Welcome back. We’ll now do another tension

problem and this one is just a slight increment harder than the

previous one just because we have to take out slightly

more sophisticated algebra tools than we did

in the last one. But it’s not really

any harder. But you should actually see this

type of problem because you’ll probably see

it on an exam. So let’s figure out the

tension in the wire. So first of all, we know

that this point right here isn’t moving. So the tension in this little

small wire right here is easy. It’s trivial. The force of gravity is pulling

down at this point with 10 Newtons because you

have this weight here. And of course, since this

point is stationary, the tension in this wire has to

be 10 Newtons upward. That’s an easy one. So let’s just figure out the

tension in these two slightly more difficult wires to figure

out the tensions of. So once again, we know that this

point right here, this point is not accelerating

in any direction. It’s not accelerating in the

x direction, nor is it accelerating in the vertical

direction or the y direction. So we know that the net forces

in the x direction need to be 0 on it and we know the

net forces in the y direction need to be 0. So what are the net forces

in the x direction? Well they’re going to be the x

components of these two– of the tension vectors of

both of these wires. I guess let’s draw the tension

vectors of the two wires. So this T1, it’s pulling. The tension vector pulls in

the direction of the wire along the same line. So let’s say that this is the

tension vector of T1. If that’s the tension vector,

its x component will be this. Let me see how good

I can draw this. It’s intended to be a straight

line, but that would be its x component. And its x component, let’s

see, this is 30 degrees. This is 30 degrees right here. And hopefully this is a bit

second nature to you. If this value up here

is T1, what is the value of the x component? It’s T1 cosine of 30 degrees. And you could do your

SOH-CAH-TOA. You know, cosine is adjacent

over hypotenuse. So the cosine of 30 degrees is

equal to– This over T1 one is equal to the x component

over T1. And if you multiply both sides

by T1, you get this. This should be a little bit of

second nature right now. That the x component is going to

be the cosine of the angle between the hypotenuse and

the x component times the hypotenuse. And similarly, the x component

here– Let me draw this force vector. So if this is T2, this would

be its x component. And very similarly, this is 60

degrees, so this would be T2 cosine of 60. Now what do we know about

these two vectors? We know that their

net force is 0. Or that you also know that the

magnitude of these two vectors should cancel each other out

or that they’re equal. I mean, they’re pulling in

opposite directions. That’s pretty obvious. And so you know that their

magnitudes need to be equal. So we know that T1 cosine

of 30 is going to equal T2 cosine of 60. So let’s write that down. T1 cosine of 30 degrees is

equal to T2 cosine of 60. And then we could bring the

T2 on to this side. And actually, let’s also– I’m

trying to save as much space as possible because I’m guessing

this is going to take up a lot of room,

this problem. What’s the cosine

of 30 degrees? If you haven’t memorized

it already, it’s square root of 3 over 2. So this becomes square root

of 3 over 2 times T1. That’s the cosine

of 30 degrees. And then I’m going to bring

this on to this side. So the cosine of 60

is actually 1/2. You could use your calculator

if you forgot that. So this is 1/2 T2. Bring it on this side so

it becomes minus 1/2. I’m skipping more steps than

normal just because I don’t want to waste too much space. And this equals 0. But if you seen the other

videos, hopefully I’m not creating too many gaps. And this is relatively

easy to follow. So we have the square root of

3 times T1 minus 1/2 T2 is equal to 0. So that gives us an equation. One equation with two unknowns,

so it doesn’t help us much so far. But let’s square that away

because I have a feeling this will be useful. Now what’s going to be happening

on the y components? So let’s say that this is the y

component of T1 and this is the y component of T2. What do we know? What what do we know about

the two y components? I could’ve drawn them here too

and then just shift them over to the left and the right. We know that their combined

pull upwards, the combined pull of the two vertical tension

components has to offset the force of gravity

pulling down because this point is stationary. So we know these two y

components, when you add them together, the combined tension

in the vertical direction has to be 10 Newtons. Because it’s offsetting

this force of gravity. So what’s this y component? Well, this was T1

of cosine of 30. This should start to become a

little second nature to you that this is T1 sine of 30, this

y component right here. So T1– Let me write it here. T1 sine of 30 degrees plus this

vector, which is T2 sine of 60 degrees. You could review your

trigonometry and your SOH-CAH-TOA. Frankly, I think, just seeing

what people get confused on is the trigonometry. But you can review the trig

modules and maybe some of the earlier force vector modules

that we did. And hopefully, these

will make sense. I’m skipping a few steps. And these will equal

10 Newtons. And let’s rewrite this up here

where I substitute the values. So what’s the sine of 30? Actually, let me do

it right here. What’s the sine of 30 degrees? The sine of 30 degrees is 1/2 so

we get 1/2 T1 plus the sine of 60 degrees, which is square

root of 3 over 2. Square root of 3 over

2 T2 is equal to 10. And then I don’t like

this, all these 2’s and this 1/2 here. So let’s multiply this

whole equation by 2. So 2 times 1/2, that’s 1. So you get T1 plus the square

root of 3 T2 is equal to, 2 times 10 , is 20. Similarly, let’s take this

equation up here and let’s multiply this equation by 2

and bring it down here. So this is the original

one that we got. So if we multiply this whole

thing by 2– I’ll do it in this color so that

you know that it’s a different equation. So if you multiply square root

of 3 over 2 times 2– I’m just doing this to get rid of the

2’s in the denominator. So you get square root of 3 T1

minus T2 is equal to 0 because 0 times 2 is 0. And let’s see what

we could do. What if we take this top

equation because we want to start canceling out some terms.

Let’s take this top equation and let’s multiply

it by– oh, I don’t know. Let’s multiply it by the

square root of 3. So you get the square

root of 3 T1. I’m taking this top equation

multiplied by the square root of 3. This is just a system

of equations that I’m solving for. And the square root of 3

times this right here. Square root of 3 times square

root of 3 is 3. So plus 3 T2 is equal to

20 square root of 3. And now what I want to do is

let’s– I know I’m doing a lot of equation manipulation here. But this is just hopefully, a

review of algebra for you. Let’s subtract this equation

from this equation. So you can also view it as

multiplying it by negative 1 and then adding the 2. So when you subtract this from

this, these two terms cancel out because they’re the same. And so then you’re left with

minus T2 from here. Minus this, minus 3 T2 is equal

to 0 minus 20 square roots of 3. And so this becomes minus 4 T2

is equal to minus 20 square roots of 3. And then, divide both sides by

minus 4 and you get T2 is equal to 5 square roots

of 3 Newtons. So that’s the tension

in this wire. And now we can substitute

and figure out T1. Let’s use this formula

right here because it looks suitably simple. So we have the square root

of 3 times T1 minus T2. Well T2 is 5 square

roots of 3. 5 square roots of

3 is equal to 0. So we have the square root of

3 T1 is equal to five square roots of 3. Divide both sides by square

root of 3 and you get the tension in the first wire

is equal to 5 Newtons. So this is pulling with a force

or tension of 5 Newtons. Or a force. And this is pulling– the second

wire –with a tension of 5 square roots

of 3 Newtons. So this wire right

here is actually doing more of the pulling. It’s actually more of the force

of gravity is ending up on this wire. That makes sense because

it’s steeper. So since it’s steeper,

it’s contributing more to the y component. It’s good whenever you do these

problems to kind of do a reality check just to make sure

your numbers make sense. And if you think about it,

their combined tension is something more than

10 Newtons. And that makes sense because

some of the force that they’re pulling with is wasted against

pulling each other in the horizontal direction. Anyway, I’ll see you all

in the next video.

If you were to reverse the diagram and place the 10 N force on top the system would be in compression. It would be like a truss. Would the force in the diagonal member be the same, just in compression?

FUCK DUDE i got lost when he started using radicals and stuff

Yes it is right, as T2 and T1 are not simply vertical but rather "diagonal" in a sense, as some of the tension is "wasted" horizontally.

The only forces that must equal 10N are the vertical variables, which he demonstrates by setting both adding the vertical aspects and setting them equal to 10N.

In my Physics class I thought the same thing. Now I have been enlightened and i don't need my senile grumpy old teacher anymore ðŸ™‚

thank you soooo, soooo much for this one.

thank you very much! this problem used to piss me off…….not anymore ðŸ˜€ great videos!

@ paulojunior201:

first of all, you should be appreciative that this man is spending his time making videos to help others. Second, have you ever thought about the other people in the world, who don't understand certain concepts as quickly?

Besides, his accomplishments in education will be better than you can ever achieve. So think before you speak you ignoramus.

Phenominal! Thank you so much

you are awesome keep up the good work

helped me out alot way better then my physics teacher!!

wasn't happy about the way I said it in my first comment ^^ I just want to ask why do you call it tension when there is no area involved for the dimension of tension [N/m^2]?

In Singapore We learnt Design N Technology and will do questions like these. Thanks for helping me out for this^^LURVE U LOTS now I can let my Teacher now relax. HEY the homework involving this question, I am was the only student that bothered to tell the teacher that I will not be able to hand in the homework and no one even handed it in.

If you did the sum of the forces in the x and the sum of the forces in the y, it'd be a lot easier.

x=T2 cos 30 – T1 cos 60. Then solve for T2.

T2 = T1(cos 60/cos 30)

y= T1 sin 60 + T2 sin 30 -10n = 0

Then plug in T2 into the second equation giving you:

T1 (cos 30) + T1(cos 60/cos30)(sin 30) -10=0 Solve for T1 giving you T1=8.66N

Since we already know T2 = T1(cos60/cos30), just plug in T1 making it 8.66 (or 5 square root of three) multiplied by (cos 60/cos 30). This gives 4.99N.

@paulojunior201 He should also stop using words and math as well. Just an empty video, that would be your preference?

1:39 why is that 30 degrees?

@paulojunior201 apparently he needs to since you haven't grasped its soh cah toa not soh cah toah…

@zRemify yeah but if thats 30 degrees shouldnt angle y be 30 degrees as well? because its alternate angles??

u are great!

I had a problem.. what about the wire that connected the 10N block with T1 and T2? Shouldn't that also be taken into account when offsetting the force of gravity on that block?

damn, why this guy is so good at teaching?! Love him!

mind blown…. this really helps a lot! thank you very much!!! ðŸ˜€

You are an awesome teacher. Thanks to the Youtube app on my Iphone you're always with me when I'm trying to figure out my physics homework.

Thanks so much

this was extremely helpful to me!!

Warm regards

thanks!

At the very end time 7:55, why are you subtracting one equation from the other. in my head i know this makes sense but can't put it into words so im having trouble with why were doing it and relating to other problems???

@rosyred158 There's a Kahn Academy app so you can have all his videos arranged on convenient playlists wherever you have internet connection. I've been using it all day to study for my physics final until my battery died haha

thanks alot for the help, your an excellent teacher, easy to comprehend and good explanations

@fishysmell321 Looking at: y= T1 sin 60 + T2 sin 30 -10n = 0

Shouldn't this be y= T1 sin 30 + T2 sin 60 – 10 N = 0?

@tolsonw Yes. I didn't realize he chose the right wire as T1 and the left wire as T2.

Dude, thank you so much!

its difficult to follow

I t is amazing thank u sooooooooooooooooooo much

replace my kpark physics honors teacher? cool, thanks. ðŸ˜€

2 times 10 equals 20?

can tension be negative because of gravity? Please comply. ðŸ™‚

wow never new simultaneous equations would be involved

i have an A-level physics exam tomorrow morning. if i get an A, it'll be because of this dude ðŸ™‚

is this for a level or as level

because i used it for as level questions – but some reason it did not work

Thanks for the video man but i dont really understand the part of the angles of the triangles its a little confusing. I hope you can help me out a bit.

tension was always something I got wrong but now I understand it completely thanks so much

THANKS!!!!

thanks so much. that was so helpful

can we solve it by using trig functions since i knw the angle ??

THANK YOU SO MUCH. Not to be a disrespectful student but my teacher literally has no idea what she's doing

I signed in just to give this a thumbs up. Thanks so much Sal! You taught me in like 3 minutes (thank you YouTube for the ability to skip around) what I could not figure out after over 30 minutes of searching the internet. When I get into a good college thanks to you and get a lot of money, I will be sure to donate some back to Khan Academy! Haha

where did he get 30 degress from??

My professor allowed us to make an "equation sheet" to use with our physics exam, this morning. I wrote this tension problem down, and sure enough, there was a problem exactly like this on the exam. Along with the rest of your videos, I can't thank you enough for helping me prepare; and if I wasn't broke, I'd try and donate. Maybe I can buy you a drink sometime! Thanks again Sal!

i know right!! btw i love your username xD

thank you!!!!! this is such a great tutorial ðŸ˜€ im now ready to go to my physics class!!

thanks a lot! this really helped! btw, did you use paint to do this? and did you have a tablet laptop as well?

So I was sitting here doing physics homework, and I just couldn't get it. So I decided to look online for some help. As soon as I watched your video, something just clicked in my head and I got it! Thank you so much for everything you do, and have a great day.

Thank you so much for everything you do for us! I finally understand forces. Once you understand it, it becomes quite easy.

God bless you really worked after 12 years of my school finally got the point….thanks ðŸ™‚

i feel stupid, i still cant do it x(

You really are my saviour ðŸ™‚

I like creating a triangle of forces and using the sine rule. It's faster.

This is easy peasy. The part I'm having trouble with is a lorry pulling a car. Teacher never explained this to us. So flipping confused cuz I can't find ANY explanations anywhere

MY god….compared to my physics teacher in school, you are so much better.

YOU HAVE SAVED MY LIFE. well, my life. BUT my grades are my life at this point so thank you. I appreciate your videos very much

*well, my grades

Honestly you made that way more confusing than it had to be all you need is two equations with two variables (t2)sin30=(t1)sin60 and (t2)cos30+(t1)cos60=10 from there it's simple algebra a lot Easier than what you just did

How would you go about solving for T1 and T2 if the hanging weight is unknown?

I get the way how to find the answer, but to tell you the truth, i dont understand why it gives the tension of the string/rope. So i wouldn't know what to do if there was 3 strings

Wait… those should be different variables. T1x, T1y, T2x, T2y

Â

Ooo

Maaan, 1 hour ago I knew nothing about tension, I watched the first video on tension and now I`ve mannaged to solve this alone without watching the video. Yaaayy :))

I'm a little confused up with the rationale behind the y-component of the two tensions. I initially set the problem with a y-component, shared by both tensions, as equaled to the weight of the hanging mass, 10N. But, that's not the case here. Is it because the weight of the mass here is supported by two strings, so the tensions of their y components are divided between the two strings? I'm just thinking back to part 1 of the video where the y component of the tension of one of the strings was equaled to the weight of the mass… Hope for some clarification here. Thank you.Â

I love these videos, but my problem is figuring the physics questions out in less than a minute

Shouldn't the cos(60) be negative? Since the second quadrant on the xy plane contains negative cos values.Â

Shouldn't the x components be opposite signs if they are opposing each other? And if added together they are equal to 0 then one of them has to be negative… Also I don't think the net force in the y direction is equal to 0, it should be equal to 10N. You stated it wrong in the beginning.

You taught me more in 2 tension videos than my teacher has in 12 weeks lol..smh

THANK YOU VERY MUCH!

keep up the good work, marvellous job this one

Would the tensions change if the load was on a pulley (to make them equal regardless of angles)?

Instead of solving for the cos and sin of whatever angle (and giving a radical answer), why not just explain this problem in simpler terms. You're unnecessarily complicating this problem by introducing square roots, and short cuts to algebra such as : multiplying both sides by two and then Â multiplying each side by the square root of three. Not everyone taking physics is fresh out of trigonometry, sir.

Not your best work ðŸ˜‰

First tell me, how the net force in that point is zero? And how did u get the value of angle y as 30 digree? Plz tell me quickly, my exams r just going to commence.

Srry, looks like angle y indeed IS 30. It looked like y in the first look but when I looked at it more deeply, it was clear. Well u still need to tell me how the net force at those two points are zero. Please its an urgent requirement.

bless you

one of the equations can be made w.r.t one variable and then substituted in the other …this is even simpler faster and takes less space ðŸ˜€

There is a proportion to check this problem or maybe lead to an easier way.( Never took physics) But I am taking trigonometry,Â so cos30/cos60= 5(rad3)/5

0.866025../0.5= 8.66025../5.0

radical 3 = radical 3

I really wish you would do it in degrees and not radians

Repeatedly press the time bar at 5:40.

I have my physics final in 2.433 hours.

u can also say +T1 cos 30-T2 cos 60 = 0( addin all the x components) and equal to 0 because it is stationery ( T2 cos 60 is negative because it is in the negative x direction) also for adding the y components u have T1 sin 30 +T2 sin 60- 10= 0 (10 is negative cos it is in the negative y direction)… -10 will move to the opposite direction and become +10 when u collect like terms… correct me if am wrong

I am going to fail physics I don't understand jack

bless up

Convoluted. Also, it's hard to make out the numbers that you wrote.

Why isn't there any effect from the force normal (Fn) here?

Are these physics videos relevant to the new MCAT?

Either there is a fly zipping by the mic, or someone in the same room as him keeps farting.

Thanks a lot Sir…

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When you set both X components together, wouldnt one have to be negative? since they cancel eachother out

I'm gonna fail this test tomorrow lol

At about 8:33 in the video, a value jumps from (10)(square root of 3) to (20)(square root of 3)

Thank you khan saab,fantastic lecture as usual.

at 5:55 how T1*Sin30 + T2*sin60 = 10 N instead of 10 * 9.8? Because W = m*g

I thought it would be T1*Sin30 + T2*sin60 = 10 * 9.8

Is it because he already said it is 10N not 10 kg?

Why did he subtract the first equation from the second one?

I don't understand why the magnitude of both x components would be same

use the lami's theorem

Brilliant vid