Introduction to tension

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.