If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Introduction to tension

## Video transcript

I will now introduce you to the concept of tension so it's tension is really just the the force that exists either within or applied by a string or a wire when it's usually lifting something or pulling on something so let's say I had a weight let's say say I have a weight here and let's say it's it's a hundred Newtons 100 Newtons and it's suspended from this wire which is right here let's at eight let's say it's attached to the ceiling right there well we already know that the force you know if we're on this planet that this weight is being pulled down by gravity right so we already know that there's a downward force on this weight which is the force of gravity force of gravity and that equals 100 Newtons right but we also know that this this weight isn't accelerating right it's actually stationary it is it also has no velocity the important thing is it's not accelerating so for the but given that we know that the net force on it must be zero by by Newton's laws so what is the counteracting force well you could you could you know you didn't have to know about tension to say well the strings pulling on it right 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 and that is also another way to think about it is that's also the force that's within within the wire and that is going to exactly offset the the force of gravity on this weight and that's what keeps this point right here that's what keeps this point right here stationary and keeps it from accelerating that's that's pretty straightforward tension it's just you know the force of a string and and just so you can conceptualize it you you you know on a guitar the more you pull on a on some of those higher was the thought you know the really thin strings that that sound higher pitch the more you pull on it the higher you the tension it actually creates a higher higher pitch note so you've dealt with tension a lot I think actually when do they sell wires or strings they'll probably tell you the tension that that wire strength is 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 you know fairly simple example Don let's create a slightly more a 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 let's make that green string there and it's attached to the ceiling up here that's the ceiling now and let's say 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 t1 and t2 well like the first problem this point right here this red point is stationary right it's not accelerating and 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 what is going to be the offset because we know that already that at this point right here there's going to be downward force right which is the force of gravity again right the force of 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 call it w-2 I guess the second wire is just pulling to the left it has its it has no y component it's not lifting up at all so all of the so it's just pulling to the left so all of the upward lifting all of that is going to occur from this first wire from from t1 so we know that we know that the Y component of t1 so let's call so if we say that this vector here doing good different color because I know when I draw these diagrams it starts to get confusing so let's let me just actually use a line tool so I have this and if you can see let me make a thicker line so we have this vector here which is t1 right and we 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 all right this is its Y component we call it we could call this t1 sub y and then of course it has an X component too and I'll do that in in let's see I'll do that in red and this is once again this is just breaking up a force into its component vectors like we've a vector force into its X&Y components like we've been doing the last several problems and these are these are just trigonometry problems right and so we have we could actually now visually see that this is T sub 1 X and this T sub 1 sub y oh and I forgot to give you a important product with the problem that you would that you needed to know before solving it is that the angle the first wire forms with the ceiling that this is 30 degrees so if that is 30 degrees we also know that what these this is this is a parallel line to this right so this is started 30 degrees this is also going to be 30 degrees right so this angle right here this 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 figured out we could have said that you know this if this angle is 30 degrees this angle 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 sohcahtoa top because this is really just trigonometry sohcahtoa 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 them the sine of 30 degrees is equal to t1 sub y over the tension in the string going in this direction and so if we solve for t1 sub y we get t1 sine 30 degrees is equal to t1 sub y right and what did we just say before we kind of dived into the math we said all of the lifting on this point all of the lifting on that point is being done by the Y component of t1 right because t2 is not doing any lifting up or down it's only lifting it's only pulling to the left right so the entire 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 so this has to equal the force of gravity that has to equal this or at this point full so that's 100 Newtons and I really want to hit this point home there because it might be a little confusing to you we just said this point is stationary it's not moving up or down it's not accelerating up or down and so we know that there is 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 by the on the first wire so given that we can now solve for the tension in this in this first wire because we have t1 t1 what's sine of 30 sine of 30 degrees in case you haven't memorized it sine of 30 degrees is one-half so t1 times 1/2 times 1/2 is equal to 100 Newtons divide both sides by 1/2 and you get t1 is equal to 200 Newtons t1 is equal to 200 Newtons so now we've got to figure out what the tension in this second wire is and 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 in this wire must be it must be being offset by attention or some other force in the opposite direction and that force in the opposite direction is the X component of the first wires tension so it's this right so t2 is equal to the X component of the first wires tension and what's the X component what is going to be the tension in the first wire 200 Newton's times the cosine of 30 degrees right 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 roots of 3 so the tension in this wire is 100 square roots of 3 which completely offsets its and it's to the left and the X component of this wire is 100 square roots of 3 Newtons to the right hopefully I didn't confuse you see in the next video