Normal force in an elevator | Forces and Newton's laws of motion | Physics | Khan Academy
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- Опубликовано: 19 июн 2011
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How the normal force changes when an elevator accelerates. Created by Sal Khan.
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am i the only who is thankful that his 2yr old son was interested in elevators
HE MUST BE 13 NOW IN 2021
@@fatimahhussaind Guess what, he is
@@fatimahhussaind 15 now
@@sulltancena7613 no
@@pinkman17_ 14 or 14 and 1/2
"We're only going to talk about vertical direction" Glad to know we're not dealing with Willy Wonka's GLASS elevator.
I love this
this comment is 9 years old. but is legendary and i have no idea how it doesnt have 20k likes
Lol
this is SOOO intuitive when trying ot udnerstand newtons second law of motion. Constant velocity = no net force, which totally explains why you feel nothing in an elevator going at constant velocity! Initially I struggled to completely grasp the idea, now it is mine, my mind has accepted it, thanks so much sal
Saved me future frustration on homework in physics class, thank you so much for this video
You have just saved my for my IB mock exams! All of your videos were very helpful and thanks a lot!!! :)
aw i wanna see sal's son
@nikola 1809 Its always good to look at these comments at different time periods
Why tho?
Bruh whenever his son needs help studying Sal will easily clutch up
@@manutebol956 lmao
hes 18 now
Simple "when the elevator accelerates your weight increase because of an increase in normal force and when the elevator decelerate normal force decreases and you feel light".
Sal: “My two-and-a-half year old son is _obsessed_ with elevators”
DieselDucy: “My true successor.”
High iq joke
my biggest problem before this video was elevator in dynamic .but after this , its easiest problem in physic for me . thanks khan🌹
The one dislike is my physics teacher
to clarify when you say how you feel are you referring to the force of the normal? the counteracting force to the downward force ( not the net force)
what an aamzing explanation!...................................thank you very much!
this was an interesting lesson!!! Thanks so much!!!!!!!
Thank you khan academy for your great video lessins.
I have a questiin pls : when the elevator was accelerating the net force was 20N upward , so shouldnt the person feels lighter since the net force is pushing him upward?? .
Thank you in advace
Thank you for the video.
I had a doubt regarding this.
If I am having an acceleration in the downward direction greater than acceleration due to gravity (> g), what will happen to the body?
According to the textbook, the body inside the elevator would start to fly inside the elevator.
But if we consider the forces in a situation where the lift accelerates with 2g in the downward direction, the normal force comes out to be N = -mg, which I am not able to visualize that easily. Kindly explain.
I love you... thank you for this video!
great explanation
thanks so much. you just helped me out a lot
Assuming the frictional force between them to be zero, what would happen if you pushed a block against a wall? How much force do you need to apply so it's stationary? Note: the force applied and the wall and everything are perpendicular to the earth's surface.
The Khan Academy voice is a core youth memory.
Sir ur great. This video helped a lot
thanks khan! amazing yet again
If you take the point (0,1) on the x-y coordinate system and connect it to the origin, that is a j unit vector. The i unit vector (not to be confused with the imaginary number) is the same thing, but for the coordinate (1,0). If you are familiar with the three-d coordinate system, the k unit vector is (0,0,1) connected to the origin
In the second example, the normal force is acting on the person at their center of gravity to move them, within the system, upward. Just to be clear, that normal force is increasing as a result of the torque on the pulley system (and the friction of the coil, and the force of gravity on the system as well) bringing the elevator upward right? So the pulley system has ungone an increase of torque to bring the system in the upward direction. The person within the system, is also subject to that same amount of force. Like in a car accident, that every object within the vehicle is subject to the same amount of force that the system (car) itself is subject to, so that if the car comes to a sudden stop, everything in the car continues to move until Newton's first law is "satisfied" and everything comes to a stop?
So technically, if an elevator was moving Upward really quickly, and the elevator had no top (think Willy Wonka minus the glass roof), and the elevator came to a sudden stop, that the person inside said elevator would be subject to the same amount of torque the pulley system was subject to, in an upward motion?
If I am correct, even in a super simplified view, because I am pretty new to this, that is wild 8O
WOW, that's all I can say. I now understand why I don't feel anything while am on the elevator at my school.
Lol when I was little i wasn't really obsessed with the elevator, I was obsessed with pressing the buttons on the elevator.
the j unit vector tell us we movin up to the verticle ligne,to emphasize it more. the special vectors i, j , k, are assigned to the x,y, and z axes. example:a vecor 3i represent a three unit vector in the +xdirection, while -5k represent a five unit vector in the -z direction.
Thank you Khan!
@Makro80 :
With all you respect I did not try to critize you excellent and educated videos; I just try to be more helpful than anything else.
I apreciated your effort and the good intension of your videos.
Thanks.
That's a subject i discussed with a friend. If the elevator goes down with 1 km/h and will suddenly hit a surface which immediately stops the motion: how many newtons are necessary to prevent the mass from crushing your bones?
If you jump in an elevator you would still fall down with the elevator's velocity minus your own jumping velocity.
Then immediately after that you will be subject to gravity again and eventually reach normal falling velocity.
why is the tendency to topple an object less when the applied force is more collinear with the frictional force against it
:OOOOOOOOOOOOOOOOOOOO scinenceeeeeeeeeeeeee i ma just starting to scratch you and i love itttttttttt.
what you have studied is g positive in downward direction and velocity is taken always positve..but here we always take g negative and positive v if it is in upward direction and negative v if it is downward direction...It does matter which method you use because finally negative sign will cancel.
I will be understand more , If you were our teacher!!
Yea I will to
I'm having some trouble with a Calculus 2 Problem involving Force and Work, and I was wondering if anyone could help me. The problem asks “How much work is done in lifting a 1.2 kg book off the floor to put it on a desk that is 0.7 m high? Use the fact that the acceleration due to gravity is g = 9.8 m/s^2.”
It says that the force is equal to the mass times acceleration, which is just mass times gravity, but wouldn't there be a moment of acceleration upwards like in the video? I'm having trouble wrapping my head around the force needed to lift the book being exactly equal to the books weight at all times...
hey use the fact that all energy contributes to work and calculate potential energy mgh which is your work
Thank you!
Thanks! But I don't get one part . Moving up is ?negative
In second case normal reaction force has to be greater than downward force for the toddler to accelerate. Had they been the same they would have cancelled each other and toddler wouldn't have moved fn>fg Fn - Fg=20N
Omg this is awesome!! My physics teacher is so incompetent this will SAVE me! Thanks so much.
same here
They all are
yes it should, he just couldnt be asked to write the negative sign down, besides, diagrams should always have a positive side, for example, when on a diagram there is a positive side to the east and a car is moving towards east it is moving towards the positive side and when the car suddenly changes direction to west then it is going to the negative side that is when you put the negative numbers, hope i helped, and yhmm correct me if im wrong, so that i might learn something too :)
i thought g should be positive as its going downwards? someone pls enlighten me.
what is the programe used here ???
please send me info about it plzzzzzzzzz
thanks man
shouldnt u take into account the tension in the string??
At 1:25, it should k hat, because j hat is in the direction of the y axis, and if you want the up and down direction, the k hat is the correct one.
I was obsessed with elevators when I was two too... or so ive been told
I actually calculated the weight that the child would be during the positive and negative acceleration, as if they were standing on a scale during these accelerations. Accelerating up the child would weigh 118kg and during the negative acceleration the child's weight would read 78kg. Is this a correct way to view this?
Wait, a son? I feel like if he wants anything, he will grow up not necessarily being given it, but know how to get it. This the same reason I like your videos so much, you don't just say that normal force is cosine theta, but you teach us how to get there and why it works.
It would be more interesting if you use powerpoint or other type of media presentation than those scribles that you are using.
In that way you would concentrate more on the concepts than the distrated movements of the mouse. JUST A SIMPLE OBSERVATION.
Wow, I also want a transparent elevator
Helped out a lot, thx!
Yeah it helps out now that quarantine has ended, if we were still in quarantine it would help-in
I have a question about the last example. Say that the toddler is accelerating at -10 m/s^2.
The net force would then be -100 N. And since the force of gravity on the toddler is -98 N, wouldn’t the normal force be -2N? How is this possible? What would happen in this situation? Would the toddler just be in free fall, not in contact with the ceiling nor the ground?
Assuming the negative means that it is accelerating downwards, the baby cannot accelerate downwards at more than -9.8 m/s^2 as it and the elevator would then be in free fall. At which point the elevator would no longer support the baby, so the normal force shrinks to 0 N. Hence, Fg = ma, mg=ma and the inertial and gravitational masses cancel resulting in a=g.
What would the normal force be if the elevator is moving downwards and decelerating?
+Maria MacArdle The normal force would be greater than the force due to gravity. A good way to think about it is essentially the same question, but with the direction you consider to be positive in the opposite direction, so that the force due to gravity (your weight) now acts in the positive direction and if you are decelerating then the force acting upwards must be greater than your weight and hence normal force > weight
just to clarify. the last scenario in this example shows an elevator accelerating downwards at -2m/s/s right? that's why normal force is less than gravitational pull...? because if it's moving upward and decelerating then 20N must be subtracted from the force due to acceleration before it started decelerating. am i right?
@@thelastcipher9135 -2 m/s2 j means that the vector head would be poinitng downwards....so its basically 2 m/s2 accl. downwards...and the second part of your doubt, had the elevator been at rest and start moving downwards with an |a|=2m/s2 then too the normal force would be 20 less than the weight....to better understand this....look at it like this, earth is always pullin everything towards it with an acceleration of -9.8 m/s2 j (minus sign implies downwards) normal forces are the reason why we are able to stand without being sucked up into the the earth....because they balance this pull of the earth..a person always has to stand on the floor of the lift, which can apply contact(normal) force. At the same time earth is pulling the person....so when elevator is moving upwards with an acceleration the person has to accelerate with the same accl and direction I.e. upwards and therefore it has to make the body move at same accl upwards with lift....for this it has to overcome the force of gravity mg and then apply a force ma f=mg+ma. Now no forces will act if the lift is in motion or moving with constant v, so it will judt balance the earths pull mg and f= mg. Now if the motion is of constant velocity upward and the lift starts acceleratin downwards then again, to keep person on floor the force will balance the force of gravity, then to keep the person accelerating in the same direction, will be lesser than weight since the weight helps in the case to accelerate the person....
How can the third scenario from the left happen without acceleration ? If there is velocity and there is displacement in J direction , there must be acceleration ??
Acceleration is the rate of change of velocity or how quickly your velocity is changing. At constant velocity, there is no acceleration. Things only accelerate when their velocity changes, hence when speeding up, slowing down, or changing direction.
My physics instructor had spend to the entire class to let us think through the the concept, which only takes eleven minutes long.
F = ma + mg (where a = acceleration and g = gravity. m is ofc. = mass)
Yay i remember your channel before the test, i suck at physics
Grade ?ur class?
i love you khan. that's about it.
thanks
Well it is v2=2as+u2 where bot u and a is 0 hence in bot cases therefore 1st. and 3rd case arr same
What about when mass is not given, and you want to find acceleration and which direction that acceleration is in?
You can determine the direction of the acceleration from the direction of the net force as Newton's 2nd says that they are always in the same direction, however, in this case, without the mass, you will not be able to calculate the magnitude of the acceleration.
5:48
in the first 0.6s . a passenger in the elevator is holding a 3kg package by a vertical string . what is the tension in the string during the acceleration provest.
27N?
what is J direction??
@koreankid28 Actually i = X axis j = Y axis k = Z axis. Since this problem only functions in 1 dimension (up, down a.k.a. y axis) then all you would need is the j notation.
Sir plzz explain monkey climbing on pulley concept
so in this case do you mean that you feel 20 newtons lighter when the system is declereating etc
And nothing when u fall
The balance is it measuring the weight or the reaction
And the force due to acceleration is it upward or downward
I think it should be up when the acceleration is downward
The balance measure the normal force acting
on the person
@loonycassarole Ah, my mistake. I see. Thank you for the correction!
I wonder if a really fat person got into an elevator can have his legs broken. I.e. if the his legs can buckle due to high compression force combining the affect of acceleration and weight.
MilitanT07 in the name of science you should test that
your two year old son is going to be a genius.
I would very much appreciate if someone could give me an explanation of how is it physically
possible that the net force is 0 when the elevator goes upwards in a constant velocity. obviously when the elevator starts to accelerate force is being applied to it according to newton's second law, however when the velocity is constant (2m/s in this case) shouldn't force still be applied to the elevator constantly in order for the elevator to move up as the elevator is not in a vacuum but on earth? for example newton's second law would make sense to me in
a vacuum since there is no friction, air resistance nor the force of gravity in a true vacuum thus there are no opposing forces to the object's constant velocity so the object will remain in constant velocity. However if the elevator is here on earth there are constant forces that are being applied on it so in order for it to move in constant velocity it
has to overcome these forces, as these forces are obviously being applied on it always in constant velocity as well. These forces are gravity, and the friction that the elevator has with the cable it is attached to. (and perhaps other forces that I'm not aware of). So in order for the elevator to move in a constant velocity the elevator must overcome these forces that are being applied on it so a force must be applied constantly on the elevator in order for it to move constantly? I've just started to learn Physics and as a beginner I sometimes find myself
wondering and bothered about Newton's laws and if they are being applied correctly in a scenario like the one given in this video. Isn't Newton's second law only be true if we assume that the object is in a true vacuum but if the object is here on earth for example constant velocity doesn't automatically mean no force is being applied on it? Or am I missing something? I know It's a very long comment but i wanted to give my full perspective and assumptions on the matter to get a comprehensive answer. Anyhow if anyone would actually read this and can just affirm or disprove my assumption i would be very grateful, thanks in advance!
The scenario posed is definitely simplistic but according to Newton's Second Law (F=ma) so if a force was applied to the elevator, the velocity would change (not remain constant) as the elevator would accelerate due to the force applied.
Hello Stav, you're confusing Force with NET FORCE. Newton's 2nd doesn't say that acceleration will occur when any force force is applied, only if a NET FORCE is applied. So in this case of the elevator, when it is moving at constant velocity and therefore not accelerating, the downward force of gravity (as well as any frictional effects) are balanced by the upward force acting upon the elevator so that there is no NET FORCE and therefore no acceleration. Newton's 1st Law which builds on Galileo's Law of Inertia says that an object will continue in its same state of motion UNLESS acted upon by a net force. Forces may be acting, but as long as they are balanced, no acceleration occurs.
why we dont add -20N to the downward direction why we take it from normal force????
In this scenario, the elevator is massless. Do not worry about the weight of the elevator since it becomes more complicated. Just know that the 98 newtons represents the force of gravity only acting on you.The force of gravity on an object is always the same. If the elevator is going up but slowing to a stop, that means the downward force on you from the elevator must be greater than the upward forced exerted by the elevator on you so that acceleration would point opposite the direction of the elevator’s or your velocity. Therefore, you would slow down if the acceleration points down. The reason why the normal force is less is simply because the elevator literally is pushing you up with less force so that you would slow down to a stop as you go up.
Look at the elevator as someone pulling a box up. To lift it, they'll need to pull at it with a force 20N greater than weight force upwards. Now if they want to let the box (elevator) go down, they wouldn't need to push it 20N downwards, but let go of it slightly (apply 20N less than weight force upwards)
EXACTLY! when i first get into the elevator and it accelerates downwards towards the ground floor for that short moment i feel lighter, and then when i decelerate as i come to a stop i feel heavier lol idk i guess I've confused myself
What would the normal force be if they both drop at the same time, should it be 0N?
If they are both in Free Fall, then yes, the normal force would be 0 N as the elevator floor is not supporting the rider as they free fall.
What if the elevator has weight?
Why would anyone even dislike this wth :0
@lubime10 I look at it as more of something that puts ur eye on what he is talking about at that exact moment :)
If anyone knows the answer, please, explain it
@ChFe95 thx alot m8 now it makes sense
in case you haven't got an answer already; It just means 1 unit in the y direction.
omg his son!!
Sals son is probably the smartest kid ever
His 2yrs old toddler is 14 now 😮
what is the thing app/software he uses to draw stuff
Jacloch Plays right I have no idea how he's able to write and draw so well with a mouse
How does he draw that good? I mean, it's not by a mouse, isn't it?
It's on a drawing tablet with a stylus... pretty much exactly like writing with a pen on a paper
Thanks! Just like I thought)
Reeta Singh okay good I thought he was some sort of inhuman creature drawing with a mouse
Sarcalogos Tortolero he used to in the beginning, after he became rich enough, he got himself a better system
whats that j thingy
i just dont understand when is normal force perpendicular and when it is equal to weight?...please answer me
From what I understand, the normal force (N) will always be perpendicular on all surfaces (flat and incline). As for when it equals to weight, it isn't always like that. If you have a book on a flat surface which is not accelerating in the y direction, there will be only two forces acting on it making the weight equal to the book which makes N=Fgravity=W=mg=ma. But if you push the same book down there will be another force acting on it but N will be the same but it will not be equal to the gravity force and the force that's pushing [N is not equal to Fgravity(W)+F]. I hope that's not confusing. If I got it wrong everyone can correct my explanation so that other people don't get it wrong.
ibrahim amanullah the normal force is equal to the weight (in magnitude) if the object is on a flat surface let's say a table. But if the surface has an inclination, then the normal force would be perpendicular to the surface, while the weight or gravitational force would still be pointing downwards almost parallel to the inclined surface (depending on the angle of inclination ) in this last case, the normal force and the gravitational force wouldnt be the same. If you were to draw a cartesian plane taking the object as the origin, and decompose the vectors of all of the forces acting on the object, you would normally get to the conclusion that the normal = mg cos A
But why i feel heavier when i accelerate upward ??
i know the force of the elevator on me will be greater than the force of the gravity on me, but i am still confused why i feel heavier. can u show in mathematical method ?
billy chau to accelerate upwards you need a resultant force right - Newton’s first law. The fact that youre accelerating upwards means there is a resultant force upwards.
Therefore the normal force and weight are no longer in equilibrium. Your weight stays the same but the normal contact force on you increases.
You ‘feel’ heavier when the normal force is greater than your weight. You feel lighter when the normal force is less than your weight. If the normal force on you is zero than youll feel weightless.
remember that Flift - Fg = ma
Does this mean that weight is equal to normal force?
only when in stationary
So the elevator it self has 0 mass?
Well, I guess that you make the evelator mass 0 to keep it simple
What is the "j unit vector"? I understood everything besides that.
Miguel Soria he explained it in the video it's not important
How y get th 78 N at th end please!?
🤔🤔
His son is now 12 years 😂😂❤
Hi! Thanks for the video. It was for me clear almost till the end. But then you said: "If the elevator is accellerating, you feel heavier and if the elevator is decelerating, you feel lighter". This would be against the idea of feeling weightless in an free falling elevator, right?
Why can someone put a scale in an elevator and let the elevator accelerate and measure the weight? So we could confirm that...
why haven't we took the force ,(applied by the machine or whatsoever we would have used to pull up the elevator) ,into consideration?? kindly help😑😑😭😭
Because Sal is talking about the forces on the rider, not on the elevator itself.
y would you leave a todler in an elevator ALONE!?
you are the equivalent of a god.