Introduction to tension (part 2) | Forces and Newton’s laws of motion | Physics | Khan Academy

Introduction to tension (part 2) | Forces and Newton’s laws of motion | Physics | Khan Academy


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.

100 Replies to “Introduction to tension (part 2) | Forces and Newton’s laws of motion | Physics | Khan Academy”

  1. 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?

  2. 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 🙂

  3. @ 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.

  4. 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]?

  5. 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.

  6. 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.

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

  8. 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?

  9. 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.

  10. 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???

  11. @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

  12. @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?

  13. 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.

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

  15. 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

  16. 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!

  17. 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.

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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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 :))

  24. 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. 

  25. 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.

  26. 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  multiplying each side by the square root of three. Not everyone taking physics is fresh out of trigonometry, sir.

  27. 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.

  28. 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.

  29. 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 😀

  30. There is a proportion to check this problem or maybe lead to an easier way.( Never took physics) But I am taking trigonometry, so cos30/cos60= 5(rad3)/5
    0.866025../0.5= 8.66025../5.0
    radical 3 = radical 3

  31. 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

  32. 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

  33. 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?

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