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Teacher/Instructor Jonathan Osbourne
Jonathan Osbourne

PhD., University of Maryland
Published author

Jonathan is a published author and recently completed a book on physics and applied mathematics.

Tension in physics is the intermolecular force that exists in inside a rope, string, or any kind of elastic materials. The purpose of tension is to maintain integrity of the system. The foundation idea about tension in the system is strings can only pull.

Alright so let's talk about tension, what is tension? Tension is the intermolecular force that exists inside of ropes and strings and other kind of materials that you can pull on but that also are flexible. So here's the idea the purpose of tension is to maintain the integrity of the string or the rope. No stretchy no breaky, okay same length and it's not going to break. So that's the idea, now I can't push you with a string alright, if I've got a string I tried to push on it well it's going to go limp. I can only pull with a string, so that means that whenever you see a string or rope tied to something it's always pulling it cannot push. And that seems like a really simple idea but you wouldn't believe the number of mistakes that I've where people aren't thinking about it that way and they this other kind of idea that goes into their minds and then everything is just wrong.

Alright so let's see how this works, so first off I want to imagine that I've got a rock that I'm swinging in a circle a vertical circle like this alright and I want to determine what is the tension in this rope when the rock is at the bottom and when the rock is at the top of the circle. Alright so let's do the bottom first, well of course we're going to start with a free body diagram I mean jeez what are we idiots? So we'll start with the free body diagram and we'll say alright what about the acceleration? What acceleration does this rock experience at the bottom? Well it's got to have an acceleration towards the center of the circle that's v squared over r. And that's because it's moving in a circle what do you want. So you got an acceleration that's going up in this case and it's v squared over r alright. So that means that the tension minus mg must be ma, m v squared over r so that gives us a tension of the weight plus mv squared over r, so at the bottom of the circle the tension is actually larger than the weight.

Alright, what happens at the top? Alright now this is where people get all messed up. Alright at the top, where's the string, well it's like that so what direction is the tension force in? Strings can only pull, it's pulling down, it can't do anything else, that string can't push it up alright the string will just go limp before it did that. So at the top we have a tension force acting down and and what about the weight? Does the weight act up now? No the weight always acts down towards the center of the earth. So this is a kind of a strange looking free body diagram, everybody is pulling down, well the acceleration is down too. V squared over r just like it was before except in this case it's down because it's got to be directed towards the center of the circle. Alright so what have we got, well we've got tension plus mg has to equal ma. So that means that the tension is equal to mv squared over r minus mg alright now this is an issue, this minus sign means that it could be the case that this tension was negative.

But what would that mean? would that mean I'm trying to push with a string? You're not allowed to push with a string you can pull with string, so that means that this tension got to be positive. So that means that the speed has to be bigger than the square root of r times g. This is the minimum speed at which you can get the rock to go in this circle, if you try to go slower than that it'll get up to a certain point and then the string will just go just fall alright and then it won't make it to the top because there's not enough speed to do it. Alright so that's the idea, let's look at another problem involving a string. So here I've got a mass m and a mass one half m, mass m is sitting on a frictionless counter top and mass a half m is pulling down like this and the 2 masses are connected by a string which runs over a pulley.

Alright now what's the purpose of a pulley? an ideal pulley does nothing more than change the direction of the tension that's the only purpose, so here the tension is pulling that way because the string's there you got to pull right. Here which direction is the tension in? Alright this is another place that I've seen a lot of mistakes people look at the string and they're like jeez the tension is going in that direction so the tension must always go in that direction, it must be pushing down. But no you're not allowed to push with a string. You can only pull so the tension must be pulling up on this guy, opposite directions but you can't argue with a string can't push, it can only pull. Alright so let's draw some free body diagrams try to determine the acceleration of this system. Alright well we've got 2 objects here, we've got mass m and mass a half m alright so let's look at mass m. Well what are the forces acting on it well jeez it's got a weight like everybody does, it's also on the table now the table doesn't want it to fall through it so the table is going to put a normal force on it to push it up. Remember that the purpose of a normal force is to maintain the integrity of a solid object so you're pushing on it and it fights back.

Alright so we'll have a normal force here which is however big it needs to be to ensure that there's no acceleration into the table and then we've got the force acting this way tension t alright so we've got this tension that's inside of this string. So how is this tension t associated with the tension that's acting on a half m? Well let's see alright on a half m we've certainly got our weight 1 half mg and then what about the tension force. Well that's going to certainly be acting up and what's the relationship between this tension and this tension? Well they're obviously in different directions but they're associated with the same strength. Now the way that ideal strings go is that the tension throughout them has to be the same. Now this isn't actually true in real strings because the tension will be a little bit higher at the bottom here I'm sorry the top here because tension has to support the weight of the string.

But you know what strings don't weigh very much let's just ignore that. And so what we'll do is we'll say that this is the same magnitude force t as this one. Alright what about the acceleration, well what do we expect, I mean we expect this mass to go the right, this mass to go down. So we'll have acceleration here and we'll have acceleration here. Okay now what's the relationship between those two accelerations? Well the string isn't allowed to change it's length, the string is what we call inextensible. Alright so it can't change it's length, so that means that if these guys are accelerating they got to accelerate at the same rate. So these accelerations have to have the same magnitude alright, so let's see what's going on.

Alright Newton's second law these two forces cancel so I might as well not even have written them down. I wanted to be correct so I wrote them, but at normal forces just going to cancel the weight, because there can't be any acceleration vertically. Alright what about horizontally? Well we'll have t equal ma alright that's easy enough and then here we'll have one half mg minus t equals one half m times a. Alright these two equations are very easy to solve, I'm going to show you a nice technique, all we're going to do is we're gong to take these two equations and we're going to add them together, we can of course do that. I mean if this one's true and that one's true then the sum of them is going to be true too. Well look what's beautiful about this, the tension is going to cancel so that means that we'll have one half mg equals and let's add these 2 together, we'll this is going to give a half plus one. Well that's 3 halves ma but look how beautiful that is m cancels the 2 cancels a is one third of the acceleration due to gravity. So this system is going to accelerate at one third of the acceleration due to gravity.

Now if we wanted to we can determine the tension just by saying well jeez the tension is m times a third the acceleration due to gravity so the tension is going to be 1 third mg, 1 third of the rate of that mass. Alright let's do one more situation and this is a standard situation with pulleys. We got a pulley system here, we've got 2 pulleys, we've attached this one to the ceiling and then we're winded a string around, we attached it to the bottom of that pulley and we run it around that pulley. Then we attached this pulley to a mass alright, we want to know what force do I have to pull with in order to support this mass. Alright so let's go ahead and determine first off what's going to be the tension in this string?

Now remember this string is all 1 string that means it's tension has to be the same throughout. Well I'm pulling with f here so that means the tension in this string has to be f all the way through. Alright let's draw a free body diagram of this pulley alright well what force is pulling down on that pulley? Well it looks to me tension 2 is, alright now tension 2 of course just equals mg because that's got to support the weight of that guy. So this is just mg, alright here's the tricky part. What force is acting up on this pulley? Well, the string is here and strings can only pull so it must be pulling up but the string is also here. Now this is the weird part that throws people off and they're thinking well the string can't pull twice, yes it can the string can pull twice and it will. It's attached here and it's attached here, so it's pulling up both sides. So that means I've got f and f again alright, so that means 2 times f in equilibrium has to equal mg. So that means the force that I need to supply is only a half of the weight of the thing that I'm trying to support. So essentially by using this pulley system I have doubled my force, right? So I'm supporting something that's twice as heavy as the force that I need to apply to support it.

Alright and that's the idea of a pulley system in general you'll multiply your force by the number of pulleys, here we've got 2 pulleys [IB] multiply it by 2. If we had 50 pulleys we would multiply it by 50 alright. One other real quick thing, let's determine what this tension is right here. Alright so that tension we're going to draw a free body diagram of this up top of our pulley, so pulling up tension 1, what's pulling down? Alright again whenever you see a string it is pulling, alright it's pulling. Well I see a string, I see 1 there, 1 there and, there's 3 tensions pulling down on this guy, so this will be 1, 2 and 3, f, f and f. So that means tension 1 has to be 3f which is 3 halves of mg.

And that's essentially how every single pulley problem goes, they're all really really really easy once you recognize that strings can only pull and that's tension.