Deal with problem of force

[Primarygun]

Hello, I am here to seek for an answer for my question.

In general, if a lift is disconnected with the rope, it falls with the acceleration about 9.8 m/s2. If a man is inside, he will fall at that acceleration too.

I know that calculation

[MATH] F_n = ma+mg [/MATH]

That states that when a=g, the two objects cannot touch each other, so there are no normal force.

1. What is observed if a>g?

It is known that the normal force and the apparent weight are negative, what does it mean?

 

2. What's the normal force? I do not totally understand this word.

I think it's action-reaction force. The reaction force but I suspected why there are no movement by the objects.

 

3. What does the [MATH] \Sigma [/MATH] mean?

4. Is the net force not included the frictional force?

 

Seek for every great answer

[pulkit]

1. What is observed if a>g?

It is known that the normal force and the apparent weight are negative, what does it mean?

 

Since force equation is vectorial, you need to see direction and can't just apply F=mg+ma, it can also be F=mg-ma.

There is nothing like negative weight, in case of acc > g, the man still falls with acc g, henece he will lose contact with the floor and move upwards, eventually hitting the cieling. The very concept of apparent weight is redundant, as the man in his refrence frame will not feel any force pulling him down, instead rather he will feel being pushed up.

 

2. What's the normal force? I do not totally understand this word.

 

When two bodies are in contact, the force they feel due to each other perpendicular to contact plane is called normal force. It is equal and opposite on both bodies. Normal is used becoz of direction (perpendicular to contact plane).

 

3. What does the "sigma" mean?

 

Sigma is a symbol that means sum of.

 

4. Is the net force not included the frictional force?

 

I can't understand your question here.

 

I have one little question of my own, how do u put all those formulas and symbols into your post ?

[JaKiri]

Read the sticky on LaTex.

[pulkit]

thanx for the help ....

[Primarygun]

Since force equation is vectorial' date=' you need to see direction and can't just apply F=mg+ma, it can also be F=mg-ma.

There is nothing like negative weight, in case of acc > g, the man still falls with acc g, henece he will lose contact with the floor and move upwards, eventually hitting the cieling. The very concept of apparent weight is redundant, as the man in his refrence frame will not feel any force pulling him down, instead rather he will feel being pushed up.

 

 

 

When two bodies are in contact, the force they feel due to each other perpendicular to contact plane is called normal force. It is equal and opposite on both bodies. Normal is used becoz of direction (perpendicular to contact plane).

 

 

 

Sigma is a symbol that means sum of.

 

 

 

I can't understand your question here.

 

I have one little question of my own, how do u put all those formulas and symbols into your post ?

 

I used latex.

I have one more question.

When a<g, the man does not float. This is because the acceleration just only reduce the effect of gravitational force from earth.

When a=g, the effect of gravity is totally reduced.

But when a>g, the man and the lift should fall into the same speed and same acceleration.

1.How come he can move upward in the lift and finally hit the ceiling if the acceleration continue to increase?This is when the calculation came negative.

[pulkit]

When a>g, it is the list that would be falling with this acceleration. As the man has no physical link with the lift, he will not fall with the same acceleration, he would feel literally as if the floor has been pulled away from his feet, and as you can very well imagine in such a case, he will fall only with an accleration g. Remember that gravity is the only force acting on him in the refrence frame of the Earth, so he can't go any faster than g. It is only once his head hits the cieling that the lift starts to exhert on him a fore and then forces him to fall with accleration greter then a. In the intermediate period of leaving floor and reaching cieling, to an observer on the ground it will appear as if he is falling down with an acceleration of g.

 

Do you know about frames of refrence and more importantly about pseudo forces ?

[Primarygun]

When a>g' date=' it is the list that would be falling with this acceleration. As the man has no physical link with the lift, he will not fall with the same acceleration, he would feel literally as if the floor has been pulled away from his feet, and as you can very well imagine in such a case, he will fall only with an accleration g. Remember that gravity is the only force acting on him in the refrence frame of the Earth, so he can't go any faster than g. It is only once his head hits the cieling that the lift starts to exhert on him a fore and then forces him to fall with accleration greter then a. In the intermediate period of leaving floor and reaching cieling, to an observer on the ground it will appear as if he is falling down with an acceleration of g.

 

Do you know about frames of refrence and more importantly about pseudo forces ?[/quote']

 

Very sorry, I am going to learn the newton's force.

Oh, so the man is all the time with a=g except when pushing from the ceiling.

But, why does he has this gravity force which is the same as the earth's gravity force? Is the lift give him?

I'm totally immersed in this frame of refrence problem.

Hope you can help me!

[Primarygun]

oh maybe I get some of it.

Is it like when we sit in a car.

Regardless the direction of y forces, when the car a=0, (no matter v is 1000 or 0 or others), we can sit inside safely and happily without the change in position.

Oh, I got into a trap.

Notice a but not v, I get why we are unchange in the car.

But, I don't know why I can't bring this concept into the lift.

Why do we stay at the middle part of the lift when a=g?

I know the calculation, maybe still confused in the normal force.

Why don't our body fall with the speed a=g? And thus, we won't leave the base of the lift.

Can anyone tell me by concept?

[JaKiri]

We'll fall at g. The lift may not.

[pulkit]

Do not get confused.

 

Analyse the man when he is not in contact with the cieling yet, but he has left the floor.

 

The only force acting on him is the grvity of Earth, i.e., mg so he is acctualy falling down with acc g (slower than the lifts, which is why he hits the cieling) for some observer on Earth.

 

Why I ask about your knowledge of frames of refrence is because all this stuff about apparent weight comes into picture if you analyse the problem with respect to the lift. It is easier to understand with respect to lift if you know what pseudo forces are (The formula you mentioned rite in the beginning then just becomes obvious).

 

If you are still confused, forget about all the apparent weight stuff, just look at the problem as if you are observing the guy while sitting on the ground. The ONLY force on him in that case is mg, hence the acc of g.

 

About the car thing, its the same concept in a diffrent orientation. When there is no acceleration, there is no force, When it accelerates forward lets say, you feel a backward push (exactly like the lift, when it falls down with an acceleration, you feel an upward push and vice versa)

[Primarygun]

ya sorry, I get confused because I was not very clear in concept with weight and normal force.

Are they vector measure?Now I don't think so, right?

I get troubled in

the [MATH]mg[/MATH]

because I always thought it was the weight and should act towards the earth.

However, in the equation, it's the normal force, so act towards the sky and the [MATH]ma[/MATH] act towards the core, so the equation is correct in concept.

However, one thing I am not very sure is the normal force.

Why can scientist think a force which is present that in opposite direction as the [MATH]W[/MATH]?It seems it doesn't act like reaction force, but it does. Why?

 

After I thought again, the highlighed words are wrong.

The ma is the force that lift by the acceleration.

That means when the lift fall with a=g,

the normal force for him is the normal force for his actual weight

minus the force by the acceleration.

 

 

I think I am currently trap in the reference frames. I need to learn it shortly first

When we pull a thing up, its normal force decreases.

When we push a things down, its normal force increases.

But, why it is reversed for the man in the lift?

When the lift falls, the lift is like being pushing down.

But, why its normal forces decreases?

Why do the rope pull the lift make the lift heavier but make the man inside lighter?

[Primarygun]

Does the lift have weight while it's falling?

Does it have the normal force? ( with air?)

[pulkit]

Does the lift have weight while it's falling?

Weight is defined as force with which earth pulls an object towards it, so yes it does have weight. Technically, weight is measured in Newtons. We express it in Kgs, it is acctualy not Kilograms but KilogramWieght (the unit).x Kilogramweight meaning x * 9.8 newtons.Maybe that would make the concept of weight clearer. Remember wight is a force and not a mass.

 

Does it have the normal force? ( with air?)

 

Any normal force with air is neglected for 2 reasons :-

1) It is inconsequentially small

2)It acts on all four sides and cancels itself out

[Primarygun]

oh I see. This acts like body in car. acceleration (+) make him behind