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Current time:0:00Total duration:14:09

Video transcript

so I found that when students have to do problems involving the force of tension they get annoyed maybe more than any other kind of force problem and we'll talk about why in a minute but so we'll go over some examples involving tension to try to demystify this force make you more comfortable with finding it so it's not so annoying when you get one of these problems and at the same time we'll talk about the misconceptions that a lot of people have about tension so that you don't make those mistakes as well all right so to make this a little more clear what is tension that's the first good question is what is this tension well tension is the force exerted by a rope or a string or a cable or any rope like object so if you had a box of cheese snacks and we tied a rope to it so we tie a rope over to here and we figure out how much force do I have to pull with since the force is being transmitted through a rope we'd call that tension so I mean it's a force just like any other force I'm going to call this t1 because we're going to add more ropes in a minute so it's a force exerted just like any other force you treat it just like any other force it just is a force that happens to be transmitted by a rope so what's happening here in this rope ropes are typically composed of fibers that have been braided together or wound around each other so that when I exert a force at this end right when I exert a force down here and this end I pull on this end of the rope that force gets transmitted through the rope all the way to the other end and it'll exert a force on this box the way that works is I pull on the fibers here they're braided around each other so these fibers will now pull on these fibers here which pull on the fibers here and this could be parts of the same fiber but the same idea holds once this force is exerted here it pulls on the ones behind them keeps pulling on the ones behind them eventually that force gets transmitted all the way to this other end and so if I pull on this end with a force this end gets pulled with force so tension is useful ropes are useful because they allow us to transmit a force over some large distance and so what you might hear a typical problem might say this typical problem might say alright so let's say there's a force of tension on this box and it causes the box to accelerate with some acceleration a zero and the question might say how much tension is required in order to accelerate this box of mass M with this acceleration a zero what tension is required for that now a lot of the problems will say assuming the rope is massless and you might be like what how can you help first of all how can you have a massless rope second of all why would you ever want one well the reason physics problems say that a lot of times is because imagine if the rope was not massless imagine the Rope is heavy very massive one of those thick massive ropes well then these fibers here at this end would not only have to be pulling the box it also be pulling all of the rope in between them so they don't you pulling all this heavy rope in between them whereas the fibers right here would have to be pulling this amount of rope half of it right the fibers here would have you pulling this amount of rope which is heavy not as heavy as all of the rope and the box and then so the tension here would be a little less than the tension of this end and then the tension over this end well these fibers would only be pulling the box they don't have to pull any heavy rope behind them so not since they're not dragging any heavy rope behind them the tension over here would be less than the tension over here you'd have a tension gradient or attention vary a varying amount of tension where the tension is big at this end smaller smaller smaller smaller that's complicated we don't have to deal with that most of these physics problems don't want to have to deal with that so what we say is that the Rope is massless but we don't really mean the Rope is massless the ropes got to be made out of something what we mean is that the ropes mass is negligible it's small compared to any mass here so that even though there is some small variation within this rope of the tension it really doesn't matter much in other words maybe the tension at this end is 50 Newtons and the tension at this end is like forty nine point nine nine nine eight so okay yeah they're a little different the tension here is a little greater but it's insignificant that's what we're saying so let's try to do this problem let's get rid of that let me get rid of all of this and let's ask what what is this required tension right here in order to pull this box with an acceleration of a zero so to do these problems we're going to draw a force diagram the way you draw any force diagram you draw the forces on the object so we can say the force of gravity is equal to mg force of gravity is wonderful force of gravity has its own formula mg I just plug right and I get the force of gravity and if you're near the earth there's always a force of gravity pulling down since this box is in contact with the floor there's going to be a normal force because the floor is a surface the box is a surface when two surfaces are in contact you'll have this normal force and then you're going to have this tension here's the first big misconception people look at this line of this rope and they say well it looks like an arrow pushing that way so they think that this ropes pushing on this box and that doesn't make any sense you can't push with a rope if you don't believe me go right now go taut pause this video tie a rope to something and try to push on it and you'll realize oh yeah if I try that the Rope just goes slack and I can't really push but you can pull on things with a rope ropes or ropes cause this tension which is a pulling force tension is a pulling force because this rope gets taught it gets tight and now I can pull on things normal force is a pushing force right forces the two surfaces push on each other the ground pushes out to keep the box out of the ground but ropes tension is a pulling force so that I have to draw this force this way so this tension t1 them in my force diagram would point to the right I'd call this t1 those are all my forces I could have friction here but let's say this cheese snack manufacturing plant has made transporting their cheese snacks as efficient as possible they've made frictionless ground and if that sounds unbelievable maybe there's ball bearings under here to prevent their from being basically any friction we'll just keep it simple to start we'll make it more complicated here in a minute but let's just keep this simple to start now how do I solve for tension the reason people don't like solving for tension I think is because tension doesn't have a nice formula like gravity look at gravity's formula force of gravity is just mg you can just find it right away it's so nice but define the force of tension there's no corresponding formula that's like t equals and then something analogous to mg the way you find tension in almost all problems is by using Newton's second law so Newton's second law says that the acceleration equals the net force over the mass now if you don't like Newton's second law that's probably why you don't like solving for tension because this is what you have to do to find the tension since there's no formula dedicated to just tension itself so what's the acceleration oh I have to pick a direction first so do I want to treat the vertical direction or the horizontal direction I'm going to treat the horizontal direction because my tension that I want to find is in the horizontal direction so my acceleration in this horizontal direction is a zero that's going to equal my net force and in the X direction I only have one four so I've got this tension force so only force I have is T one in the X direction and since that points right and I'm going to consider rightward is positive I'm going to call this positive T one even though that's pretty much implied but positive because it points to the right and I'm going to assume rightward is positive you could call leftward positive you if you really wanted to be kind of weird in this case now I divide by the mass I can solve for T one now so I just do a little algebra I get the T one the tension in this first rope right here it's going to equal the mass times whatever the acceleration of this box is the causing with this rope so don't draw acceleration as a force this is a no-no people try to draw this sometimes acceleration is not a force acceleration is caused by a force acceleration itself is not a force so don't ever draw that okay so we found tension not too bad but this is probably the easiest imaginable tension problem you could ever come up with so let's step it up you're probably gonna face problems that are more difficult than this let's say we made it harder let's say over here someone's pulling on this side with another rope so let's say there's another t2 so people are fighting over these cheese snacks people are hungry and someone's pulling on this end what would that change let's say I still this person over here like ah you're not going to get my cheese snacks say they pull with a force to maintain this acceleration to be the same right they're going to pull with whatever they need to even with this new force here so that the acceleration just remains a zero to the right what would that change up here in my calculation well my force diagram I've got another force but I can't I don't draw this force pushing on the box again you can't push with tension you can only pull with tension so this rope can pull to the left so I'm going to draw that as t2 and how do I include that here well that's a force to the left so I'd subtract it because the leftward forces we're going to consider negative and now I do my algebra I multiply both sides by M to get MA but then I have to add T 2 to both sides and this makes sense if I'm going to pull over here if I want my T 1 to compensate and make it so that this box still accelerates with a zero even though these people over here are pulling to the left this tension has to increase in order to maintain the same acceleration to the right let's step it up even more let's say it's about two wars about to break out over these cheese snacks right here so let's say someone pulls this way some pulls that way with a force T three so we'll call this T three someone's pulling at an angle this time let's say this force is at an angle theta now what is that change again let's say this T one has to be such that you get the same acceleration what would that change up here you're going to have a tension force up in to the right in my force diagram so you get a tension force this way this is T three at an angle theta now I can't plug all of T three into this formula this formula was just for the horizontal direction so I have to plug only the horizontal component of this T three four so this component right here only that component of T three do I plug in that's come I'm going to call that T three X T three X is what I plug in up here T three why this isn't going to get plugged into this formula at all the t3 Y does not affect the horizontal acceleration it will only affect the vertical acceleration and maybe any forces that are exerted vertically so I'll call this T three why how do I find T three X I have to use trigonometry the way you find components of these vectors is always trigonometry so I'm going to say cosine theta and I'm going to use cosine because I know this side t3 X is adjacent to this angle that I'm given so since this is adjacent to the angle I'm given I use cosine because the definition of cosine is adjacent over hypotenuse and my adjacent side is t 3x my hypotenuse is this side here which is the entire tension t3 so if this tension was like 50 Newtons at an angle of 30 degrees plug the hole 50 Newtons in here I'd say that 50 Newtons times if I solve this 43 X I get t3 x equals so I'm going to multiply both sides by t3 so that would be like our 50 Newtons so t3 the entire magnitude of the tension force times the cosine theta whatever that theta is if it was 30 degrees at plug in 30 degrees this is what I can plug in up here so now I can plug this into my Newton's second law I couldn't plug the hole force in because the entire force was not in the x-direction the entire force was composed of this vertical and horizontal component and the vertical component does not affect the acceleration in the horizontal direction only the horizontal component of this tension which is this amount so if I plug this in up here I'll put a plus because this horizontal component points to the right plus t3 times cosine theta and again the way it solve for t1 is I'd multiply both sides by M so I get M a naught and then I'd add T 2 to both sides and then I have to subtract t3 cosine theta by both from both sides in order to solve for this algebraically and this makes sense my tension t1 doesn't have to be as big anymore because it's got a force helping it pull to the right there's someone on its side pulling to the right so it doesn't have to exert as much force that's why this ends up subtracting up here t1 decreases if you give it a helper force to pull in the same direction that it's pulling so conceptually that's why this tension might increase decrease that's how you would deal with it if there were forces involved you can keep adding forces here even friction if you had a frictional force to the left you would just have to include that as a force up here you'd keep doing it using Newton's second law and then solve algebraically for the tension that you wanted to find so to recap remember the way you solve for tension is by using Newton's second law carefully getting all the signs right and doing your algebra to solve for that tension that you want to find also remember the force of tension is not a pushing force the force of tension is a pulling force you can pull with a rope but you can't push with a rope and in this problem the tension throughout the rope was the same because we assumed that either the rope was massless or the mass of the rope was so insignificant compared to the mass of this box that any variation didn't matter and in other words the tension at every point in this rope was the same and that could have made a difference because if we if we were asking the question how hard does this person over here have to pull on this rope to cause this acceleration if the Rope itself was massive this person would have to not only pull on this massive box but also on this massive rope and there'd be a variation in tension to hear that honestly we often don't want to have to deal with so we assume the rope is massless and then we could just assume that whatever force this person pulls with because tensions of pulling force is also transmitted here undiluted if this person pulled with 50 Newtons then this point on the rope would also pull on the box with 50 Newtons