"We're only going to talk about vertical direction" Glad to know we're not dealing with Willy Wonka's GLASS elevator.

am i the only who is thankful that his 2yr old son was interested in elevators

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

Sal: “My two-and-a-half year old son is _obsessed_ with elevators” DieselDucy: “My true successor.”

You have just saved my for my IB mock exams! All of your videos were very helpful and thanks a lot!!! :)

Saved me future frustration on homework in physics class, thank you so much for this video

aw i wanna see sal's son

The one dislike is my physics teacher

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".

The Khan Academy voice is a core youth memory.

I love you... thank you for this video!

this was an interesting lesson!!! Thanks so much!!!!!!!

my biggest problem before this video was elevator in dynamic .but after this , its easiest problem in physic for me . thanks khan🌹

great explanation

what an aamzing explanation!...................................thank you very much!

thanks khan! amazing yet again

thanks so much. you just helped me out a lot

Thank you Khan!

WOW, that's all I can say. I now understand why I don't feel anything while am on the elevator at my school.

Sir ur great. This video helped a lot

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)

I will be understand more , If you were our teacher!!

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.

Omg this is awesome!! My physics teacher is so incompetent this will SAVE me! Thanks so much.

Lol when I was little i wasn't really obsessed with the elevator, I was obsessed with pressing the buttons on the elevator.

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!

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.

@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.

:OOOOOOOOOOOOOOOOOOOO scinenceeeeeeeeeeeeee i ma just starting to scratch you and i love itttttttttt.

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

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.

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

thanks man

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 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.

Wow, I also want a transparent elevator

Helped out a lot, thx!

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?

What would the normal force be if the elevator is moving downwards and decelerating?

I was obsessed with elevators when I was two too... or so ive been told

your two year old son is going to be a genius.

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 :)

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.

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

i love you khan. that's about it.

Yay i remember your channel before the test, i suck at physics

thanks

Thanks! But I don't get one part . Moving up is ?negative

Sals son is probably the smartest kid ever

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?

F = ma + mg (where a = acceleration and g = gravity. m is ofc. = mass)

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...

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.

why is the tendency to topple an object less when the applied force is more collinear with the frictional force against it

5:48

@ChFe95 thx alot m8 now it makes sense

shouldnt u take into account the tension in the string??

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?

My physics instructor had spend to the entire class to let us think through the the concept, which only takes eleven minutes long.

@loonycassarole Ah, my mistake. I see. Thank you for the correction!

why we dont add -20N to the downward direction why we take it from normal force????

so in this case do you mean that you feel 20 newtons lighter when the system is declereating etc

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.

@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.

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

The balance is it measuring the weight or the reaction And the force due to acceleration is it upward or downward

Why would anyone even dislike this wth :0

What about when mass is not given, and you want to find acceleration and which direction that acceleration is in?

How does he draw that good? I mean, it's not by a mouse, isn't it?

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 is J direction??

Sir plzz explain monkey climbing on pulley concept

His son is now 12 years 😂😂❤

in case you haven't got an answer already; It just means 1 unit in the y direction.

His 2yrs old toddler is 14 now 😮

i thought g should be positive as its going downwards? someone pls enlighten me.

omg his son!!

@lubime10 I look at it as more of something that puts ur eye on what he is talking about at that exact moment :)

What would the normal force be if they both drop at the same time, should it be 0N?

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.

If anyone knows the answer, please, explain it

What if the elevator has weight?

what is the programe used here ??? please send me info about it plzzzzzzzzz

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 ??

remember that Flift - Fg = ma

So the elevator it self has 0 mass? Well, I guess that you make the evelator mass 0 to keep it simple

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.

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!

What is the "j unit vector"? I understood everything besides that.

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?

you are the equivalent of a god.

Does this mean that weight is equal to normal force?

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 ?

How y get th 78 N at th end please!?

Thanks a lot

what is the thing app/software he uses to draw stuff

when I found out that Sal does physics too *insert Pikachu meme here*

y would you leave a todler in an elevator ALONE!?

i just dont understand when is normal force perpendicular and when it is equal to weight?...please answer me