I will now introduce you to

the concept of tension. So tension is really just the

force that exists either within or applied by

a string or wire. It’s usually lifting something

or pulling on something. So let’s say I had a weight. Let’s say I have

a weight here. And let’s say it’s

100 Newtons. And it’s suspended from this

wire, which is right here. Let’s say it’s attached to

the ceiling right there. Well we already know that the

force– if we’re on this planet that this weight is being

pull down by gravity. So we already know that there’s

a downward force on this weight, which is

a force of gravity. And that equals 100 Newtons. But we also know that this

weight isn’t accelerating, it’s actually stationary. It also has no velocity. But the important thing is

it’s not accelerating. But given that, we know that the

net force on it must be 0 by Newton’s laws. So what is the counteracting

force? You didn’t have to know about

tension to say well, the string’s pulling on it. The string is what’s keeping

the weight from falling. So the force that the string or

this wire applies on this weight you can view as

the force of tension. Another way to think about it

is that’s also the force that’s within the wire. And that is going to exactly

offset the force of gravity on this weight. And that’s what keeps this point

right here stationery and keeps it from

accelerating. That’s pretty straightforward. Tension, it’s just the

force of a string. And just so you can

conceptualize it, on a guitar, the more you pull on some of

those higher– what was it? The really thin strings that

sound higher pitched. The more you pull on it,

the higher the tension. It actually creates a

higher pitched note. So you’ve dealt with

tension a lot. I think actually when they sell

wires or strings they’ll probably tell you the tension

that that wire or string can support, which is important if

you’re going to build a bridge or a swing or something. So tension is something that

should be hopefully, a little bit intuitive to you. So let’s, with that fairly

simple example done, let’s create a slightly more

complicated example. So let’s take the same weight. Instead of making the

ceiling here, let’s add two more strings. Let’s add this green string. Green string there. And it’s attached to the

ceiling up here. That’s the ceiling now. And let’s see. This is the wall. And let’s say there’s another

string right here attached to the wall. So my question to you is, what

is the tension in these two strings So let’s call

this T1 and T2. Well like the first problem,

this point right here, this red point, is stationary. It’s not accelerating in

either the left/right directions and it’s not

accelerating in the up/down directions. So we know that the net forces

in both the x and y dimensions must be 0. My second question to

you is, what is going to be the offset? Because we know already that

at this point right here, there’s going to be a downward

force, which is the force of gravity again. The weight of this

whole thing. We can assume that the wires

have no weight for simplicity. So we know that there’s going

to be a downward force here, this is the force of

gravity, right? The whole weight of this entire

object of weight plus wire is pulling down. So what is going to be the

upward force here? Well let’s look at each

of the wires. This second wire, T2, or we

could call it w2, I guess. The second wire is just

pulling to the left. It has no y components. It’s not lifting up at all. So it’s just pulling

to the left. So all of the upward lifting,

all of that’s going to occur from this first wire, from T1. So we know that the y component

of T1, so let’s call– so if we say that

this vector here. Let me do it in a

different color. Because I know when I draw these

diagrams it starts to get confusing. Let me actually use

the line tool. So I have this. Let me make a thicker line. So we have this vector

here, which is T1. And we would need to figure

out what that is. And then we have the other

vector, which is its y component, and I’ll draw

that like here. This is its y component. We could call this T1 sub y. And then of course, it has an

x component too, and I’ll do that in– let’s see. I’ll do that in red. Once again, this is just

breaking up a force into its component vectors like we’ve–

a vector force into its x and y components like we’ve been

doing in the last several problems. And these are just

trigonometry problems, right? We could actually now, visually

see that this is T sub 1 x and this is

T sub 1 sub y. Oh, and I forgot to give you an

important property of this problem that you needed to

know before solving it. Is that the angle that the

first wire forms with the ceiling, this is 30 degrees. So if that is 30 degrees, we

also know that this is a parallel line to this. So if this is 30 degrees,

this is also going to be 30 degrees. So this angle right here is also

going to be 30 degrees. And that’s from our– you know,

we know about parallel lines and alternate

interior angles. We could have done

it the other way. We could have said that if this

angle is 30 degrees, this angle is 60 degrees. This is a right angle,

so this is also 30. But that’s just review

of geometry that you already know. But anyway, we know that this

angle is 30 degrees, so what’s its y component? Well the y component,

let’s see. What involves the hypotenuse

and the opposite side? Let me write soh cah toa at the

top because this is really just trigonometry. soh cah toa in blood red. So what involves the opposite

and the hypotenuse? So opposite over hypotenuse. So that we know the sine– let

me switch to the sine of 30 degrees is equal to T1 sub y

over the tension in the string going in this direction. So if we solve for T1 sub y we

get T1 sine of 30 degrees is equal to T1 sub y. And what did we just say

before we kind of dived into the math? We said all of the lifting on

this point is being done by the y component of T1. Because T2 is not doing any

lifting up or down, it’s only pulling to the left. So the entire component that’s

keeping this object up, keeping it from falling

is the y component of this tension vector. So that has to equal the force

of gravity pulling down. This has to equal the

force of gravity. That has to equal this

or this point. So that’s 100 Newtons. And I really want to hit this

point home because it might be a little confusing to you. We just said, this point

is stationery. It’s not moving up or down. It’s not accelerating

up or down. And so we know that there’s a

downward force of 100 Newtons, so there must be an upward force

that’s being provided by these two wires. This wire is providing

no upward force. So all of the upward force must

be the y component or the upward component of this force

vector on the first wire. So given that, we can now solve

for the tension in this first wire because we have

T1– what’s sine of 30? Sine of 30 degrees, in case you

haven’t memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal

to 100 Newtons. Divide both sides by 1/2

and you get T1 is equal to 200 Newtons. So now we’ve got to figure out

what the tension in this second wire is. And we also, there’s

another clue here. This point isn’t moving left

or right, it’s stationary. So we know that whatever the

tension in this wire must be, it must be being offset by a

tension or some other force in the opposite direction. And that force in the opposite

direction is the x component of the first wire’s tension. So it’s this. So T2 is equal to the

x component of the first wire’s tension. And what’s the x component? Well, it’s going to be the

tension in the first wire, 200 Newtons times the cosine

of 30 degrees. It’s adjacent over hypotenuse. And that’s square root

of 3 over 2. So it’s 200 times the square

root of 3 over 2, which equals 100 square root of 3. So the tension in this wire is

100 square root of 3, which completely offsets to the left

and the x component of this wire is 100 square root of

3 Newtons to the right. Hopefully I didn’t

confuse you. See you in the next video.

Wow, wow, I never knew all these concepts are so logical and I guess that`s the problem with most of nowadays teachers, they just fill the table wih formulas without telling us where it all came from… But you sir, you mannage to do this and I must say: "Thank you a lot, sir!" 🙂

woooondeeeerfuuul >> Bless you!

this video is courtesy of potato cams

Very well explained. Tension can seem like some abstract thing, but it's really just a force, and often students aren't made aware of that until they see real-life examples.

Thanks a lot for your good work

I dont understand how you got T1 = 200 N from T1 * 1/2 = 100 N … shouldn't it be 150N?

That does NOT look like 30 degrees, it looks more like 60 degrees!

How is T2 +T1x=100N ?

Thanks, this lifted my stress as my teacher didn't explain it very well, and this explains it in such a simple way, thanks

Could you make a video with a problem that has a string with mass? Thanks

Audio's pretty bad. In middle volume, I can hardly hear anything.

It's so confusing

You kind of have a similar voice to Sparc Mac

is there a videos that does not involve trigonometry

you are good

Since the body is stationary, the sum of all forces is equal to zero? Is it so?

I want to learn physics but it's just confusing :'(

240p. Really? (/.-)

Haha!

SOH CAH TOA

in blood red.

Like the blood in my nose when I couldn't get this topic when I was in high school

hey!!! soh cah toa was a method i used and made when i was in high school.!!!

isnt force of gravity equal to mass times 9.8 so the fg would be 980N in his case and not 100

good

Thank you for this! made perfect sense for once

Appreciate all this but I must say it is SO PAINFUL listening to Sal STUTTER throughout. I think: "How does this man expect to supposed to get me through Physics when he struggles to form each and every effing sentence during the entire attempt???"

how is that angle 30 degree's? I always thought that the angle closest to the 90 degree angle is 60 degrees and the angle farthest makes the 30 degree angle.

A string is stretched by two equal and opposite forces 10N each. The tension in the string is…..?

shouldn't the Fg be: -100N ?

because the Normal force will be: +100N

@Sal Khan. Sir what application do you use to create these videos?

bro your videos are the best

How could a weight of 100N possibly cause 200N of tension on a string? I don't think this is correct…

its very helpfull for students

~◇~ I'm a grade 8 student and I still can't get this topic…………also it is confusing ;-;

recording quality uhh what happened

ur drawing is pathetic , ur concepts are good but ur explanation`s poor as hell

What about spring tension

Great video! Sal is the real man 🐉

8:50

My new PHYSICS SOLVING APP.More then 150+ formulas,Solves for any variable you want,Covers up all physics.download now.https://play.google.com/store/apps/details?id=com.physics.lenovo.myapplication

why cos 30 only was taken in the last step…

This video made me understand this concept a lot better. I do have a question though. Sal used a triangle to the right of the T_1y vector. Couldn't he have used the same triangle on the other side of the vector where the side parallel to the y component would've been straight up from the weight? The angle is the same as he stated, and the use of sin is still the same because the y component is still the opposite side from the angle.

So if the angle of the T_1 wire was 45 degrees, it would be holding a force of 141N. This is because the math is 100 / sin(45). If it was 60 degrees, it would have a force of 115N. If it was, lets say 89 degrees, it would have an upward force of 100.02N. That doesn't seem right. How come the tension on the rope grows as the angle grows from 0 to 30 degrees, but then decreases once it exceeds 30 degrees?

U r so boring

AP Physics got me like “A student is going to hang himself because of the overwhelming topic that he fails every test and paper in. He is trying to find a rope that can support his weight. The student weighs 50kg. What is the tension of the rope?”

many thankkkkss

This video is only 240p?

Poor

tysm sir

Could u plz tell me what is SOH CAH TOA…

its been 10 years, and im still watching this

how did he get square root(3)

Can anyone explain how is that angle 30 degrees?

WARNING, this is completely incorrect for 3 reasons. First of all, theta is 60, deg. not 30 deg. T1 = 1/sin Theta * 100n = 115N T2 = (90 – theta) tan * 100N = 57.7N

thnx sir

I was never able to solve complicated tension problems befor watching this👍

Great intro

my calculator was in radian and i was about to commit because i kept getting different answers

for some reason, Im not rlly fond of khan acdemy

9,81kg/cm^2

Hey, can anyone explain why when something is accelerating down, you subtract that acceleration from gravity in the equation of tension?

LOL. that is definitely not 30 degrees that is at least 60 or 70

of course tyler1 has to carry everything

11 years ago , wow thats older than many people and things lol

I’m heading over right now to fix that gap in the green line

When I start making money with the knowledge I have gained from watching your channel, I will donate to Khan Academy.

when the string breaks will the particle move a distance before hitting the ground, please I need a quick answer

anybody else notice that the intro to this video compared to the ending of the last sounded like Sal was recording in his closet trying not to wake up his wife and kid XD. This guy is seriously incredible though.

Thx teacher khan

Or By gravity and anti Gravity haha, Gravity is a product of tension, Tension is not only the force of a string but of a plane also the plane of space and time

this video made me realize this was really the easiest thing ever but why did it seem like such a messy and horrible thing to learn during class…. lol

I'm watching this and my midterm exam is tomorrow

2019 ANYONE??

Why can't we say mg cos 60° = T1?

Dear god…. this quality is so bad lol Thats how you know it's from 2008.