I Why normal force is not equal to gravity?

[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..

 

[nasu]

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.

 

[user079622]

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

 

[phinds]

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?

 

[A.T.]

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.

 

[Dale]

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.

 

[user079622]

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

 

[user079622]

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?

 

[user079622]

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..

 

[kuruman]

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.

 

[A.T.]

I dont agree that normal foce is real force, that force dont even exist...it is math construction.

Who taught you that nonsense?

 

[user079622]

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?

 

[jbriggs444]

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.

Not really worth worrying about.

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.

 

[kuruman]

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.

 

[user079622]

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..

 

[user079622]

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?

 

[A.T.]

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

 

[user079622]

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?

 

[kuruman]

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.

 

[user079622]

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?

 

[A.T.]

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.

 

[user079622]

No, if they would cancel, the Earth would not stay in orbit, but move along a line.

So different frames get different results?

 

[kuruman]

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.

 

[user079622]

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)?

 

[kuruman]

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]

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?

 

[kuruman]

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?

 

[nasu]

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.

 

[A.T.]

Why object not accelerate, because frame accelerate together with object?

Yes

why we need introduce fictive forces if object dont accelerate?

To make Newtons 2nd Law work.

 

[A.T.]

So different frames get different results?

Different coordinate accelerations thus different net forces.

 

[user079622]

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.

 

[user079622]

@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?

 

[jbriggs444]

@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.

 

[user079622]

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?

 

[kuruman]

I would say centrifugal force, from nowhere.

Exactly. You asked

why we need introduce fictive forces if object dont accelerate?

and you answered your own question.

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.

 

[user079622]

here is an "action" force on the mass but the mass can exert no "reaction" force on anything because it doesn't exist. .

action force is centripetal?

What you say about my post #34?

 

[kuruman]

action force is centripetal?

What you say about my post #34?

I say that's a different physical situation. I am considering a simpler case to answer your question

why we need introduce fictive forces if object dont accelerate?

and I have answered that.

 

[jbriggs444]

But if moon rotate very fast,(so no synchronous rotation) we have now two sources of centripetal acceleration, revolution and rotation?
Why not then?

You would still be double dipping. The two sources are still not independent.

 

[user079622]

You would still be double dipping. The two sources are still not independent.

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

 

[A.T.]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal force as ball that is only spinning 1000rpm?

This sound impossible.

What do mean by "has same centripetal force"? Maybe draw a diagram and label is properly.

 

[user079622]

What do mean by "has same centripetal force"? Maybe draw a diagram and label is properly.

I mean centripetal acceleration,

Point A and B has same centripetal acceleration?

 

[jbriggs444]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

Be careful. Are you specifying a "sidereal" rotation rate? Or a rotation rate relative to a frame that rotates with the orbit?

Let us assume sidereal. Let us speak of a bug perched on a rigidly spinning ball in a circular orbit about the center of a carousel.

We calculate the centripetal acceleration required to keep the bug on the surface of the spinning ball without accounting for the orbital motion. Simple. That's $\omega^2r$ where r is the ball's radius and $\omega$ is the rotation rate of the ball.

Now we go to add in the orbital motion.

The center of the ball is at one orbital radius. The outer surface of the ball is at a larger orbital radius. The acceleration of the bug relative to the ball due to the orbital motion would seem to be the difference of the two. But wait a minute... If we used this calculation, it would be like assuming that the bug is moving in lock step with the carousel. But it is not. The ball is spinning. The bug is moving with the ball. And we've already accounted for the acceleration associated with that spin.

So to do the calculation correctly, we would need to do the orbital calculation as if the ball has zero sidereal rotation. In that case, the bug's acceleration due to orbital motion would exactly match the acceleration of the center of the ball.

 

[user079622]

The bug is moving with the ball. And we've already accounted for the acceleration associated with that spin.

Tangential velocity and radius of point A and B are different.

why you calculate like there are same?

it could be 2 turn table,frame earth, inertial

small table r=1m, 1000rpm
big table r=10m, 500rpm

point A at small table, ac=10966 m/s2 , tangential velocity=104 m/s

now I am first find what is tangetianl velocity at point B if table is lock to table, so r=11m tan. velocity=576 m/s

I add both tang. velocity 104+576=680 m/s is in point B,when table spin unlock from big table
ac=42036 m/s2

I am not sure if I correct calculate tangential velocity at point B...I strugle in relative motion.
Velocity 104 and 576 are both relative to earth, so I think cant add them,
I turn out that velocity at B due to just "revolution"(lock small table to big one) is faster then velocity of A, I must get velocity at B when small table is spinning

 

[user079622]

@jbriggs444
I post task to Homework help, where it belong.

 

[jbriggs444]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

What is providing the centripetal acceleration? Gravity or a carousel?

 

[user079622]

What is providing the centripetal acceleration? Gravity or a carousel?

carousel

 

[jbriggs444]

So not relevant to the normal force on the moon.

 

[user079622]

So not relevant to the normal force on the moon.

But I didnt mention normal force for this task.

 

[PeterO]

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..

When an object is placed on a horizontal bench in the class room, we have several pairs of forces, which are reaction forces as mentioned in Newtons 3rd law.
1: The object pushes down on the surface -- the surface pushes up on the object.
2: The Earth pulls down on the object -- the object pulls up on the Earth (effect of gravity)
3: The bench pushes down on the Earth -- the Earth pushes up on the bench.

Each force in those pairs are equal in magnitude and opposite in direction.

If you were to place a 1 kg mass on top of your object, the pair of forces mentioned in #2 do not change, but the upward force on the object MUST increase, as it now supports both the object AND the 1 kg mass

 

[PeterO]

16:23

The presenter made a very misleading statement.
He said the force of gravity had to be slightly more than the normal (reaction) force.
The size of the force of gravity is determined by the mass of the two objects (Earth and Person) and the separation of their centres of mass. We cannot alter any of those, so the force of gravity does not change.
He should have said "the reaction force is slightly less than the force of gravity" (giving the net downward force)
Note: another way to reduce the Normal force, would be to attach a Helium balloon to you - but not one big enough to lift you off the round. (Perhaps a couple of those balloons people take to parties).