moon's orbital velocity in its path around the Earth and count rotation only. Of course, rotation is also negligible as estimated above.
If moon and man accelerate at same rate in direction of the earth, does it mean I must get same result if I calculate from rotation or from revolution?
Of course assume that revolution path is circle..

kuruman said:
The Earth experiences a non-zero net force of attraction in the direction of the Sun.
Isnt earth centrifugal force equal sun gravity that pull earth toward him?just like tension in rope is same for stone and arm..

You mean net force is non zero because path is elipse? or?

user079622 said:
Isnt earth centrifugal force equal sun gravity that pull earth toward him?just like tension in rope is same for stone and arm..
In the rotating rest frame of both, yes. In the inertial frame there is no centrifugal force

A.T. said:
In the rotating rest frame of both, yes. In the inertial frame there is no centrifugal force
Yes I know that , but he say it is net force non zero toward sun..
in inertial frame forces must cancel out too?
isnt it?

user079622 said:
You mean net force is non zero because path is elipse? or?
Yes, because the path is an ellipse. The Earth's velocity is changing direction continuously. That means non-zero acceleration.

kuruman said:
Yes, because the path is an ellipse. The Earth's velocity is changing direction continuously. That means non-zero acceleration.
But when I rotate stone on string, net force is zero in radial direction, even stone accelerate toward center all the time?

user079622 said:
in inertial frame forces must cancel out too?
No, if they would cancel, the Earth would not stay in orbit, but move along a line.

A.T. said:
No, if they would cancel, the Earth would not stay in orbit, but move along a line.
So different frames get different results?

user079622 said:
But when I rotate stone on string, net force is zero in radial direction, even stone accelerate toward center all the time?
No, the net force in the radial direction is not zero. The net force in the radial direction provides an acceleration in the radial direction according to Newton's second law. That radial acceleration changes the direction of the velocity but does not change the speed so the stone goes around in a circle at constant speed.

kuruman said:
No, the net force in the radial direction is not zero. The net force in the radial direction provides an acceleration in the radial direction according to Newton's second law. That radial acceleration changes the direction of the velocity but does not change the speed so the stone goes around in a circle at constant speed.
This is in inertial frame.
In non inertial they cancel out or net force is non zero , outward(centrifugal)?

In the inertial frame the tension in the string is mass times acceleration and you write
##T=m\dfrac{v^2}{R}.##
In the non-inertial frame you move ##m\dfrac{v^2}{R}## to the left side of the equation and change its sign to get
##T-m\dfrac{v^2}{R}=0.## Then you rename ##m\dfrac{v^2}{R}## "centrifugal force" and you write
##T-F_{\text{centrifugal}}=0.##
In the non-inertial frame the net force is zero and the object does not accelerate.

The professor in the video calculates the acceleration of the person standing on the Earth in the inertial frame in which the person is rotating at constant speed around the center of the Earth just like the stone at the end of the string.

user079622
kuruman said:
In the non-inertial frame the net force is zero and the object does not accelerate.
Why object not accelerate, because frame accelerate together with object?
why we need introduce fictive forces if object dont accelerate?

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user079622 said:
Why object not accelerate, because frame accelerate together with object?
why we need introduce fictive forces if object dont accelerate?
I explained that to you in post #25. Suppose you are on a rotating platform with closed walls all around and you hold a string with a mass at the other end. The mass is on the frictionless floor of the platform and you are the center of the platform. You know that you are holding a mass at the end of a string which is under tension and you see that the mass is at rest with respect to you. You conclude that the net force on the mass is zero. You also know that one of the forces on the mass is the tension in the string and that it is directed towards you. What would you say is the other force that balances the tension and points away from you? Where does that force come from?

user079622 said:
In static normal force has equal magnitude as mg.
No, it does not. For a body siting on an inclined plane (static) the normal force is not equal to mg. For someone pushing horizontaly against a wall, the normal force is not equal to any mg. There is only one case out of many others when the normal foce may be equal to mg (body at rest on horizontal surface) This is not the rule.

user079622 said:
Why object not accelerate, because frame accelerate together with object?
Yes
user079622 said:
why we need introduce fictive forces if object dont accelerate?
To make Newtons 2nd Law work.

user079622 said:
So different frames get different results?
Different coordinate accelerations thus different net forces.

kuruman said:
You conclude that the net force on the mass is zero. You also know that one of the forces on the mass is the tension in the string and that it is directed towards you. What would you say is the other force that balances the tension and points away from you? Where does that force come from?
I would say centrifugal force, from nowhere.

@jbriggs444

centrp. acc for moon from revolution (r= moon radius +distanace to earth) = 0.0027 m/s2 and from rotation= 0.0000124 m/s2
Its weird that these result are not the same,because they rotate at same rate.

I dont need to added this two numbers to get total centripetal acceleration on dark side of moon?

Do I need add this two numbers if moon rotate around itself faster, 50rotation/per 1month?

centr. force due to revolution at visible side act away from moon surface and on dark side act toward surface?

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user079622 said:
@jbriggs444

centrp. acc for moon from revolution (r= moon radius +distanace to earth) = 0.0027 m/s2 and from rotation= 0.0000124 m/s2
Its weird that these result are not the same,because they rotate at same rate.

I dont need to added this two numbers to get total centripetal acceleration on dark side of moon?

Do I need add this two numbers if moon rotate around itself faster, 50rotation/per 1month?

centr. force due to revolution at visible side act away from moon surface and on dark side act toward surface?
No. Do not add them. That would be double dipping. If you consider the revolution of the moon about the earth as a rigid body, you have already accounted for all of the motion.

Dale
jbriggs444 said:
No. Do not add them. That would be double dipping. If you consider the revolution of the moon about the earth as a rigid body, you have already accounted for all of the motion.
But if moon rotate very fast,(so no synchronous rotation) we have now two sources of centripetal acceleration, revolution and rotation?
Why not then?

user079622 said:
I would say centrifugal force, from nowhere.
user079622 said:
why we need introduce fictive forces if object dont accelerate?

In the non-inertial rotating frame the mass at the end of the string is at rest so you say that a "a centrifugal from nowhere" must be acting on it. That force "from nowhere" is fictitious. You made it up because you wanted to satisfy Newton's first law because if the mass is at rest there must be a second force to cancel it. However in so doing, you violated Newton's third law. There is an "action" force on the mass but the mass can exert no "reaction" force on anything because it doesn't exist. Moving mass times acceleration to the left-hand side of Newton's second law does not make it a real force.

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jbriggs444

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