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Introduction to tension

An introduction to tension. Solving for the tension(s) in a set of wires when a weight is hanging from them. Created by Sal Khan.

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  • blobby green style avatar for user turn.profoundly
    If the block is not moving, and the forces are balanced, why doesn't the tension of both the ropes equal 100N? How can there be a tension of 200N in one of the ropes, if only 100N is pulling down on it?
    (67 votes)
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    • blobby green style avatar for user Robert
      Imagine holding a something heavy (a physics textbook) in your hand with your arm straight down, by your side. Then raise your arm, still completely straight, up so it is at at an angle with you body. You will notice that it is much more difficult to hold the book in that position than if you let your arm hang straight down. It is the same principle.
      (140 votes)
  • blobby green style avatar for user TANVI TAVARNA
    At he says he's adding 2 more strings and then considers the tension in only those 2 strings but what about the first string?
    (6 votes)
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  • blobby green style avatar for user Fahim
    When we've attached the two new wires, does the original blue wire not provide any force to support the block anymore? I mean is it now just extending the downward force of the block to the two new wires and otherwise just hanging around? And if this is the case could you please explain why? I can set up and do a problem like this but I just don't understand why that original wire no longer supports the weight anymore in any way. Thanks.
    (4 votes)
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    • leaf green style avatar for user Derek Edrich
      The original blue wire will still provide the same amount of force to the block, and it still has the same amount of tension on it, but is still transferring that force to the block.

      To put it another way, if you were to cut the blue wire, the block would fall because you removed the force that the blue wire supplies to the block.
      (4 votes)
  • blobby green style avatar for user limmarcus_13
    Is the T1x = T2 an application of Newton's 3rd law? some thing liddat?
    (2 votes)
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  • leaf blue style avatar for user shreyachand66
    Why do we need to make the x and the y component of T1 wire?
    (4 votes)
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    • leaf green style avatar for user Hrit Roy
      The fundamental reason behind this is that two mutually perpendicular vectors do not affect one another.Now, any vector can be resolved into two mutually perpendicular vectors, right? In this case T₁ᵪ and T₁ᵧ are those components.Now since T₁ᵪ is perpendicular to the 100N force acting downwards, it will not balance it, it will not add to it; it will not affect it in anyway.But the T₁ᵧ vector is parallel to the 100N force.Now, this T₁ᵧ MUST BE balancing the 100N downward force as the point, as Sal mentions, is not accelerating in any direction.So the 100N downward force is actually balanced by this T₁ᵧ=100N upwards force.
      This is why we first MUST resolve the T₁ vector into two components, one perpendicular to the 100N force and one parallel to it.Hope I was clear enough :)
      (2 votes)
  • blobby green style avatar for user a.harif13
    At he says 'sine of 30 degrees is one half' what does he mean by that? That kinda made me confused :(
    (2 votes)
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  • blobby green style avatar for user Khalid Anjum
    If a string is stretched by two opposite & equal forces 10 N each.what will be tension in string
    (1 vote)
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    • male robot hal style avatar for user Andrew M
      If the string is attached to the wall and you pull on it with 10 N of force, the string will have 10 N of tension, right? That's easy.

      Now, ask yourself, how hard is the wall pulling on the string? It must also be pulling with a force of 10 N, otherwise the string would be accelerating toward you.

      Now, if we replace the wall with another force that applies 10 N, does the string know the difference? No, right?
      (3 votes)
  • blobby green style avatar for user Whitney
    So if T1 was directly above T2 instead of off to the side a bit, would you solve it the same way or would it just offset the force of tension in T2?
    (2 votes)
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    • leaf red style avatar for user Andy Petsch
      Hey Whitney! You're right. In the case, that each wire is perpendicular to the walls, all of the force counteracting against the force of gravity would just be in the string going straight up from the mass.
      That's just because of Newton's First Law, stating that the opposing force has to have the same magnitude AND has to point in the opposite direction (if their is no acceleration!)
      (2 votes)
  • aqualine ultimate style avatar for user super00bird
    I still get confuse that the original string provides 100N upward to balance the block with 100N downward ,but after he added 2 more strings and he proofed T1y =100N, doesn't it mean that there are 200N upward and 100N downward. So, I don't get it.
    (2 votes)
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    • male robot hal style avatar for user Ben Willetts
      No, adding more strings does not add an upward force, it just balances out the total upward force amongst the strings. So in this example one string would supply a 100N upward force, but if you added a second string, the two strings might supply a 50N force each (or one 40N and one 60N, or whatever). This is the same reason that if you try to lift a heavy rock by yourself, you can't do it, but if ten people try all together, they can lift it.
      (2 votes)
  • female robot grace style avatar for user Mariam Varkey
    how will the numerical part of the tension problems be afftected if the string has mass?
    (2 votes)
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Video transcript

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.